Generalized Fractional Calculus: New Advancements and Applications [1st ed.] 9783030569617, 9783030569624

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Studies in Systems, Decision and Control 305

George A. Anastassiou

Generalized Fractional Calculus New Advancements and Applications

Studies in Systems, Decision and Control Volume 305

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

The series “Studies in Systems, Decision and Control” (SSDC) covers both new developments and advances, as well as the state of the art, in the various areas of broadly perceived systems, decision-making, and control–quickly, up to date and with a high quality. The intent is to cover the theory, applications, and perspectives on the state of the art and future developments relevant to systems, decisionmaking, control, complex processes and related areas, as embedded in the fields of engineering, computer science, physics, economics, social and life sciences, as well as the paradigms and methodologies behind them. The series contains monographs, textbooks, lecture notes, and edited volumes in systems, decision-making, and control spanning the areas of Cyber-Physical Systems, Autonomous Systems, Sensor Networks, Control Systems, Energy Systems, Automotive Systems, Biological Systems, Vehicular Networking and Connected Vehicles, Aerospace Systems, Automation, Manufacturing, Smart Grids, Nonlinear Systems, Power Systems, Robotics, Social Systems, Economic Systems and other. Of particular value to both the contributors and the readership are the short publication timeframe and the worldwide distribution and exposure which enables both wide and rapid dissemination of research output. ** Indexing: The books of this series are submitted to ISI, SCOPUS, DBLP, Ulrichs, MathSciNet, Current Mathematical Publications, Mathematical Reviews, Zentralblatt Math: MetaPress and Springerlink.

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

George A. Anastassiou

Generalized Fractional Calculus New Advancements and Applications

123

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

ISSN 2198-4182 ISSN 2198-4190 (electronic) Studies in Systems, Decision and Control ISBN 978-3-030-56961-7 ISBN 978-3-030-56962-4 (eBook) https://doi.org/10.1007/978-3-030-56962-4 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Dedicated to my wife Koula and to all doctors fighting Covid—19

Preface

Computational and Fractional Analysis nowadays are more and more in the center of mathematics and of other related sciences either by themselves because of their rapid development, which is based on very old foundations, or because they cover a great variety of applications in the real world. In this monograph, all presented is original work by the author given at a very general level to cover a maximum number of cases in various applications. We apply generalized fractional differentiation techniques of Caputo, Canavati, and conformable types to various kinds of inequalities such as Ostrowski, Grüss, Iyengar, Polya, Poincaré, Sobolev, Hilbert-Pachpatte, and Opial type inequalities. Some of these are extended to Banach space valued functions. Many of these are studying the average of a function as compared to values of the on hand function. These inequalities also have a great impact on numerical analysis, stochastics, and fractional differential equations. We continue with generalized fractional approximation by positive sublinear operators including the case of iterated one which derives from the newly presented Korovkin type inequalities. The linear case is extended to Banach space valued functions. It follows for the first time multivariate complex Korovkin quantitative approximation theory. We also present M-fractional integral inequalities of Ostrowski and Polya types. The M-fractional derivative is based on Mittag-Leffler special function and possesses all nice properties of ordinary derivative. Our results are weighted so they provide a great variety of cases and applications. The last eight chapters of the book deal for the first time with the quantitative fractional Korovkin type approximation of stochastic processes. We lay there the foundations of stochastic fractional calculus. We consider both Caputo and Conformable fractional directions and we derive regular and trigonometric results. Our positive linear operators can be expectation operator commutative or not. The case of non-commutative operators is especially elaborate and interesting. This monograph is the natural and expected evolution of recent author’s research work put in a book form for the first time. The presented approaches are original, and the 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 vii

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the maximum our computational, estimating and approximation methods to all possible directions. The motivation to write this monograph came by the following: various issues related to the modeling and analysis of ordinary and fractional order systems have gained an increased popularity, as witnessed by many books and volumes in Springer’s program: http://www.springer.com/gp/search?query=fractional&submit=Prze%C5%9Blij and the purpose of our book is to capture at a very general level a deeper formal analysis on some issues that are relevant to many areas for instance: decision making, complex processes, systems modeling and control, and related areas. 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: Mixed Caputo weighted-fractional Ostrowski type Inequalities, Caputo weighted-fractional Ostrowski and Grüss inequalities for several functions, Weighted fractional Iyengar type Inequalities in the Caputo direction, Generalized Canavati type g-fractional Iyengar and Ostrowski type inequalities, Generalized Canavati type g-fractional Polya type inequalities, Caputo generalized weighted-fractional integral Inequalities, Generalized weighted-fractional Approximation by Sublinear Operators, Generalized g-iterated fractional Approximations by Sublinear Operators, Generalized g-Fractional vector Representation Formula and integral Inequalities for Banach space valued functions, Iterated g-Fractional vector Representation Formulae and Inequalities for Banach space valued functions, Vectorial generalized g-fractional direct and iterated approximations by linear operators, Quantitative Multivariate Complex Korovkin Theory, M-fractional integral inequalities, Foundation of Stochastic Fractional Calculus with Fractional Approximation of Stochastic Processes, Trigonometric Fractional Approximation of Stochastic Processes, Conformable Fractional Approximation of Stochastic Processes, Trigonometric Conformable Fractional Quantitative Approximation of Stochastic Processes, Commutative Caputo fractional Korovkin inequalities for Stochastic Processes, Trigonometric Commutative Caputo fractional Korovkin theory for Stochastic Processes, Commutative Conformable fractional Korovkin properties for Stochastic Processes, Trigonometric Commutative Conformable fractional Korovkin properties for Stochastic Processes. An extensive list of references is given per chapter.

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The book’s results are expected to find applications in many areas of pure and applied mathematics, especially in approximation theory, stochastic processes, inequalities and differential equations as they connect to fractionality. Thus this monograph is suitable for researchers, graduate students, and seminars of the above disciplines, also to be in all science and engineering libraries. The preparation of the book took place during 2019–2020 at the University of Memphis and finished during the Coronavirus outbreak keeping alive and in control the author! The author likes to thank Professor Alina Lupas of University of Oradea, Romania, for checking and reading the manuscript. Memphis, TN, USA June 2020

George A. Anastassiou

Contents

1

Caputo ˆ-Fractional Ostrowski Inequalities . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . 1.2 Background . . . . . . . . . . . . . . . . . . . . . 1.3 Main Results . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Caputo ˆ-Fractional Ostrowski and Involving Several Functions . . . . . . 2.1 Introduction . . . . . . . . . . . . . . 2.2 Background . . . . . . . . . . . . . . 2.3 Main Results . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . .

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Weighted Caputo Fractional Iyengar Type Inequalities 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Main Results—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Main Results—II . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Generalized Canavati g-Fractional Polya Inequalities . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Grüss Inequalities . . . . .

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Caputo Generalized ˆ-Fractional Integral Type Inequalities 6.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Generalized ˆ-Fractional Quantitative Approximation by Sublinear Operators . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Generalized g-Iterated Fractional By Sublinear Operators . . . . . . . 8.1 Background . . . . . . . . . . . . 8.2 Main Results . . . . . . . . . . . 8.3 Applications . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . .

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Quantitative Approximation . . . . .

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Generalized g-Fractional Vector Representation Formula And Bochner Integral Type Inequalities for Banach Space Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 Iterated g-Fractional Vector Bochner Integral Representation Formulae and Inequalities for Banach Space Valued Functions 10.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11 Vectorial Generalized g-Fractional Direct and Iterated Quantitative Approximation by Linear Operators . . . . 11.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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12 Quantitative Multivariate Complex Korovkin Approximation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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13 M-Fractional Integral Type Inequalities 13.1 Introduction . . . . . . . . . . . . . . . . . 13.2 Background . . . . . . . . . . . . . . . . . 13.3 Main Results . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . .

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14 Principles of Stochastic Caputo Fractional Calculus with Fractional Approximation of Stochastic Processes 14.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Foundation of Stochastic Fractional Calculus . . . . . 14.3 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 Stochastic Korovkin Results . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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15 Trigonometric Caputo Fractional Approximation of Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.2 Foundation of Stochastic Fractional Calculus ([10]) . . . . . . 15.3 Background (See Also [10]) . . . . . . . . . . . . . . . . . . . . . . . 15.4 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.6 Trigonometric Stochastic Korovkin Results . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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16 Conformable Fractional Quantitative Approximation of Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . 16.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . 16.3 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.6 Stochastic Korovkin Results . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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17 Trigonometric Conformable Fractional Approximation of Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . . 17.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.2 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.3 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . . 17.4 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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17.6 Trigonometric Conformable Fractional Stochastic Korovkin Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 18 Commutative Caputo Fractional Korovkin Approximation for Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.6 Caputo Fractional Stochastic Korovkin Theory . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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19 Trigonometric Commutative Caputo Fractional Korovkin Approximation for Stochastic Processes . . . . . . . . . . . . . . . . . 19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.4 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.5 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.6 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.7 Commutative Trigonometric Caputo Fractional Stochastic Korovkin Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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20 Commutative Conformable Fractional Korovkin Approximation for Stochastic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.5 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.6 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.7 Conformable Fractional Stochastic Korovkin Theory . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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21 Trigonometric Commutative Conformable Fractional Korovkin Approximation for Stochastic Processes . . . . . . . . . . . . . . . . . . . 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Background—I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Background—II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.6 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

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21.7 Commutative Trigonometric Conformable Fractional Stochastic Korovkin Properties . . . . . . . . . . . . . . . . . . . . . . . . 495 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 22 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497

Chapter 1

Caputo ψ-Fractional Ostrowski Inequalities

Very general univariate mixed Caputo ψ-fractional Ostrowski type inequalities are presented. Estimates are with respect to · p , 1 ≤ p ≤ ∞. We give also applications. This chapter relies on [4].

1.1 Introduction In 1938, Ostrowski [5] proved the following important inequality. Theorem 1.1 Let f : [a, b] → R be continuous on [a, b] and differentiable on   (a, b) whose derivative f  : (a, b) → R is bounded on (a, b), i.e.,  f  ∞ := sup t∈(a,b)     f (t) < +∞. Then    1  b − a

a

b

 2       x − a+b 1 2  + f (t) dt − f (x) ≤ · (b − a)  f  ∞ , 2 4 (b − a)

for any x ∈ [a, b]. The constant

1 4

(1.1)

is the best possible.

Since then there has been a lot of activity around these inequalities with important applications to numerical analysis and probability. In this chapter we are greatly motived and inspired by Theorem 1.1, see also [2]. Here we present various ψ-fractional Ostrowski type inequalities and we give interesting applications.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_1

1

1 Caputo ψ -Fractional Ostrowski Inequalities

2

1.2 Background Here we follow [1]. Let α > 0, [a, b] ⊂ R, f : [a, b] → R which is integrable and ψ ∈ C 1 ([a, b]) an increasing function such that ψ  (x) = 0, for all x ∈ [a, b]. Consider n = α, the ceiling of α. The left and right fractional integrals are defined, respectively, as follows:  x 1 α,ψ Ia+ f (x) := ψ  (t) (ψ (x) − ψ (t))α−1 f (t) dt, (1.2)  (α) a and α,ψ

Ib− f (x) :=

1  (α)



b

ψ  (t) (ψ (t) − ψ (x))α−1 f (t) dt,

(1.3)

x

for any x ∈ [a, b], where  is the gamma function. The following semigroup property is valid for fractional integrals: if α, β > 0, then α,ψ β,ψ α+β,ψ α,ψ β,ψ α+β,ψ f (x) , and Ib− Ib− f (x) = Ib− f (x) . Ia+ Ia+ f (x) = Ia+ We mention Definition 1.2 ([1]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. The left ψCaputo fractional derivative of f of order α is given by C

α,ψ Da+

f (x) :=

n−α,ψ Ia+

1 d  ψ (x) d x

n f (x) ,

(1.4)

and the right ψ-Caputo fractional derivative of f is given by C

α,ψ Db−

f (x) :=

n−α,ψ Ib−

−

1 d  ψ (x) d x

n f (x) .

(1.5)

To simplify notation, we will use the symbol f ψ[n]

(x) :=

1 d ψ  (x) d x

n f (x) ,

(1.6)

with f ψ[0] (x) = f (x) . By the definition, when α = m ∈ N, we have α,ψ

Da+ f (x) = f ψ[m] (x) and C α,ψ Db− f (x) = (−1)m f ψ[m] (x) . C

(1.7)

1.2 Background

3

If α ∈ / N, we have C

α,ψ

Da+ f (x) =

1  (n − α)



x

a

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

(1.8)

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

(1.9)

and C

α,ψ Db−

(−1)n f (x) =  (n − α)



b x

∀ x ∈ [a, b] . In particular, when α ∈ (0, 1), we have α,ψ

Da+ f (x) = and C α,ψ Db− f (x) = C

1 (1−α) −1 (1−α)

x a

b x

(ψ (x) − ψ (t))−α f  (t) dt, (1.10) (ψ (t) − ψ (x))

−α



f (t) dt

∀ x ∈ [a, b] . Clearly the above is a generalization of left and right Caputo fractional derivatives. For more see [1]. Still we need from [1] the following left and right fractional Taylor’s formulae: Theorem 1.3 ([1]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. Then, the left fractional Taylor formula follows, f (x) =

n−1 f ψ[k] (a)

k!

k=0

α,ψ C

(ψ (x) − ψ (a))k + Ia+

α,ψ

Da+ f (x) ,

(1.11)

and the right fractional Taylor formula follows, f (x) =

n−1 k=0

(−1)

k

f ψ[k] (b) k!

α,ψ C

(ψ (b) − ψ (x))k + Ib−

α,ψ

Db− f (x) ,

(1.12)

∀ x ∈ [a, b] . In particular, given α ∈ (0, 1), we have α,ψ

α,ψ

f (x) = f (a) + Ia+ C Da+ f (x) , and α,ψ α,ψ f (x) = f (b) + Ib− C Db− f (x) , ∀ x ∈ [a, b] .

(1.13)

1 Caputo ψ -Fractional Ostrowski Inequalities

4

Remark 1.4 For convenience we can rewrite (1.11)–(1.13) as follows: f (x) =

n−1 f ψ[k] (a) k=0

1  (α)



x

(ψ (x) − ψ (a))k +

ψ  (t) (ψ (x) − ψ (t))α−1

C

(1.14)

α,ψ

Da+ f (t) dt,

a

and

n−1 (−1)k f ψ[k] (b)

f (x) =

k!

k=0

1  (α)

k!



b

(ψ (b) − ψ (x))k +

ψ  (t) (ψ (t) − ψ (x))α−1

C

x

(1.15)

α,ψ

Db− f (t) dt,

∀ x ∈ [a, b] . When α ∈ (0, 1), we get: f (x) = f (a) + and f (x) = f (b) +

1 (α) 1 (α)

x a

b x

α,ψ

ψ  (t) (ψ (x) − ψ (t))α−1

C

Da+ f (t) dt,

ψ  (t) (ψ (t) − ψ (x))α−1

C

Db− f (t) dt,

α,ψ

(1.16)

∀ x ∈ [a, b] . Again from [1] we have the following: Consider the norms ·∞ : C ([a, b]) → R and ·C [n] : C n ([a, b]) → R, where ψ  n 

 [k]   f C [n] :=  fψ  . ψ

k=0

∞

We have Theorem 1.5 ([1]) The ψ-Caputo fractional derivatives are bounded operators. For all α > 0 (n = α)   C α,ψ  (1.17)  Da+  ≤ K  f C [n] ∞

and

  C α,ψ   Db− 

∞

where K =

ψ

≤ K  f C [n] , ψ

(ψ (b) − ψ (a))n−α > 0.  (n + 1 − α)

(1.18)

(1.19)

1.3 Main Results

5

1.3 Main Results We present the following ψ-fractional Ostrowski type inequalities: Theorem 1.6 Let α > 0, n ∈ N : n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. Let x0 ∈ [a, b] and assume that f ψ[k] (x0 ) = 0, for k = 1, . . . , n − 1. Then    1   (ψ (b) − ψ (a))

a

b

  f (x) ψ (x) d x − f (x0 ) ≤ 

(1.20)

   1  α,ψ  (ψ (x0 ) − ψ (a))α+1 C Dx0 − f  ∞,[a,x0 ] (ψ (b) − ψ (a))  (α + 2)    α,ψ  + (ψ (b) − ψ (x0 ))α+1 C Dx0 + f 

 ≤

∞,[x0 ,b]

     1  α,ψ   α,ψ  max C Dx0 − f  , C Dx0 + f  (ψ (b) − ψ (a))α . (1.21) ∞,[a,x0 ] ∞,[x0 ,b]  (α + 2) In case of 0 < α ≤ 1, (1.20)–(1.21) are still valid without any initial conditions. Proof By Theorem 1.3 we have that f (x) − f (x0 ) =

1  (α)



x

α,ψ

ψ  (t) (ψ (x) − ψ (t))α−1

C

Dx0 + f (t) dt,

ψ  (t) (ψ (t) − ψ (x))α−1

C

Dx0 − f (t) dt,

x0

(1.22)

∀ x ∈ [x0 , b] , and f (x) − f (x0 ) =

1  (α)



x0 x

α,ψ

(1.23)

∀ x ∈ [a, x0 ] . Hence | f (x) − f (x0 )| ≤ 1  (α)



x

1  (α)



x x0

   α,ψ  ψ  (t) (ψ (x) − ψ (t))α−1 C Dx0 + f (t) dt ≤

ψ  (t) (ψ (x) − ψ (t))α−1 dt

x0

  C α,ψ   Dx0 + f 

∞,[x0 ,b]

 (α + 1)

  C α,ψ   Dx0 + f 

(ψ (x) − ψ (x0 ))α .

∞,[x0 ,b]

=

(1.24)

1 Caputo ψ -Fractional Ostrowski Inequalities

6

  C α,ψ   Dx0 + f 

That is | f (x) − f (x0 )| ≤

∞,[x0 ,b]

(ψ (x) − ψ (x0 ))α ,

 (α + 1)

(1.25)

∀ x ∈ [x0 , b] . Similarly, it holds 1 | f (x) − f (x0 )| ≤  (α) 1  (α)



x0



x0 x

   α,ψ  ψ  (t) (ψ (t) − ψ (x))α−1 C Dx0 − f (t) dt ≤ α−1



ψ (t) (ψ (t) − ψ (x))

x

  C α,ψ   Dx0 − f 

∞,[a,x0 ]

 (α + 1) That is | f (x) − f (x0 )| ≤

   α,ψ  dt C Dx0 − f 

∞,[a,x0 ]

=

(ψ (x0 ) − ψ (x))α .

  C α,ψ   Dx0 − f 

∞,[a,x0 ]

(ψ (x0 ) − ψ (x))α ,

 (α + 1)

(1.26)

∀ x ∈ [a, x0 ] . We observe that    b   1   f (x) ψ (x) d x − f (x0 ) =  (ψ (b) − ψ (a)) a    1   (ψ (b) − ψ (a))

b

a

   1   (ψ (b) − ψ (a))

  f (x) ψ  (x) d x − f (x0 ) (ψ (b) − ψ (a))  = b





f (x) ψ (x) d x −

a

a

   1   (ψ (b) − ψ (a))

b

a

1 (ψ (b) − ψ (a)) 1 (ψ (b) − ψ (a))



x0

b



b

  f (x0 ) ψ (x) d x  = 

  ( f (x) − f (x0 )) ψ  (x) d x  ≤ | f (x) − f (x0 )| ψ  (x) d x =

(1.27)





a

| f (x) − f (x0 )| ψ  (x) d x +

b x0

a (by (1.25), (1.26))

≤

| f (x) − f (x0 )| ψ  (x) d x

1.3 Main Results

7

1 (ψ (b) − ψ (a))  (α + 1)

 +

b x0

 

x0

a

   α,ψ  (ψ (x0 ) − ψ (x))α ψ  (x) d x C Dx0 − f 

   α,ψ  (ψ (x) − ψ (x0 )) ψ (x) d x C Dx0 + f  α

∞,[a,x0 ]





∞,[x0 ,b]

=

(1.28)

  1 C α,ψ  (ψ (x0 ) − ψ (a))α+1  Dx0 − f  ∞,[a,x0 ] (ψ (b) − ψ (a))  (α + 2)    α,ψ  + C Dx0 + f 

∞,[x0 ,b]

 (ψ (b) − ψ (x0 ))α+1 ≤

     1  α,ψ   α,ψ  max C Dx0 − f  , C Dx0 + f  (ψ (b) − ψ (a))α . (1.29) ∞,[a,x0 ] ∞,[x0 ,b]  (α + 2)  We make Remark 1.7 In our setting, clearly, it is f ψ[n] ∈ C ([a, b]). Given f ∈ C ([a, b]), by α,ψ

Theorem 4.10, p. 98 of [3], we get that Ia+ f ∈ C ([a, b]), and by Theorem 4.11, p. α,ψ 101 of [3], we get that Ib− f ∈ C ([a, b]). α,ψ α,ψ Therefore, we obtain that C Da+ f , C Db− f ∈ C ([a, b]). We continue with Theorem 1.8 All as in Theorem 1.6 and α ≥ 1. Then    1   (ψ (b) − ψ (a))

a

b

  f (x) ψ (x) d x − f (x0 ) ≤ 

(1.30)

   1  α,ψ  (ψ (x0 ) − ψ (a))α C Dx0 − f  L 1 ([a,x0 ],ψ) (ψ (b) − ψ (a))  (α + 1)    α,ψ  + (ψ (b) − ψ (x0 ))α C Dx0 + f 

 L 1 ([x0 ,b],ψ)

≤

     1  α,ψ   α,ψ  max C Dx0 − f  , C Dx0 + f  (ψ (b) − ψ (a))α−1 . L 1 ([a,x0 ],ψ) L 1 ([x0 ,b],ψ)  (α + 1) (1.31) Proof By (1.22) we obtain: (ψ (x) − ψ (x0 ))α−1 | f (x) − f (x0 )| ≤  (α)



x x0

  C α,ψ   Dx0 + f (t) dψ (t) ≤

1 Caputo ψ -Fractional Ostrowski Inequalities

8

(ψ (x) − ψ (x0 ))α−1  (α)



b x0

  C α,ψ   Dx0 + f (t) dψ (t) =

 (ψ (x) − ψ (x0 ))α−1  C α,ψ  .  Dx0 + f  L 1 ([x0 ,b],ψ)  (α)

(1.32)

That is, we get | f (x) − f (x0 )| ≤

 (ψ (x) − ψ (x0 ))α−1  C α,ψ  ,  Dx0 + f  L 1 ([x0 ,b],ψ)  (α)

(1.33)

∀ x ∈ [x0 , b] . Similarly, by (1.23), we get: | f (x) − f (x0 )| ≤

(ψ (x0 ) − ψ (x))α−1  (α)

(ψ (x0 ) − ψ (x))α−1  (α)



x0

a



x0 x

  C α,ψ   Dx0 − f (t) dψ (t) ≤

  C α,ψ   Dx0 − f (t) dψ (t) =

(1.34)

 (ψ (x0 ) − ψ (x))α−1  C α,ψ  .  Dx0 − f  L 1 ([a,x0 ],ψ)  (α) That is, we derive | f (x) − f (x0 )| ≤

 (ψ (x0 ) − ψ (x))α−1  C α,ψ  ,  Dx0 − f  L 1 ([a,x0 ],ψ)  (α)

(1.35)

∀ x ∈ [a, x0 ] . As in the proof of Theorem 1.6, we have    1   (ψ (b) − ψ (a))

a

1 (ψ (b) − ψ (a))



x0

b

  f (x) ψ  (x) d x − f (x0 ) ≤ 

| f (x) − f (x0 )| ψ (x) d x +

a



b





| f (x) − f (x0 )| ψ (x) d x

x0

(1.36)

(by (1.33), (1.35))

≤

1 (ψ (b) − ψ (a))  (α)

  a

x0

   α,ψ  (ψ (x0 ) − ψ (x))α−1 dψ (x) C Dx0 − f 

L 1 ([a,x0 ],ψ)

1.3 Main Results

9

 +

b

(ψ (x) − ψ (x0 ))

α−1

x0

   α,ψ  dψ (x) C Dx0 + f 

 L 1 ([x0 ,b],ψ)

=

   1  α,ψ  (ψ (x0 ) − ψ (a))α C Dx0 − f  L 1 ([a,x0 ],ψ) (ψ (b) − ψ (a))  (α + 1)    α,ψ  + (ψ (b) − ψ (x0 ))α C Dx0 + f 

 ≤

L 1 ([x0 ,b],ψ)

(1.37)

     1  α,ψ   α,ψ  max C Dx0 − f  , C Dx0 + f  (ψ (b) − ψ (a))α−1 . L 1 ([a,x0 ],ψ) L 1 ([x0 ,b],ψ)  (α + 1)  Next we present Theorem 1.9 All as in Theorem 1.6. Let p, q > 1 :    1   (ψ (b) − ψ (a))

b

a

1 p

+

1 q

= 1, and α ≥ 1. Then

  f (x) ψ (x) d x − f (x0 ) ≤ 

1

  1 (ψ (b) − ψ (a))  (α) ( p (α − 1) + 1) p α + 1p    1  α,ψ  (ψ (x0 ) − ψ (a))α+ p C Dx0 − f    1  α,ψ  (ψ (b) − ψ (x0 ))α+ p C Dx0 + f 

L q ([a,x0 ],ψ)

(1.38)

+

 L q ([x0 ,b],ψ)

≤

1

  1  (α) ( p (α − 1) + 1) p α + 1p    α,ψ  max C Dx0 − f 

L q ([a,x0 ],ψ)

   α,ψ  , C Dx0 + f 

 L q ([x0 ,b],ψ)

(1.39)

(ψ (b) − ψ (a))α− q . 1

Proof By (1.22) and Hölder’s inequality we have | f (x) − f (x0 )| ≤ 1  (α)



x x0

1  (α)



x x0

   α,ψ  (ψ (x) − ψ (t))α−1 C Dx0 + f (t) dψ (t) ≤

(ψ (x) − ψ (t)) p(α−1) dψ (t)

1p 

x x0

q1 q   C α,ψ ≤  Dx0 + f (t) dψ (t) (1.40)

1 Caputo ψ -Fractional Ostrowski Inequalities

10

1  1 (ψ (x) − ψ (x0 ))α−1+ p  C α,ψ  D f .  x0 +  L q ([x0 ,b],ψ)  (α) ( p (α − 1) + 1) 1p

That is 1  1 (ψ (x) − ψ (x0 ))α−1+ p  C α,ψ  | f (x) − f (x0 )| ≤ ,  Dx0 + f  1 L q ([x0 ,b],ψ)  (α) ( p (α − 1) + 1) p

(1.41)

∀ x ∈ [x0 , b] . By (1.23) and Hölder’s inequality we have 1 | f (x) − f (x0 )| ≤  (α) 1  (α)



x0

(ψ (t) − ψ (x))



   α,ψ  (ψ (t) − ψ (x))α−1 C Dx0 − f (t) dψ (t) ≤

x0 x

p(α−1)

dψ (t)

1p 

x

x0 x

q1  q C α,ψ  ≤  Dx0 − f (t) dψ (t) (1.42)

1  1 (ψ (x0 ) − ψ (x))α−1+ p  C α,ψ  D f .  x0 −  L q ([a,x0 ],ψ)  (α) ( p (α − 1) + 1) 1p

That is 1  1 (ψ (x0 ) − ψ (x))α−1+ p  C α,ψ  | f (x) − f (x0 )| ≤ ,  Dx0 − f  1 L q ([a,x0 ],ψ)  (α) ( p (α − 1) + 1) p

(1.43)

∀ x ∈ [a, x0 ] . As in the proof of Theorem 1.6, we have    1   (ψ (b) − ψ (a))

a

1 (ψ (b) − ψ (a))



x0

b

  f (x) ψ  (x) d x − f (x0 ) ≤

| f (x) − f (x0 )| ψ  (x) d x +

x0

a

 +

b x0

| f (x) − f (x0 )| ψ  (x) d x

1

(ψ (b) − ψ (a))  (α) ( p (α − 1) + 1) p

  1  α,ψ  (ψ (x0 ) − ψ (x))α−1+ p dψ (x) C Dx0 − f 

(ψ (x) − ψ (x0 ))

α−1+ 1p

   α,ψ  dψ (x) C Dx0 + f 



(1.44)

1

(by (1.41), (1.43))

 

b x0

a

≤



L q ([a,x0 ],ψ)

 L q ([x0 ,b],ψ)

=

1.3 Main Results

11

1

  1 (ψ (b) − ψ (a))  (α) ( p (α − 1) + 1) p α + 1p    1  α,ψ  (ψ (x0 ) − ψ (a))α+ p C Dx0 − f    1  α,ψ  (ψ (b) − ψ (x0 ))α+ p C Dx0 + f  ≤    α,ψ  max C Dx0 − f 

L q ([a,x0 ],ψ)

+ 

L q ([x0 ,b],ψ)

1

  1  (α) ( p (α − 1) + 1) p α + 1p

L q ([a,x0 ],ψ)

   α,ψ  , C Dx0 + f 

 L q ([x0 ,b],ψ)

(1.45)

(ψ (b) − ψ (a))α− q . 1

 Corollary 1.10 (to Theorem 1.9) All as in Theorem 1.6. Here p = q = 2 and α ≥ 1. Then    b   1    f ψ d x − f (x) (x) (x ) 0 ≤  (ψ (b) − ψ (a)) a 1   √ (ψ (b) − ψ (a))  (α) (2α − 1) α + 21    1  α,ψ  (ψ (x0 ) − ψ (a))α+ 2 C Dx0 − f    1  α,ψ  (ψ (b) − ψ (x0 ))α+ 2 C Dx0 + f  ≤    α,ψ  max C Dx0 − f 

L 2 ([a,x0 ],ψ)

(1.46)

+ 

L 2 ([x0 ,b],ψ)

1   √  (α) (2α − 1) α + 21

L 2 ([a,x0 ],ψ)

   α,ψ  , C Dx0 + f 

 L 2 ([x0 ,b],ψ)

(1.47)

(ψ (b) − ψ (a))α− 2 . 1

Some applications of Theorem 1.6 follow. In the case of ψ (x) = e x we get: Proposition 1.11 Let α > 0, n ∈ N : n = α, [a, b] ⊂ R, f ∈ C n ([a, b]). Let x0 ∈ [a, b] and assume that f e[k] x (x 0 ) = 0, k = 1, . . . , n − 1. Then

1 Caputo ψ -Fractional Ostrowski Inequalities

12

   b   1 1   x   f e d x − f (x) (x ) ≤  b  b 0 a a   e −e e − e  (α + 2) a    x0 α+1  C α,ex  e − ea  Dx0 − f 

∞,[a,x0 ]

 α+1   C α,ex  + eb − e x0  Dx0 + f 

(1.48) 

∞,[x0 ,b]

      b α x 1   C α,ex  e − ea . max C Dxα,e f , D f    x0 + 0− ∞,[a,x0 ] ∞,[x0 ,b]  (α + 2)

≤

(1.49)

In case of 0 < α ≤ 1, (1.48)–(1.49) are still valid without any initial conditions. In case of ψ (x) = ln x we obtain: Proposition 1.12 Let α > 0, n ∈ N : n = α, [a, b] ⊂ (0, +∞), f ∈ C n ([a, b]). Let x0 ∈ [a, b] and assume that f ln[k]x (x0 ) = 0, k = 1, . . . , n − 1. Then     1  b f (x) 1   d x − f (x0 ) ≤  b   b   ln a a x ln a  (α + 2) 

   x α+1  b α+1  0 C α,ln x  C α,ln x  ln + ln  Dx0 − f   Dx0 + f  ∞,[a,x0 ] ∞,[x0 ,b] a x0

 

   1 b α  C α,ln x  x  ln max C Dxα,ln f , D f .    x0 + 0− ∞,[a,x0 ] ∞,[x0 ,b]  (α + 2) a

(1.50)  ≤

(1.51)

In case of 0 < α ≤ 1, (1.50)–(1.51) are still valid without any initial conditions. Note. Many other interesting examples of our theorems could follow but we skip this task.

References 1. Almeida, R.: A Caputo fractional derivative of a function with respect to another function. Commun. Nonlinear Sci. Numer. Simul. 44, 460–481 (2017) 2. Anastassiou, G.A.: Ostrowski type inequalities. Proc. AMS 123, 3775–3781 (1995) 3. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 4. Anastassiou, G.: Mixed Caputo ψ-fractional Ostrowski type inequalities (2019). Submitted 5. Ostrowski, A.: Über die Absolutabweichung einer differtentiebaren Funktion von ihrem Integralmittelwert. Comment. Math. Helv. 10, 226–227 (1938)

Chapter 2

Caputo ψ-Fractional Ostrowski and Grüss Inequalities Involving Several Functions

Very general univariate mixed Caputo ψ-fractional Ostrowski and Grüss type inequalities for several functions are presented. Estimates are with respect to · p , 1 ≤ p ≤ ∞. We give also applications. See also [4].

2.1 Introduction In 1938, Ostrowski [6] proved the following important inequality. Theorem 2.1 Let f : [a, b] → R be continuous on [a, b] and differentiable on   (a, b) whose derivative f  : (a, b) → R is bounded on (a, b), i.e.,  f  ∞ := sup t∈(a,b)     f (t) < +∞. Then    1  b − a

a

b

 2       x − a+b 1 2  + f (t) dt − f (x) ≤ · (b − a)  f  ∞ , 2 4 (b − a)

for any x ∈ [a, b]. The constant

1 4

(2.1)

is the best possible.

Since then there has been a lot of activity around these inequalities with important applications to numerical analysis and probability. In this chapter we are greatly motivated and inspired by Theorem 2.1, see also [2]. Here we present various ψ-fractional Ostrowski and Grüss type inequalities for several functions and we give interesting applications.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_2

13

2 Caputo ψ-Fractional Ostrowski and Grüss …

14

2.2 Background Here we follow [1]. Let α > 0, [a, b] ⊂ R, f : [a, b] → R which is integrable and ψ ∈ C 1 ([a, b]) an increasing function such that ψ  (x) = 0, for all x ∈ [a, b]. Consider n = α, the ceiling of α. The left and right fractional integrals are defined, respectively, as follows:  x 1 α,ψ Ia+ f (x) := ψ  (t) (ψ (x) − ψ (t))α−1 f (t) dt, (2.2)  (α) a and α,ψ

Ib− f (x) :=

1  (α)



b

ψ  (t) (ψ (t) − ψ (x))α−1 f (t) dt,

(2.3)

x

for any x ∈ [a, b], where  is the gamma function. The following semigroup property is valid for fractional integrals: if α, β > 0, then α,ψ β,ψ

α+β,ψ

Ia+ Ia+ f (x) = Ia+

α,ψ β,ψ

α+β,ψ

f (x) , and Ib− Ib− f (x) = Ib−

f (x) .

We mention Definition 2.2 ([1]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. The left ψCaputo fractional derivative of f of order α is given by C



α,ψ

n−α,ψ

Da+ f (x) := Ia+

1 d  ψ (x) d x

n f (x) ,

(2.4)

and the right ψ-Caputo fractional derivative of f is given by C

α,ψ Db−

f (x) :=

n−α,ψ Ib−

−

1 d  ψ (x) d x

n f (x) .

(2.5)

To simplify notation, we will use the symbol f ψ[n]

(x) :=

1 d ψ  (x) d x

n f (x) ,

(2.6)

with f ψ[0] (x) = f (x) . By the definition, when α = m ∈ N, we have α,ψ

Da+ f (x) = f ψ[m] (x) and C α,ψ Db− f (x) = (−1)m f ψ[m] (x) . C

(2.7)

2.2 Background

15

If α ∈ / N, we have C

α,ψ

Da+ f (x) =

1  (n − α)



x

a

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

(2.8)

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

(2.9)

and C

α,ψ Db−

(−1)n f (x) =  (n − α)



b x

∀ x ∈ [a, b] . In particular, when α ∈ (0, 1), we have α,ψ

Da+ f (x) = and C α,ψ Db− f (x) = C

1 (1−α) −1 (1−α)

x a

b x

(ψ (x) − ψ (t))−α f  (t) dt, (2.10) (ψ (t) − ψ (x))

−α



f (t) dt

∀ x ∈ [a, b] . Clearly the above is a generalization of left and right Caputo fractional derivatives. For more see [1]. Still we need from [1] the following left and right fractional Taylor’s formulae: Theorem 2.3 ([1]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. Then, the left fractional Taylor formula follows, f (x) =

n−1 f ψ[k] (a)

k!

k=0

α,ψ C

(ψ (x) − ψ (a))k + Ia+

α,ψ

Da+ f (x) ,

(2.11)

and the right fractional Taylor formula follows, f (x) =

n−1 k=0

(−1)

k

f ψ[k] (b) k!

α,ψ C

(ψ (b) − ψ (x))k + Ib−

α,ψ

Db− f (x) ,

(2.12)

∀ x ∈ [a, b] . In particular, given α ∈ (0, 1), we have α,ψ

α,ψ

f (x) = f (a) + Ia+ C Da+ f (x) , and α,ψ α,ψ f (x) = f (b) + Ib− C Db− f (x) , ∀ x ∈ [a, b] .

(2.13)

2 Caputo ψ-Fractional Ostrowski and Grüss …

16

Remark 2.4 For convenience we can rewrite (2.11)–(2.13) as follows: f (x) =

n−1 f ψ[k] (a) k=0

1  (α)



x

(ψ (x) − ψ (a))k +

ψ  (t) (ψ (x) − ψ (t))α−1

C

(2.14)

α,ψ

Da+ f (t) dt,

a

and

n−1 (−1)k f ψ[k] (b)

f (x) =

k!

k=0

1  (α)

k!



b

(ψ (b) − ψ (x))k +

ψ  (t) (ψ (t) − ψ (x))α−1

C

x

(2.15)

α,ψ

Db− f (t) dt,

∀ x ∈ [a, b] . When α ∈ (0, 1), we get: f (x) = f (a) + and f (x) = f (b) +

1 (α) 1 (α)

x a

b x

α,ψ

ψ  (t) (ψ (x) − ψ (t))α−1

C

Da+ f (t) dt,

ψ  (t) (ψ (t) − ψ (x))α−1

C

Db− f (t) dt,

α,ψ

(2.16)

∀ x ∈ [a, b] . Again from [1] we have the following: Consider the norms ·∞ : C ([a, b]) → R and ·C [n] : C n ([a, b]) → R, where ψ  n 

 [k]   f C [n] :=  fψ  . ψ

k=0

∞

We have Theorem 2.5 ([1]) The ψ-Caputo fractional derivatives are bounded operators. For all α > 0 (n = α)   C α,ψ  (2.17)  Da+  ≤ K  f C [n] ∞

and

  C α,ψ   Db− 

∞

where K =

ψ

≤ K  f C [n] , ψ

(ψ (b) − ψ (a))n−α > 0.  (n + 1 − α)

(2.18)

(2.19)

2.3 Main Results

17

2.3 Main Results At first we present the following ψ-fractional Ostrowski type inequalities for several functions: Theorem 2.6 Let α > 0, n ∈ N : n = α, [a, b] ⊂ R and f i , ψ ∈ C n ([a, b]), i = 1, . . . , r ; with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. Let x0 ∈ [a, b] [k] and assume that f iψ (x0 ) = 0, for k = 1, . . . , n − 1; i = 1, . . . , r . Set   ( f 1 , . . . , fr ) (x0 ) := r

b



a

⎡ r i=1

⎢ ⎢ f i (x0 ) ⎣



r 

a

⎜ ⎜ ⎝

f k (x) dψ (x) − ⎤

⎞ r  j=1 j =i

(2.20)

k=1

⎛ b



⎥ ⎟ ⎥ f j (x)⎟ ⎠ dψ (x)⎦ .

Then ⎡⎡ | ( f 1 , . . . , fr ) (x0 )| ≤

α+1,ψ

∞,[a,x0 ]

i=1

⎡

Ia+

α+1,ψ

∞,[x0 ,b]

Ib−

⎜ ⎜ ⎝

r  j=1 j =i

 ⎟⎥  f j (x0 )⎟⎥ ⎠⎦

⎞⎤⎤

⎛

 ⎢ C α,ψ  +⎢ D f   i + x 0 ⎣

⎞⎤

⎛

 ⎢⎢ C α,ψ  ⎢⎢ D f  x0 − i  ⎣⎣

r

(2.21)

r  ⎟⎥⎥ ⎜ ⎜  f j (x0 )⎟⎥⎥ . ⎠⎦⎦ ⎝ j=1 j =i

If 0 < α ≤ 1, then (2.21) is valid without any initial conditions. Proof By Theorem 2.3 we have that 1 f i (x) − f i (x0 ) =  (α)



x

α,ψ

ψ  (t) (ψ (x) − ψ (t))α−1

C

Dx0 + f i (t) dt,

ψ  (t) (ψ (t) − ψ (x))α−1

C

Dx0 − f i (t) dt,

x0

(2.22)

∀ x ∈ [x0 , b] , and 1 f i (x) − f i (x0 ) =  (α) ∀ x ∈ [a, x0 ] ; for all i = 1, . . . , r.



x0 x

α,ψ

(2.23)

2 Caputo ψ-Fractional Ostrowski and Grüss …

18

⎛ ⎜ Multiplying (2.22) and (2.23) by ⎝

⎞ r  j=1 j =i

⎟ f j (x)⎠ we obtain, respectively, ⎞

⎛ r  k=1

⎛ ⎜ ⎝

⎜ f k (x) − ⎜ ⎝

r  j=1 j =i

⎟ f j (x)⎟ ⎠ f i (x0 ) =

⎞ r  j=1 j =i

⎟ f j (x)⎠



 (α)

x

ψ  (t) (ψ (x) − ψ (t))α−1

C

x0

∀ x ∈ [x0 , b] ,

k=1

⎛

(2.24)

⎞

⎛ r 

α,ψ

Dx0 + f i (t) dt,

r ⎟ ⎜ f k (x) − ⎜ f j (x)⎟ ⎠ f i (x0 ) = ⎝ j=1 j =i

⎞

r ⎜ ⎟ f j (x)⎠ ⎝ j=1 j =i



 (α)

x0

ψ  (t) (ψ (t) − ψ (x))α−1

α,ψ

C

Dx0 − f i (t) dt,

⎞

⎤

x

(2.25)

∀ x ∈ [a, x0 ] ; for all i = 1, . . . , r. Adding (2.24) and (2.25), separately, we obtain

r

 r 

f k (x) −

k=1

r r ⎟ ⎜ 1 ⎢ ⎢⎜ f j (x)⎟ ⎠  (α) i=1 ⎣⎝ j=1 j =i

∀ x ∈ [x0 , b] , and

r i=1

⎞

⎡⎛

⎡⎛





⎢⎜ ⎢⎜ ⎣⎝

r  j=1 j =i

⎟ ⎥ ⎥ f j (x)⎟ ⎠ f i (x0 )⎦ = ⎤

x x0

ψ  (t) (ψ (x) − ψ (t))α−1

C

⎥ α,ψ Dx0 + f i (t) dt ⎥ ⎦ , (2.26)

2.3 Main Results

19

r

 r 

f k (x) −

k=1

1  (α)

i=1

⎢⎜ ⎢⎜ ⎣⎝

r  j=1 j =i

r i=1

⎞

⎡⎛ r

⎟ f j (x)⎟ ⎠

⎞

⎡⎛



⎢⎜ ⎢⎜ ⎣⎝

r  j=1 j =i

⎤

⎟ ⎥ ⎥ f j (x)⎟ ⎠ f i (x0 )⎦ = ⎤



x0

ψ  (t) (ψ (t) − ψ (x))α−1

C

x

⎥ α,ψ Dx0 − f i (t) dt ⎥ ⎦ , (2.27)

∀ x ∈ [a, x0 ] . Next we integrate (2.26) and (2.27) with respect to ψ (x), x ∈ [a, b]. We have 

b

 r 

x0

k=1

r

⎡

⎡

 f k (x) dψ (x) −

r

⎢ ⎢ f i (x0 ) ⎣

i=1

⎛

b x0

⎜ ⎜ ⎝

r 

⎥ ⎟ ⎥ f j (x)⎟ ⎠ dψ (x)⎦ =

j=1 j =i

⎞

 x  b  r ⎜ r ⎟ 1 ⎢ ⎢ ⎜ ⎟ f ψ  (t) (ψ (x) − ψ (t))α−1 (x) j ⎣ ⎝ ⎠  (α) x0 x0 i=1

⎤

⎞

⎛



⎤



C

j=1 j =i

⎥ α,ψ Dx0 + f i (t) dt dψ (x)⎥ ⎦,

(2.28) and

r

 x0  r a

⎡

 f k (x) dψ (x) −

k=1

⎛

⎞

 x0  r ⎢ ⎜ r ⎟ ⎥ ⎢ f i (x0 ) ⎜ f j (x)⎟ dψ (x)⎥ ⎣ ⎝ ⎠ ⎦= a i=1

j=1 j =i

⎡ ⎛ ⎞  x0   x0 r r ⎢ ⎜ ⎟ 1 ⎢ ⎜ ⎟[ f ψ  (t) (ψ (t) − ψ (x))α−1 (x) j ⎣ ⎝ ⎠  (α) a x i=1

⎤

j=1 j =i

⎤ C

⎥ α,ψ Dx0 − f i (t) dt]dψ (x)⎥ ⎦.

(2.29) Adding (2.28) and (2.29) we derive the identity:   ( f 1 , . . . , fr ) (x0 ) := r a

⎡ r i=1

⎢ ⎢ f i (x0 ) ⎣

 a

⎛ b

⎜ ⎜ ⎝

b



r 

 f k (x) dψ (x) −

k=1

⎞ r  j=1 j =i

⎤

⎟ ⎥ ⎥ f j (x)⎟ ⎠ dψ (x)⎦ =

2 Caputo ψ-Fractional Ostrowski and Grüss …

20 ⎡⎡

⎛

⎞

 x0  r r ⎢ x0 ⎜  ⎟ 1 ⎢ ⎢⎢ ⎜ ⎟ f ψ  (t) (ψ (t) − ψ (x))α−1 (x) j ⎣⎣ ⎝ ⎠  (α) a x i=1

j=1 j =i

⎤



C

⎥ α,ψ Dx0 − f i (t) dt dψ (x)⎥ ⎦

(2.30) ⎡ ⎢ +⎢ ⎣



⎛ b x0

⎞

r ⎜ ⎟ ⎜ f j (x)⎟ ⎝ ⎠



⎤⎤

x

ψ  (t) (ψ (x) − ψ (t))α−1

C

x0

j=1 j =i

⎥⎥ α,ψ ⎥ Dx0 + f i (t) dt dψ (x)⎥ ⎦⎦ .

Hence it holds | ( f 1 , . . . , fr ) (x0 )| ≤ ⎡⎡ 1  (α)

⎞

⎛

⎤

 x 

⎥  ⎟ 0  ⎥  α,ψ  f j (x)⎟ ψ  (t) (ψ (t) − ψ (x))α−1 C Dx0 − f i (t) dt dψ (x)⎥ ⎟ ⎠ x ⎦

r ⎢⎢ x0 ⎜  r  ⎢⎢ ⎜ i=1

⎢⎢ ⎣⎣ a

⎜ ⎝

j=1 j =i

⎞ ⎤⎤ ⎛   r  ⎟  ⎥⎥ ⎢ b ⎜   x   ⎥⎥ ⎢  ⎜ α,ψ  f j (x)⎟ +⎢ ψ  (t) (ψ (x) − ψ (t))α−1 C Dx0 + f i (t) dt dψ (x)⎥⎥ =: (∗) . ⎟ ⎜ ⎠ x0 ⎦⎦ ⎣ x0 ⎝ ⎡

j=1 j =i

(2.31) We observe that ⎡⎡

⎡



 ⎢ C α,ψ  +⎢ D f  x0 + i  ⎣

∞,[x0 ,b]

⎞

⎛

 x r ⎢⎢  0 1 ⎢⎢C α,ψ  (∗) ≤ ⎢⎢ Dx0 − f i  ∞,[a,x0 ] a ⎣⎣  (α + 1) i=1

x0

⎜ ⎜ ⎝

j=1 j =i

⎞

⎛ b

r  j=1 j =i

⎤⎤

 ⎟ ⎥⎥  f j (x)⎟ (ψ (x) − ψ (x0 ))α dψ (x)⎥⎥ = (2.32) ⎠ ⎦⎦

⎡⎡ r i=1

⎤

r  ⎜ ⎥ ⎟ ⎜ ⎥  f j (x)⎟ ⎟ (ψ (x0 ) − ψ (x))α dψ (x)⎥ ⎜ ⎠ ⎝ ⎦

⎞⎤

⎛

 ⎢⎢ C α,ψ  ⎢⎢ D f  x0 − i  ⎣⎣

α+1,ψ

∞,[a,x0 ]

⎡  ⎢ C α,ψ  ⎢ D f   i + x 0 ⎣

∞,[x0 ,b]

Ia+

⎜ ⎜ ⎝

r  j=1 j =i

⎞⎤⎤

⎛ α+1,ψ

Ib−

⎜ ⎜ ⎝

 ⎟⎥  f j (x0 )⎟⎥ + ⎠⎦

r  j=1 j =i

 ⎟⎥⎥  f j (x0 )⎟⎥⎥ . ⎠⎦⎦

2.3 Main Results

21

⎛ α+1,ψ

By Theorem 4.10, p. 98 of [3], we get that Ia+

⎜ ⎝

r  j=1 j =i

⎞  ⎟  f j ⎠ ∈ C ([a, b]) and so

at any x0 ∈ [a, b] is⎛finite, i = ⎞ 1, . . . , r . Similarly, by Theorem 4.11, p. 101 of [3], r α+1,ψ ⎜   ⎟ we get that Ib− ⎝  f j ⎠ ∈ C ([a, b]) and so at any x0 ∈ [a, b] is finite, i = j=1 j =i

α,ψ

α,ψ

1, . . . , r . Arguing similarly, we get that C Da+ f i , C Db− f i ∈ C ([a, b]), for all i = 1, . . . , r . The theorem is proved.  We continue with Theorem 2.7 All as in Theorem 2.6 with α ≥ 1. Then ⎡ r  ⎢ C α,ψ  ⎢ | ( f 1 , . . . , fr ) (x0 )| ≤ D f   i − x 0 ⎣

α,ψ ⎜ Ia+ ⎜ ⎝ L 1 ([a,x0 ],ψ)

i=1

r  j=1 j =i

⎟   f j (x0 )⎟ ⎠

⎞⎤

⎛    α,ψ  + C Dx0 + f i 

⎞

⎛

r  ⎟⎥  α,ψ ⎜  f j (x0 )⎟⎥ . Ib− ⎜ ⎠⎦ ⎝ L 1 ([x0 ,b],ψ)

(2.33)

j=1 j =i

Proof From (2.31) we get 1 (∗) ≤  (α) i=1 ⎞ ⎛ r

⎛ ⎜ ⎜ ⎝



x0

a

⎜ ⎜ ⎝

⎡

r  j=1 j =i

  C α,ψ   Dx0 − f i 

L 1 ([a,x0 ],ψ)

⎞⎤

⎟  ⎟⎥  f j (x)⎟ (ψ (x0 ) − ψ (x))α−1 dψ (x)⎟⎥ + ⎠ ⎠⎦ ⎛

⎞⎤⎤

⎞

⎛

r  ⎟⎥⎥  ⎢ ⎜ b ⎜  ⎟ ⎟⎥⎥ ⎢C α,ψ  ⎜ ⎜  f j (x)⎟ ⎟ (ψ (x) − ψ (x0 ))α−1 dψ (x)⎟⎥⎥ ⎢ D x 0 + f i  ⎜ ⎜ L 1 ([x0 ,b],ψ) ⎝ x0 ⎝ ⎠⎦⎦ ⎠ ⎣ j=1 j =i

⎡ =

⎞

⎛

r r    ⎟ ⎢ α,ψ ⎜ C α,ψ   f j (x0 )⎟ + ⎢ ⎜ D f I   x0 − i a+ ⎝ ⎠ ⎣ L 1 ([a,x0 ],ψ) i=1

j=1 j =i

(2.34)

2 Caputo ψ-Fractional Ostrowski and Grüss …

22

⎞⎤

⎛   C α,ψ   Dx0 + f i 

α,ψ ⎜ Ib− ⎜ ⎝ L 1 ([x0 ,b],ψ)

r  j=1 j =i

 ⎟⎥  f j (x0 )⎟⎥ , ⎠⎦

(2.35)



proving the theorem. We continue with Theorem 2.8 All as in Theorem 2.6 with p, q > 1 :

| ( f 1 , . . . , fr ) (x0 )| ≤

i=1

+

 α+

1 q

1 p

= 1, α ≥ 1. Then

! 1

 (α) ( p (α − 1) + 1) p

⎡⎡ r

1 p

⎞⎤

⎛

 ⎢⎢ C α,ψ  ⎢⎢ D f  i x0 −  ⎣⎣

α+ 1p ,ψ

L q ([a,x0 ],ψ)

Ia+

⎡

⎜ ⎜ ⎝

r  j=1 j =i

⎟ ⎥   f j (x0 )⎟⎥ ⎠⎦ ⎞⎤⎤

⎛

 ⎢ C α,ψ  +⎢ D f  x0 + i  ⎣

⎜ ⎜ ⎝

α+ 1p ,ψ

Ib−

L q ([x0 ,b],ψ)

(2.36)

r  j=1 j =i

 ⎟⎥⎥  f j (x0 )⎟⎥⎥ . ⎠⎦⎦

Proof From (2.31) we obtain ⎡⎡

⎞

⎛

 r r  ⎟ ⎢ x0 ⎜  1 ⎢  f j (x)⎟ ⎢⎢ ⎜ (∗) ≤ ⎠ ⎣ ⎣ ⎝  (α) a i=1



x0

(ψ (t) − ψ (x))

p(α−1)

j=1 j =i

1  p dψ (t)

x

x

⎡ ⎢ ⎢ ⎣



x x0



⎛ b x0



1  q q C α,ψ  dψ (x) +  Dx0 − f i (t) dψ (t)

⎞

 r ⎜  ⎟ ⎜  f j (x)⎟ ⎝ ⎠ j=1 j =i

x

(ψ (x) − ψ (t))

a

⎛ x0

p(α−1)

1 p dψ (t)

(2.37)

x0

1  q q C α,ψ  D f dψ dψ (x) (t)  x0 + i (t)

⎡⎡  r  ⎢⎢ C α,ψ  ⎢⎢ D f   i x0 − ⎣⎣ L q ([a,x 0 ],ψ) i=1

x0

 ≤ ⎞

1 1

( p (α − 1) + 1) p  (α) ⎤

r  ⎜ ⎟ ⎥ 1  f j (x)⎟ (ψ (x0 ) − ψ (x))α−1+ p dψ (x)⎥ ⎜ ⎝ ⎠ ⎦ j=1 j =i

2.3 Main Results

23

⎡  ⎢ C α,ψ  +⎢ ⎣ D x 0 + f i 

 L q ([x 0 ,b],ψ)

 α+

1 p

⎛ b x0

⎞

⎤⎤

r  ⎜ ⎟ ⎥⎥ 1  f j (x)⎟ (ψ (x) − ψ (x0 ))α−1+ p dψ (x)⎥⎥ = ⎜ ⎝ ⎠ ⎦⎦

⎡⎡

!

r

 (α) ( p (α − 1) + 1)

1 p

(2.38)

j=1 j =i

i=1

⎛

⎞⎤

 ⎢⎢ α+ 1p ,ψ ⎜ C α,ψ  ⎢⎢ ⎜ ⎣⎣ Dx0 − f i  L q ([a,x0 ],ψ) Ia+ ⎝

⎡  ⎢ C α,ψ  +⎢ ⎣ Dx0 + f i 

⎛ α+ I L q ([x 0 ,b],ψ) b−

1 p ,ψ

⎜ ⎜ ⎝

r  j=1 j =i

 ⎟⎥  f j (x0 )⎟⎥ ⎠⎦

⎞ ⎤⎤ r  j=1 j =i

  ⎟ ⎥⎥  f j (x0 )⎟⎥⎥ , ⎠ ⎦⎦

(2.39)



proving the theorem. We mention as motivation for Grüss type inequalities the following:

ˇ Theorem 2.9 (1882, Cebyšev [5]) Let f, g : [a, b] → R be absolutely continuous   functions with f , g ∈ L ∞ ([a, b]). Then    1  b − a

b

f (x) g (x) d x −

a

≤

1 b−a





b

f (x) d x a

1 b−a



b

a

  g (x) d x 

    1 (b − a)2  f  ∞ g  ∞ . 12

(2.40)

The above integrals are assumed to exist. Next follow ψ-Caputo fractional Grüss type inequalities for several functions. Theorem 2.10 Let 0 < α ≤ 1, [a, b] ⊂ R and f i , ψ ∈ C 1 ([a, b]), i = 1, . . . , r ∈  N − {1};  with ψ being increasing and ψ (x)  = 0, for all x ∈ [a, b]. Assume that C α,ψ   α,ψ  sup  Dx0 − f i  < ∞, and sup C Dx0 + f i  < ∞, i = 1, . . . , r. ∞,[a,x0 ]

x0 ∈[a,b]

x0 ∈[a,b]

Set ψ

∞,[x0 ,b]



 ( f 1 , . . . , fr ) := r (ψ (b) − ψ (a))

 r b 

a

⎡ r i=1

Then

⎢ ⎢ ⎣

 a

b

⎛

 ⎜ f i (x) dψ (x) ⎜ ⎝

a

⎜ ⎜ ⎝

r  j=1 j =i



f k (x) dψ (x) −

k=1

⎞

⎛ b



⎞⎤

⎟⎥ ⎟ ⎟⎥ . dψ f j (x)⎟ (x) ⎠⎦ ⎠

 ψ   ( f 1 , . . . , fr ) ≤ (ψ (b) − ψ (a))

(2.41)

2 Caputo ψ-Fractional Ostrowski and Grüss …

24

⎧ ⎞⎤ ⎡⎡ ⎛ ⎪ ⎪ r r   ⎨  ⎟⎥ ⎢⎢ α,ψ  α+1,ψ ⎜  f j (x0 )⎟⎥ ⎢⎢ sup  sup Ia+ ⎜ C Dx0 − f i  ⎠⎦ ⎣ ⎝ ⎣ ⎪ ∞,[a,x0 ] x0 ∈[a,b] x0 ∈[a,b] ⎪ j=1 ⎩ i=1 j =i

⎞⎤⎤⎫ ⎪ ⎪   ⎢ ⎟⎥⎥⎬ ⎜    α,ψ α+1,ψ C ⎢   ⎟ ⎥ ⎥ ⎜ + ⎣ sup  Dx0 + f i  f j (x0 ) ⎠⎦⎦ . sup Ib− ⎝ ⎪ ∞,[x0 ,b] x0 ∈[a,b] x0 ∈[a,b] ⎪ j=1 ⎭ ⎡

⎛

r 

(2.42)

j =i

Proof Here 0 < α ≤ 1, i.e. n = 1. Then (2.30) is valid without any initial conditions. Clearly  ( f 1 , . . . , fr ) ∈ C ([a, b]). Thus, by integrating (2.30) against ψ we obtain 

ψ

b

 ( f 1 , . . . , fr ) =

 ( f 1 , . . . , fr ) (x0 ) dψ (x0 ) =

a

 r (ψ (b) − ψ (a))

b

a

⎡ r i=1



⎢ ⎢ ⎣

b

a



 r 



f k (x) dψ (x) −

k=1

⎞⎤ ⎛ ⎞ ⎛

 b  r ⎟⎥ ⎜ ⎟ ⎜ ⎟⎥ ⎜ f i (x) dψ (x) ⎜ f j (x)⎟ ⎝ a ⎝ ⎠ dψ (x)⎠⎦ = j=1 j =i

⎧ ⎞ ⎡⎡ ⎛ ⎪ ⎪  r r b ⎨ ⎟ ⎢⎢ x0 ⎜ 1 ⎢⎢ ⎜ f j (x)⎟ ⎠ ⎣⎣ a ⎝  (α) a ⎪ ⎪ j=1 ⎩ i=1 

j =i



x0

α−1 C



ψ (t) (ψ (t) − ψ (x))

x

⎡  ⎢ ⎢ ⎣

⎛ b x0

⎞

r ⎜ ⎟ ⎜ f j (x)⎟ ⎝ ⎠ j=1 j =i



x x0

α,ψ Dx0 − f i

ψ  (t) (ψ (x) − ψ (t))α−1





(t) dt dψ (x) +

(2.43)

⎤⎤⎫ ⎪ ⎪ ⎥⎥⎬ C α,ψ ⎥ dψ (x0 ) . Dx0 + f i (t) dt dψ (x)⎥ ⎦⎦⎪ ⎪ ⎭

Hence it holds ⎧ ⎞ ⎡⎡ ⎛ ⎪ ⎪  b ⎨  r r   ⎟  ψ ⎢⎢ x0 ⎜  ( f 1 , . . . , fr ) ≤ 1 ⎢⎢ ⎜  f j (x)⎟ ⎠ ⎣ ⎝ ⎣  (α) a ⎪ a ⎪ j=1 ⎩ i=1 j =i

2.3 Main Results

 x0 x



25

⎞ ⎛ ⎡   r   ⎢ b ⎜  ⎟ ⎜   ⎢ α,ψ  f j (x)⎟ ψ  (t) (ψ (t) − ψ (x))α−1 C Dx0 − f i (t) dt dψ (x) + ⎢ ⎟ ⎜ ⎠ ⎣ x0 ⎝

j=1 j =i

x

(2.44) )  

 α,ψ dψ (x0 ) =: (∗∗) .  Dx0 + f i (t) dt dψ (x)

α−1 C



ψ (t) (ψ (x) − ψ (t))

x0

Using (2.21) we derive (∗∗) ≤ (ψ (b) − ψ (a)) ⎧ ⎞⎤ ⎡⎡ ⎛ ⎪ ⎪ r r   ⎨   ⎟⎥ ⎢⎢ α+1,ψ ⎜ C α,ψ  ⎢⎢ sup  ⎜  f j (x0 )⎟⎥ D f sup I   i − x a+ 0 ⎠⎦ ⎣⎣x ∈[a,b] ⎝ ⎪ ∞,[a,x ] 0 x 0 ∈[a,b] ⎪ 0 j=1 ⎩ i=1 j =i

⎞⎤⎤⎫ ⎪ ⎪ r    ⎢  ⎟⎥⎥⎬ C α,ψ  α+1,ψ ⎜ ⎢   ⎟ ⎥ ⎥ ⎜ + ⎣ sup  Dx0 + f i  f j (x0 ) ⎠⎦⎦ , sup Ib− ⎝ ⎪ ∞,[x0 ,b] x0 ∈[a,b] x0 ∈[a,b] ⎪ j=1 ⎭ ⎡

⎛

(2.45)

j =i



proving the theorem. We make

Remark 2.11 Let α > 0, [a, b] ⊂ R, f ∈ C ([a, b]) and ψ ∈ C 1 ([a, b]) an increasing function such that ψ  (x) = 0, for all x ∈ [a, b]. Let x0 ∈ [a, b]. We observe the following   (2.2)  α,ψ  Ia+ f (x0 ) ≤  f ∞,[a,x0 ]  (α) That is



x0

1  (α)



x0

ψ  (t) (ψ (x0 ) − ψ (t))α−1 | f (t)| dt ≤

a

(ψ (x0 ) − ψ (t))α−1 dψ (t) =

a

 f ∞,[a,x0 ] (ψ (x0 ) − ψ (a))α .  (α + 1) (2.46)

  f  α,ψ  ∞,[a,x0 ] (ψ (x0 ) − ψ (a))α .  Ia+ f (x0 ) ≤  (α + 1)

Similarly, we obtain

(2.47)

2 Caputo ψ-Fractional Ostrowski and Grüss …

26

  (2.3)  α,ψ  Ib− f (x0 ) ≤  f ∞,[x0 ,b]  (α)



b

1  (α)



b

ψ  (t) (ψ (t) − ψ (x0 ))α−1 | f (t)| dt ≤

x0

(ψ (t) − ψ (x0 ))α−1 dψ (t) =

x0

That is

 f ∞,[x0 ,b] (ψ (b) − ψ (x0 ))α .  (α + 1) (2.48)

 f    α,ψ ∞,[x0 ,b] (ψ (b) − ψ (x0 ))α . Ib− f (x0 ) ≤  (α + 1)

(2.49)

We make Remark 2.12 Let α ≥ 1, the rest as in Remark 2.11. We observe that   (2.2) 1  x0  α,ψ  ψ  (t) (ψ (x0 ) − ψ (t))α−1 | f (t)| dt ≤  Ia+ f (x0 ) ≤  (α) a (ψ (x0 ) − ψ (a))α−1  (α) That is



x0

| f (t)| dψ (t) =

a

(ψ (x0 ) − ψ (a))α−1  f  L 1 ([a,x0 ],ψ) .  (α) (2.50)

  (ψ (x ) − ψ (a))α−1 0  α,ψ   f  L 1 ([a,x0 ],ψ) . Ia+ f (x0 ) ≤  (α)

(2.51)

Similarly, we get   (2.3)  α,ψ  Ib− f (x0 ) ≤ (ψ (b) − ψ (x0 ))α−1  (α) That is



1  (α) b



b

ψ  (t) (ψ (t) − ψ (x0 ))α−1 | f (t)| dt ≤

x0

| f (t)| dψ (t) =

x0

(ψ (b) − ψ (x0 ))α−1  f  L 1 ([x0 ,b],ψ) .  (α) (2.52)

  (ψ (b) − ψ (x ))α−1 0  α,ψ   f  L 1 ([x0 ,b],ψ) . Ib− f (x0 ) ≤  (α)

Next, we simplify our main theorems: Proposition 2.13 All as in Theorem 2.6. Then | ( f 1 , . . . , fr ) (x0 )| ≤

1  (α + 2)

(2.53)

2.3 Main Results

27

      r     f j  ∞,[a,x0 ]   j=1  j =i 

⎡⎡ r i=1

⎤

 ⎢⎢ C α,ψ  ⎢⎢ D f  x0 − i  ⎣⎣

⎥ (ψ (x0 ) − ψ (a))α+1 ⎥ ⎦

∞,[a,x0 ]

       r   fj   ∞,[x0 ,b]    j=1  j =i

⎤⎤

⎡  ⎢ C α,ψ  +⎢ D f   i + x 0 ⎣

⎥⎥ ⎥ (ψ (b) − ψ (x0 ))α+1 ⎥ ⎦⎦ .

(2.54)

∞,[x0 ,b]

If 0 < α ≤ 1, then (2.54) is valid without any initial conditions. 

Proof By (2.21), (2.47) and (2.49). Next comes Proposition 2.14 All as in Theorem 2.6 with α ≥ 1. Then | ( f 1 , . . . , fr ) (x0 )| ≤       r r     ⎢⎢   α,ψ  ⎢⎢C Dx − f i  fj 0  ⎣⎣ L 1 ([a,x0 ],ψ)   j=1  i=1  j =i 

1  (α)

⎡⎡

⎡  ⎢ C α,ψ  +⎢ D f  x0 + i  ⎣

      r     f j  L 1 ([x0 ,b],ψ)   j=1  j =i 

⎤ ⎥ (ψ (x0 ) − ψ (a))α−1 ⎥ ⎦ L 1 ([a,x0 ],ψ)

⎤⎤ ⎥⎥ ⎥ (ψ (b) − ψ (x0 ))α−1 ⎥ ⎦⎦ .

(2.55)

L 1 ([x0 ,b],ψ)



Proof By (2.33), (2.51) and (2.53). Next follows Proposition 2.15 All as in Theorem 2.6 with p, q > 1 : | ( f 1 , . . . , fr ) (x0 )| ≤

 (α) α +

      r r     ⎢⎢ C α,ψ   ⎢⎢ fj  ⎣⎣ Dx0 − f i  L q ([a,x0 ],ψ)   j=1  i=1  j =i 

1 p

!

⎡⎡

∞,[a,x0 ]

1 p

+

1 q

= 1, α ≥ 1. Then

1 1

( p (α − 1) + 1) p ⎤ 1 ⎥ (ψ (x0 ) − ψ (a))α+ p ⎥ ⎦

2 Caputo ψ-Fractional Ostrowski and Grüss …

28

⎡  ⎢ C α,ψ  +⎢ D f  x0 + i  ⎣

      r     f j  L q ([x0 ,b],ψ)   j=1  j =i 

⎤⎤ 1 ⎥⎥ ⎥ (ψ (b) − ψ (x0 ))α+ p ⎥ ⎦⎦ .

(2.56)

∞,[x0 ,b]



Proof By (2.36), (2.47) and (2.49). We continue with Proposition 2.16 All as in Theorem 2.10. Then

⎧ ⎪  ⎪ r ⎨ ⎪ ⎪ ⎩ i=1

α+2  ψ   ( f 1 , . . . , fr ) ≤ (ψ (b) − ψ (a))  (α + 2)

      r        C α,ψ  C α,ψ   + sup  Dx0 + f i  fj sup  Dx0 − f i    ∞,[a,x0 ] ∞,[x0 ,b]  x0 ∈[a,b] x0 ∈[a,b]    j=1 j =i

⎫ ⎪ ⎪ ⎬

∞,[a,b]

⎪ ⎪ ⎭

.

(2.57) 

Proof By (2.42), (2.47) and (2.49). Next we make some applications of our main results. We need Remark 2.17 We have that (r = 2)   ( f 1 , f 2 ) (x0 ) = 2

b

f 1 (x) f 2 (x) dψ (x) −

(2.58)

a



b

f 1 (x0 )



b

f 2 (x) dψ (x) − f 2 (x0 )

a

f 1 (x) dψ (x) ,

a

and (r = 3)  ( f 1 , f 2 , f 3 ) (x0 ) = 3

 − f 2 (x0 ) a

etc.

b

 b a

f 1 (x) f 2 (x) f 3 (x) dψ (x) − f 1 (x0 )

 f 1 (x) f 3 (x) dψ (x) − f 3 (x0 ) a

b

 b a

f 2 (x) f 3 (x) dψ (x)

f 1 (x) f 2 (x) dψ (x) ,

(2.59)

2.3 Main Results

29

Furthermore we derive (r = 2) 

ψ



b

 ( f 1 , f 2 ) = 2 (ψ (b) − ψ (a))

f 1 (x) f 2 (x) dψ (x) −

a





b



b

f 1 (x) dψ (x)

f 2 (x) dψ (x)

a

,

(2.60)

a

and (r = 3) ψ ( f 1 , f 2 , f 3 ) = 3 (ψ (b) − ψ (a))



b

f 1 (x) f 2 (x) f 3 (x) dψ (x) −

a





b

b

f 1 (x) dψ (x) a



f 2 (x) f 3 (x) dψ (x) −

a



b

b

f 2 (x) dψ (x) a



f 1 (x) f 3 (x) dψ (x) −

a



b

f 3 (x) dψ (x) a

b

f 1 (x) f 2 (x) dψ (x) ,

(2.61)

a

etc. We give the special cases of fractional Ostrowski type inequalities. Proposition 2.18 Let α > 0, n ∈ N : n = α, [a, b] ⊂ R and f 1 , f 2 ∈ C n ([a, b]). Let x0 ∈ [a, b] and assume that f 1e[k]x (x0 ) = f 2e[k]x (x0 ) = 0, for k = 1, . . . , n − 1. Then    b  b  b   x x x ≤ 2 f d x − f f d x − f f d x f e e e (x) (x) (x ) (x) (x ) (x) 1 2 1 0 2 2 0 1   a

a

⎡⎡ 1  (α + 2)

2 i=1

⎢⎢ C α,e x ⎢⎢ ⎣⎣ Dx0 −

a

      2        fi  f j  ∞,[a,x0 ]   j=1  j =i 

⎤  x0 α+1 ⎥ ⎥ e − ea ⎦

∞,[a,x0 ]

⎡ ⎢ C α,ex +⎢ ⎣ Dx0 +

      2       fi  fj   ∞,[x0 ,b]    j=1  j =i

⎤⎤  b α+1 ⎥⎥ ⎥⎥ . e − e x0 ⎦⎦

∞,[x0 ,b]

If 0 < α ≤ 1, then (2.62) is valid without any initial conditions.

(2.62)

2 Caputo ψ-Fractional Ostrowski and Grüss …

30

Proof Case of ψ (x) = e x , apply Proposition 2.13 for r = 2.



We continue with Proposition 2.19 Let α > 0, n ∈ N : n = α, [a, b] ⊂ (0, +∞) and f 1 , f 2 , f 3 ∈ C n ([a, b]). Let x0 ∈ [a, b] and assume that f i[k] ln x (x 0 ) = 0, for k = 1, . . . , n − 1; i = 1, 2, 3. Then    3 

f 1 (x) f 2 (x) f 3 (x) d x − f 1 (x0 ) x

b

a



b

f 2 (x0 ) a

f 1 (x) f 3 (x) d x − f 3 (x0 ) x

⎡⎡ 1  (α + 2)

3 i=1

⎢⎢ C α,ln x ⎢⎢ ⎣⎣ Dx0 −



b

a



f 2 (x) f 3 (x) d x− x  f 1 (x) f 2 (x)  dx ≤ x

b

a

      3       fi  fj   ∞,[a,x0 ]    j=1  j =i

⎤ x0 !!α+1 ⎥ ⎥ ⎦ a

ln

∞,[a,x0 ]

⎡ ⎢ C α,ln x +⎢ ⎣ Dx0 +

      3       fi  fj   ∞,[x0 ,b]    j=1  j =i

⎤⎤

α+1 ⎥⎥ b ⎥⎥ . ln ⎦⎦ x0

(2.63)

∞,[x0 ,b]

If 0 < α ≤ 1, then (2.63) is valid without any initial conditions. Proof Case of ψ (x) = ln x, apply Proposition 2.13 for r = 3.



Next we present the special cases of fractional Grüss type inequality: 1 Proposition 2.20 Let 0 < α ≤ 1, [a, b] ⊂ R   and fx1 , f 2 ∈ C ([a, b]). Assume that C α,ex  C α,e  sup  Dx0 − f i  < ∞, and sup  Dx0 + f i  < ∞, i = 1, 2. ∞,[a,x0 ]

x0 ∈[a,b]

∞,[x0 ,b]

x0 ∈[a,b]

Then    b  a  2 e −e

b





f 1 (x) f 2 (x) e d x −

a



b

f 1 (x) e d x

x

x

a

b

  f 2 (x) e d x  ≤ x

a

 b α+2 e − ea  (α + 2) ⎧ ⎪  ⎪ 2 ⎨ ⎪ ⎪ ⎩ i=1

  x   f sup C Dxα,e  i − 0

x0 ∈[a,b]

∞,[a,x0 ]

 x  + sup C Dxα,e 0+ x0 ∈[a,b]

      2       fi  fj   ∞,[x0 ,b]     j=1 j =i

⎫ ⎪ ⎪ ⎬

∞,[a,b]

⎪ ⎪ ⎭

.

(2.64)

2.3 Main Results

31

Proof Apply Proposition 2.16, for ψ (x) = e x , r = 2.



We finish with another ψ-fractional Grüss type inequality: Proposition 2.21 Assume that sup

1 Let  0 < α ≤1, [a, b] ⊂ (0, ∞) and  f i ∈ C ([a,  b]), i = 1, 2, 3. C α,ln x  C α,ln x  < ∞, sup  Dx0 + f i  < ∞, i =  Dx0 − f i  ∞,[a,x0 ]

x0 ∈[a,b]

∞,[x0 ,b]

x0 ∈[a,b]

1, 2, 3. Then  

 b

 b

 b f (x) f (x) f (x) f 1 (x) f 2 (x) f 3 (x) b 3 1 2 3  dx − dx dx − ln  a x x x a a a  a

b

f 2 (x) dx x

 a

b

 b

 b

  f 1 (x) f 3 (x) f 3 (x) f 1 (x) f 2 (x) dx − dx d x  (2.65) x x x a a

  b α+2 * 3    ln a  x  ≤ fi  + sup C Dxα,ln 0− ∞,[a,x 0 ]  (α + 2) x 0 ∈[a,b] i=1

  x sup C Dxα,ln 0+

x 0 ∈[a,b]

      3       fi  f j ∞,[x 0 ,b]   j=1   j =1 

Proof By Proposition 2.16, for ψ (x) = ln x, r = 3.

⎫ ⎪ ⎪ ⎪ ⎬ ∞,[a,b]

⎪ ⎪ ⎪ ⎭

.



References 1. Almeida, R.: A Caputo fractional derivative of a function with respect to another function. Commun. Nonlinear Sci. Numer. Simul. 44, 460–481 (2017) 2. Anastassiou, G.A.: Ostrowski type inequalities. Proc. AMS 123, 3775–3781 (1995) 3. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 4. Anastassiou, G.: Caputo ψ-fractional Ostrowski and Grüss inequalities for several functions. J. Comput. Anal. Appl. 29(6), 1097–1114 (2021) ˇ 5. Cebyšev, P.L.: Sur les expressions approximatives des intégrales définies par les aures prises entre les mêmes limites. Proc. Math. Soc. Charkov 2, 93–98 (1882) 6. Ostrowski, A.: Über die Absolutabweichung einer differtentiebaren Funktion von ihrem Integralmittelwert. Comment. Math. Helv. 10, 226–227 (1938)

Chapter 3

Weighted Caputo Fractional Iyengar Type Inequalities

Here we present weighted fractional Iyengar type inequalities with respect to L p norms, with 1 ≤ p ≤ ∞. Our employed fractional calculus is of Caputo type defined with respect to another function. Our results provide quantitative estimates for the approximation of the Lebesgue-Stieljes integral of a function, based on its values over a finite set of points including at the endpoints of its interval of definition. Our method relies on the right and left generalized fractional Taylor’s formulae. The iterated generalized fractional derivatives case is also studied. We give applications at the end. See also [3].

3.1 Introduction 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

  M (b − a) 2 ( f (b) − f (a))2 1 . f (x) d x − (b − a) ( f (a) + f (b)) ≤ − 2 4 4M (3.1)

We need Definition 3.2 ([1]) Let α > 0, α = n, · the ceiling of the number. Here g ∈ AC ([a, b]) (absolutely continuous functions) and strictly increasing. We assume that  (n) f ◦ g −1 ◦ g ∈ L ∞ ([a, b]). We define the left generalized g-fractional derivative of f of order α as follows:

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_3

33

34



3 Weighted Caputo Fractional Iyengar Type Inequalities

 α f (x) := Da+;g

1  (n − α)



x

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

a

(3.2)

x ≥ a. α f ∈ C ([a, b]). If α ∈ / N, by [2], pp. 360–361, we have that Da+;g We see that     α  (n) n−α Ia+;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , x ≥ a. We set n f (x) := Da+;g

When g = id, then



f ◦ g −1

(n)

 ◦ g (x) ,

(3.3)

(3.4)

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

(3.5)

α α α f = Da+;id f = D∗a f, Da+;g

(3.6)

the usual left Caputo fractional derivative. We mention the following g-left fractional generalized Taylor’s formula: Theorem 3.3 ([1]) Let g be strictly increasing function and g ∈ AC ([a, b]).  (n−1)   ∈ AC ([g (a) , We assume that f ◦ g −1 ∈ AC n ([g (a) , g (b)]), i.e. f ◦ g −1  (n) g (b)]) , where N n = α, α > 0. Also we assume that f ◦ g −1 ◦g ∈ L ∞ ([a, b]). Then (k) n−1   f ◦ g −1 (g (a)) f (x) = f (a) + (g (x) − g (a))k + k! k=1 1  (α)

 a

x

  α f (t) dt, ∀ x ∈ [a, b] . (g (x) − g (t))α−1 g  (t) Da+;g

(3.7)

Calling Rn (a, x) the remainder of (3.7), we get that Rn (a, x) =

 g(x)    1 α f ◦ g −1 (z) dz, ∀ x ∈ [a, b] . (3.8) (g (x) − z)α−1 Da+;g  (α) g(a)

We need Definition 3.4 ([1]) Here g ∈ AC ([a, b]) and strictly increasing. We assume that  (n) f ◦ g −1 ◦ g ∈ L ∞ ([a, b]), where N n = α, α > 0. We define the right generalized g-fractional derivative of f of order α as follows: 

α Db−;g



(−1)n f (x) :=  (n − α)



b x

(n)  (g (t) − g (x))n−α−1 g  (t) f ◦ g −1 (g (t)) dt, (3.9)

3.1 Introduction

35

all x ∈ [a, b] .   α f ∈ C ([a, b]). If α ∈ / N, by [2], p. 378, we get that Db−;g We see that    (n)   α n−α ◦ g (x) = Db−;g f (x) , a ≤ x ≤ b. Ib−;g (−1)n f ◦ g −1 We set n f (x) = (−1)n Db−;g



f ◦ g −1

(n)

 ◦ g (x) ,

(3.10)

(3.11)

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

When g = id, then

α α α f (x) = Db−;id f (x) = Db− f, Db−;g

(3.12)

the usual right Caputo fractional derivative. We mention the g-right generalized fractional Taylor’s formula: Theorem 3.5 ([1]) Let  g be strictly increasing function and g ∈ AC ([a, b]). We assume that f ◦ g −1 ∈ AC n ([g (a) , g (b)]), where N n = α, α > 0. Also we (n)  assume that f ◦ g −1 ◦ g ∈ L ∞ ([a, b]). Then (k) n−1   f ◦ g −1 (g (b)) f (x) = f (b) + (g (x) − g (b))k + k! k=1 1  (α)



b x

  α f (t) dt, all a ≤ x ≤ b. (g (t) − g (x))α−1 g  (t) Db−;g

(3.13)

Calling Rn (b, x) the remainder in (3.13), we get that 1 Rn (b, x) =  (α)



g(b) g(x)

(z − g (x))α−1



  α Db−;g f ◦ g −1 (z) dz, ∀ x ∈ [a, b] . (3.14)

Denote by nα α α α Db−;g := Db−;g Db−;g . . . Db−;g (n-times), n ∈ N.

(3.15)

We mention the following g-right generalized modified Taylor’s formula: kα Theorem 3.6 ([1]) Suppose that Fk := Db−;g f , for k = 0, 1, . . . , n + 1, fulfill: Fk ◦   g −1 ∈ AC ([c, d]) , where c = g (a), d = g (b), and Fk ◦ g −1 ◦ g ∈ L ∞ ([a, b]) , where 0 < α ≤ 1. Then

36

3 Weighted Caputo Fractional Iyengar Type Inequalities

f (x) =

n  (g (b) − g (x))iα 

 (iα + 1)

i=0

1  ((n + 1) α)



b x

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

n  (g (b) − g (x))iα  i=0

 iα Db−;g f (b) +

 (iα + 1)

 

iα Db−;g f (b) +

 (n+1)α Db−;g f (ψx )

 ((n + 1) α + 1)

(3.16)

(g (b) − g (x))(n+1)α , (3.17)

where ψx ∈ [x, b], any x ∈ [a, b] . Denote by nα α α α Da+;g := Da+;g Da+;g . . . Da+;g (n-times), n ∈ N.

(3.18)

We mention the following g-left generalized modified Taylor’s formula: kα Theorem 3.7 ([1]) Suppose that Fk := Da+;g f , for k = 0, 1, . . . , n + 1, fulfill: Fk ◦   −1 g ∈ AC ([c, d]), where c = g (a), d = g (b), and Fk ◦ g −1 ◦ g ∈ L ∞ ([a, b]) , where 0 < α ≤ 1. Then

f (x) =

n  (g (x) − g (a))iα 

 (iα + 1)

i=0

1  ((n + 1) α)



x

a

n  (g (x) − g (a))iα  i=0

 (iα + 1)

 iα Da+;g f (a) +

(3.19)

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

iα Da+;g f (a) +

 (n+1)α Da+;g f (ψx )

 ((n + 1) α + 1)

(g (x) − g (a))(n+1)α , (3.20)

where ψx ∈ [a, x], any x ∈ [a, b] . Next we present generalized fractional Iyengar type inequalities.

3.2 Main Results We present the following Caputo type generalized g-fractional Iyengar type inequality: Theorem 3.8 Let g be strictly increasing function and g ∈ AC ([a, b]). We assume  that f ◦ g −1 ∈ AC n ([g (a) , g (b)]), where N n = α, α > 0. We also assume  (n) that f ◦ g −1 ◦ g ∈ L ∞ ([a, b]) (clearly here it is f ∈ C ([a, b])). Then

3.2 Main Results

37

(i)  n−1  b   (k) 1  f ◦ g −1 f dg − (x) (x) (g (a)) (g (t) − g (a))k+1   a (k + 1)! k=0

  (k)  + (−1)k f ◦ g −1 (g (b)) (g (b) − g (t))k+1  ≤  α max Da+;g f

L ∞ ([a,b])

α , Db−;g f

L ∞ ([a,b])

 (α + 2)   (g (t) − g (a))α+1 + (g (b) − g (t))α+1 ,

(3.21)

∀ t ∈ [a, b] , , the right hand side of (3.21) is minimized, and we get: (ii) at g (t) = g(a)+g(b) 2  n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

 α max Da+;g f

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤

L ∞ ([a,b])

α , Db−;g f

L ∞ ([a,b])

 (α + 2)

(g (b) − g (a))α+1 , 2α

(3.22)

 (k) (k)  (iii) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 0, 1, . . . , n − 1, we obtain   b   ≤  f dg (x) (x)   a

 α max Da+;g f L

∞

α , Db−;g f L ([a,b])

 (g (b) − g (a))α+1 ∞ ([a,b])

 (α + 2) 2α

,

(3.23)

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds    n−1  b  g (b) − g (a) k+1 1  f dg − (x) (x)   a N (k + 1)! k=0



  (k)  (k)  j k+1 f ◦ g −1 (g (a)) + (−1)k (N − j)k+1 f ◦ g −1 (g (b))  ≤

38

3 Weighted Caputo Fractional Iyengar Type Inequalities

 α max Da+;g f

L ∞ ([a,b])

α , Db−;g f

L ∞ ([a,b])

 (α + 2) 

g (b) − g (a) N

α+1



 j α+1 + (N − j)α+1 ,

(3.24)

(k) (k)   (v) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 1, . . . , n − 1, from (3.24) we obtain    

b

 f (x) dg (x) −

a

   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

 α max Da+;g f

L ∞ ([a,b])

α , Db−;g f

L ∞ ([a,b])

 (α + 2) 

g (b) − g (a) N

α+1



 j α+1 + (N − j)α+1 ,

(3.25)

j = 0, 1, 2, . . . , N , (vi) when N = 2, j = 1, (3.25) turns to    

b

 f (x) dg (x) −

a

 α max Da+;g f

L ∞ ([a,b])

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

α , Db−;g f

 (α + 2)

L ∞ ([a,b])

(g (b) − g (a))α+1 . 2α

(3.26)

(vii) when 0 < α ≤ 1, inequality (3.26) is again valid without any boundary conditions. Proof We have by (3.7) that f (x) − 1  (α)



(k) n−1   f ◦ g −1 (g (a)) (g (x) − g (a))k = k! k=0 x

a

∀ x ∈ [a, b] . Also by (3.13) we get

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

(3.27)

3.2 Main Results

39

(k) n−1   f ◦ g −1 (g (b)) f (x) − (g (x) − g (b))k = k! k=0 1  (α)



b x

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

(3.28)

∀ x ∈ [a, b] . By (3.27) we derive (by [6], p. 107)   (k) n−1     f ◦ g −1 (g (a))  k (g (x) − g (a))  ≤  f (x) −   k! k=0

α Da+;g f

L ∞ ([a,b])

 (α + 1)

(g (x) − g (a))α ,

(3.29)

and by (3.28) we obtain   (k) n−1     f ◦ g −1 (g (b))  k (g (x) − g (b))  ≤  f (x) −   k! k=0

α Db−;g f

L ∞ ([a,b])

 (α + 1) ∀ x ∈ [a, b] . Call ϕ1 := and ϕ2 :=

(g (b) − g (x))α ,

α Da+;g f

L ∞ ([a,b])

 (α + 1) α Db−;g f

L ∞ ([a,b])

 (α + 1)

(3.30)

,

(3.31)

.

(3.32)

Set ϕ := max {ϕ1 , ϕ2 } .

(3.33)

That is   (k) n−1     f ◦ g −1 (g (a))  k (g (x) − g (a))  ≤ ϕ (g (x) − g (a))α ,  f (x) −   k! k=0

40

3 Weighted Caputo Fractional Iyengar Type Inequalities

and   (k) n−1     f ◦ g −1 (g (b))  k (g (x) − g (b))  ≤ ϕ (g (b) − g (x))α , (3.34)  f (x) −   k! k=0

∀ x ∈ [a, b] . Equivalently, we have (k) n−1   f ◦ g −1 (g (a)) (g (x) − g (a))k − ϕ (g (x) − g (a))α ≤ k! k=0

(3.35)

(k) n−1   f ◦ g −1 (g (a)) f (x) ≤ (g (x) − g (a))k + ϕ (g (x) − g (a))α , k! k=0 and (k) n−1   f ◦ g −1 (g (b)) (g (x) − g (b))k − ϕ (g (b) − g (x))α ≤ k! k=0

(3.36)

(k) n−1   f ◦ g −1 (g (b)) f (x) ≤ (g (x) − g (b))k + ϕ (g (b) − g (x))α , k! k=0 ∀ x ∈ [a, b] . Let any t ∈ [a, b], then by integration against g over [a, t] and [t, b], respectively, we obtain (k) n−1   f ◦ g −1 (g (a)) ϕ (g (t) − g (a))α+1 (g (t) − g (a))k+1 − + 1)! + 1) (k (α k=0  ≤

t

f (x) dg (x) ≤

a

(k) n−1   f ◦ g −1 (g (a)) ϕ (g (t) − g (a))α+1 , (3.37) (g (t) − g (a))k+1 + + 1)! + 1) (k (α k=0 and (k) n−1   f ◦ g −1 (g (b)) ϕ − (g (b) − g (t))α+1 (g (t) − g (b))k+1 − + 1)! + 1) (k (α k=0

3.2 Main Results

41



b

≤

f (x) dg (x) ≤

t

(k) n−1   f ◦ g −1 (g (b)) ϕ − (g (b) − g (t))α+1 . (g (t) − g (b))k+1 + + 1)! + 1) (k (α k=0 (3.38) Adding (3.37) and (3.38), we obtain  n−1  k=0

 (k) 1 f ◦ g −1 (g (a)) (g (t) − g (a))k+1 − (k + 1)! 

f ◦ g −1

(k)

(g (b)) (g (t) − g (b))k+1



−

  ϕ (g (t) − g (a))α+1 + (g (b) − g (t))α+1 (α + 1)  ≤

b

f (x) dg (x) ≤

a

 n−1  k=0

 (k) 1 f ◦ g −1 (g (a)) (g (t) − g (a))k+1 − (k + 1)! 

f ◦ g −1

(k)

(g (b)) (g (t) − g (b))k+1



+

  ϕ (g (t) − g (a))α+1 + (g (b) − g (t))α+1 , (α + 1)

(3.39)

∀ t ∈ [a, b] . Consequently we derive:  n−1  b   (k) 1  f ◦ g −1 f (x) dg (x) − (g (a)) (g (t) − g (a))k+1   a (k + 1)! k=0

+ (−1)k



f ◦ g −1

(k)

  (g (b)) (g (b) − g (t))k+1  ≤

  ϕ (g (t) − g (a))α+1 + (g (b) − g (t))α+1 , (α + 1) ∀ t ∈ [a, b] . Let us consider

(3.40)

42

3 Weighted Caputo Fractional Iyengar Type Inequalities

θ (z) := (z − g (a))α+1 + (g (b) − z)α+1 , ∀ z ∈ [g (a) , g (b)] . That is θ (g (t)) = (g (t) − g (a))α+1 + (g (b) − g (t))α+1 , ∀ t ∈ [a, b] . We have that   θ (z) = (α + 1) (z − g (a))α − (g (b) − z)α = 0, giving (z − g (a))α = (g (b) − z)α and z − g (a) = g (b) − z, that is z = g(a)+g(b) 2 the only critical number of that θ (g (a)) = θ (g (b)) = (g (b) −  θ. We have α+1 = (g(b)−g(a)) , which is the minimum of θ over g (a))α+1 , and θ g(a)+g(b) 2 2α [g (a) , g (b)] . , Consequently the right hand side of (3.40) is minimized when g (t) = g(a)+g(b) 2 with value

ϕ (g(b)−g(a))α+1 . 2α (α+1)   −1 (k)

(k)  Assuming f ◦ g (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 0, 1, . . . , n − 1, then we obtain that    

a

b

  f (x) dg (x) ≤

ϕ (g (b) − g (a))α+1 , 2α (α + 1)

(3.41)

which is a sharp inequality. , then (3.40) becomes When g (t) = g(a)+g(b) 2  n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤

ϕ (g (b) − g (a))α+1 . 2α (α + 1)

(3.42)

    , that is Next let N ∈ N, j = 0, 1, 2, . . . , N and g t j = g (a) + j g(b)−g(a) N g (t0 ) = g (a), g (t1 ) = g (a) + (g(b)−g(a)) , . . . , g (t N ) = g (b) . N Hence it holds         g (b) − g (a) g (b) − g (a) , g (b) − g t j = (N − j) , g t j − g (a) = j N N (3.43) j = 0, 1, 2, . . . , N .

3.2 Main Results

43

We notice

α+1   α+1    + g (b) − g t j = g t j − g (a) 

g (b) − g (a) N

α+1



 j α+1 + (N − j)α+1 ,

(3.44)

j = 0, 1, 2, . . . , N , and (for k = 0, 1, . . . , n − 1) 

f ◦ g −1

(k)

k+1    + (g (a)) g t j − g (a)

 (k)   k+1

= (−1)k f ◦ g −1 (g (b)) g (b) − g t j 

g (b) − g (a) N

k+1 

f ◦ g −1

(k)

(g (a)) j k+1 +

 (k) (−1)k f ◦ g −1 (g (b)) (N − j)k+1 ,

(3.45)

j = 0, 1, 2, . . . , N . By (3.40) we get    n−1  b  g (b) − g (a) k+1 1  f (x) dg (x) −   a N (k + 1)! k=0



f ◦ g −1

(k)



  (k)  (g (a)) j k+1 + (−1)k f ◦ g −1 (g (b)) (N − j)k+1  ≤

ϕ α+1



g (b) − g (a) N

α+1



 j α+1 + (N − j)α+1 ,

(3.46)

j = 0, 1, 2, . . . , N .  (k) (k)  If f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, k = 1, . . . , n − 1, then (3.46) becomes  b      g (b) − g (a)  ≤ f dg − j f + − j) f [ (x) (x) (a) (N (b)]   N a 

ϕ α+1



g (b) − g (a) N

α+1



 j α+1 + (N − j)α+1 ,

j = 0, 1, 2, . . . , N . When N = 2 and j = 1, then (3.47) becomes

(3.47)

44

3 Weighted Caputo Fractional Iyengar Type Inequalities

   

b

 f (x) dg (x) −

a



   g (b) − g (a) ( f (a) + f (b)) ≤ 2

   ϕ ϕ (g (b) − g (a))α+1 (g (b) − g (a))α+1 2 = . α+1 2α+1 α+1 2α

(3.48)

Let 0 < α ≤ 1, then n = α = 1. In that case, without any boundary conditions, we derive from (3.48) again that    

b

 f (x) dg (x) −

a



ϕ α+1

   g (b) − g (a) ( f (a) + f (b)) ≤ 2



(g (b) − g (a))α+1 . 2α

(3.49) 

We have proved theorem in all possible cases. Next we give modified g-fractional Iyengar type inequalities:

Theorem 3.9 Let g be strictly increasing function and g ∈ AC ([a, b]), and f ∈ kα C ([a, b]). Let 0 < α ≤ 1, and Fk := Da+;g f , for k = 0, 1, . . . , n + 1; n ∈ N. We   −1 assume that Fk ◦ g ∈ AC ([g (a) , g (b)]) and Fk ◦ g −1 ◦ g ∈ L ∞ ([a, b]). Also kα let Fk := Db−;g f , for k = 0, 1, . . . , n + 1, they fulfill Fk ◦ g −1 ∈ AC ([g (a) , g (b)])   −1  and Fk ◦ g ◦ g ∈ L ∞ ([a, b]) . Then (i)   n  b   iα  1  Da+;g f (a) (g (t) − g (a))iα+1 f (x) dg (x) −   a  (iα + 2) i=0

  iα  + Db−;g f (b) (g (b) − g (t))iα+1  ≤  (n+1)α max Da+;g f

∞,[a,b]

(n+1)α , Db−;g f

∞,[a,b]

 ((n + 1) α + 2)   (g (t) − g (a))(n+1)α+1 + (g (b) − g (t))(n+1)α+1 ,

(3.50)

∀ t ∈ [a, b] , , the right hand side of (3.50) is minimized, and we get: (ii) at g (t) = g(a)+g(b) 2   n  b  1 (g (b) − g (a))iα+1  f (x) dg (x) −   a  (iα + 2) 2iα+1 i=0

3.2 Main Results

45

  (n+1)α max Da+;g f

  iα   iα Da+;g f (a) + Db−;g f (b)  ≤ (n+1)α , Db−;g f

(g (b) − g (a))(n+1)α+1 ,  ((n + 1) α + 2) 2(n+1)α (3.51)     iα iα f (a) = Db−;g f (b) = 0, for i = 0, 1, . . . , n, we obtain (iii) assuming Da+;g ∞,[a,b]

   

b

a

 (n+1)α max Da+;g f

∞,[a,b]

∞,[a,b]

  f (x) dg (x) ≤

(n+1)α , Db−;g f

∞,[a,b]

 ((n + 1) α + 2)

(g (b) − g (a))(n+1)α+1 , 2(n+1)α (3.52)

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds   n    b  g (b) − g (a) iα+1 1  f (x) dg (x) −   a  (iα + 2) N i=0



  iα  iα Da+;g f (a) j iα+1 + Db−;g f (b) (N − j)iα+1  ≤ 

 (n+1)α max Da+;g f

∞,[a,b]

(n+1)α , Db−;g f

∞,[a,b]

 ((n + 1) α + 2) 

g (b) − g (a) N

(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 ,

(3.53)

    iα iα (v) if Da+;g f (a) = Db−;g f (b) = 0, for i = 1, . . . , n, from (3.53) we get:    

b

 f (x) dg (x) −

a

   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

 (n+1)α max Da+;g f

∞,[a,b]

(n+1)α , Db−;g f

∞,[a,b]

 ((n + 1) α + 2) 

g (b) − g (a) N

(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 ,

(3.54)

46

3 Weighted Caputo Fractional Iyengar Type Inequalities

for j = 0, 1, 2, . . . , N , (vi) when N = 2, j = 1, (3.54) becomes    



b

f (x) dg (x) −

a

 (n+1)α max Da+;g f

∞,[a,b]

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

(n+1)α , Db−;g f

∞,[a,b]

 ((n + 1) α + 2)

(g (b) − g (a))(n+1)α+1 . 2(n+1)α (3.55)

Proof We have by (3.19) that f (x) =

n  (g (x) − g (a))iα  i=0

1  ((n + 1) α)



x

a

 (iα + 1)

 iα Da+;g f (a) +

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

(3.56)

∀ x ∈ [a, b] . Also by (3.16) we get f (x) =

n  (g (b) − g (x))iα  i=0

1  ((n + 1) α)



b x

 (iα + 1)

 iα Db−;g f (b) +

  (n+1)α f (t) dt, (g (t) − g (x))(n+1)α−1 g  (t) Db−;g

(3.57)

∀ x ∈ [a, b] . (n+1)α (n+1)α f, Db−;g f ∈ C ([a, b]) . Clearly here it is Da+;g By (3.56) we derive (by [6], p. 107)   n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤  f (x) −    (iα + 1) i=0

(n+1)α Da+;g f

∞,[a,b]

(g (x) − g (a))(n+1)α ,  ((n + 1) α + 1)

and by (3.57) we obtain   n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤  f (x) −    (iα + 1) i=0

(3.58)

3.2 Main Results

47

(n+1)α Db−;g f

∞,[a,b]

∀ x ∈ [a, b] . Call γ1 := and γ2 :=

(g (b) − g (x))(n+1)α ,  ((n + 1) α + 1)

(n+1)α Da+;g f

∞,[a,b]

 ((n + 1) α + 1) (n+1)α Db−;g f

∞,[a,b]

 ((n + 1) α + 1)

(3.59)

,

(3.60)

.

(3.61)

Set γ := max {γ1 , γ2 } .

(3.62)

That is   n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤ γ (g (x) − g (a))(n+1)α , (3.63)  f (x) −    (iα + 1) i=0

and   n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤ γ (g (b) − g (x))(n+1)α , (3.64)  f (x) −    (iα + 1) i=0

∀ x ∈ [a, b] . Equivalently, we have n  (g (x) − g (a))iα  i=0

 (iα + 1)

 iα Da+;g f (a) − γ (g (x) − g (a))(n+1)α ≤ f (x) ≤

n  (g (x) − g (a))iα  i=0

 (iα + 1)

 iα Da+;g f (a) + γ (g (x) − g (a))(n+1)α ,

(3.65)

and n  (g (b) − g (x))iα  i=0

 (iα + 1)

 iα Db−;g f (b) − γ (g (b) − g (x))(n+1)α ≤ f (x) ≤

n  (g (b) − g (x))iα  i=0

 (iα + 1)

 iα Db−;g f (b) + γ (g (b) − g (x))(n+1)α ,

(3.66)

48

3 Weighted Caputo Fractional Iyengar Type Inequalities

∀ x ∈ [a, b] . Let any t ∈ [a, b], then by integration against g over [a, t] and [t, b], respectively, we obtain n   i=0

 γ (g (t) − g (a))iα+1 iα − Da+;g f (a) (g (t) − g (a))(n+1)α+1  (iα + 2) ((n + 1) α + 1)  ≤

t

f (x) dg (x) ≤

a n   i=0

 γ (g (t) − g (a))iα+1 iα + Da+;g f (a) (g (t) − g (a))(n+1)α+1 ,  (iα + 2) ((n + 1) α + 1) (3.67)

and n  (g (b) − g (t))iα+1 

 (iα + 2)

i=0

 iα Db−;g f (b) − 

b

≤

γ (g (b) − g (t))(n+1)α+1 ((n + 1) α + 1)

f (x) dg (x) ≤

t n  (g (b) − g (t))iα+1 

 (iα + 2)

i=0

 iα Db−;g f (b) +

γ (g (b) − g (t))(n+1)α+1 . ((n + 1) α + 1) (3.68)

Adding (3.67) and (3.68), we obtain ⎧ n ⎨ ⎩

i=0

⎫   

⎬  1 iα iα+1 iα iα+1 + Db−;g f (b) (g (b) − g (t)) Da+;g f (a) (g (t) − g (a)) ⎭  (iα + 2)

−

  γ (g (t) − g (a))(n+1)α+1 + (g (b) − g (t))(n+1)α+1 ((n + 1) α + 1)  ≤

b

f (x) dg (x) ≤

a

⎧ n ⎨ ⎩

i=0

+

⎫

⎬     1 iα iα+1 iα iα+1 Da+;g f (a) (g (t) − g (a)) + Db−;g f (b) (g (b) − g (t)) ⎭  (iα + 2)

  γ (g (t) − g (a))(n+1)α+1 + (g (b) − g (t))(n+1)α+1 , ((n + 1) α + 1)

∀ t ∈ [a, b] .

(3.69)

3.2 Main Results

49

Consequently, we derive:   n  b   iα  1  Da+;g f (a) (g (t) − g (a))iα+1 f (x) dg (x) −   a  (iα + 2) i=0

  iα  + Db−;g f (b) (g (b) − g (t))iα+1  ≤   γ (g (t) − g (a))(n+1)α+1 + (g (b) − g (t))(n+1)α+1 , ((n + 1) α + 1)

(3.70)

∀ t ∈ [a, b] . Let us consider φ (z) := (z − g (a))(n+1)α+1 + (g (b) − z)(n+1)α+1 , ∀ z ∈ [g (a) , g (b)] . That is φ (g (t)) = (g (t) − g (a))(n+1)α+1 + (g (b) − g (t))(n+1)α+1 , ∀ t ∈ [a, b] . We have that   φ (z) = ((n + 1) α + 1) (z − g (a))(n+1)α − (g (b) − z)(n+1)α = 0, giving (z − g (a))(n+1)α = (g (b) − z)(n+1)α and z − g (a) = g (b) − z, that is z = g(a)+g(b) the only critical number of φ. We have that 2 φ (g (a)) = φ (g (b)) = (g (b) − g (a))(n+1)α+1 , and



g (a) + g (b) φ 2

 =

(g (b) − g (a))(n+1)α+1 , 2(n+1)α

which is the minimum of φ over [g (a) , g (b)]. Consequently, the right hand side of (3.70) is minimized when g (t) = (n+1)α+1

g(a)+g(b) , 2

(g(b)−g(a)) γ for some t ∈ [a, b], with value ((n+1)α+1) . 2(n+1)α     iα iα Assuming Da+;g f (a) = Db−;g f (b) = 0, i = 0, 1, . . . , n, then we obtain that   b   γ (g (b) − g (a))(n+1)α+1 ≤  f dg , (3.71) (x) (x)  ((n + 1) α + 1)  2(n+1)α a

50

3 Weighted Caputo Fractional Iyengar Type Inequalities

which is a sharp inequality. , then (3.70) becomes When g (t) = g(a)+g(b) 2   n  b  1 (g (b) − g (a))iα+1  f (x) dg (x) −   a  (iα + 2) 2iα+1 i=0



 iα   iα Da+;g f (a) + Db−;g (b)  ≤ 

γ (g (b) − g (a))(n+1)α+1 . 2(n+1)α ((n + 1) α + 1)

(3.72)

    , that is Next let N ∈ N, j = 0, 1, 2, . . . , N and g t j = g (a) + j g(b)−g(a) N g (t0 ) = g (a) , g (t1 ) = g (a) + (g(b)−g(a)) , . . . , g (t N ) = g (b) . N Hence it holds         g (b) − g (a) g (b) − g (a) , g (b) − g t j = (N − j) , g t j − g (a) = j N N (3.73) j = 0, 1, 2, . . . , N . We notice (n+1)α+1   (n+1)α+1    + g (b) − g t j = g t j − g (a) 

g (b) − g (a) N

(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 ,

(3.74)

j = 0, 1, 2, . . . , N , and (for i = 0, 1, . . . , n)  

iα+1  iα        iα+1

iα Da+;g = f (a) g t j − g (a) + Db−;g f (b) g (b) − g t j

g (b) − g (a) N

iα+1



   iα  iα Da+;g f (b) (N − j)iα+1 , (3.75) f (a) j iα+1 + Db−;g

for j = 0, 1, 2, . . . , N . By (3.70) we get   n    b  g (b) − g (a) iα+1 1  f (x) dg (x) −   a  (iα + 2) N i=0



  iα  iα Da+;g f (a) j iα+1 + Db−;g f (b) (N − j)iα+1  ≤ 

3.2 Main Results

51

γ ((n + 1) α + 1)



g (b) − g (a) N

(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 , (3.76)

j = 0,  1, 2, . . ., N .   iα iα If Da+;g f (a) = Db−;g f (b) = 0, i = 1, . . . , n, then (3.76) becomes    

b

 f (x) dg (x) −

a

γ ((n + 1) α + 1)



   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

g (b) − g (a) N

(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 , (3.77)

j = 0, 1, 2, . . . , N . When N = 2 and j = 1, then (3.77) becomes    

b

 f (x) dg (x) −

a

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

γ 2 (g (b) − g (a))(n+1)α+1 = 2(n+1)α+1 ((n + 1) α + 1) γ (g (b) − g (a))(n+1)α+1 . 2(n+1)α ((n + 1) α + 1)

(3.78) 

We have proved theorem in all possible cases. We give L 1 variants of last theorems:

Theorem 3.10 All as in Theorem 3.8 with α ≥ 1. If α = n ∈ N, we assume that  (n) f ◦ g −1 ◦ g ∈ C ([a, b]). Then (i)  n−1  b   (k) 1  f ◦ g −1 f (x) dg (x) − (g (a)) (g (t) − g (a))k+1   a (k + 1)! k=0

  (k)  + (−1)k f ◦ g −1 (g (b)) (g (b) − g (t))k+1  ≤  α max Da+;g f

L 1 ([a,b],g)

α , Db−;g f

L 1 ([a,b],g)

 (α + 1)   (g (t) − g (a))α + (g (b) − g (t))α ,

(3.79)

52

3 Weighted Caputo Fractional Iyengar Type Inequalities

∀ t ∈ [a, b] , , the right hand side of (3.79) is minimized, and we get: (ii) at g (t) = g(a)+g(b) 2  n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

 α max Da+;g f

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤

L 1 ([a,b],g)

α , Db−;g f

(g (b) − g (a))α , 2α−1

L 1 ([a,b],g)

 (α + 1)

(3.80)

 (k) (k)  (iii) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 0, 1, . . . , n − 1, we obtain   b   ≤  f dg (x) (x)   a

 α max Da+;g f

L 1 ([a,b],g)

α , Db−;g f

L 1 ([a,b],g)

 (α + 1)

(g (b) − g (a))α , 2α−1

(3.81)

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds    n−1  b  1 g (b) − g (a) k+1  f dg − (x) (x)   a N (k + 1)! k=0



  (k)  (k)  j k+1 f ◦ g −1 (g (a)) + (−1)k (N − j)k+1 f ◦ g −1 (g (b))  ≤  α max Da+;g f

L 1 ([a,b],g)

α , Db−;g f

L 1 ([a,b],g)

 (α + 1) 

g (b) − g (a) N

α



 j α + (N − j)α ,

(3.82)

(k) (k)   (v) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 1, . . . , n − 1, from (3.82) we obtain    

a

b

 f (x) dg (x) −

   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

3.2 Main Results

53

 α max Da+;g f

α , Db−;g f

L 1 ([a,b],g)

L 1 ([a,b],g)

 (α + 1) 

g (b) − g (a) N

α



 j α + (N − j)α ,

(3.83)

j = 0, 1, 2, . . . , N , (vi) when N = 2, j = 1, (3.83) turns to    

b

 f (x) dg (x) −

a

 α max Da+;g f

L 1 ([a,b],g)

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

α , Db−;g f

L 1 ([a,b],g)

 (α + 1)

(g (b) − g (a))α . 2α−1

(3.84)

Proof From (3.27) we get   (k) n−1     f ◦ g −1 (g (a))  k (g (x) − g (a))  ≤  f (x) −   k! k=0

1  (α)

 a

x

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

(g (x) − g (a))α−1  (α) (g (x) − g (a))α−1  (α)



x

a



(g (x) − g (a))α−1  (α) α Da+;g f

b

a



 α   g  (t)  Da+;g f (t) dt ≤  α   g  (t)  Da+;g f (t) dt = b

a

L 1 ([a,b],g)

 (α) ∀ x ∈ [a, b] .

 α  D

a+;g

  f (t) dg (t) =

(g (x) − g (a))α−1 ,

(3.85)

54

3 Weighted Caputo Fractional Iyengar Type Inequalities

Similarly, from (3.28) we obtain   (k) n−1     f ◦ g −1 (g (b))   (g (x) − g (b))k  ≤  f (x) −   k! k=0

1  (α)



b x

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

(g (b) − g (x))α−1  (α) α Db−;g f



b

b−;g

x

L 1 ([a,b],g)

 (α) ∀ x ∈ [a, b] . Call

 α  D

 α f L δ := max Da+;g

1

  f (t) dg (t) ≤

(3.86)

(g (b) − g (x))α−1 ,

α , Db−;g f L ([a,b],g)

 1 ([a,b],g)

.

(3.87)

We have proved that   (k) n−1     f ◦ g −1 (g (a))   (g (x) − g (a))k  ≤  f (x) −   k! k=0

δ (g (x) − g (a))α−1 ,  (α) and

(3.88)

  (k) n−1     f ◦ g −1 (g (b))   (g (x) − g (b))k  ≤  f (x) −   k! k=0

δ (g (b) − g (x))α−1 ,  (α)

(3.89)

∀ x ∈ [a, b] . The rest of the proof is as in Theorem 3.8.



It follows Theorem 3.11 All as in Theorem 3.9, with  (n+1)α ρ := max Da+;g f

1 n+1

L 1 ([a,b],g)

≤ α ≤ 1. Call

(n+1)α , Db−;g f

L 1 ([a,b],g)

.

(3.90)

3.2 Main Results

55

Then (i)   n  b   iα  1  Da+;g f (a) (g (t) − g (a))iα+1 f (x) dg (x) −   a  (iα + 2) i=0

  iα  + Db−;g f (b) (g (b) − g (t))iα+1  ≤   ρ (g (t) − g (a))(n+1)α + (g (b) − g (t))(n+1)α ,  ((n + 1) α + 1)

(3.91)

∀ t ∈ [a, b] , (ii) at g (t) = g(a)+g(b) , the right hand side of (3.91) is minimized, and we get: 2   n  b  1 (g (b) − g (a))iα+1  f (x) dg (x) −   a  (iα + 2) 2iα+1 i=0





 iα   iα Da+;g f (a) + Db−;g f (b)  ≤

ρ (g (b) − g (a))(n+1)α ,  ((n + 1) α + 1) 2(n+1)α−1

(3.92)

    iα iα f (a) = Db−;g f (b) = 0, i = 0, 1, . . . , n, we obtain (iii) assuming Da+;g    

b

a

  f (x) dg (x) ≤

ρ (g (b) − g (a))(n+1)α ,  ((n + 1) α + 1) 2(n+1)α−1

(3.93)

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds   n    b  g (b) − g (a) iα+1 1  f dg − (x) (x)   a  (iα + 2) N i=0



   iα  iα Da+;g f (a) j iα+1 + Db−;g f (b) (N − j)iα+1  ≤

ρ  ((n + 1) α + 1)



g (b) − g (a) N

(n+1)α



 j (n+1)α + (N − j)(n+1)α , (3.94)

56

3 Weighted Caputo Fractional Iyengar Type Inequalities

    iα iα (v) if Da+;g f (a) = Db−;g f (b) = 0, i = 1, . . . , n, from (3.94) we get:    

b

 f (x) dg (x) −

a

ρ  ((n + 1) α + 1)



   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

g (b) − g (a) N

(n+1)α



 j (n+1)α + (N − j)(n+1)α , (3.95)

for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (3.95) becomes    

b

 f (x) dg (x) −

a

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

ρ (g (b) − g (a))(n+1)α .  ((n + 1) α + 1) 2(n+1)α−1

(3.96)

Proof By (3.56) we get   n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤  f (x) −    (iα + 1) i=0

1  ((n + 1) α)

 a

x

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

(g (x) − g (a))(n+1)α−1  ((n + 1) α) (g (x) − g (a))(n+1)α−1  ((n + 1) α)

 a



(g (x) − g (a))(n+1)α−1  ((n + 1) α) (n+1)α Da+;g f

∀ x ∈ [a, b] .

b

a



L 1 ([a,b],g)

 ((n + 1) α)

x

     (n+1)α g  (t)  Da+;g f (t) dt ≤      (n+1)α g  (t)  Da+;g f (t) dt = b

a

     (n+1)α  Da+;g f (t) dg (t) =

(g (x) − g (a))(n+1)α−1 ,

(3.97)

3.2 Main Results

57

Similarly, from (3.57) we derive   n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤  f (x) −    (iα + 1) i=0

1  ((n + 1) α)



b x

     (n+1)α f (t) dt ≤ (g (t) − g (x))(n+1)α−1 g  (t)  Db−;g 

(g (b) − g (x))(n+1)α−1  ((n + 1) α) (n+1)α Db−;g f

b x

L 1 ([a,b],g)

 ((n + 1) α)

    (n+1)α   Db−;g f (t) dg (t) ≤

(3.98)

(g (b) − g (x))(n+1)α−1 ,

∀ x ∈ [a, b] . We have proved that   n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤  f (x) −    (iα + 1) i=0

ρ (g (x) − g (a))(n+1)α−1 ,  ((n + 1) α) and

(3.99)

  n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤  f (x) −    (iα + 1) i=0

ρ (g (b) − g (x))(n+1)α−1 ,  ((n + 1) α)

(3.100)

∀ x ∈ [a, b] . The rest of the proof is as in Theorem 3.9.



Next follow L p variants of Theorems 3.8, 3.9. Theorem 3.12 All as in Theorem 3.8 with α ≥ 1, and p, q > 1 : (n)  ◦ g ∈ C ([a, b]). Set n ∈ N, we assume that f ◦ g −1  α f L μ := max Da+;g

q

α , Db−;g f L ([a,b],g)

1 p

+

 q ([a,b],g)

.

1 q

= 1. If α =

(3.101)

58

3 Weighted Caputo Fractional Iyengar Type Inequalities

Then (i)  n−1  b   (k) 1  f ◦ g −1 f dg − (x) (x) (g (a)) (g (t) − g (a))k+1   a (k + 1)! k=0

  (k)  + (−1)k f ◦ g −1 (g (b)) (g (b) − g (t))k+1  ≤ μ   1 1  (α) α + p ( p (α − 1) + 1) p

1 1 (g (t) − g (a))α+ p + (g (b) − g (t))α+ p ,

(3.102)

∀ t ∈ [a, b] , (ii) at g (t) = g(a)+g(b) , the right hand side of (3.102) is minimized, and we get: 2  n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤ (g (b) − g (a))α+ p 1

μ

  1  (α) α + 1p ( p (α − 1) + 1) p

2α− q

1

,

(3.103)

(k) (k)   (iii) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 0, 1, . . . , n − 1, we obtain   b    f (x) dg (x) ≤  a

μ

  1  (α) α + 1p ( p (α − 1) + 1) p

(g (b) − g (a))α+ p 1

2α− q

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds

1

,

  b   n−1   g (b) − g (a) k+1 1  f dg − (x) (x)  N (k + 1)!  a k=0

(3.104)

3.2 Main Results



59

  (k)  (k)  j k+1 f ◦ g −1 (g (a)) + (−1)k (N − j)k+1 f ◦ g −1 (g (b))  ≤ 

μ

  1  (α) α + 1p ( p (α − 1) + 1) p

 1   g (b) − g (a) α+ p α+ 1p α+ 1p j , + (N − j) N

(3.105) (k) (k)   (v) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 1, . . . , n − 1, from (3.105) we obtain      b  g (b) − g (a)   f (x) dg (x) − [ j f (a) + (N − j) f (b)] ≤   a  N







μ

1  (α) α + 1p ( p (α − 1) + 1) p

 1   g (b) − g (a) α+ p α+ 1p α+ 1 j + (N − j) p , N

(3.106) j = 0, 1, 2, . . . , N , (vi) when N = 2, j = 1, (3.106) turns to    



b

f (x) dg (x) −

a

μ

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

  1  (α) α + 1p ( p (α − 1) + 1) p

(g (b) − g (a))α+ p 1

2α− q

1

.

(3.107)

Proof From (3.27) we get   (k) n−1     f ◦ g −1 (g (a))  k (g (x) − g (a))  ≤  f (x) −   k! k=0

1  (α)



x

a

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

(by [7], p. 439) 1  (α)

 a

x

 α   f (t) dg (t) ≤ (g (x) − g (t))α−1  Da+;g

(3.108)

(by [4]) 1  (α)

 a

x

 1p  (g (x) − g (t)) p(α−1) dg (t) a

x

 q1  q  dg (t) f ≤ (t) a+;g

 α  D

60

3 Weighted Caputo Fractional Iyengar Type Inequalities

1 (g (x) − g (a))α− q Dα f . 1 a+;g L q ([a,b],g)  (α) ( p (α − 1) + 1) p 1

  (k) n−1     f ◦ g −1 (g (a))   (g (x) − g (a))k  ≤  f (x) −   k!

That is

k=0

α Da+;g f

L q ([a,b],g)

 (α) ( p (α − 1) + 1)

(g (x) − g (a))α− q , 1

1 p

(3.109)

∀ x ∈ [a, b] . Similarly, from (3.28) we obtain   (k) n−1     f ◦ g −1 (g (b))   (g (x) − g (b))k  ≤  f (x) −   k! k=0

1  (α)



b x

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

(by [7], p. 439) 1  (α)



b x

 α   f (t) dg (t) ≤ (g (t) − g (x))α−1  Db−;g

(by [4]) 1  (α)



b

(g (t) − g (x))

p(α−1)

 1p  dg (t)

x

b x

q  b−;g f (t) dg (t)

 α  D



 q1

≤

(3.110)

1 (g (b) − g (x))α− q Dα f . 1 b−;g L q ([a,b],g)  (α) ( p (α − 1) + 1) p 1

That is

  (k) n−1     f ◦ g −1 (g (b))   (g (x) − g (b))k  ≤  f (x) −   k! k=0

α Db−;g f

L q ([a,b],g)

 (α) ( p (α − 1) + 1) ∀ x ∈ [a, b] .

(g (b) − g (x))α− q , 1

1 p

(3.111)

3.2 Main Results

61

We have proved that   (k) n−1     f ◦ g −1 (g (a))   (g (x) − g (a))k  ≤  f (x) −   k! k=0

μ  (α) ( p (α − 1) + 1) and

(g (x) − g (a))α− q , 1

1 p

(3.112)

  (k) n−1     f ◦ g −1 (g (b))   (g (x) − g (b))k  ≤  f (x) −   k! k=0

μ

(g (b) − g (x))α− q , 1

 (α) ( p (α − 1) + 1)

1 p

(3.113)

∀ x ∈ [a, b] . The rest of the proof is as in Theorem 3.8.



We continue with 1 Theorem 3.13 All as in Theorem 3.9, with n+1 ≤ α ≤ 1, and p, q > 1 : Set

 (n+1)α (n+1)α . , Db−;g f θ := max Da+;g f L q ([a,b],g)

1 p

L q ([a,b],g)

+

1 q

= 1.

(3.114)

Then (i)   n  b   iα  1  Da+;g f (a) (g (t) − g (a))iα+1 f (x) dg (x) −   a  (iα + 2) i=0

  iα + Db−;g f (b) (g (b) − g (t))iα+1  ≤ 

θ   1  ((n + 1) α) (n + 1) α + 1p ( p ((n + 1) α − 1) + 1) p

1 1 (g (t) − g (a))(n+1)α+ p + (g (b) − g (t))(n+1)α+ p ,

(3.115)

∀ t ∈ [a, b] , (ii) at g (t) = g(a)+g(b) , the right hand side of (3.115) is minimized, and we get: 2 ⎧   b n ⎨  1 (g (b) − g (a))iα+1  f (x) dg (x) −  ⎩  (iα + 2) 2iα+1  a i=0

62

3 Weighted Caputo Fractional Iyengar Type Inequalities



    iα iα f (b)  ≤ Da+;g f (a) + Db−;g  θ

  1  ((n + 1) α) (n + 1) α + 1p ( p ((n + 1) α − 1) + 1) p

(g (b) − g (a)) 2

(n+1)α+ 1p

(n+1)α− q1

,

(3.116) 







iα iα f (a) = Db−;g f (b) = 0, i = 0, 1, . . . , n, we obtain (iii) assuming Da+;g

    b   f (x) dg (x) ≤    a

θ



(g (b) − g (a))



1  ((n + 1) α) (n + 1) α + 1p ( p ((n + 1) α − 1) + 1) p

2

(n+1)α+ 1p

(n+1)α− q1

,

(3.117) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds   n    b  g (b) − g (a) iα+1 1  f dg − (x) (x)   a  (iα + 2) N i=0



   iα  iα Da+;g f (a) j iα+1 + Db−;g f (b) (N − j)iα+1  ≤

θ   1  ((n + 1) α) (n + 1) α + 1p ( p ((n + 1) α − 1) + 1) p 

g (b) − g (a) N

(n+1)α+ 1p

1 1 j (n+1)α+ p + (N − j)(n+1)α+ p ,

(3.118)

    iα iα (v) if Da+;g f (a) = Db−;g f (b) = 0, i = 1, . . . , n, from (3.118) we get:    

b

 f (x) dg (x) −

a

   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

θ   1  ((n + 1) α) (n + 1) α + 1p ( p ((n + 1) α − 1) + 1) p 

g (b) − g (a) N

(n+1)α+ 1p

1 1 j (n+1)α+ p + (N − j)(n+1)α+ p ,

(3.119)

3.2 Main Results

63

j = 0, 1, 2, . . . , N , (vi) when N = 2, j = 1, (3.119) turns to

     b  g (b) − g (a)   f (x) dg (x) − ( f (a) + f (b)) ≤   a  2 θ

  1  ((n + 1) α) (n + 1) α + 1p ( p ((n + 1) α − 1) + 1) p

(g (b) − g (a)) 2

(n+1)α+ 1p

(n+1)α− q1

.

(3.120) Proof By (3.56) we get   n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤  f (x) −    (iα + 1)

(3.121)

i=0

1  ((n + 1) α)



x

a

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

(by [7]) 1  ((n + 1) α)

 a

x

    (n+1)α  f (t) dg (t) ≤ (g (x) − g (t))(n+1)α−1  Da+;g

(by [4]) 1  ((n + 1) α) 

x

a



x

(g (x) − g (t))

p((n+1)α−1)

dg (t)

 1p

a

 q1   q   (n+1)α ≤  Da+;g f (t) dg (t)

p((n+1)α−1)+1 p 1 (g (x) − g (a)) (n+1)α f . D a+;g L q ([a,b],g)  ((n + 1) α) ( p ((n + 1) α − 1) + 1) 1p

That is

  n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤  f (x) −    (iα + 1) i=0

(n+1)α Da+;g f

L q ([a,b],g)

 ((n + 1) α) ( p ((n + 1) α − 1) + 1) ∀ x ∈ [a, b] .

(g (x) − g (a))(n+1)α− q , 1

1 p

(3.122)

64

3 Weighted Caputo Fractional Iyengar Type Inequalities

Similarly, from (3.57) we derive   n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤  f (x) −    (iα + 1) i=0

1  ((n + 1) α)



b x

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

(by [7]) 1  ((n + 1) α)



b x

    (n+1)α  f (t) dg (t) ≤ (g (t) − g (x))(n+1)α−1  Db−;g

(by [4]) 1  ((n + 1) α) 



b x

b

(g (t) − g (x))

p((n+1)α−1)

 1p dg (t)

x

 q1   q  (n+1)α  ≤  Db−;g f (t) dg (t)

(3.123)

p((n+1)α−1)+1 p 1 (g (b) − g (x)) (n+1)α f . 1 Db−;g L q ([a,b],g)  ((n + 1) α) ( p ((n + 1) α − 1) + 1) p

That is

  n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤  f (x) −    (iα + 1) i=0

(n+1)α Db−;g f

L q ([a,b],g)

 ((n + 1) α) ( p ((n + 1) α − 1) + 1)

(g (b) − g (x))(n+1)α− q , 1

1 p

(3.124)

∀ x ∈ [a, b] . We have proved that   n     (g (x) − g (a))iα  iα  Da+;g f (a) ≤  f (x) −    (iα + 1) i=0

θ  ((n + 1) α) ( p ((n + 1) α − 1) + 1) and

(g (x) − g (a))(n+1)α− q , 1

1 p

(3.125)

3.2 Main Results

65

  n     (g (b) − g (x))iα  iα  Db−;g f (b) ≤  f (x) −    (iα + 1) i=0

θ

(g (b) − g (x))(n+1)α− q , 1

 ((n + 1) α) ( p ((n + 1) α − 1) + 1)

1 p

(3.126)

∀ x ∈ [a, b] . The rest of the proof is as in Theorem 3.9.



Applications follow:

  Proposition 3.14 We assume that ( f ◦ ln x) ∈ AC n ea , eb , where N n = α, α > 0. We also assume that ( f ◦ ln x)(n) ◦ e x ∈ L ∞ ([a, b]), f ∈ C ([a, b]). Set  α T1 := max Da+;e x f L

∞

α , Db−;e x f L ([a,b])

 ∞ ([a,b])

.

(3.127)

Then (i)

 n−1  b    k+1 1  f (x) e x d x − ( f ◦ ln x)(k) ea et − ea   a (k + 1)! k=0

  k+1  (−1)k ( f ◦ ln x)(k) eb eb − et ≤  α+1  b α+1

T1 et − ea , + e − et  (α + 2)

(3.128)

∀ t ∈ [a,  b] ,  a b , the right hand side of (3.128) is minimized, and we get: (ii) at t = ln e +e 2  k+1  b n−1  b  e − ea 1  x f (x) e d x −   a 2k+1 (k + 1)! k=0

     ( f ◦ ln x)(k) ea + (−1)k ( f ◦ ln x)(k) eb  ≤ α+1  b e − ea T1 ,  (α + 2) 2α

(3.129)

  (iii) if ( f ◦ ln x)(k) (ea ) = ( f ◦ ln x)(k) eb = 0, for k = 0, 1, . . . , n − 1, we obtain    

a

b

 b α+1   e − ea  f (x) e d x  ≤ T1 ,  (α + 2) 2α x

(3.130)

66

3 Weighted Caputo Fractional Iyengar Type Inequalities

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds  k+1  b n−1  b  e − ea 1  x f e d x − (x)   a N (k + 1)! k=0



    j k+1 ( f ◦ ln x)(k) ea + (−1)k (N − j)k+1 ( f ◦ ln x)(k) eb  ≤ T1  (α + 2)



eb − ea N

α+1



 j α+1 + (N − j)α+1 ,

(3.131)

  (v) if ( f ◦ ln x)(k) (ea ) = ( f ◦ ln x)(k) eb = 0, for k = 1, . . . , n − 1, from (3.131) we obtain  b    b   e − ea x  f (x) e d x − [ j f (a) + (N − j) f (b)] ≤  N a

T1  (α + 2)



eb − ea N

α+1



 j α+1 + (N − j)α+1 ,

(3.132)

j = 0, 1, 2, . . . , N , (vi) when N = 2, j = 1, (3.132) turns to    

b

 f (x) e x d x −

a

eb − ea 2



  ( f (a) + f (b)) ≤

α+1  b e − ea T1 ,  (α + 2) 2α

(3.133)

(vii) when 0 < α ≤ 1, inequality (3.133) is again valid without any boundary conditions. Proof By Theorem 3.8, for g (x) = e x .



We continue with Proposition 3.15 Here f ∈ C ([a, b]), where [a, b] ⊂ (0, +∞). Let 0 < α ≤ 1, kα x and G k := Da+;ln x f , for k = 0, 1, . . . , n + 1; n ∈ N. We assume that G k ◦ e ∈ kα AC ([ln a, ln b]) and (G k ◦ e x ) ◦ ln x ∈ L ∞ ([a, b]). Also let G k := Db−;ln f , for   x x k = 0, 1, . . . , n + 1, they fulfill G k ◦ e ∈ AC ([ln a, ln b]) and G k ◦ e x ◦ ln x ∈ L ∞ ([a, b]). Set  (n+1)α T2 := max Da+;ln x f

∞,[a,b]

(n+1)α , Db−;ln x f

∞,[a,b]

.

(3.134)

3.2 Main Results

67

Then (i)

  n     b f (x)    iα t iα+1 1  dx − Da+;ln x f (a) ln   a x  (iα + 2) a i=0 +



T2  ((n + 1) α + 2)

iα Db−;ln x



t ln a

   b iα+1 f (b) ln t 

(n+1)α+1

   ≤ 

   b (n+1)α+1 + ln , t

(3.135)

∀ t ∈ [a, b] , ln ab (ii) at t = e( 2 ) , the right hand side of (3.135) is minimized, and we get:   n  b iα+1  b f (x)  ln a 1  dx −   a x  (iα + 2) 2iα+1 i=0 

  iα   iα ≤ Da+;ln x f (a) + Db−;ln x f (b)  b (n+1)α+1 ln a T2 ,  ((n + 1) α + 2) 2(n+1)α

(3.136)

  iα   iα (iii) assuming Da+;ln x f (a) = Db−;ln x f (b) = 0, i = 0, 1, . . . , n, we obtain    

b

a

 b (n+1)α+1  ln a T2 f (x)  dx ≤ , x  ((n + 1) α + 2) 2(n+1)α

(3.137)

which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds ⎧  ! "iα+1  b n ⎨  ln ab f 1 (x)  dx −  ⎩ x  (iα + 2) N  a i=0 

   iα  iα iα+1 iα+1  ≤ Da+;ln + Db−;ln x f (a) j x f (b) (N − j)

T2  ((n + 1) α + 2)

!

ln ab N

"(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 , (3.138)

68

3 Weighted Caputo Fractional Iyengar Type Inequalities

 iα   iα  (v) if Da+;ln x f (a) = Db−;ln x f (b) = 0, i = 1, . . . , n, from (3.138) we get:   ! "   b f (x) ln ab   dx − ( j f (a) + (N − j) f (b)) ≤    a x N T2  ((n + 1) α + 2)

!

ln ab N

"(n+1)α+1



 j (n+1)α+1 + (N − j)(n+1)α+1 , (3.139)

for j = 0, 1, 2, . . . , N , (vi) if N = 2 and j = 1, (3.139) becomes   ! "  b f (x)  ln ab   dx − ( f (a) + f (b)) ≤   a  x 2  b (n+1)α+1 ln a T2 .  ((n + 1) α + 2) 2(n+1)α Proof By Theorem 3.9, for g (x) = ln x.

(3.140) 

We could give many other interesting applications that are based in our other theorems, due to lack of space we skip this task.

References 1. Anastassiou, G.: Advanced fractional Taylor’s formulae 2. Anastassiou, G., Argyros, I.: Intelligent Numerical Methods: Applications to Fractional Calculus. Springer, Heidelberg (2016) 3. Anastassiou, G.: Weighted fractional Iyengar type Inequalities in the Caputo direction. Mathematics 7, 1119 (2019). https://doi.org/10.3390/MATH7111119 4. Dragomir, S.S.: Inequalities for the Riemann-Stieljes integral of ( p, q) − H −Dominated integrators with applications. Appl. Math. E-Notes 15, 243–260 (2015). http://www.math.nthu. edu.tw/~amen 5. Iyengar, K.S.K.: Note on an inequality. Math. Student 6, 75–76 (1938) 6. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co. Inc., New York (1968) 7. Royden, H.L., Fitzpatrick, P.M.: Real Analysis, 4th edn. Pearson, New York (2010)

Chapter 4

Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

We present here generalized Canavati type g-fractional Iyengar and Ostrowski type inequalities. Our inequalities are with respect to all L p norms: 1 ≤ p ≤ ∞. We finish with applications. See also [6].

4.1 Background—I We are motivated by the following famous Iyengar inequality (1938), [8].   Theorem 4.1 Let f be a differentiable function on [a, b] and  f  (x) ≤ M. Then    

a

b

f (x) d x −

  M (b − a)2 1 ( f (b) − f (a))2 − . (b − a) ( f (a) + f (b)) ≤ 2 4 4M (4.1)

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,  ∞ −tb]), 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,

(4.2)

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

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_4

(4.3)

69

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

70

α For f ∈ Ca+ ([a, b]), we define the left generalized α-fractional derivative of f over [a, b] as



a+ (m) α Da+ f := J1−β f ,

(4.4)

see [1, p. 24]. Canavati first in [7] introduced the above over [0, 1]. n f = f (n) ; n ∈ N. We have that Da+ α f ∈ C ([a, b]). Notice that Da+ Furthermore we need: Let again α > 0, m = [α], β = α − m, f ∈ C ([a, b]), call the right RiemannLiouville fractional integral operator by 

 α f (x) := Jb−

1  (α)



b

(t − x)α−1 f (t) dt,

(4.5)

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]) .

(4.6)

Define the right generalized α-fractional derivative of f over [a, b] as

 1−β α Db− f = (−1)m−1 Jb− f (m) ,

(4.7)

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 [5] we have the following Canavati fractional Iyengar type inequalities: ν ν Theorem 4.2 Let ν ≥ 1, n = [ν] and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then (i)

   b n−1

   1  f (k) (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1  ≤ f (x) d x −  (k + 1)!  a  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 , the right hand side of (4.8) is minimized, and we get: 2

(4.8)

4.1 Background—I

71

  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 ∞,([a,b]) ,  Db− f ∞,([a,b]) (b − a)ν+1 2ν

 (ν + 2)

,

(4.9)

(iii) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1, we obtain     ν   max   Dν f  ,  Db− f ∞,([a,b]) (b − a)ν+1  a+ ∞,([a,b]) f (x) d x  ≤ ,  (ν + 2) 2ν a (4.10) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds    

b

    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 f

max  Da+ ,  Db− f ∞,([a,b])  b − a ν+1 ∞,([a,b]) ≤ j ν+1 + (N − j)ν+1 ,  (ν + 2) N

(4.11)

(v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (4.11) we get:      b b−a   f (x) d x − [ j f (a) + (N − j) f (b)] ≤    a N

    ν   ν  max  Da+ f ∞,([a,b]) , Db− f

 

∞,([a,b])

 (ν + 2)



b − a ν+1 ν+1 j + (N − j)ν+1 , N

(4.12)

j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (4.12) turns to    

a

b

 f (x) d x −

b−a 2



  f + f ( (a) (b)) ≤

    ν  ν max  Da+ f ∞,([a,b]) ,  Db− f ∞,([a,b]) (b − a)ν+1  (ν + 2)

2ν

.

(4.13)

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

72

We continue with L 1 estimates: ν ν Theorem 4.3 ([5]) Let ν ≥ 1, n = [ν], and f ∈ Ca+ ([a, b]) ∩ Cb− ([a, b]). Then

(i)

   b n−1

   1 k+1 k k+1 (k) (k) ≤  f (a) (t − a) f (x) d x − + (−1) f (b) (b − t)   (k + 1)!   a k=0      ν   ν  max  Da+ f  L ([a,b]) , Db− f 1  L 1 ([a,b])  (t − a)ν + (b − t)ν ,  (ν + 1)

(4.14)

∀ t ∈ [a, b] , (ii) when ν = 1, from (4.14), 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] ,

(4.15)

(iii) from (4.15), 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) , (4.16)

ν > 1, the right hand side of (4.14) 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  (ν + 1)

2ν−1

,

(4.17)

(v) if f (k) (a) = f (k) (b) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (4.17), 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  ≤ ,   (ν + 1) 2ν−1 a (4.18) which is a sharp inequality,

4.1 Background—I

73

(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)  f (x) d x − f f j (a) + (−1) (N − j) (b)   N (k + 1)!   a k=0      ν   ν    max  Da+ f  L ([a,b]) , Db− f  1 b−a ν ν L 1 ([a,b]) ≤ j + (N − j)ν ,  (ν + 1) N

(4.19)

(vii) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (4.19) we get:    

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 , (4.20)

j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (4.20) 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

,

(4.21)

We continue with L p estimates: Theorem 4.4 ([5]) Let p, q > 1 : ν Cb− ([a, b]). Then

1 p

+

1 q

ν = 1, ν ≥ 1, n = [ν] ; f ∈ Ca+ ([a, b]) ∩

(i)    b n−1

   1  f (k) (a) (t − a)k+1 + (−1)k f (k) (b) (b − t)k+1  ≤ f (x) d x −  (k + 1)!  a  k=0      ν   ν    max  Da+ f  L ([a,b]) , Db− f q L q ([a,b]) ν+ 1p ν+ 1p , + (b − t) (t − a)

1  (ν) ν + 1p ( p (ν − 1) + 1) p

(4.22) ∀ t ∈ [a, b] , , the right hand side of (4.22) is minimized, and we get: (ii) at t = a+b 2

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

74

  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

(4.23)

(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 ([a,b]) (b − a)ν+ 1p  a+ L q ([a,b]) q f (x) d x  ≤ ,

1 ν− 1  (ν) ν + 1p ( p (ν − 1) + 1) p 2 q

(4.24)

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 f

max  Da+ ,  Db− f  L ([a,b])  b − a ν+ 1p L q ([a,b]) q ν+ 1 ν+ 1 ≤ j p + (N − j) p ,

1 N  (ν) ν + 1p ( p (ν − 1) + 1) p

(4.25) (v) if f (k) (a) = f (k) (b) = 0, k = 1, . . . , n − 1, from (4.25) we get:    

b

 f (x) d x −

a

b−a N



  [ j f (a) + (N − j) f (b)] ≤

    ν  ν f

max  Da+ ,  Db− f  L ([a,b])  b − a ν+ 1p L q ([a,b]) q ν+ 1 ν+ 1 j p + (N − j) p ,

1 N  (ν) ν + 1p ( p (ν − 1) + 1) p

(4.26) for j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (4.26) turns to    

a

b

 f (x) d x −

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 Next we follow [4].

(4.27)

4.1 Background—I

75

Let g : [a, b] → R be a strictly increasing function. Let f ∈ C n ([a, b]), n ∈ N. Assume that g ∈ C 1 ([a, b]), and g −1 ∈ C n ([a, b]). Call l := f ◦ g −1 : [g (a) , g (b)] → R. It is clear that l, l  , . . . , l (n) are continuous functions from [g (a) , g (b)] into f ([a, b]) ⊆ R. Let ν ≥ 1 such that [ν] = n, n ∈ N as above, where [·] is the integral part of the number. Clearly when 0 < ν < 1, [ν] = 0. Next we follow [1, pp. 7–9]. (I) Let h ∈ C ([g (a) , g (b)]), we define the left Riemann-Liouville fractional integral as  z  z0  1 (4.28) Jν h (z) := (z − t)ν−1 h (t) dt,  (ν) z0 g (a) ≤ z 0 ≤ z ≤ g (b), where  is the gamma function;  (ν) = for ∞ −t ν−1 e t dt. 0 We set J0z0 h = h. ν Let α := ν − [ν] (0 < α < 1). We define the subspace Cg(x ([g (a) , g (b)]) of 0) [ν] C ([g (a) , g (b)]), where x0 ∈ [a, b] :   g(x ) ν Cg(x ([g (a) , g (b)]) := h ∈ C [ν] ([g (a) , g (b)]) : J1−α0 h ([ν]) ∈ C 1 ([g (x0 ) , g (b)]) . ) 0

(4.29)

ν So let h ∈ Cg(x ([g (a) , g (b)]); we define the left g-generalized fractional derivative 0) of h of order ν, of Canavati type, over [g (x0 ) , g (b)] as

 g(x0 ) ([ν]) ν h := J h . Dg(x ) 1−α 0

(4.30)

ν Clearly, for h ∈ Cg(x ([g (a) , g (b)]), there exists 0)



 ν Dg(x h (z) = 0)

d 1  (1 − α) dz



z

g(x0 )

(z − t)−α h ([ν]) (t) dt,

(4.31)

for all g (x0 ) ≤ z ≤ g (b) . ν In particular, when f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]) we have that 0) 

  ν Dg(x f ◦ g −1 (z) = 0)

d 1  (1 − α) dz 



z g(x0 )

 ([ν]) (z − t)−α f ◦ g −1 (t) dt,

  (n) n f ◦ g −1 = f ◦ g −1 and for all g (x0 ) ≤ z ≤ g (b) . We have Dg(x ) 0   0 −1 −1 Dg(x0 ) f ◦ g = f ◦g .

(4.32)

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

76

(II) Next we follow [3, pp. 345–348]. Let h ∈ C ([g (a) , g (b)]), we define the right Riemann-Liouville fractional integral as  z0  ν  1 Jz0 − h (z) := (4.33) (t − z)ν−1 h (t) dt,  (ν) z for g (a) ≤ z ≤ z 0 ≤ g (b). We set Jz00 − h = h. ν Let α := ν − [ν] (0 < α < 1). We define the subspace Cg(x ([g (a) , g (b)]) of 0 )− [ν] C ([g (a) , g (b)]), where x0 ∈ [a, b] : ν Cg(x ([g (a) , g (b)]) := 0 )−



 1−α 1 ([ν]) h ∈ C [ν] ([g (a) , g (b)]) : Jg(x h ∈ C , g . ([g (a) (x )]) 0 0 )−

(4.34)

ν So let h ∈ Cg(x ([g (a) , g (b)]); we define the right g-generalized fractional deriva0 )− tive of h of order ν, of Canavati type, over [g (a) , g (x0 )] as



 1−α ν h := (−1)n−1 Jg(x h ([ν]) . Dg(x 0 )− 0 )−

(4.35)

ν Clearly, for h ∈ Cg(x ([g (a) , g (b)]), there exists 0 )−



 (−1)n−1 d ν Dg(x h = (z) 0 )−  (1 − α) dz



g(x0 )

(t − z)−α h ([ν]) (t) dt,

(4.36)

z

for all g (a) ≤ z ≤ g (x0 ) ≤ g (b) . ν In particular, when f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]) we have that 0 )−

 g(x0 )



([ν]) (−1)n−1 d ν −1 Dg(x f ◦ g = (z) (t − z)−α f ◦ g −1 (t) dt, 0 )−  (1 − α) dz z

(4.37)

for all g (a) ≤ z ≤ g (x0 ) ≤ g (b) . We get that    (n)  n Dg(x0 )− f ◦ g −1 (z) = (−1)n f ◦ g −1 (z)

(4.38)

 

  0 −1 f ◦ g and Dg(x (z) = f ◦ g −1 (z) , all z ∈ [g (a) , g (x0 )] . 0 )− Let g be strictly increasing and continuous over [a, b], and f ∈ C ([a, b]). We have that  b a

f (x) dg (x) =

 b a

 g(b)



f ◦ g −1 (g (x)) dg (x) = f ◦ g −1 (z) dz.

Here it is f ◦ g −1 ∈ C ([g (a) , g (b)]) .

g(a)

(4.39)

4.2 Main Results—I

77

4.2 Main Results—I Next we present generalized g-fractional Iyengar type inequalities: Theorem 4.5 Let ν ≥ 1 such that [ν] = n ∈ N, and g : [a, b] → R be a strictly increasing function. Assume that f ∈ C n ([a, b]), g ∈ C 1 ([a, b]), g −1 ∈ C n ([g (a) , g (b)]). ν ν Assume further that f ◦ g −1 ∈ Cg(a) ([g (a) , g (b)]) ∩ Cg(b)− ([g (a) , g (b)]). Set    ν     ν f ◦ g −1 ∞,[g(a),g(b)] ,  Dg(b)− f ◦ g −1 ∞,[g(a),g(b)] . M1 ( f, g) := max  Dg(a) (4.40) Then (i)

 n−1  b   (k) 1  f ◦ g −1 f (x) dg (x) − (g (a)) (z − g (a))k+1   a (k + 1)! k=0

  (k)  +(−1)k f ◦ g −1 (g (b)) (g (b) − z)k+1  ≤  M1 ( f, g)  (z − g (a))ν+1 + (g (b) − z)ν+1 ,  (ν + 2)

(4.41)

∀ z ∈ [g (a) , g (b)] , (ii) at z = g(a)+g(b) , the right hand side of (4.41) is minimized, and we get: 2  n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤

M1 ( f, g) (g (b) − g (a))ν+1 ,  (ν + 2) 2ν

(4.42)

 (k) (k)  (iii) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for all k = 0, 1, . . . , n − 1, we obtain  b  ν+1     ≤ M1 ( f, g) (g (b) − g (a)) , f dg (4.43) (x) (x)    (ν + 2) 2ν a which is a sharp inequality,

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

78

(iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds    n−1  b  1 g (b) − g (a) k+1  f (x) dg (x) −   a N (k + 1)! k=0



  (k)  (k)  j k+1 f ◦ g −1 (g (a)) + (−1)k (N − j)k+1 f ◦ g −1 (g (b))  M1 ( f, g) ≤  (ν + 2)



g (b) − g (a) N

ν+1



 j ν+1 + (N − j)ν+1 ,

(4.44)

(k) (k)   (v) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, k = 1, . . . , n − 1, from (4.44) we get:    

b

 f (x) dg (x) −

a

M1 ( f, g)  (ν + 2)



   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N

g (b) − g (a) N

ν+1



 j ν+1 + (N − j)ν+1 ,

(4.45)

j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (4.45) turns to    

b

 f (x) dg (x) −

a

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

M1 ( f, g) (g (b) − g (a))ν+1 .  (ν + 2) 2ν

(4.46)

Proof Apply Theorem 4.2 for f ◦ g −1 over [g (a) , g (b)] and take into account (4.39).  Next come L 1 estimates: Theorem 4.6 All as in Theorem 4.5. Set   ν     ν M2 ( f, g) := max  Dg(a) f ◦ g −1  L [g(a),g(b)] ,  Dg(b)− f ◦ g −1  L 1

Then



. (4.47)

1 [g(a),g(b)]

4.2 Main Results—I

(i)

79

 n−1  b   (k) 1  f ◦ g −1 f dg − (x) (x) (g (a)) (z − g (a))k+1   a (k + 1)! k=0

  (k)  +(−1)k f ◦ g −1 (g (b)) (g (b) − z)k+1  ≤  M2 ( f, g)  (z − g (a))ν + (g (b) − z)ν ,  (ν + 1)

(4.48)

∀ z ∈ [g (a) , g (b)] , (ii) when ν = 1, from (4.48), we have    

  f (x) dg (x) − [ f (a) (z − g (a)) + f (b) (g (b) − z)] ≤

b

a

      f ◦ g −1 

L 1 ([g(a),g(b)])

(g (b) − g (a)) ,

(4.49)

∀ z ∈ [g (a) , g (b)] , (iii) from (4.49) we obtain (ν = 1 case)    

b

 f (x) dg (x) −

a

      f ◦ g −1  (iv) at z =

g(a)+g(b) , 2

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

L 1 ([g(a),g(b)])

(g (b) − g (a)) ,

(4.50)

ν > 1, the right hand side of (4.48) is minimized, and we get:

 n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤ M2 ( f, g) (g (b) − g (a))ν ,  (ν + 1) 2ν−1

(4.51)

 (k) (k)  (v) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for all k = 0, 1, . . . , n − 1; ν > 1, from (4.51), we obtain

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

80

   

b

a

  M2 ( f, g) (g (b) − g (a))ν f (x) dg (x) ≤ ,  (ν + 1) 2ν−1

(4.52)

which is a sharp inequality, (vi) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds    n−1  b  g (b) − g (a) k+1 1  f (x) dg (x) −   a N (k + 1)! k=0



  (k)  (k)  j k+1 f ◦ g −1 (g (a)) + (−1)k (N − j)k+1 f ◦ g −1 (g (b))  ≤

M2 ( f, g)  (ν + 1)



g (b) − g (a) N

ν



 j ν + (N − j)ν ,

(4.53)

(k) (k)   (vii) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, k = 1, . . . , n − 1, from (4.53) we get:    

b

 f (x) dg (x) −

a

M2 ( f, g)  (ν + 1)



   g (b) − g (a) [ j f (a) + (N − j) f (b)] ≤ N g (b) − g (a) N

ν



 j ν + (N − j)ν ,

(4.54)

j = 0, 1, 2, . . . , N , (viii) when N = 2 and j = 1, (4.54) turns to    

b

 f (x) dg (x) −

a

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

M2 ( f, g) (g (b) − g (a))ν .  (ν + 1) 2ν−1

(4.55)

Proof Application of Theorem 4.3 for f ◦ g −1 over [g (a) , g (b)] and take into account (4.39).  We continue with L p estimates: Theorem 4.7 All as in Theorem 4.5. Let p, q > 1 :    ν M3 ( f, g) := max  Dg(a) f ◦ g −1  L

q

1 p

+

1 q

= 1. Set

 ν   −1  D , g(b)− f ◦ g L [g(a),g(b)]



. (4.56)

q [g(a),g(b)]

4.2 Main Results—I

81

Then (i)

 n−1  b   (k) 1  f ◦ g −1 f (x) dg (x) − (g (a)) (z − g (a))k+1   a (k + 1)! k=0

  (k)  +(−1)k f ◦ g −1 (g (b)) (g (b) − z)k+1  ≤

1 1 M3 ( f, g)

(z − g (a))ν+ p + (g (b) − z)ν+ p , 1  (ν) ν + 1p ( p (ν − 1) + 1) p (4.57) ∀ z ∈ [g (a) , g (b)] , (ii) at z = g(a)+g(b) , the right hand side of (4.57) is minimized, and we get: 2

 n−1  b  1 (g (b) − g (a))k+1  f (x) dg (x) −   a 2k+1 (k + 1)! k=0



f ◦ g −1

(k)

  (k)  (g (a)) + (−1)k f ◦ g −1 (g (b))  ≤

M3 ( f, g) (g (b) − g (a))ν+ p

, 1 1 2ν− q  (ν) ν + 1p ( p (ν − 1) + 1) p 1

(4.58)

(k) (k)   (iii) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for all k = 0, 1, . . . , n − 1, we obtain    

  f (x) dg (x) ≤

M3 ( f, g) (g (b) − g (a))ν+ p

, 1 1 a 2ν− q  (ν) ν + 1p ( p (ν − 1) + 1) p (4.59) which is a sharp inequality, (iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds b

1

  b   n−1   g (b) − g (a) k+1 1  f dg − (x) (x)  N (k + 1)!  a k=0





(k)

(k)  j k+1 f ◦ g −1 (g (a)) + (−1)k (N − j)k+1 f ◦ g −1 (g (b))  ≤

M3 ( f, g)

1  (ν) ν + 1p ( p (ν − 1) + 1) p



  1  g (b) − g (a) ν+ p ν+ 1p ν+ 1 j + (N − j) p , N

(4.60)

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

82

(k) (k)   (v) if f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, k = 1, . . . , n − 1, from (4.60) we get:  b      g (b) − g (a)  ≤ j f + − j) f f dg − [ (a) (N (b)] (x) (x)   N a M3 ( f, g)

1  (ν) ν + 1p ( p (ν − 1) + 1) p



g (b) − g (a) N

ν+ 1 p j

ν+ 1p

ν+ 1p

+ (N − j)

,

(4.61) j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (4.61) turns to    

b

 f (x) dg (x) −

a

   g (b) − g (a) ( f (a) + f (b)) ≤ 2

M3 ( f, g) (g (b) − g (a))ν+ p

. 1 1 2ν− q  (ν) ν + 1p ( p (ν − 1) + 1) p 1

(4.62)

Proof Application of Theorem 4.4 to f ◦ g −1 over [g (a) , g (b)] and using (4.39). 

4.3 Background—II In 1938, Ostsrowski [9] proved the following important inequality. Theorem 4.8 Let f : [a, b] → R be continuous on [a, b] and differentiable   on (a, b)  whose derivative f  : (a, b) → R is bounded on (a, b), i.e.,  f  ∞ := sup  f  (t) < +∞. Then t∈(a,b)

   1  b − a

a

b

 2       x − a+b 1 2  + f (t) dt − f (x) ≤ · (b − a)  f  ∞ , 2 4 (b − a)

for any x ∈ [a, b]. The constant

1 4

(4.63)

is the best possible.

Since then there has been a lot of activity around these inequalities with important applications to numerical analysis and probability. We mention and need the following left generalized g-fractional, of Canavati type, Taylor’s formula: ν Theorem 4.9 ([4]) Let f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]), where x0 ∈ [a, b] is fixed, 0) ν ≥ 1. Then

4.3 Background—II

83

f (x) − f (x0 ) =

[ν]−1 



f ◦ g −1



g(x)

g(x0 )

(g (x0 ))

k!

k=1

1  (ν)

(k)

(g (x) − g (x0 ))k +

 ν   f ◦ g −1 (t) dt, all x ∈ [a, b] : x ≥ x0 . (g (x) − t)ν−1 Dg(x 0) (4.64)

We also mention and need the following right generalized g-fractional, of Canavati type, Taylor’s formula: ν Theorem 4.10 ([4]) Let f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]), where x0 ∈ [a, b] is fixed, 0 )− ν ≥ 1. Then

f (x) − f (x0 ) =

[ν]−1 



f ◦ g −1



g(x0 )

g(x)

(g (x0 ))

k!

k=1

1  (ν)

(k)

(g (x) − g (x0 ))k +

 ν   f ◦ g −1 (t) dt, all a ≤ x ≤ x0 . (t − g (x))ν−1 Dg(x 0 )−

(4.65)

4.4 Main Results—II Next we present generalized g-fractional Ostrowski type inequalities: Theorem 4.11 Let g : [a, b] → R be a strictly increasing function, ν ≥ 1, [ν] = n ∈ N, f ∈ C n ([a, b]) . Assume that g ∈ C 1 ([a, b]), and g −1 ∈ C n ([g (a) , g (b)]). ν For x0 ∈ [a, b], assume that f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]) and f ◦ g −1 ∈ 0) (k)  ν Cg(x ([g (a) , g (b)]). Furthermore assume that f ◦ g −1 (g (x0 )) = 0, all k = 0 )− 1, . . . , n − 1. Then    1   (g (b) − g (a))

b

a

  f (x) dg (x) − f (x0 ) ≤

   1 −1   Dν (g (x0 ) − g (a))ν+1 + g(x0 )− f ◦ g ∞,[g(a),g(x0 )] (g (b) − g (a))  ν D



g(x0 )

 f ◦ g −1 ∞,[g(x

0

 ν+1 ≤ − g (g (b) (x )) 0 ),g(b)]

   1 ν max  Dg(x f ◦ g −1 ∞,[g(a),g(x )] , 0 )− 0  (ν + 2)  ν D

g(x0 )



 f ◦ g −1 ∞,[g(x

 0 ),g(b)]

(g (b) − g (a))ν .

(4.66)

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

84

(k)  Proof By Theorem 4.9, when f ◦ g −1 (g (x0 )) = 0, for k = 1, . . . , n − 1, we get 1  (ν)

f (x) − f (x0 ) =



g(x) g(x0 )

 ν   f ◦ g −1 (t) dt, (g (x) − t)ν−1 Dg(x 0)

(4.67)

∀ x ∈ [x0 , b] .  (k) By Theorem 4.10, when f ◦ g −1 (g (x0 )) = 0, for k = 1, . . . , n − 1, we get 1 f (x) − f (x0 ) =  (ν)



g(x0 )

g(x)

 ν   f ◦ g −1 (t) dt, (t − g (x))ν−1 Dg(x 0 )−

(4.68)

∀ x ∈ [a, x0 ] . Hence 1  (ν)

| f (x) − f (x0 )| ≤

≤

1  (ν) =



g(x)

g(x0 )



g(x) g(x0 )

 ν    f ◦ g −1 (t) dt (g (x) − t)ν−1  Dg(x 0)

  ν   f ◦ g −1 ∞,[g(x (g (x) − t)ν−1 dt  Dg(x 0)

0 ),g(b)]

  1 (g (x) − g (x0 ))ν  −1   Dν g(x0 ) f ◦ g ∞,[g(x0 ),g(b)]  (ν) ν =

  (g (x) − g (x0 ))ν  −1   Dν , g(x0 ) f ◦ g ∞,[g(x0 ),g(b)]  (ν + 1)

(4.69)

∀ x ∈ [x0 , b] . That is

| f (x) − f (x0 )| ≤

   ν    Dg(x0 ) f ◦ g −1 

∞,[g(x0 ),g(b)]

 (ν + 1)

(g (x) − g (x0 ))ν ,

(4.70)

∀ x ∈ [x0 , b] . Similarly, it holds | f (x) − f (x0 )| ≤ 

1  (ν)

g(x0 )



g(x0 )

g(x)

 ν    f ◦ g −1 (t) dt (t − g (x))ν−1  Dg(x 0 )−

  ν   f ◦ g −1 ∞,[g(a),g(x (t − g (x))ν−1 dt  Dg(x 0 )−

≤

1  (ν)

=

 ν   1 f ◦ g −1 ∞,[g(a),g(x )] , (g (x0 ) − g (x))ν  Dg(x 0 )− 0  (ν + 1)

g(x)

0 )]

(4.71)

4.4 Main Results—II

85

∀ x ∈ [a, x0 ] . That is

| f (x) − f (x0 )| ≤

    ν  Dg(x0 )− f ◦ g −1 

∞,[g(a),g(x0 )]

 (ν + 1)

(g (x0 ) − g (x))ν ,

(4.72)

∀ x ∈ [a, x0 ] . We observe that    b   1   f (x) g (x) d x − f (x0 ) =  (g (b) − g (a)) a    1   (g (b) − g (a))

  f (x) g (x) d x − f (x0 ) (g (b) − g (a)) =

b



a

   1   (g (b) − g (a))

b





f (x) g (x) d x −

a

a

   1   (g (b) − g (a))

b

a

1 (g (b) − g (a)) 1 (g (b) − g (a))



x0



b

≤

  f (x0 ) g (x) d x  = 

  ( f (x) − f (x0 )) g  (x) d x  ≤

(4.73)

| f (x) − f (x0 )| g  (x) d x =

a 

| f (x) − f (x0 )| g (x) d x +



b





| f (x) − f (x0 )| g (x) d x

x0

a

(by (4.70), (4.72))

b

⎧    ν −1  ⎪ f ◦ g  ⎨ Dg(x )− 0

1 (g (b) − g (a)) ⎪ ⎩ 

x0

∞,[g(a),g(x0 )]

 (ν + 1)

(g (x0 ) − g (x))ν g  (x) d x+

a

   ν    Dg(x0 ) f ◦ g −1 

∞,[g(x0 ),g(b)]

 (ν + 1)



b

(g (x) − g (x0 ))ν g  (x) d x

x0

⎫ ⎪ ⎬ ⎪ ⎭

=

(4.74)

   1 −1   Dν (g (x0 ) − g (a))ν+1 g(x0 )− f ◦ g ∞,[g(a),g(x0 )] (g (b) − g (a))  (ν + 2)  ν   f ◦ g −1 ∞,[g(x +  Dg(x 0)

0

 ν+1 ≤ − g (g (b) (x )) 0 ),g(b)]

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

86

   1 ν max  Dg(x f ◦ g −1 ∞,[g(a),g(x )] , 0 )− 0  (ν + 2)  ν D



g(x0 )

 f ◦ g −1 ∞,[g(x

 0 ),g(b)]

(g (b) − g (a))ν .

(4.75) 

We continue with an L 1 -estimate: Theorem 4.12 All as in Theorem 4.11. Then    b   1  f (x) dg (x) − f (x0 ) ≤  (g (b) − g (a)) a    1 −1   Dν (g (x0 ) − g (a))ν g(x0 )− f ◦ g L 1 ([g(a),g(x0 )]) (g (b) − g (a))  (ν + 1)  ν   f ◦ g −1  L +  Dg(x 0)

1 [g(x 0

 ν ≤ − g (g (b) (x )) 0 ),g(b)]

   1 ν max  Dg(x f ◦ g −1  L ([g(a),g(x )]) , 0 )− 1 0  (ν + 1)  ν D

g(x0 )



 f ◦ g −1  L

 1 ([g(x 0 ),g(b)])

(g (b) − g (a))ν−1 .



Proof Similar to the proof of Theorem 4.11. We continue with an L p -estimate: Theorem 4.13 All as in Theorem 4.11, and let p, q > 1 :    1   (g (b) − g (a))

a

b

1 p

+

1 q

= 1. Then

  f (x) dg (x) − f (x0 ) ≤ 1



1 (g (b) − g (a))  (ν) ( p (ν − 1) + 1) p ν + 1p    1  ν f ◦ g −1  L (g (x0 ) − g (a))ν+ p  Dg(x 0 )−   1  ν f ◦ g −1  L + (g (b) − g (x0 ))ν+ p  Dg(x 0)

(4.76)

q ([g(a),g(x 0 )])

 q [g(x 0 ),g(b)]

≤

4.4 Main Results—II

87

   −1   Dν

, max g(x0 )− f ◦ g 1 L q ([g(a),g(x0 )])  (ν) ( p (ν − 1) + 1) p ν + 1p 1

 ν D



g(x0 )

 f ◦ g −1  L

 q ([g(x 0 ),g(b)])

(g (b) − g (a))ν− q . 1

Proof Similar to the proof of Theorem 4.11.

(4.77) 

Applications follow: b]), where Proposition 4.14 Let ν ≥ 1 such that [ν] = n ∈ N, and f ∈  C n ([a, ν ν a b [a, b] ⊂ (0, +∞). Assume that f ◦ ln x ∈ Cea e , e ∩ Ceb − ea , eb . Set        M1 f, e x := max  Deνa ( f ◦ ln x)∞,[ea ,eb ] ,  Deνb − ( f ◦ ln x)∞,[ea ,eb ] . (4.78) Then (i)

 n−1  b    k+1 1  f (x) e x d x − ( f ◦ ln x)(k) ea z − ea   a (k + 1)! k=0

k+1    +(−1)k ( f ◦ ln x)(k) eb eb − z ≤ ν+1  b ν+1 M1 ( f, e x )  z − ea , + e −z  (ν + 2)

(4.79)

  ∀ z ∈ ea , eb , a b (ii) at z = e +e , the right hand side of (4.79) is minimized, and we get: 2  k+1  b n−1  b  e − ea 1  x f (x) e d x −   a 2k+1 (k + 1)! k=0

     ( f ◦ ln x)(k) ea + (−1)k ( f ◦ ln x)(k) eb  ≤ ν+1  M1 ( f, e x ) eb − ea ,  (ν + 2) 2ν

(4.80)

  (iii) if ( f ◦ ln x)(k) (ea ) = ( f ◦ ln x)(k) eb = 0, for all k = 0, 1, . . . , n − 1, we obtain    b    M1 ( f, e x ) eb − ea ν+1 x  f (x) e d x  ≤ , (4.81)   (ν + 2) 2ν a

which is a sharp inequality,

4 Generalized Canavati g-Fractional Iyengar and Ostrowski Inequalities

88

(iv) more generally, for j = 0, 1, 2, . . . , N ∈ N, it holds  k+1  b n−1  b  1 e − ea  x f (x) e d x −   a N (k + 1)! k=0



    j k+1 ( f ◦ ln x)(k) ea + (−1)k (N − j)k+1 ( f ◦ ln x)(k) eb  M1 ( f, e x ) ≤  (ν + 2)



eb − ea N

ν+1



 j ν+1 + (N − j)ν+1 ,

(4.82)

  (v) if ( f ◦ ln x)(k) (ea ) = ( f ◦ ln x)(k) eb = 0, k = 1, . . . , n − 1, from (4.82) we get:  b    b   e − ea x  f (x) e d x − [ j f (a) + (N − j) f (b)] ≤  N a

M1 ( f, e x )  (ν + 2)



eb − ea N

ν+1



 j ν+1 + (N − j)ν+1 ,

(4.83)

j = 0, 1, 2, . . . , N , (vi) when N = 2 and j = 1, (4.83) turns to    

b

 f (x) e d x − x

a

eb − ea 2



  ( f (a) + f (b)) ≤

ν+1  M1 ( f, e x ) eb − ea .  (ν + 2) 2ν

(4.84)

Proof By application of Theorem 4.5 for g (x) = e x .



We finish with Proposition 4.15 Let ν ≥ 1, [ν] = n ∈ N, f ∈ C n ([a, b]), [a, b] ⊂ (0, +∞) . For x0 ∈ [a, b], assume that f ◦ e x ∈ Clnν x0 ([ln a, ln b]) , and f ◦ e x ∈ Clnν x0 − ([ln a, ln b]). Furthermore assume that ( f ◦ e x )(k) (ln x0 ) = 0, all k = 1, . . . , n − 1. Then     1  b f (x)   d x − f (x0 ) ≤  b   ln a a x  x ν+1  ν   1 0 x  D  b ln + ln x0 − f ◦ e ∞,[ln a,ln x0 ] a ln a  (ν + 2)  ν D

ln x0



 f ◦ e ∞,[ln x x

  b ν+1 ln 0 ,ln b] x0

≤

(4.85)

4.4 Main Results—II

89

   1 ν x  max  Dln , x0 − f ◦ e ∞,[ln a,ln x0 ]  (ν + 2)  ν D

ln x0



 f ◦ e x ∞,[ln x

Proof By Theorem 4.11, for g (x) = ln x.

  b ν ln . 0 ,ln b] a 

References 1. Anastassiou, G.A.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009) 2. Anastassiou, G.A.: On right fractional calculus. Chaos Solitons Fractals 42, 365–376 (2009) 3. Anastassiou, G.A.: Inteligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 4. Anastassiou, G.A.: Generalized Canavati type Fractional Taylor’s formulae. J. Comput. Anal. Appl. 21(7), 1205–1212 (2016) 5. Anastassiou, G.A.: Canavati fractional Iyengar type Inequalities. Analele Universitatii Oradea, Fasc. Matematica, Tom XXVI(1), 141–151 (2019) 6. Anastassiou, G.: Generalized Canavati type g-fractional Iyengar and Ostrowski type inequalities. Ann. Commun. Math. 2(2), 57–72 (2019) 7. Canavati, J.A.: The Riemann-Liouville Integral. Nieuw Archief Voor Wiskunde 5(1), 53–75 (1987) 8. Iyengar, K.S.K.: Note on a inequality. Math. Stud. 6, 75–76 (1938) 9. Ostrowski, A.: Über die Absolutabweichung einer differtentiebaren Funktion von ihrem Integralmittelwert. Comment. Math. Helv. 10, 226–227 (1938)

Chapter 5

Generalized Canavati g-Fractional Polya Inequalities

We present here generalized Canavati type g-fractional Polya type inequalities. We cover also the iterated case. Our inequalities are with respect to all L p norms: 1 ≤ p ≤ ∞. We finish with applications. See also [6].

5.1 Introduction We are motivated by the following famous Polya’s integral inequality, see [12], [13, p. 62], [14] and [15, p. 83]. Theorem 5.1 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.

(5.1)

a

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]), ∞ [a, b] ⊂ R, x ∈ [a, b]. The gamma function  is given by  (α) = 0 e−t t α−1 dt. We define the left Riemann–Liouville integral 

 Jαa+ f (x) =

1  (α)



x

(x − t)α−1 f (t) dt,

(5.2)

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

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_5

(5.3)

91

5 Generalized Canavati g-Fractional Polya Inequalities

92

α For f ∈ Ca+ ([a, b]), we define the left generalized α-fractional derivative of f over [a, b] as



a+ (m) α Da+ f := J1−β f ,

(5.4)

see [1, p. 24]. Canavati first in [7] introduced the above over [0, 1]. α f ∈ C ([a, b]). Notice that Da+ 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 f (x) := Jb−

1  (α)



b

(t − x)α−1 f (t) dt,

(5.5)

x

x ∈ [a, b], see also [2, 8–10, 16]. Define the subspace of functions α Cb− ([a, b]) =



 1−β f ∈ C m ([a, b]) : Jb− f (m) ∈ C 1 ([a, b]) .

(5.6)

Define the right generalized α-fractional derivative of f over [a, b] as

 1−β α Db− f = (−1)m−1 Jb− f (m) ,

(5.7)

0 α see [2]. We set Db− f = f. Notice that Db− f ∈ C ([a, b]). We are also motivated by the following fractional Polya type integral inequality without any boundary conditions. α Theorem  a+b 5.2 ([4, pp.α 1–7])  a+b Let  0 < α < 1, f ∈ C ([a, b]). Assume f ∈ Ca+ a, 2 and f ∈ Cb− , b . Set 2

Then

 

α α f ∞,[a, a+b ] , Db− f ∞,[ a+b ,b] . M ( f ) = max Da+

(5.8)

   

(5.9)

b

a

   f (x) d x  ≤

b

2

2

| f (x)| d x ≤ M ( f )

(b − a)α+1 .  (α + 2) 2α

a

Inequality (5.9) is sharp, namely it is attained by  f ∗ (x) =

  (x − a)α , x ∈ a, a+b 2 , 0 < α < 1. ,b (b − x)α , x ∈ a+b 2

(5.10)

Clearly here non zero constant functions f are excluded. In this chapter we present very general g-fractional Polya type inequalities.

5.2 Background

93

5.2 Background Here we follow [5]. Let g : [a, b] → R be a strictly increasing function. Let f ∈ C n ([a, b]), n ∈ N. Assume that g ∈ C 1 ([a, b]), and g −1 ∈ C n ([g (a) , g (b)]). Call l := f ◦ g −1 : [g (a) , g (b)] → R. It is clear that l, l  , . . . , l (n) are continuous functions from [g (a) , g (b)] into f ([a, b]) ⊆ R. Let ν ≥ 1 such that [ν] = n, n ∈ N as above, where [·] is the integral part of the number. Clearly when 0 < ν < 1, [ν] = 0 and n = 0. Next we follow [1, pp. 7–9]. (I) Let h ∈ C ([g (a) , g (b)]), we define the left Riemann–Liouville fractional integral as  z  z0  1 (5.11) Jν h (z) := (z − t)ν−1 h (t) dt,  (ν) z0 ∞ for g (a) ≤ z 0 ≤ z ≤ g (b), where  is the gamma function;  (ν) = 0 e−t t ν−1 dt. We set J0z0 h = h. ν Let α := ν − [ν] (0 < α < 1). We define the subspace Cg(x ([g (a) , g (b)]) of 0) [ν] C ([g (a) , g (b)]), where x0 ∈ [a, b] :   g(x ) ν Cg(x ([g (a) , g (b)]) := h ∈ C [ν] ([g (a) , g (b)]) : J1−α0 h ([ν]) ∈ C 1 ([g (x0 ) , g (b)]) . 0)

(5.12) ν So let h ∈ Cg(x , g we define the left g-generalized fractional derivative ([g (a) (b)]); 0) of h of order ν, of Canavati type, over [g (x0 ) , g (b)] as

 g(x0 ) ([ν]) ν h := J h . Dg(x ) 1−α 0

(5.13)

ν Clearly, for h ∈ Cg(x ([g (a) , g (b)]), there exists 0)



 ν Dg(x h (z) = 0)

d 1  (1 − α) dz



z

g(x0 )

(z − t)−α h ([ν]) (t) dt,

(5.14)

for all g (x0 ) ≤ z ≤ g (b) . ν In particular, when f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]) we have that 0) 

  ν Dg(x f ◦ g −1 (z) = 0)

d 1  (1 − α) dz 



z g(x0 )

 ([ν]) (z − t)−α f ◦ g −1 (t) dt,

  (n) n f ◦ g −1 = f ◦ g −1 and for all g (x0 ) ≤ z ≤ g (b) . We have Dg(x ) 0   0 −1 −1 Dg(x0 ) f ◦ g = f ◦g .

(5.15)

We mention and need the following left generalized g-fractional, of Canavati type, Taylor’s formula:

5 Generalized Canavati g-Fractional Polya Inequalities

94

ν Theorem 5.3 Let f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]), where x0 ∈ [a, b] is fixed. 0)

(i) if ν ≥ 1, then f (x) − f (x0 ) =

[ν]−1 



k=1

1  (ν)



g(x) g(x0 )

f ◦ g −1

(k)

(g (x0 ))

k!

(g (x) − g (x0 ))k +

 ν   f ◦ g −1 (t) dt, all x ∈ [a, b] : x ≥ x0 , (g (x) − t)ν−1 Dg(x 0) (5.16)

(ii) if 0 < ν < 1, we get f (x) =



1  (ν)

g(x)

g(x0 )

 

ν f ◦ g −1 (t) dt, all x ∈ [a, b] : x ≥ x0 . (g (x) − t)ν−1 Dg(x 0)

(5.17)

By the change of variable method, see [11], we may rewrite the remainder of (5.16), (5.17), as 1  (ν) 1  (ν)



x x0



g(x) g(x0 )

 ν   f ◦ g −1 (t) dt = (g (x) − t)ν−1 Dg(x 0)

(5.18)

 ν   f ◦ g −1 (g (s)) g  (s) ds, (g (x) − g (s))ν−1 Dg(x 0)

all x ∈ [a, b] : x ≥ x0 . We may rewrite (5.17) as follows: if 0 < ν < 1, we have  ν    f ◦ g −1 (g (x)) , f (x) = Jνg(x0 ) Dg(x 0)

(5.19)

all x ∈ [a, b] : x ≥ x0 . (II) Next we follow [3, pp. 345–348]. Let h ∈ C ([g (a) , g (b)]), we define the right Riemann–Liouville fractional integral as  z0  ν  1 Jz0 − h (z) := (5.20) (t − z)ν−1 h (t) dt,  (ν) z for g (a) ≤ z ≤ z 0 ≤ g (b). We set Jz00 − h = h. ν Let α := ν − [ν] (0 < α < 1). We define the subspace Cg(x ([g (a) , g (b)]) of 0 )− [ν] C ([g (a) , g (b)]), where x0 ∈ [a, b] : ν Cg(x ([g (a) , g (b)]) := 0 )−

5.2 Background



95

 1−α 1 ([ν]) . h ∈ C [ν] ([g (a) , g (b)]) : Jg(x h ∈ C , g ([g (a) (x )]) 0 )− 0

(5.21)

ν So let h ∈ Cg(x ([g (a) , g (b)]); we define the right g-generalized fractional deriva0 )− tive of h of order ν, of Canavati type, over [g (a) , g (x0 )] as



 1−α ν n−1 ([ν]) J Dg(x h := h . (−1) g(x0 )− 0 )−

(5.22)

ν Clearly, for h ∈ Cg(x ([g (a) , g (b)]), there exists 0 )−



ν Dg(x h 0 )−



(−1)n−1 d (z) =  (1 − α) dz



g(x0 )

(t − z)−α h ([ν]) (t) dt,

(5.23)

z

for all g (a) ≤ z ≤ g (x0 ) ≤ g (b) . ν In particular, when f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]) we have that 0 )− 

  (−1)n−1 d ν Dg(x f ◦ g −1 (z) = 0 )−  (1 − α) dz



g(x0 )

 ([ν]) (t − z)−α f ◦ g −1 (t) dt,

z

(5.24)

for all g (a) ≤ z ≤ g (x0 ) ≤ g (b) . We get that 

   (n) n f ◦ g −1 (z) = (−1)n f ◦ g −1 Dg(x (z) 0 )−

(5.25)

 

  0 −1 f ◦ g and Dg(x (z) = f ◦ g −1 (z) , all z ∈ [g (a) , g (x0 )] . 0 )− We mention and need the following right generalized g-fractional, of Canavati type, Taylor’s formula: ν Theorem 5.4 Let f ◦ g −1 ∈ Cg(x ([g (a) , g (b)]), where x0 ∈ [a, b] is fixed. 0 )− (i) if ν ≥ 1, then

f (x) − f (x0 ) =

[ν]−1  k=1

1  (ν)



g(x0 )

g(x)



f ◦ g −1

(k) k!

(g (x0 ))

(g (x) − g (x0 ))k +

 ν   f ◦ g −1 (t) dt, all a ≤ x ≤ x0 , (t − g (x))ν−1 Dg(x 0 )−

(5.26)

(ii) if 0 < ν < 1, we get f (x) =

1  (ν)



g(x0 )

g(x)

 ν   f ◦ g −1 (t) dt, all a ≤ x ≤ x0 . (t − g (x))ν−1 Dg(x 0 )− (5.27)

5 Generalized Canavati g-Fractional Polya Inequalities

96

By change of variable, see [11], we may rewrite the remainder of (5.26), (5.27), as 1  (ν) 1  (ν)



x0 x



g(x0 )

g(x)

 ν   f ◦ g −1 (t) dt = (t − g (x))ν−1 Dg(x 0 )−

(5.28)

 ν   f ◦ g −1 (g (s)) g  (s) ds, (g (s) − g (x))ν−1 Dg(x 0 )−

all a ≤ x ≤ x0 . We may rewrite (5.27) as follows: if 0 < ν < 1, we have  ν    ν Dg(x0 )− f ◦ g −1 (g (x)) , f (x) = Jg(x 0 )−

(5.29)

all a ≤ x ≤ x0 ≤ b. (III) Denote by mν ν ν ν = Dg(x Dg(x . . . Dg(x (m-times), m ∈ N. Dg(x 0) 0) 0) 0)

(5.30)

g(x0 ) = Jνg(x0 ) Jνg(x0 ) . . . Jνg(x0 ) (m-times), m ∈ N. Jmν

(5.31)

Also denote by

mν an iterated fractional derivative. We call Dg(x 0) We mention and need the following g-modified and generalized left fractional Taylor’s formula of Canavati type:  

iν ν Theorem 5.5 Let 0 < ν < 1. Assume that Dg(x f ◦ g −1 ∈ Cg(x ([g (a) , 0) 0)  

(m+1)ν f ◦ g −1 ∈ g (b)]), x0 ∈ [a, b], for i = 0, 1, . . . , m. Assume also that Dg(x0 ) C ([g (x0 ) , g (b)]). Then

 g(x) 

1 (m+1)ν  f ◦ g −1 (z) dz (g (x) − z)(m+1)ν−1 Dg(x 0)  ((m + 1) ν) g(x0 ) (5.32)  x 

1 (m+1)ν  −1  (m+1)ν−1 Dg(x0 ) = f ◦g (g (s)) g (s) ds, (g (x) − g (s))  ((m + 1) ν) x0 f (x) =

all x0 ≤ x ≤ b. (IV) Denote by mν ν ν ν Dg(x = Dg(x Dg(x . . . Dg(x (m-times), m ∈ N. 0 )− 0 )− 0 )− 0 )−

(5.33)

5.2 Background

97

Also denote by mν ν ν = Jg(x Jν . . . Jg(x (m-times), m ∈ N. Jg(x 0 )− 0 )− g(x 0 )− 0 )−

(5.34)

mν an iterated fractional derivative. We call Dg(x 0 )− We mention and need the following g-modified and generalized right fractional Taylor’s formula of Canavati type:  

iν ν −1 f ◦ g ∈ Cg(x Theorem 5.6 Let 0 < ν < 1. Assume that Dg(x ([g (a) , 0 )− 0 )− 

(m+1)ν  g (b)]), x0 ∈ [a, b], for all i = 0, 1, . . . , m. Assume also that Dg(x0 )− f ◦ g −1 ∈ C ([g (a) , g (x0 )]). Then

 g(x0 ) 

1 (m+1)ν  −1 f (x) = f ◦ g (z) dz (z − g (x))(m+1)ν−1 Dg(x 0 )−  ((m + 1) ν) g(x) (5.35)  x0

  1 (m+1)ν = (g (s) − g (x))(m+1)ν−1 Dg(x0 )− f ◦ g −1 (g (s)) g  (s) ds,  ((m + 1) ν) x all a ≤ x ≤ x0 ≤ b. In what follows next it is based on this background.

5.3 Main Results We present the following generalized g-fractional Polya type integral inequalities without any boundary conditions. Theorem 5.7 Let 0 < ν < 1, f ∈ C ([a, b]), g : [a, b] → R be strictly increasν g (a) , g(a)+g(b) , and f ◦ g −1 ∈ ing and g ∈ C 1 ([a, b]). Assume f ◦ g −1 ∈ Cg(a) 2  

g(a)+g(b) ν , g (b) . Set Cg(b)− 2 

M ( f, g) := max D ν



g(a)

ν

D



g(b)−

Then

   

a

b

   f dg  ≤

a

b

 f ◦ g −1 ∞,g(a), g(a)+g(b)  , 2

   

. f ◦g ∞, g(a)+g(b) ,g(b) −1

(5.36)

2

| f | dg ≤ M ( f, g)

(g (b) − g (a))ν+1 .  (ν + 2) 2ν

Inequality (5.37) is sharp, namely it is attained by f ∗ such that

(5.37)

5 Generalized Canavati g-Fractional Polya Inequalities

98



f∗ ◦ g

 −1

⎧ ⎫  ⎨ (z − g (a))ν , z ∈ g (a) , g(a)+g(b) ⎬ 2   , 0 < ν < 1. (5.38) (z) = ⎩ (g (b) − z)ν , z ∈ g(a)+g(b) , g (b) ⎭ 2

Clearly here non zero constant functions f ◦ g −1 are excluded. Proof Notice that    

b

a

    f dg  = 

a

b

   f g d x  ≤

b







| f | g dx =

a

b

| f | dg.

a



Let x0 ∈ [a, b] be such that g (x0 ) = g(a)+g(b) . , that is x0 = g −1 g(a)+g(b) 2 2  

ν Let f ◦ g −1 ∈ Cg(a) g (a) , g(a)+g(b) , 0 < ν < 1. By Theorem 5.3 we have 2 that  g(x)  ν   1 f (x) = f ◦ g −1 (t) dt, (5.39) (g (x) − t)ν−1 Dg(a)  (ν) g(a) all x ∈ [a, x0 ] .  

g(a)+g(b) ν , g , by Theorem 5.4 we have that Assume also f ◦ g −1 ∈ Cg(b)− (b) 2 f (x) =



1  (ν)

g(b)

g(x)

 ν   f ◦ g −1 (t) dt, (t − g (x))ν−1 Dg(b)−

(5.40)

all x ∈ [x0 , b] . By (5.39) we get 1 | f (x)| ≤  (ν)



g(x) g(a)

ν−1

(g (x) − t)



ν   f ◦ g −1 ∞,g(a), g(a)+g(b)  dt Dg(a) 2

ν   (g (x) − g (a))ν . f ◦ g −1 ∞,g(a), g(a)+g(b)  = Dg(a) 2  (ν + 1)

(5.41)

ν   (g (x) − g (a))ν | f (x)| ≤ Dg(a) , f ◦ g −1 ∞,g(a), g(a)+g(b)  2  (ν + 1)

(5.42)

That is

for any x ∈ [a, x0 ] . Similarly, by (5.40) we get 1 | f (x)| ≤  (ν)



g(b)

g(x)

(t − g (x))

ν−1



ν   dt Dg(b)− f ◦ g −1 ∞, g(a)+g(b) ,g(b) 2

5.3 Main Results

99

  (g (b) − g (x))ν −1  .

Dν g(b)− f ◦ g ∞, g(a)+g(b) ,g(b) 2  (ν + 1)

(5.43)

ν   (g (b) − g (x))ν | f (x)| ≤ Dg(b)− , f ◦ g −1 ∞, g(a)+g(b) ,g(b) 2  (ν + 1)

(5.44)

= That is

for any x ∈ [x0 , b] . Thus, we have    

b

a



x0

   f dg  ≤

b

 | f | dg =

a

| f (x)| g  (x) d x +

b

| f | g d x =

a



b

| f (x)| g  (x) d x

(by (5.42), (5.44))

≤

x0

a



ν 

Dg(a) f ◦ g −1

  ∞, g(a), g(a)+g(b) 2

 (ν + 1)

 

ν

Dg(b)− f ◦ g −1



x0

(g (x) − g (a))ν g  (x) d x+

(5.45)

a   ∞, g(a)+g(b) ,g(b) 2

 (ν + 1)



ν 

Dg(a) f ◦ g −1



b

(g (b) − g (x))ν g  (x) d x =

x0   ∞, g(a), g(a)+g(b) 2

 (ν + 1)

 

ν

Dg(b)− f ◦ g −1  g(a)+g(b) ∞,

2

,g(b)

 (ν + 1)



(g (x0 ) − g (a))ν+1 + (ν + 1)

(5.46)

(g (b) − g (x0 ))ν+1 = (ν + 1)

  

ν   1 g (b) − g (a) ν+1 −1  

D f ◦ g + g(a) ∞, g(a), g(a)+g(b) 2  (ν + 2) 2

ν

D

g(b)−



 f ◦ g ∞, g(a)+g(b) ,g(b) −1

2



g (b) − g (a) 2

ν+1  =

   g (b) − g (a)ν+1 ν  1 −1  

D g(a) f ◦ g ν+1 ∞, g(a), g(a)+g(b) 2  (ν + 2) 2

(5.47)

5 Generalized Canavati g-Fractional Polya Inequalities

100



ν   + Dg(b)− f ◦ g −1 ∞, g(a)+g(b) ,g(b) ≤ 2

M ( f, g) (g (b) − g (a))ν+1 .  (ν + 2) 2ν

(5.48)

We have proved inequality (5.37). Next we prove sharpness of (5.37). Here [ν] = 0 and α = ν. We see that



g(a)  J1−α f ∗ ◦ g −1 (z) = =



1  (1 − α)

z g(a)

1  (1 − α)



z g(a)

  (z − t)−α f ∗ ◦ g −1 (t) dt

(z − t)(1−α)−1 (t − g (a))(ν+1)−1 dt =

(5.49)

(by [17, p. 256])  (1 − α)  (ν + 1) 1 (z − g (a)) =  (ν + 1) (z − g (a)) .  (1 − α)  (2) Therefore 

    g (a) + g (b) ν . f ∗ ◦ g −1 (z) =  (ν + 1) , for all z ∈ g (a) , Dg(a) 2

That is

ν

D



g(a)

 f ◦ g −1 ∞,g(a), g(a)+g(b)  =  (ν + 1) .

(5.50)

(5.51)

2

Furthermore we have −



1−α Jg(b)−



−

f∗ ◦ g

−1



1  (1 − α)

−1 (z) =  (1 − α)



g(b)



g(b)

(t − z)−α (g (b) − t)ν dt =

z

(g (b) − t)(ν+1)−1 (t − z)(1−α)−1 dt =

z

(by [17, p. 256]) −

 (ν + 1)  (1 − α) 1 (g (b) − z) =  (ν + 1) (z − g (b)) ,  (1 − α)  (2)

for all z ∈ Hence



g(a)+g(b) , g (b) 2





(5.52)

.

  ν f ∗ ◦ g −1 (z) =  (ν + 1) , Dg(b)−

(5.53)

5.3 Main Results

101

and

ν

D



g(b)−

 f e ◦ g −1 ∞, g(a)+g(b) ,g(b) =  (ν + 1) .

(5.54)

2

Consequently we get that M ( f ∗ , g) =  (ν + 1) .

(5.55)

Applying f ∗ into (5.37) we obtain: R.H.S (5.37) for f ∗ =

 (ν + 1) (g (b) − g (a))ν+1 (g (b) − g (a))ν+1 = . (5.56)  (ν + 2) 2ν (ν + 1) 2ν

We get the same result when we apply f ∗ in the   L.H.S (5.37) = 

  f ∗ (x) dg (x) =

b

a

   

x0



b

f ∗ (x) dg (x) +

x0

a

   

x0



f∗ ◦ g

−1





b

(g (x)) dg (x) +

  f ∗ (x) dg (x) = 

f∗ ◦ g

−1

x0

a

   

g(x0 )



g(a)

   

g(x0 )

g(a)

 f ∗ ◦ g −1 (z) dz + (z − g (a))ν dz +



g(b)



g(x0 )



g(b)

g(x0 )



  (g (x)) dg (x) =

   f ∗ ◦ g −1 (z) dz  =

(5.57)

  (g (b) − z)ν dz  =

(g (x0 ) − g (a))ν+1 (g (b) − g (x0 ))ν+1 + = ν+1 ν+1 2 (g (b) − g (a))ν+1 (g (b) − g (a))ν+1 = . ν+1 2ν+1 (ν + 1) 2ν

(5.58)

So equality holds in (5.37) for f ∗ . Furthermore we notice that 

f∗ ◦ g

−1





g (a) + g (b) 2

  −

 =



= f∗ ◦ g

−1

g (b) − g (a) 2





g (a) + g (b) 2

  (5.59) +

ν ,

thus f ∗ ◦ g −1 is continuous on [g (a) , g (b)] and hence f ∗ ∈ C ([a, b]) .

5 Generalized Canavati g-Fractional Polya Inequalities

102

Sharpness of (5.37) has been proved. The theorem is proved completely.   (k) Remark 5.8 When ν ≥ 1, thus n = [ν] ≥ 1, and by assuming that f ◦ g −1   (k) (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 0, 1, . . . , n − 1, we can prove inequality (5.37) again. Here also f ∈ C n ([a, b]) and g −1 ∈ C n ([g (a) , g (b)]). (n)  cannot be a constant different than zero, equivalently, The function f ◦ g −1 f ◦ g −1 cannot be a non-trivial polynomial of degree n. We continue with a fractional iterated Polya type inequality: Theorem 5.9 Let 0 < ν < 1, f ∈ C ([a, b]), g : [a, b] → R  be strictly increas

  iν ν g (a) , g(a)+g(b) , f ◦ g −1 ∈ Cg(a) ing and g ∈ C 1 ([a, b]). Assume that Dg(a) 2 

   (m+1)ν for i = 0, 1, . . . , m ∈ N, and Dg(a) . Also f ◦ g −1 ∈ C g (a) , g(a)+g(b) 2 



  g(a)+g(b) iν ν assume that Dg(b)− f ◦ g −1 ∈Cg(b)− , g (b) , for i = 0, 1, . . . , m ∈ 2

 

 (m+1)ν  g(a)+g(b) N, and Dg(b)− f ◦ g −1 ∈ C , g (b) . Set 2 



(m+1)ν  M ∗ ( f, g) := max Dg(a) f ◦ g −1



(m+1)ν 

Dg(b)− f ◦ g −1

  ∞, g(a), g(a)+g(b) 2

  ∞, g(a)+g(b) ,g(b) 2

,

.

(5.60)

Then    

b

a

   f dg  ≤

b

| f | dg ≤ M ∗ ( f, g)

a

(g (b) − g (a))(m+1)ν+1 .  ((m + 1) ν + 2) 2(m+1)ν

(5.61)

Proof By Theorem 5.5 we have 1 f (x) =  ((m + 1) ν)



g(x)

g(a)

all x ∈ [a, x0 ], where x0 := g By Theorem 5.6 we have f (x) =

1  ((m + 1) ν)



−1

g(b) g(x)



(m+1)ν  f ◦ g −1 (z) dz, (g (x) − z)(m+1)ν−1 Dg(a)

g(a)+g(b) 2



(5.62) .



(m+1)ν  f ◦ g −1 (z) dz, (z − g (x))(m+1)ν−1 Dg(b)− (5.63)

all x ∈ [x0 , b] . By (5.62) we get | f (x)| ≤

1  ((m + 1) ν)

5.3 Main Results



g(x)

g(a)

=

103

 

(m+1)ν  f ◦ g −1 dz Dg(a)

(m+1)ν−1

(g (x) − z)

  ∞, g(a), g(a)+g(b) 2

 (g (x) − g (a))(m+1)ν

(m+1)ν 

, f ◦ g −1 

Dg(a) ∞, g(a), g(a)+g(b)  ((m + 1) ν + 1) 2

(5.64)

for all x ∈ [a, x0 ] . Similarly, by (5.63), we obtain 1  ((m + 1) ν)

| f (x)| ≤ 

g(b) g(x)

=

 

(m+1)ν  f ◦ g −1 dz Dg(b)−

(m+1)ν−1

(z − g (x))

  ∞, g(a)+g(b) ,g(b) 2

 (g (b) − g (x))(m+1)ν

(m+1)ν 

,

Dg(b)− f ◦ g −1  g(a)+g(b) ∞, ,g(b)  ((m + 1) ν + 1) 2

(5.65)

for all x ∈ [x0 , b] . Thus, we have    

b

a



x0

   f dg  ≤

b

 | f | dg =

a

| f | g d x =

a





b

| f (x)| g (x) d x +

b

| f (x)| g  (x) d x

(by (5.64), (5.65))

≤

x0

a



(m+1)ν 

f ◦ g −1

Dg(a)

  ∞, g(a), g(a)+g(b) 2

 ((m + 1) ν + 1)



(m+1)ν 

Dg(b)− f ◦ g −1  g(a)+g(b) ∞,

2

 ((m + 1) ν + 1)



x0

(g (x) − g (a))(m+1)ν g  (x) d x+ (5.66)

a

,g(b)





b

(g (b) − g (x))(m+1)ν g  (x) d x =

x0



 1

(m+1)ν   (g (x 0 ) − g (a))(m+1)ν+1 f ◦ g −1 

Dg(a) ∞, g(a), g(a)+g(b)  ((m + 1) ν + 2) 2



(m+1)ν  f ◦ g −1 + Dg(b)−

  ∞, g(a)+g(b) ,g(b) 2

(by g (x0 ) =

g(a)+g(b) ) 2

 (g (b) − g (x0 ))

(m+1)ν+1

=

5 Generalized Canavati g-Fractional Polya Inequalities

104

  1 (g (b) − g (a))(m+1)ν+1

(m+1)ν  −1   D f ◦ g

g(a) ∞, g(a), g(a)+g(b)  ((m + 1) ν + 2) 2(m+1)ν+1 2 (5.67) 

 

(m+1)ν  ≤ f ◦ g −1  g(a)+g(b) + Dg(b)− ∞,

2

,g(b)

M ∗ ( f, g) (g (b) − g (a))(m+1)ν+1 ,  ((m + 1) ν + 2) 2(m+1)ν

(5.68) 

proving inequality (5.61). Next we give an L 1 generalized g-fractional Polya inequality:

Theorem 5.10 Let ν ≥ 1, n = [ν] , and f ∈ C n ([a, b]), g : [a, b] → R be a −1 n strictly increasing function. Assume  that g∈C 1 ([a, b])  and g ∈C ([g (a) , ν ν g (b)]). Also assume f ◦ g −1 ∈ Cg(a) g (a) , g(a)+g(b) , and f ◦ g −1 ∈ Cg(b)− 2 

     (k) (k) g(a)+g(b) , g (b) , and that f ◦ g −1 (g (a)) = f ◦ g −1 (g (b)) = 0, for k = 2 0, 1, . . . , n − 1. Set 

ν   f ◦ g −1 L g(a), g(a)+g(b)  , M1 ( f, g) := max Dg(a) 1

ν

D



g(b)−

Then

   

a

b

   f dg  ≤

b

a

 f ◦ g −1 L

2

  1

| f | dg ≤ M1 ( f, g)



.

(5.69)

(g (b) − g (a))ν .  (ν + 1) 2ν−1

(5.70)

g(a)+g(b) ,g(b) 2

Here f ◦ g −1 cannot be a non-trivial polynomial of degree n.  

 (k) ν g (a) , g(a)+g(b) , ν ≥ 1, and f ◦ g −1 Proof Let f ◦ g −1 ∈ Cg(a) (g (a)) = 0, 2 for k = 0, 1, . . . , n − 1, n = [ν] . By Theorem 5.3 we have that 1 f (x) =  (ν)



g(x) g(a)

 ν   f ◦ g −1 (t) dt, (g (x) − t)ν−1 Dg(a)

(5.71)



. all x ∈ [a, x0 ], where x0 := g −1 g(a)+g(b) 2  

(k)  g(a)+g(b) ν Let f ◦ g −1 ∈ Cg(b)− , g (b) , and f ◦ g −1 (g (b)) = 0, for k = 2 0, 1, . . . , n − 1. By Theorem 5.4 we have that f (x) =

1  (ν)



g(b)

g(x)

 ν   f ◦ g −1 (t) dt, (t − g (x))ν−1 Dg(b)−

(5.72)

5.3 Main Results

105

all x ∈ [x0 , b] . By (5.71) we have 1 | f (x)| ≤  (ν)



g(x)

g(a)

 ν    f ◦ g −1 (t) dt ≤ (g (x) − t)ν−1  Dg(a)

(g (x) − g (a))ν−1  (ν)



 ν  D

g(x)



g(a)

g(a)

f ◦ g −1



 (t) dt ≤

(5.73)

  (g (x) − g (a))ν−1 −1   ,

Dν g(a) f ◦ g L 1 g(a), g(a)+g(b) 2  (ν) for all x ∈ [a, x0 ] . Similarly, by (5.72) we get | f (x)| ≤

1  (ν)



g(b)

g(x)

 ν    f ◦ g −1 (t) dt ≤ (t − g (x))ν−1  Dg(b)−

(g (b) − g (x))ν−1  (ν)



 ν  D

g(b)



g(b)−

g(x)

f ◦ g −1



 (t) dt ≤

  (g (b) − g (x))ν−1 −1   ,

Dν g(b)− f ◦ g L 1 g(a)+g(b) ,g(b) 2  (ν) for all x ∈ [x0 , b] . Thus, we have    

b

a



x0

   f dg  ≤

b

 | f | dg =

a

| f (x)| g (x) d x +

a

| f | g d x =

a





b

b

| f (x)| g  (x) d x

(by (5.73), (5.74))

≤

x0



ν 

Dg(a) f ◦ g −1

L1

 (ν)

 

ν

Dg(b)− f ◦ g −1

 

g(a), g(a)+g(b) 2

L1

 (ν)



x0

(g (x) − g (a))ν−1 g  (x) d x+

a 

g(a)+g(b) ,g(b) 2





b x0

(g (b) − g (x))ν−1 g  (x) d x =

(5.74)

5 Generalized Canavati g-Fractional Polya Inequalities

106



ν 

Dg(a) f ◦ g −1

L1

 

g(a), g(a)+g(b) 2

 (ν + 1)

 

ν

Dg(b)− f ◦ g −1  g(a)+g(b) L1

2

,g(b)

(g (x0 ) − g (a))ν +



 (ν + 1)

(g (b) − g (x0 ))ν =

(5.75)

   1 (g (b) − g (a))ν −1  

Dν g(a) f ◦ g L 1 g(a), g(a)+g(b) 2  (ν + 1) 2ν

+ Dν



g(b)−

 f ◦ g L



−1

 1

g(a)+g(b) ,g(b) 2



≤

M1 ( f, g) (g (b) − g (a))ν ,  (ν + 1) 2ν−1

(5.76) 

proving the claim. We continue with a L q generalized g-fractional Polya type inequality:

Theorem 5.11 Let p, q > 1 : 1p + q1 = 1, ν > q1 , n = [ν] , f ∈ C n ([a, b]), g : 1 −1 [a, b] → R be a strictly increasing function. Assume  that g ∈ C ([a,  b]) and g ∈ ν C n ([g (a) , g (b)]). Also assume f ◦ g −1 ∈ Cg(a) g (a) , g(a)+g(b) , and f ◦ g −1 ∈ 2  

(k) (k)   g(a)+g(b) ν Cg(b)− , g (b) , and that f ◦ g −1 (g (a)) = f ◦ g −1 2 (g (b)) = 0, for k = 0, 1, . . . , n − 1. When q1 < ν < 1, the last boundary conditions are void. Set 

ν   f ◦ g −1 L g(a), g(a)+g(b)  , Mq ( f, g) := max Dg(a) q

ν

D

g(b)−

Then  b      f dg  ≤ 

b



 f ◦ g L

2



−1

 q

g(a)+g(b) ,g(b) 2

Mq ( f, g)



.

(5.77)



(g (b) −g (a))ν+ p . 1 a a 2  (ν) ( p (ν − 1) + 1) ν + p (5.78) Again here f ◦ g −1 cannot be a non-trivial polynomial of degree n.  

 (k) ν g (a) , g(a)+g(b) , ν > 0, and f ◦ g −1 Proof Let f ◦ g −1 ∈ Cg(a) (g (a)) = 2 0, for k = 0, 1, . . . , n − 1, n = [ν] . By Theorem 5.3 we have that | f | dg≤

ν− q1

1

1 p

5.3 Main Results

107

1 f (x) =  (ν)



g(x) g(a)

 ν   f ◦ g −1 (t) dt, (g (x) − t)ν−1 Dg(a)

(5.79)



. all x ∈ [a, x0 ], where x0 := g −1 g(a)+g(b) 2  

(k)  g(a)+g(b) ν Let f ◦ g −1 ∈ Cg(b)− , g (b) , and f ◦ g −1 (g (b)) = 0, for k = 2 0, 1, . . . , n − 1. By Theorem 5.4 we have that 1  (ν)

f (x) =



g(b)

g(x)

 ν   f ◦ g −1 (t) dt, (t − g (x))ν−1 Dg(b)−

(5.80)

all x ∈ [x0 , b] . By (5.79) we have | f (x)| ≤ 1  (ν)



g(x) g(a)

1  (ν)



g(x)

g(a)

(g (x) − t)

 ν    f ◦ g −1 (t) dt ≤ (g (x) − t)ν−1  Dg(a)  1p 

p(ν−1)

g(x)

dt g(a)

 ν  D



g(a)

f ◦g

−1



q (t) dt

 q1

  1 (g (x) − g (a))ν−1+ p −1   ,

Dν 1 g(a) f ◦ g L q g(a), g(a)+g(b) 2  (ν) ( p (ν − 1) + 1) p

≤

1

(5.81)

for all x ∈ [a, x0 ] . Similarly, by (5.80) we get 1 | f (x)| ≤  (ν) 1  (ν)



g(b)

g(x)



 ν    f ◦ g −1 (t) dt ≤ (t − g (x))ν−1  Dg(b)−

g(b)

g(x)

(t − g (x))

 1p  p(ν−1)

g(b)

dt g(x)

 ν  D

g(b)−



f ◦g

−1



q (t) dt

  1 (g (b) − g (x))ν−1+ p −1   ,

Dν g(b)− f ◦ g L q g(a)+g(b) ,g(b) 2  (ν) ( p (ν − 1) + 1) 1p

 q1

≤

1

for all x ∈ [x0 , b] . Thus, we have

   

a

 a

x0

| f (x)| g  (x) d x +



b x0

b

   f dg  ≤

b

(5.82)

| f | dg =

a

| f (x)| g  (x) d x

(by (5.81), (5.82))

≤

1 1

 (ν) ( p (ν − 1) + 1) p

5 Generalized Canavati g-Fractional Polya Inequalities

108



ν

D



g(a)

ν

D



g(b)−

 f ◦ g −1 L

 f ◦ g L



−1

1 1

  

g(a), g(a)+g(b) q 2

 q

g(a)+g(b) ,g(b) 2



 (ν) ( p (ν − 1) + 1) p ν + 1p

+ Dν



g(b)−

b



x0

(g (x) − g (a))ν−1+ p g  (x) d x+ 1

a

(g (b) − g (x))

2

 f ◦ g L −1

g(a)

 f ◦ g L −1

≤

 q

g (x) d x = (5.83)





1

ν −1   (g (x ) − g (a))ν+ p

0

Dg(a) f ◦ g L q g(a), g(a)+g(b)

 q

g(a)+g(b) ,g(b) 2



g(a), g(a)+g(b) 2

(g (b) − g (x0 ))

ν+ 1p

 =

(g (b) − g (a))ν+ p 1



1  (ν) ( p (ν − 1) + 1) p ν + 1p 





x0

1



ν

D

ν−1+ 1p



+ Dν

2ν+ p 1



g(b)−

 f ◦ g L



−1

 q



g(a)+g(b) ,g(b) 2

(g (b) − g (a))ν+ p 1

Mq ( f, g)



1  (ν) ( p (ν − 1) + 1) p ν + 1p

2ν− q

1

,

(5.84)



proving the claim. Combining facts from Theorems 5.7, 5.10, 5.11 and Remark 5.8, we get:

Theorem 5.12 Let p, q > 1 : 1p + q1 = 1, ν ≥ 1, n = [ν] , f ∈ C n ([a, b]), 1 g : [a, b] → R be a strictly increasing function. Assume  that g ∈ C ([a,  b]) and g(a)+g(b) ν −1 n −1 , and f ◦ g ∈ C ([g (a) , g (b)]). Also assume f ◦ g ∈ Cg(a) g (a) , 2 

   (k)  (k) g(a)+g(b) ν f ◦ g −1 g −1 ∈ Cg(b)− , g (b) , and that (g (a)) = f ◦ g −1 2 (g (b)) = 0, for k = 0, 1, . . . , n − 1. Then  b   b    ≤ | f | dg ≤ (5.85) f dg   a

 min M ( f, g)

a

(g (b) − g (a))ν+1 (g (b) − g (a))ν , M , f, g) ( 1  (ν + 2) 2ν  (ν + 1) 2ν−1 ⎫ ⎬ 1

Mq ( f, g)

2

ν− q1



(g (b) − g (a))ν+ p , ⎭ 1  (ν) ( p (ν − 1) + 1) ν + p 1 p

where M ( f, g) as in (5.36), M1 ( f, g) as in (5.69) and Mq ( f, g) as in (5.77). Above f ◦ g −1 cannot be a non-trivial polynomial of degree n.

5.3 Main Results

109

Corollary 5.13 Here all as in Theorem 5.12. Then    1   g (b) − g (a)

b

a

  b  1  | f | dg ≤ f dg  ≤ g (b) − g (a) a

 (g (b) − g (a))ν (g (b) − g (a))ν−1 min M ( f, g) , M1 ( f, g) , ν  (ν + 2) 2  (ν + 1) 2ν−1 Mq ( f, g)



(g (b) − g (a)) 1 1 2ν− q  (ν) ( p (ν − 1) + 1) p ν + 1p

ν− q1

⎫ ⎬ ⎭

.

(5.86)

We continue with an L 1 iterated fractional Polya type inequality: 1 m+1

Theorem 5.14 All as in Theorem 5.9, with

≤ ν < 1. Set



M1∗



(m+1)ν  f ◦ g −1 ( f, g) := max Dg(a)



(m+1)ν 

Dg(b)− f ◦ g −1

 L1

L1

 

g(a), g(a)+g(b) 2

g(a)+g(b) ,g(b) 2



,

.

(5.87)

Then    

a

b

   f dg  ≤

a

b

| f | dg ≤ M1∗ ( f, g)

(g (b) − g (a))(m+1)ν .  ((m + 1) ν + 1) 2(m+1)ν−1

(5.88) 

Proof Similar to Theorem 5.10. We continue with an L p iterated fractional Polya type inequality: Theorem 5.15 All as in Theorem 5.9. Let p, q > 1 : 1p + q1 = 1, such that ν < 1. Set 



(m+1)ν  ∗  , f ◦ g −1  Mq ( f, g) := max Dg(a) g(a)+g(b) Lq



(m+1)ν 

Dg(b)− f ◦ g −1 Then

   

a

b

   f dg  ≤

 Lq

a

b

g(a)+g(b) ,g(b) 2

| f | dg ≤

g(a),



.

1 (m+1)q


0, y > 0, and let u ( j) (0) = v ( j) (0) = 0 for j ∈ {0, . . . , n − 1}. Then 

x 0



x

 0

y

 (k)   (k)  u (s) v (t) dsdt ≤ M (n, k, x, y) s 2n−2k−1 + t 2n−2k−1

 2 (x − s) u (n) (s) ds

0

 21 

y

 2 (y − t) v (n) (t) dt

 21

,

(6.5)

0

where M (n, k, x, y) :=

√ xy 1 . 2 2 [(n − k − 1)!] (2n − 2k − 1)

Next we follow [2]. Let α > 0, [a, b] ⊂ R, f : [a, b] → R which is integrable and ψ ∈ C 1 ([a, b]) an increasing function such that ψ (x) = 0, for all x ∈ [a, b]. Consider n = α, the ceiling of α. The left and right fractional integrals are defined, respectively, as follows:  x 1 α,ψ ψ (t) (ψ (x) − ψ (t))α−1 f (t) dt, (6.6) Ia+ f (x) :=  (α) a and α,ψ

Ib− f (x) :=

1  (α)



b

ψ (t) (ψ (t) − ψ (x))α−1 f (t) dt,

(6.7)

x

for any x ∈ [a, b], where  is the gamma function. The following semigroup property is valid for fractional integrals: if α, β > 0, then α,ψ β,ψ

α+β,ψ

Ia+ Ia+ f (x) = Ia+

α,ψ β,ψ

α+β,ψ

f (x) , and Ib− Ib− f (x) = Ib−

f (x) .

6.1 Background

115

We mention Definition 6.2 ([2]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ (x) = 0, for all x ∈ [a, b]. The left ψCaputo fractional derivative of f of order α is given by C

α,ψ Da+

 f (x) :=

n−α,ψ Ia+

1 d ψ (x) d x

n f (x) ,

(6.8)

and the right ψ-Caputo fractional derivative of f is given by C

α,ψ Db−

f (x) :=

n−α,ψ Ib−

 −

1 d ψ (x) d x

n f (x) .

(6.9)

To simplify notation, we will use the symbol  f ψ[n]

(x) :=

1 d ψ (x) d x

n f (x) ,

(6.10)

with f ψ[0] (x) = f (x). By the definition, when α = m ∈ N, we have α,ψ

Da+ f (x) = f ψ[m] (x) and C α,ψ Db− f (x) = (−1)m f ψ[m] (x) . C

(6.11)

If α ∈ / N, we have C

α,ψ Da+

1 f (x) =  (n − α)



x

a

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

(6.12)

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

(6.13)

and C

α,ψ

Db− f (x) =

(−1)n  (n − α)



b x

∀ x ∈ [a, b]. In particular, when α ∈ (0, 1), we have α,ψ

Da+ f (x) = and C α,ψ Db− f (x) = C

∀ x ∈ [a, b].

1 (1−α) −1 (1−α)

x a

b x

(ψ (x) − ψ (t))−α f (t) dt, (6.14) (ψ (t) − ψ (x))−α f (t) dt,

6 Caputo Generalized ψ-Fractional …

116

Clearly the above is a generalization of left and right Caputo fractional derivatives. For more see [2]. Still we need from [2] the following left and right fractional Taylor’s formulae: Theorem 6.3 ([2]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ (x) = 0, for all x ∈ [a, b]. Then, the left fractional Taylor formula follows, f (x) =

n−1 f ψ[k] (a)

k!

k=0

α,ψ C

(ψ (x) − ψ (a))k + Ia+

α,ψ

Da+ f (x) ,

(6.15)

and the right fractional Taylor formula follows, f (x) =

n−1

f ψ[k] (b)

(−1)k

k!

k=0

α,ψ C

(ψ (b) − ψ (x))k + Ib−

α,ψ

Db− f (x) ,

(6.16)

∀ x ∈ [a, b]. In particular, given α ∈ (0, 1), we have α,ψ

α,ψ

f (x) = f (a) + Ia+ C Da+ f (x) , and α,ψ α,ψ f (x) = f (b) + Ib− C Db− f (x) ,

(6.17)

∀ x ∈ [a, b]. Remark 6.4 For convenience we can rewrite (6.15)–(6.17) as follows: f (x) =

n−1 f ψ[k] (a) k=0

1  (α)



x

f (x) =

ψ (t) (ψ (x) − ψ (t))α−1

n−1 (−1)k f ψ[k] (b) k=0

∀ x ∈ [a, b].

(ψ (x) − ψ (a))k +

C

(6.18)

α,ψ

Da+ f (t) dt,

a

and

1  (α)

k!



b x

k!

(ψ (b) − ψ (x))k +

ψ (t) (ψ (t) − ψ (x))α−1

C

α,ψ

Db− f (t) dt,

(6.19)

6.1 Background

117

When α ∈ (0, 1), we get: f (x) = f (a) + and f (x) = f (b) +

1 (α) 1 (α)

x a

b x

α,ψ

ψ (t) (ψ (x) − ψ (t))α−1

C

Da+ f (t) dt

ψ (t) (ψ (t) − ψ (x))α−1

C

Db− f (t) dt

α,ψ

(6.20)

∀ x ∈ [a, b]. Again from [2] we have the following: Consider the norms ·∞ : C ([a, b]) → R and ·C [n] : C n ([a, b]) → R, where ψ  n 

 [k]   f C [n] :=  fψ  . ψ

k=0

∞

We have Theorem 6.5 ([2]) The ψ-Caputo fractional derivatives are bounded operators. For all α > 0 (n = α)   C α,ψ  (6.21)  Da+  ≤ K  f C [n] ∞

  C α,ψ   Db− 

and

∞

where K =

ψ

≤ K  f C [n] , ψ

(ψ (b) − ψ (a))n−α > 0.  (n + 1 − α)

(6.22)

(6.23)

We make Remark 6.6 In our setting, clearly, it is f ψ[n] ∈ C ([a, b]). Given f ∈ C ([a, b]), by α,ψ

Theorem 4.10, p. 98 of [4], we get that Ia+ f ∈ C ([a, b]), and by Theorem 4.11, p. α,ψ 101 of [4], we get that Ib− f ∈ C ([a, b]) . α,ψ α,ψ Therefore, we obtain that C Da+ f,C Db− f ∈ C ([a, b]) .

6.2 Main Results We present a generalized left fractional Poincaré type inequality: Theorem 6.7 Let p, q > 1 such that 1p + q1 = 1, and α > q1 , n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being strictly increasing and ψ (x) = 0, for all x ∈ [a, b]. Assume that f ψ[k] (a) = 0, for k = 0, 1, . . . , n − 1. Then  f q,[a,b]

1 1 

(ψ (b) − ψ (a))α− q (b − a) q   C α,ψ  ≤ .  Da+ f ◦ ψ −1  1 q,[ψ(a),ψ(b)]  (α) ( p (α − 1) + 1) p (6.24)

6 Caputo Generalized ψ-Fractional …

118

Proof By (6.18), since f ψ[k] (a) = 0, for k = 0, 1, . . . , n − 1, we have 1  (α)

f (x) =



x

ψ (t) (ψ (x) − ψ (t))α−1

C

α,ψ

Da+ f (t) dt,

(6.25)

a

∀ x ∈ [a, b]. Hence | f (x)| ≤

1  (α)

1  (α)



x



x

   α,ψ  ψ (t) (ψ (x) − ψ (t))α−1 C Da+ f (t) dt =

a

   α,ψ  (ψ (x) − ψ (t))α−1 C Da+ f (t) dψ (t) =

(6.26)

a

(by ψ being strictly increasing) 1  (α) 1  (α)





ψ(x) ψ(a)



   α,ψ (ψ (x) − z)α−1  C Da+ f ψ −1 (z)  dz ≤  1p 

ψ(x)

(ψ (x) − z) p(α−1) dz

ψ(a)

p(α−1)+1

1 (ψ (x) − ψ (a)) p  (α) ( p (α − 1) + 1) 1p

ψ(x)

ψ(a)

 q1  

 C α,ψ  −1 q D f ψ dz ≤ (z)   a+

 

 C α,ψ   Da+ f ◦ ψ −1 

q,[ψ(a),ψ(b)]

.

(6.27)

That is | f (x)| ≤

(ψ (x) − ψ (a))

p(α−1)+1 p

 (α) ( p (α − 1) + 1)

1 p

 

 C α,ψ   Da+ f ◦ ψ −1 

q,[ψ(a),ψ(b)]

,

(6.28)

∀ x ∈ [a, b]. Thus, we have q q

(ψ (x) − ψ (a))( p(α−1)+1) p   C α,ψ −1  | f (x)| ≤ D f ◦ ψ ,   q a+ q,[ψ(a),ψ(b)] ( (α))q ( p (α − 1) + 1) p

q

∀ x ∈ [a, b]. Therefore it holds  a

b

q

| f (x)|q d x ≤

(ψ (b) − ψ (a))( p(α−1)+1) p (b − a) q

( (α))q ( p (α − 1) + 1) p

(6.29)

6.2 Main Results

119

 q

 C α,ψ   Da+ f ◦ ψ −1 

q,[ψ(a),ψ(b)]

and



b

| f (x)| d x

 q1

q

,

(6.30)

(ψ (b) − ψ (a))α− q (b − a) q 1

≤

a

1

1

 (α) ( p (α − 1) + 1) p

 

 C α,ψ   Da+ f ◦ ψ −1 

q,[ψ(a),ψ(b)]

,

(6.31) 

proving the claim. Next we give a generalized right fractional Poincaré type inequality:

Theorem 6.8 Let p, q > 1 such that 1p + q1 = 1, and α > q1 , n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being strictly increasing and ψ (x) = 0, for all x ∈ [a, b]. Assume that f ψ[k] (b) = 0, for k = 0, 1, . . . , n − 1. Then  f q,[a,b] ≤

1 1 

(ψ (b) − ψ (a))α− q (b − a) q   C α,ψ −1  D f ◦ ψ .   1 b− q,[ψ(a),ψ(b)]  (α) ( p (α − 1) + 1) p (6.32)

Proof By (6.19), since f ψ[k] (b) = 0, for k = 0, 1, . . . , n − 1, we have 1 f (x) =  (α)



b

ψ (t) (ψ (t) − ψ (x))α−1

x

C

α,ψ

Db− f (t) dt,

(6.33)

∀ x ∈ [a, b]. Hence 1 | f (x)| ≤  (α) 1  (α)



b x



b x

   α,ψ  ψ (t) (ψ (t) − ψ (x))α−1 C Db− f (t) dt =

   α,ψ  (ψ (t) − ψ (x))α−1 C Db− f (t) dψ (t) =

(6.34)

(by ψ being strictly increasing) 1  (α) 1  (α)



ψ(b)

ψ(x)



ψ(b)

ψ(x)



   α,ψ (z − ψ (x))α−1  C Db− f ψ −1 (z)  dz ≤  1p 

(z − ψ (x))

p(α−1)

ψ(b)

dz ψ(x)

 q1  

 C α,ψ  −1 q ≤  Db− f ψ (z)  dz

6 Caputo Generalized ψ-Fractional …

120 p(α−1)+1

1 (ψ (b) − ψ (x)) p  (α) ( p (α − 1) + 1) 1p

 

 C α,ψ   Db− f ◦ ψ −1 

q,[ψ(a),ψ(b)]

.

(6.35)

That is | f (x)| ≤

(ψ (b) − ψ (x))

p(α−1)+1 p 1

 (α) ( p (α − 1) + 1) p

 

 C α,ψ   Db− f ◦ ψ −1 

q,[ψ(a),ψ(b)]

,

(6.36)

∀ x ∈ [a, b]. Thus, we have q q

(ψ (b) − ψ (x))( p(α−1)+1) p   C α,ψ  | f (x)| ≤ Db− f ◦ ψ −1  , q  q q,[ψ(a),ψ(b)] p ( (α)) ( p (α − 1) + 1)

q

(6.37)

∀ x ∈ [a, b]. Therefore it holds 

q

b

| f (x)| d x ≤ q

(ψ (b) − ψ (a))( p(α−1)+1) p (b − a) q

( (α))q ( p (α − 1) + 1) p

a

 q

 C α,ψ   Db− f ◦ ψ −1 

q,[ψ(a),ψ(b)]

and



b

| f (x)| d x

 q1

q

,

(6.38)

(ψ (b) − ψ (a))α− q (b − a) q 1

≤

1

1

 (α) ( p (α − 1) + 1) p

a

 

 C α,ψ   Db− f ◦ ψ −1 

q,[ψ(a),ψ(b)]

,

(6.39) 

proving the claim. We continue with a generalized left fractional Sobolev type inequality: Theorem 6.9 All as in Theorem 6.7, r > 0. Then  f r,[a,b]

1 1 

(ψ (b) − ψ (a))α− q (b − a) r   C α,ψ −1  ≤ D f ◦ ψ . (6.40)   a+ 1 q,[ψ(a),ψ(b)]  (α) ( p (α − 1) + 1) p

Proof From (6.28) we have

| f (x)|r ≤

(ψ (x) − ψ (a))

p(α−1)+1 p

r

( (α))r ( p (α − 1) + 1)

r p

 r

 C α,ψ   Da+ f ◦ ψ −1 

q,[ψ(a),ψ(b)]

,

(6.41)

6.2 Main Results

121

∀ x ∈ [a, b]. Hence we have  b a



p(α−1)+1 p

r

r

(b − a)   C α,ψ  ,  Da+ f ◦ ψ −1  r q,[ψ(a),ψ(b)] r p ( (α)) ( p (α − 1) + 1)

(ψ (b) − ψ (a))

| f (x)|r d x ≤

(6.42)

That is 

b

| f (x)|r d x

 r1

≤

a

1 

(b − a) r   C α,ψ  ,  Da+ f ◦ ψ −1  1 q,[ψ(a),ψ(b)] p  (α) ( p (α − 1) + 1)

(ψ (b) − ψ (a))

p(α−1)+1 p

(6.43) 

proving the claim. We continue with a generalized right fractional Sobolev type inequality: Theorem 6.10 All as in Theorem 6.8, r > 0. Then  f r,[a,b] ≤

1 1 

(ψ (b) − ψ (a))α− q (b − a) r   C α,ψ −1  D f ◦ ψ . (6.44)   1 b− q,[ψ(a),ψ(b)]  (α) ( p (α − 1) + 1) p

Proof From (6.36) we get that

| f (x)|r ≤

(ψ (b) − ψ (x))

p(α−1)+1 p

r r

( (α))r ( p (α − 1) + 1) p

 r

 C α,ψ   Db− f ◦ ψ −1 

q,[ψ(a),ψ(b)]

,

(6.45)

∀ x ∈ [a, b]. Hence we have  b a



| f (x)|r d x ≤

p(α−1)+1 p

r



(b − a)   C α,ψ −1 r D f ◦ ψ ,   r b− q,[ψ(a),ψ(b)] ( (α))r ( p (α − 1) + 1) p

(ψ (b) − ψ (a))

(6.46)

That is  b a

1 | f (x)|r d x

r

≤

p(α−1)+1 p

1 

(b − a) r   C α,ψ  ,  Db− f ◦ ψ −1  1 q,[ψ(a),ψ(b)] p  (α) ( p (α − 1) + 1)

(ψ (b) − ψ (a))

proving the claim.

(6.47) 

Next comes a generalized left fractional Hilbert–Pachpatte inequality: Theorem 6.11 Let p, q > 1 such that 1p + q1 = 1, α1 > q1 , n 1 ∈ N : n 1 = α1 ; and α2 > 1p , n 2 ∈ N : n 2 = α2 ; [ai , bi ] ⊂ R, for i = 1, 2. Let f i , ψi ∈ C ni ([ai , bi ]) with ψi being strictly increasing and ψi (xi ) = 0, for all xi ∈ [ai , bi ], for i = 1, 2. [ki ] Assume that f iψ (ai ) = 0, for ki = 0, 1, . . . , n i − 1; i = 1, 2. Then i

6 Caputo Generalized ψ-Fractional …

122





b1

a1

b2



a2

| f 1 (x1 )| | f 2 (x2 )| d x1 d x2 (ψ1 (x1 )−ψ1 (a1 )) p(α1 −1)+1 p

+

(ψ2 (x2 )−ψ2 (a2 ))q (α2 −1)+1 q

≤

(b1 − a1 ) (b2 − a2 ) 1

  C α1 ,ψ1  Da 1 +

1

 (α1 )  (α2 ) ( p (α1 − 1) + 1) p (q (α2 − 1) + 1) q   

  C α2 ,ψ2  f 1 ◦ ψ1−1   Da2 + f 2 ◦ ψ2−1  q,[ψ1 (a1 ),ψ1 (b1 )]

p,[ψ2 (a2 ),ψ2 (b2 )]

. (6.48)

[ki ] Proof Since f iψ (ai ) = 0, for ki = 0, 1, . . . , n i − 1; i = 1, 2, by (6.18) we get that: i

f i (xi ) =

1  (αi )



xi

ai

ψi (ti ) (ψi (xi ) − ψi (ti ))αi −1

C

α ,ψ

Daii+ i f i (ti ) dti ,

(6.49)

∀ xi ∈ [ai , bi ] , i = 1, 2. Hence it holds  xi   1  α ,ψ  | f i (xi )| ≤ ψi (ti ) (ψi (xi ) − ψi (ti ))αi −1 C Daii+ i f i (ti ) dti , (6.50)  (αi ) ai ∀ xi ∈ [ai , bi ] , i = 1, 2. Let p, q > 1 : 1p + q1 = 1, then 1 | f 1 (x1 )| ≤  (α1 ) 1  (α1 )



1  (α1 )

x1

a1





x1

a1

   α ,ψ  ψ1 (t1 ) (ψ1 (x1 ) − ψ1 (t1 ))α1 −1 C Da11+ 1 f 1 (t1 ) dt1 =

   α ,ψ  (ψ1 (x1 ) − ψ1 (t1 ))α1 −1 C Da11+ 1 f 1 (t1 ) dψ1 (t1 ) =

ψ1 (x1 ) ψ1 (a1 )



   α ,ψ (ψ1 (x1 ) − z 1 )α1 −1  C Da11+ 1 f 1 ψ1−1 (z 1 )  dz 1 ≤

1  (α1 ) 



ψ1 (x1 ) ψ1 (a1 )

ψ1 (x1 ) ψ1 (a1 )

(ψ1 (x1 ) − z 1 )

p(α1 −1)

 1p dz 1

 q1 

 q  C α1 ,ψ1 −1 ≤  Da1 + f 1 ψ1 (z 1 )  dz 1 p (α1 −1)+1

1 (ψ1 (x1 ) − ψ1 (a1 )) p 1  (α1 ) ( p (α1 − 1) + 1) p That is

(6.51)

 

 C α1 ,ψ1   Da1 + f 1 ◦ ψ1−1 

q,[ψ1 (a1 ),ψ1 (b1 )]

.

(6.52)

6.2 Main Results

| f 1 (x1 )| ≤

123

(ψ1 (x1 ) − ψ1 (a1 ))

p (α1 −1)+1 p

 (α1 ) ( p (α1 − 1) + 1)

1 p

 

 C α1 ,ψ1   Da1 + f 1 ◦ ψ1−1 

q,[ψ1 (a1 ),ψ1 (b1 )]

,

(6.53)

∀ x1 ∈ [a1 , b1 ], where α1 > q1 . Similarly, by assuming α2 > 1p , we get | f 2 (x2 )| ≤

(ψ2 (x2 ) − ψ2 (a2 ))

q (α2 −1)+1 q

 (α2 ) (q (α2 − 1) + 1)

1 q

 

 C α2 ,ψ2   Da2 + f 2 ◦ ψ2−1 

p,[ψ2 (a2 ),ψ2 (b2 )]

,

(6.54)

∀ x2 ∈ [a2 , b2 ] . Hence we have (by (6.53) and (6.54) multiplication) 1

| f 1 (x1 )| | f 2 (x2 )| ≤

  C α1 ,ψ1  Da 1 +

1

(ψ1 (x1 ) − ψ1 (a1 )) 

 f 1 ◦ ψ1−1 

p (α1 −1)+1 p

q,[ψ1 (a1 ),ψ1 (b1 )]

(ψ2 (x2 ) − ψ2 (a2 ))  

 C α2 ,ψ2   Da2 + f 2 ◦ ψ2−1  1

1

(by using Young’s inequality for a, b ≥ 0, a p b q ≤ ≤ 

1

 (α1 )  (α2 ) ( p (α1 − 1) + 1) p (q (α2 − 1) + 1) q

a p

q (α2 −1)+1 q

p,[ψ2 (a2 ),ψ2 (b2 )]

(6.55)

+ qb )

1 1

1

 (α1 )  (α2 ) ( p (α1 − 1) + 1) p (q (α2 − 1) + 1) q

(ψ2 (x2 ) − ψ2 (a2 ))q(α2 −1)+1 (ψ1 (x1 ) − ψ1 (a1 )) p(α1 −1)+1 + p q

 

  C α1 ,ψ1  Da1 + f 1 ◦ ψ1−1 

q,[ψ1 (a1 ),ψ1 (b1 )]

 

 C α2 ,ψ2   Da2 + f 2 ◦ ψ2−1 



p,[ψ2 (a2 ),ψ2 (b2 )]

(6.56)

∀ xi ∈ [ai , bi ] , i = 1, 2. Therefore we can write 

| f 1 (x1 )| | f 2 (x2 )| (ψ1 (x1 )−ψ1 (a1 )) p

 

 C α1 ,ψ1   Da1 + f 1 ◦ ψ1−1 

p (α1 −1)+1

q,[ψ1 (a1 ),ψ1 (b1 )]

+

(ψ2 (x2 )−ψ2 (a2 ))q (α2 −1)+1 q

 

 C α2 ,ψ2   Da2 + f 2 ◦ ψ2−1  1 p

≤

p,[ψ2 (a2 ),ψ2 (b2 )] 1

 (α1 )  (α2 ) ( p (α1 − 1) + 1) (q (α2 − 1) + 1) q ∀ xi ∈ [ai , bi ] , i = 1, 2.

, (6.57)

6 Caputo Generalized ψ-Fractional …

124

The denominator in the left hand side of (6.57) can be zero only when both x1 = a1 and x2 = a2 .  By integrating (6.57) over [a1 , b1 ] × [a2 , b2 ] we obtain (6.48). Next comes a generalized right fractional Hilbert–Pachpatte inequality: Theorem 6.12 Let p, q > 1 such that 1p + q1 = 1, α1 > q1 , n 1 ∈ N : n 1 = α1 ; and α2 > 1p , n 2 ∈ N : n 2 = α2 ; [ai , bi ] ⊂ R, for i = 1, 2. Let f i , ψi ∈ C ni ([ai , bi ]) with ψi being strictly increasing and ψi (xi ) = 0, for all xi ∈ [ai , bi ], for i = 1, 2. [ki ] Assume that f iψ (bi ) = 0, for ki = 0, 1, . . . , n i − 1; i = 1, 2. Then i 

b1

a1



b2

a2



| f 1 (x1 )| | f 2 (x2 )| d x1 d x2 (ψ1 (b1 )−ψ1 (x1 )) p(α1 −1)+1 p

+

(ψ2 (b2 )−ψ2 (x2 ))q (α2 −1)+1 q

≤

(b1 − a1 ) (b2 − a2 ) 1

  C α1 ,ψ1  Db1 −

1

 (α1 )  (α2 ) ( p (α1 − 1) + 1) p (q (α2 − 1) + 1) q   

  C α2 ,ψ2  f 1 ◦ ψ1−1   Db2 − f 2 ◦ ψ2−1  q,[ψ1 (a1 ),ψ1 (b1 )]

p,[ψ2 (a2 ),ψ2 (b2 )]

. (6.58)

Proof Similar to the proof of Theorem 6.11, but now using (6.19). We omit the details.  Preparing to give multivariate results we make   Remark 6.13 Let N ≥ 2, S N −1 := x ∈ R N : |x| = 1 the unit sphere on R N , where |·| stands for the Euclidean norm in R N . Also denote the ball   B (0, R) := x ∈ R N : |x| < R ⊆ R N , R > 0, and the spherical shell A := B (0, R2 ) − B (0, R1 ), 0 < R1 < R2 . For the following see [8, pp. 149–150] and [9, pp. 87–88]. For x ∈ R N − {0} we can write uniquely x = r ω, where r = |x| > 0, and ω = x ∈ S N −1 , |ω| = 1. r Clearly here R N − {0} = (0, ∞) × S N −1 , and

A = [R1 , R2 ] × S N −1 .

In the sequel the following related theorem will be used:

6.2 Main Results

125

Theorem 6.14 ([3], p. 458) Let f : A → R be a Lebesgue integrable function. Then 



 f (x) d x = A

S N −1

R2

f (r ω) r

N −1

 dr dω.

(6.59)

R1

So we are able to write an integral in polar form using the polar coordinates (r, ω) . We need Definition 6.15 Let α > 0, n ∈ N such that n = α, [R1 , R2 ] ⊂ R; R2 > R1 > 0, and f (·ω) , ψ ∈ C n ([R1 , R2 ]) for any ω ∈ S N −1 , with ψ being increasing and ψ (r ) = 0, for all r ∈ [R1 , R2 ]. Denote by   1 d n [n] f (r ω) , (6.60) f ψ (r ω) := ψ (r ) dr with f ψ[0] (r ω) = f (r ω), ∀ ω ∈ S N −1 . The left ψ-Caputo fractional radial derivative of f : A → R of order α is given by: (x ∈ A is written as x = r ω) for α ∈ / N, we have C

1  (n − α)



α,ψ

α,ψ

D R1 + f (x) =C D R1 + f (r ω) := r R1

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

(6.61)

for α = m ∈ N, we have C

α,ψ

α,ψ

D R1 + f (x) =C D R1 + f (r ω) = f ψ[m] (x) = f ψ[m] (r ω) .

(6.62)

And the right ψ-Caputo fractional radial derivative of f : A → R of order α is given by: α,ψ C α,ψ D R2 − f (x) =C D R2 − f (r ω) := (−1)n  (n − α)

 r

R2

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

(6.63)

for α = m ∈ N, we have C

α,ψ

α,ψ

D R2 − f (x) =C D R2 − f (r ω) = (−1)m f ψ[m] (x) = (−1)m f ψ[m] (r ω) .

(6.64)

6 Caputo Generalized ψ-Fractional …

126

We set C

0,ψ

0,ψ

D R1 + f =C D R2 − f := f.

Next we present a generalized left fractional Poincaré type inequality on the shell A. Theorem 6.16 Let p, q > 1 such that 1p + q1 = 1, and α > q1 , n ∈ N such that   n = α. Let f ∈ C n A , ψ ∈ C n ([R1 , R2 ]) with ψ being strictly increasing and ψ (r ) = 0, for all r ∈ [R1 , R2 ]. Assume that f ψ[k] (R1 ω) = 0, for k = 0, 1, . . . , n −   α,ψ 1; ∀ ω ∈ S N −1 , and C D R1 + f ∈ C A . Then

 f q,A ≤

  q1 ψ 

∞,[R1 ,R2 ] (ψ (R2 ) − ψ (R1 ))

α− q1 1

 (α) ( p (α − 1) + 1) p

1 N −1  (R2 − R1 ) q  R2  q  C α,ψ  .  D R1 + f  q,A R1

(6.65)

Proof By (6.18), since f ψ[k] (R1 ω) = 0, for k = 0, 1, . . . , n − 1, ∀ ω ∈ S N −1 , we have  r 1 α,ψ f (r ω) = ψ (t) (ψ (r ) − ψ (t))α−1 C D R1 + f (tω) dt, (6.66)  (α) R1 ∀ r ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Hence  r   1  α,ψ  | f (r ω)| ≤ ψ (t) (ψ (r ) − ψ (t))α−1 C D R1 + f (tω) dt =  (α) R1 1  (α)



r R1

    α,ψ (ψ (r ) − ψ (t))α−1 C D R1 + f (tω) dψ (t) =

(6.67)

(by ψ being strictly increasing) 1  (α) 1  (α)



ψ(r ) ψ(R1 )



ψ(r )

ψ(R1 )



   α,ψ (ψ (r ) − z)α−1  C D R1 + f ψ −1 (z) ω  dz ≤  1p 

(ψ (r ) − z)

p(α−1)

dz ψ(R1 )

p(α−1)+1 p

1 (ψ (r ) − ψ (R1 ))  (α) ( p (α − 1) + 1) 1p That is

ψ(r )



ψ(R2 )

ψ(R1 )

 q1 

 q  C α,ψ −1 ≤  D R1 + f ψ (z) ω  dz

(6.68)  q1  

 q   C α,ψ  D R1 + f ψ −1 (z) ω  dz .

6.2 Main Results

127

(ψ (r ) − ψ (R1 ))

| f (r ω)| ≤

p(α−1)+1 p



1

 (α) ( p (α − 1) + 1) p

ψ(R2 )

ψ(R1 )

 q1 

 q  C α,ψ −1 D f ψ dz , ω (z)   R1 + (6.69)

∀ r ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Thus, we have q

| f (r ω)| ≤ q

(ψ (r ) − ψ (R1 ))( p(α−1)+1) p



q

( (α))q ( p (α − 1) + 1) p

ψ(R2 )

ψ(R1 )

 

 q  C α,ψ  D R1 + f ψ −1 (z) ω  dz , (6.70)

∀ r ∈ [R1 , R2 ], ∀ ω ∈ S N −1 . Therefore it holds 

R2

q

| f (r ω)|q dr ≤

(ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) q

( (α))q ( p (α − 1) + 1) p

R1



ψ(R2 )

ψ(R1 )

 

 q  C α,ψ  D R1 + f ψ −1 (z) ω  dz = q

(ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 )



R2

q

( (α))q ( p (α − 1) + 1) p   ψ 

R1

(6.71)

  q

 C α,ψ  D f ψ dt ≤ (tω) (t)   R1 + q

∞,[R1 ,R2 ]

(ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) q

( (α))q ( p (α − 1) + 1) p 

R2 R1

(6.72)

q  

  C α,ψ  D R1 + f (r ω) dr ,

∀ ω ∈ S N −1 . We have proved that 

R2

| f (r ω)|q dr ≤

  ψ 

q

∞,[R1 ,R2 ]

(ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) q

( (α))q ( p (α − 1) + 1) p

R1



R2 R1

q  

  C α,ψ  D R1 + f (r ω) dr ,

(6.73)

∀ ω ∈ S N −1 . Because R1 ≤ r ≤ R2 , N ≥ 2, we have that R1N −1 ≤ r N −1 ≤ R2N −1 and R21−N ≤ 1−N ≤ R11−N . r By (6.73) we get

6 Caputo Generalized ψ-Fractional …

128

 R21−N

R2

r

N −1



R2

| f (r ω)| dr ≤ q

r

R1

1−N N −1

r

 | f (r ω)| dr = q

R1

  ψ 

R2

| f (r ω)|q dr ≤

R1 q

(ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 )

∞,[R1 ,R2 ]

q

( (α))q ( p (α − 1) + 1) p 

q  

  α,ψ r 1−N r N −1  C D R1 + f (r ω) dr ≤

R2 R1

  ψ 

(6.74)

q

∞,[R1 ,R2 ]

(ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) R11−N q

( (α))q ( p (α − 1) + 1) p 

 q 

  α,ψ r N −1  C D R1 + f (r ω) dr ,

R2 R1

∀ ω ∈ S N −1 . We have proved that



R2

| f (r ω)|q r N −1 dr ≤

R1

  ψ 

  (ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) R2 N −1 q

∞,[R1 ,R2 ]

q

( (α))q ( p (α − 1) + 1) p 

R2 R1

R1

  q

 C α,ψ  N −1 D f ω) r dr , (r   R1 +

(6.75)

∀ ω ∈ S N −1 . Then by integration of (6.75) on S N −1 we obtain 

 S N −1

  ψ 

R2

| f (r ω)| r q

N −1

 dr dω ≤

R1

  (ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) R2 N −1 q

∞,[R1 ,R2 ]

q

( (α))q ( p (α − 1) + 1) p 

 S N −1

That is (by (6.59))

R2 R1

  q

 C α,ψ   D R1 + f (r ω) r N −1 dr dω.  | f (x)|q d x ≤ A

R1 (6.76)

6.2 Main Results

129

  ψ 

  (ψ (R2 ) − ψ (R1 ))( p(α−1)+1) p (R2 − R1 ) R2 N −1 q

∞,[R1 ,R2 ]

q

( (α))q ( p (α − 1) + 1) p  

q  C α,ψ   D R1 + f (x) d x.

R1 (6.77)

A

Consequently we get

 | f (x)|q d x

 q1

≤

A

  q1 ψ 

(ψ (R2 ) − ψ (R1 ))α− q (R2 − R1 ) q 1

∞,[R1 ,R2 ]

1

 (α) ( p (α − 1) + 1) p

1



R2 R1

 Nq−1

 

q  q1   C α,ψ ,  D R1 + f (x) d x

(6.78)

A



proving the claim.

It follows a generalized right fractional Poincaré type inequality on the shell A. Theorem 6.17 Let p, q > 1 such that 1p + q1 = 1, and α > q1 , n ∈ N such that   n = α. Let f ∈ C n A , ψ ∈ C n ([R1 , R2 ]) with ψ being strictly increasing and ψ (r ) = 0, for all r ∈ [R1 , R2 ]. Assume that f ψ[k] (R2 ω) = 0, for k = 0, 1, . . . , n −   α,ψ 1; ∀ ω ∈ S N −1 , and C D R2 − f ∈ C A . Then  f q,A

  q1 1   N −1 α− q1  ψ  (R2 − R1 ) q R2 q  C α,ψ  ∞,[R1 ,R2 ] (ψ (R2 ) − ψ (R1 )) ≤  D R2 − f  . 1 q,A R1  (α) ( p (α − 1) + 1) p

(6.79)

Proof Similar to the proof of Theorem 6.16, now using (6.19). We skip the details.  Fractional Poincaré L 1 inequalities come next Theorem 6.18 Let α ≥ 1, n ∈ N : n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being strictly increasing and ψ (x) = 0, for all x ∈ [a, b]. Assume that f ψ[k] (a) = 0, for k = 0, 1, . . . , n − 1. Then  f  L 1 ([a,b])

   (ψ (b) − ψ (a))α−1 ψ ∞,[a,b] (b − a)  C α,ψ  ≤ . (6.80)  Da+ f  L 1 ([a,b])  (α)

Proof By (6.18), since f ψ[k] (a) = 0, for k = 0, 1, . . . , n − 1, we have

6 Caputo Generalized ψ-Fractional …

130

f (x) =

1  (α)



x

ψ (t) (ψ (x) − ψ (t))α−1

C

α,ψ

Da+ f (t) dt,

(6.81)

a

∀ x ∈ [a, b]. Hence | f (x)| ≤

1  (α)

  ψ 



∞,[a,b]

x

   α,ψ  ψ (t) (ψ (x) − ψ (t))α−1 C Da+ f (t) dt ≤

a

(ψ (b) − ψ (a))α−1   (α)

  ψ 

∞,[a,b]

x

  C α,ψ   Da+ f (t) dt ≤

a

(ψ (b) − ψ (a))α−1   (α)

b

  C α,ψ   Da+ f (t) dt,

(6.82)

a

∀ x ∈ [a, b]. Therefore it holds    b ψ   (ψ (b) − ψ (a))α−1 (b − a)  b C α,ψ  ∞,[a,b] | f (x)| d x ≤  Da+ f (x) d x,  (α) a a (6.83) proving the claim.  Theorem 6.19 Let α ≥ 1, n ∈ N : n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being strictly increasing and ψ (x) = 0, for all x ∈ [a, b]. Assume that f ψ[k] (b) = 0, for k = 0, 1, . . . , n − 1. Then  f  L 1 ([a,b])

   (ψ (b) − ψ (a))α−1 ψ ∞,[a,b] (b − a)  C α,ψ  ≤ . (6.84)  Db− f  L 1 ([a,b])  (α) 

Proof Similar to the proof of Theorem 6.18.

Next come multivariate on the shell A L 1 -fractional Poincaré type inequalities.   Theorem 6.20 Let α ≥ 1, n∈N such that n= α. Let f ∈C n A , ψ ∈ C n ([R1 , R2 ]) with ψ being strictly increasing and ψ (r ) = 0, for all r ∈ [R1 , R2 ]. Assume   that α,ψ f ψ[k] (R1 ω) = 0, for k = 0, 1, . . . , n − 1; ∀ ω ∈ S N −1 , and C D R1 + f ∈ C A . Then  f  L 1 (A)

   (ψ (R2 ) − ψ (R1 ))α−1 ψ ∞,[R1 ,R2 ] (R2 − R1 )  R2  N −1  C α,ψ  ≤ .  D R1 + f  L 1 (A)  (α) R1

(6.85)

Proof By (6.83) we get:  R21−N

R2 R1

r N −1 | f (r ω)| dr ≤



R2 R1

r 1−N r N −1 f (r ω) dr =



R2 R1

| f (r ω)| dr ≤

6.2 Main Results

131

  ψ 

∞,[R1 ,R2 ]

(ψ (R2 ) − ψ (R1 ))α−1 (R2 − R1 )  (α)



R2 R1

 

  α,ψ r 1−N r N −1  C D R1 + f (r ω) dr ≤

(6.86)

  ψ  (ψ (R2 ) − ψ (R1 ))α−1 (R2 − R1 ) R11−N ∞,[R1 ,R2 ]  (α) 

R2 R1

∀ ω ∈ S N −1 . Therefore it holds

 

  α,ψ r N −1  C D R1 + f (r ω) dr,



R2

| f (r ω)| r N −1 dr ≤

R1

    ψ  (ψ (R2 ) − ψ (R1 ))α−1 (R2 − R1 ) R2 N −1 ∞,[R1 ,R2 ]  (α) 

R2 R1

∀ ω ∈ S N −1 . Thus we obtain

 

 C α,ψ   D R1 + f (r ω) r N −1 dr,





R1

S N −1

R2

| f (r ω)| r

N −1

(6.87)

 dr dω ≤

R1

    ψ  (ψ (R2 ) − ψ (R1 ))α−1 (R2 − R1 ) R2 N −1 ∞,[R1 ,R2 ]  (α) 

 S N −1

R2 R1

R1

  

 C α,ψ   D R1 + f (r ω) r N −1 dr dω,

which means (by using (6.59)) that  | f (x)| d x ≤ A

    ψ  (ψ (R2 ) − ψ (R1 ))α−1 (R2 − R1 ) R2 N −1 ∞,[R1 ,R2 ]  (α)

R1

(6.88)

6 Caputo Generalized ψ-Fractional …

132

    C α,ψ  D R1 + f (x) d x,

(6.89)

A

   Theorem 6.21 Let α ≥ 1, n∈N such that n= α. Let f ∈C n A , ψ ∈ C n ([R1 , R2 ]) with ψ being strictly increasing and ψ (r ) = 0, for all r ∈ [R1 , R2 ]. Assume   that α,ψ f ψ[k] (R2 ω) = 0, for k = 0, 1, . . . , n − 1; ∀ ω ∈ S N −1 , and C D R2 − f ∈ C A . Then proving the claim.

 f  L 1 (A)

   (ψ (R2 ) − ψ (R1 ))α−1 ψ ∞,[R1 ,R2 ] (R2 − R1 )  R2  N −1  C α,ψ  ≤ .  D R2 − f  L 1 (A)  (α) R1

(6.90) 

Proof As similar to the proof of Theorem 6.20 is omitted. Applications follow:

Proposition 6.22 Let p, q > 1 such that 1p + q1 = 1, α1 > q1 , n 1 ∈ N : n 1 = α1 ; and α2 > 1p , n 2 ∈ N : n 2 = α2 ; [a1 , b1 ] ⊂ R, [a2 , b2 ] ⊂ (0, ∞). Let f i ∈ C ni ([ai , bi ]), i = 1, 2. Assume that f 1e[kx11] (a1 ) = 0, for k1 = 0, 1, . . . , n 1 − 1; f 2[kln2 ]x2 (a2 ) = 0, for k2 = 0, 1, . . . , n 2 − 1. Then 

b1

a1



b2

a2



| f 1 (x1 )| | f 2 (x2 )| d x1 d x2 (e x1 −ea1 ) p(α1 −1)+1 p

+



q (α2 −1)+1  x ln a2

≤

2

q

(b1 − a1 ) (b2 − a2 ) 1

1

 (α1 )  (α2 ) ( p (α1 − 1) + 1) p (q (α2 − 1) + 1) q    

 C α1 ,ex1    α ,ln x .  Da1 + f 1 ◦ ln x1  a b  C Da22+ 2 f 2 ◦ e x2  1 1 q,[e ,e ] q,[ln a2 ,ln b2 ] Proof By Theorem 6.11, for ψ1 (x1 ) = e x1 and ψ2 (x2 ) = ln x2 .

(6.91) 

Proposition 6.23 Let p, q > 1 such that 1p + q1 = 1, and α > q1 , n ∈ N : n = α.   Let f ∈ C n A and assume that f e[k] r (R1 ω) = 0, for k = 0, 1, . . . , n − 1; ∀ ω ∈   r N −1 C α,e S , and D R1 + f ∈ C A . Then  f q,A ≤

e

R2 q

 R α− q1 1   N −1  e 2 − e R1 (R2 − R1 ) q R2 q  C α,er   D R1 + f  . 1 q,A R1  (α) ( p (α − 1) + 1) p

Proof By Theorem 6.16, for ψ (r ) = er , r ∈ [R1 , R2 ] .

(6.92) 

References

133

References 1. Acosta, G., Durán, R.G.: An optimal Poincaré inequality in L for convex domains. Proc. AMS 132(1), 195–202 (2003) 2. Almeida, R.: A Caputo fractional derivative of a function with respect to another function. Commun. Nonlinear Sci. Numer. Simul. 44, 460–481 (2017) 3. Anastassiou, G.A.: Fractional Differentiation Inequalities. Research Monograph. Springer, New York (2009) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 5. Anastassiou, G.: Caputo generalized ψ-fractional integral inequalities (2019). Submitted 6. Evans, L.C.: Partial Differential Equations. Graduate studies in Mathematics, vol. 19. American Mathematical Society, Providence, R.I. (1998) 7. Pachpatte, B.G.: Inequalities similar to the integral analogue of Hilbert’s inequalities. Tamkang J. Math. 30(1), 139–146 (1999) 8. Rudin, W.: Real and Complex Analysis, International Student edn. Mc Graw Hill, London (1970) 9. Stroock, D.: A Concise Introduction to the Theory of Integration, 3rd edn. Birkäuser, Boston (1999)

Chapter 7

Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

Here we consider the approximation of functions by sublinear positive operators with applications to several Max-Product operators under generalized fractional differentiability. Our study is based on our generalized fractional results about positive sublinear operators. We derive Jackson type inequalities under iterated initial conditions. So our approach is quantitative by producing inequalities with their right hand sides involving the modulus of continuity of generalized fractional derivative of the function under approximation. See also [4].

7.1 Background Here we study the approximation properties of positive sublinear operators converging the unit operator related to the generalized fractional smoothness of the involved function under approximation. This is a quantitative study using inequalities. We need some background: We mention Definition 7.1 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]) ,

(7.1)

(ii) if f, g ∈ C+ ([a, b]) : f ≤ g, then L N ( f ) ≤ L N (g) , ∀ N ∈ N,

(7.2)

L N ( f + g) ≤ L N ( f ) + L N (g) , ∀ f, g ∈ C+ ([a, b]) .

(7.3)

(iii)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_7

135

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

136

We call {L N } N ∈N positive sublinear operators. We need a Hölder’s type inequality, see next: Theorem 7.2 (see [2, 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 (·) g (·)) (s∗ ) ≤ L ( f (·)) p (s∗ ) p L (g (·))q (s∗ ) q .

(7.4)

We make Remark 7.3 By [5, p. 17], we get: let f, g ∈ C+ ([a, b]), then |L N ( f ) (x) − L N (g) (x)| ≤ L N (| f − g|) (x) , ∀ x ∈ [a, b] .

(7.5)

Furthermore, we also have that |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) + | f (x)| |L N (e0 ) (x) − 1| , (7.6) ∀ x ∈ [a, b]; e0 (t) = 1. From now on we assume that L N (1) = 1. Hence it holds |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) , ∀ x ∈ [a, b] .

(7.7)

In [5, p. 10], the authors introduced the basic Max-product Bernstein operators, B N(M) ( f ) (x) =

N k=0

p N ,k (x) f

N

k=0

where



k N

p N ,k (x) 

stands for maximum, and p N ,k (x) =

N k



, N ∈ N,

(7.8)

x k (1 − x) N −k and f : [0, 1] →

R+ = [0, ∞). These are nonlinear and piecewise rational operators, in particular they are positive sublinear operators. The authors in [5] studied similar such nonlinear operators such as: the Maxproduct Favard–Szász–Mirakjan operators and their truncated version, the Maxproduct Baskakov operators and their truncated version, also many other similar specific operators. These Max-product operators tend to converge faster to the on hand function.

7.1 Background

137

So we mention from [5, p. 30], that for f : [0, 1] → R+ continuous, we have the estimate     1  (M)  , for all N ∈ N, x ∈ [0, 1] , B N ( f ) (x) − f (x) ≤ 12ω1 f, √ N +1 (7.9) Also from [5, p. 36], we mention that for f : [0, 1] → R+ being concave function we get that     1  (M)  , for all x ∈ [0, 1] , (7.10) B N ( f ) (x) − f (x) ≤ 2ω1 f, N a much faster convergence. Above ω1 is the first modulus of continuity. Inequalities (7.9) and (7.10) motivate our work. Some fractional background follows: Next we use [1]. Let α > 0, [a, b] ⊂ R, f : [a, b] → R which is integrable and ψ ∈ C 1 ([a, b]) an increasing function such that ψ  (x) = 0, for all x ∈ [a, b]. Consider n = α, the ceiling of α. The left and right fractional integrals are defined, respectively, as follows:  x 1 α,ψ ψ  (t) (ψ (x) − ψ (t))α−1 f (t) dt, (7.11) Ia+ f (x) :=  (α) a and α,ψ Ib−

1 f (x) :=  (α)



b

ψ  (t) (ψ (t) − ψ (x))α−1 f (t) dt,

(7.12)

x

for any x ∈ [a, b], where  is the gamma function. The following semigroup property is valid for fractional integrals: if α, β > 0, then α,ψ β,ψ

α+β,ψ

Ia+ Ia+ f (x) = Ia+

α,ψ β,ψ

α+β,ψ

f (x) , and Ib− Ib− f (x) = Ib−

f (x) .

We mention Definition 7.4 ([1]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. The left ψCaputo fractional derivative of f of order α is given by C

α,ψ Da+

 f (x) :=

n−α,ψ Ia+

1 d ψ  (x) d x

n f (x) ,

(7.13)

and the right ψ-Caputo fractional derivative of f is given by C

α,ψ Db−

f (x) :=

n−α,ψ Ib−

 −

1 d ψ  (x) d x

n f (x) .

(7.14)

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

138

To simplify notation, we will use the symbol  f ψ[n]

(x) :=

1 d ψ  (x) d x

n f (x) ,

(7.15)

with f ψ[0] (x) = f (x) . By the definition, when α = m ∈ N, we have α,ψ

Da+ f (x) = f ψ[m] (x) and C α,ψ Db− f (x) = (−1)m f ψ[m] (x) . C

(7.16)

If α ∈ / N, we have C

α,ψ Da+

1 f (x) =  (n − α)



x

a

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

(7.17)

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

(7.18)

and C

α,ψ

Db− f (x) =

(−1)n  (n − α)



b x

∀ x ∈ [a, b] . In particular, when α ∈ (0, 1), we have α,ψ

Da+ f (x) = and C α,ψ Db− f (x) = C

1 (1−α) −1 (1−α)

x a

b x

(ψ (x) − ψ (t))−α f  (t) dt, (7.19) (ψ (t) − ψ (x))−α f  (t) dt

∀ x ∈ [a, b] . Clearly the above is a generalization of left and right Caputo fractional derivatives. For more see [1]. Still we need from [1] the following left and right fractional Taylor’s formulae: Theorem 7.5 ([1]) Let α > 0, n ∈ N such that n = α, [a, b] ⊂ R and f, ψ ∈ C n ([a, b]) with ψ being increasing and ψ  (x) = 0, for all x ∈ [a, b]. Then, the left fractional Taylor formula follows, f (x) =

n−1

f ψ[k] (a) k=0

k!

α,ψ C

(ψ (x) − ψ (a))k + Ia+

α,ψ

Da+ f (x) ,

(7.20)

7.1 Background

139

and the right fractional Taylor formula follows, f (x) =

n−1

(−1)

f ψ[k] (b)

k

k!

k=0

α,ψ C

(ψ (b) − ψ (x))k + Ib−

α,ψ

Db− f (x) ,

(7.21)

∀ x ∈ [a, b] . In particular, given α ∈ (0, 1), we have α,ψ

α,ψ

f (x) = f (a) + Ia+ C Da+ f (x) , and α,ψ α,ψ f (x) = f (b) + Ib− C Db− f (x) ,

(7.22)

∀ x ∈ [a, b] . Remark 7.6 For convenience we can rewrite (7.20)–(7.22) as follows: f (x) =

n−1

f ψ[k] (a) k=0

1  (α)



x

(ψ (x) − ψ (a))k +

ψ  (t) (ψ (x) − ψ (t))α−1

C

(7.23)

α,ψ

Da+ f (t) dt,

a

and

n−1

(−1)k f ψ[k] (b)

f (x) =

k=0

1  (α)

k!



b

k!

(ψ (b) − ψ (x))k +

ψ  (t) (ψ (t) − ψ (x))α−1

x

C

(7.24)

α,ψ

Db− f (t) dt,

∀ x ∈ [a, b] . When α ∈ (0, 1), we get: f (x) = f (a) + and f (x) = f (b) +

1 (α) 1 (α)

x a

b x

α,ψ

ψ  (t) (ψ (x) − ψ (t))α−1

C

Da+ f (t) dt,

ψ  (t) (ψ (t) − ψ (x))α−1

C

Db− f (t) dt,

α,ψ

(7.25)

∀ x ∈ [a, b] . Again from [1] we have the following: Consider the norms ·∞ : C ([a, b]) → R and ·C [n] : C n ([a, b]) → R, where ψ n  [k]  f C [n] := fψ . ψ

k=0

We have

∞

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

140

Theorem 7.7 ([1]) The ψ-Caputo fractional derivatives are bounded operators. For all α > 0 (n = α) C α,ψ (7.26) Da+ ≤ K  f C [n] ∞

C α,ψ Db−

and

∞

where K =

ψ

≤ K  f C [n] ,

(7.27)

ψ

(ψ (b) − ψ (a))n−α > 0.  (n + 1 − α)

(7.28)

We make Remark 7.8 In our setting, clearly, it is f ψ(n) ∈ C ([a, b]). Given f ∈ C ([a, b]), by α,ψ

Theorem 4.10, p. 98 of [3], we get that Ia+ f ∈ C ([a, b]), and by Theorem 4.11, p. α,ψ 101 of [3], we get that Ib− f ∈ C ([a, b]). α,ψ α,ψ Therefore, we obtain that C Da+ , C Db− ∈ C ([a, b]). We mention Definition 7.9 Let f ∈ C ([a, b]), we define ω1 ( f, δ) :=

| f (x) − f (y)| , where 0 < δ ≤ b − a.

sup x,y∈[a,b]: |x−y|≤δ

7.2 Main Results We need Definition 7.10 Let

C

ψ −1 , and δ > 0. We set





 α,ψ α,ψ α,ψ Dx0 f ◦ ψ −1 denote any of C Dx0 + f ◦ ψ −1 , C Dx0 − f ◦ ω1

C



  α,ψ max ω1 C Dx0 − f ◦ ψ −1 , δ

  Dxα,ψ f ◦ ψ −1 , δ := 0

[ψ(a),ψ(x0 )]

, ω1



C

  α,ψ Dx0 + f ◦ ψ −1 , δ

(7.29)  [ψ(x0 ),ψ(b)]

,

where x0 ∈ [a, b]. Here the moduli of continuity are considered over [ψ (a) , ψ (x0 )] and [ψ (x0 ) , ψ (b)], respectively. We present Theorem 7.11 Let α > 0, α ∈ / N, n ∈ N such that n = α, [a, b] ⊂ R, and f, ψ ∈ C n ([a, b]) with ψ being strictly increasing. Let a fixed x 0 ∈ [a, b] such that f ψ[k] (x0 ) = 0, for k = 1, . . . , n − 1. Then

7.2 Main Results

141

| f (x) − f (x0 )| ≤ 

ω1



C

  α,ψ Dx0 f ◦ ψ −1 , δ  (α + 1)

 |ψ (x) − ψ (x0 )|α+1 |ψ (x) − ψ (x0 )| + , δ (α + 1) α

(7.30)

δ > 0, ∀ x ∈ [a, b]. If 0 < α < 1, then we do not need any initial conditions. Proof By (7.23) we have f (x) =

n−1

f ψ[k] (x0 ) k=0

1  (α)



x

k!

(ψ (x) − ψ (x0 ))k +

ψ  (t) (ψ (x) − ψ (t))α−1

C

x0

(7.31)

α,ψ

Dx0 + f (t) dt,

∀ x ∈ [x0 , b] . And, by (7.24) we have f (x) =

n−1

(−1)k f ψ[k] (x0 )

k!

k=0

1  (α)



x0

(ψ (x0 ) − ψ (x))k +

ψ  (t) (ψ (t) − ψ (x))α−1

x

C

(7.32)

α,ψ

Dx0 − f (t) dt,

∀ x ∈ [a, x0 ] . By the assumption f ψ[k] (x0 ) = 0, k = 1, . . . , n − 1, we get f (x) − f (x0 ) =

1  (α)



x

α,ψ

ψ  (t) (ψ (x) − ψ (t))α−1

C

Dx0 + f (t) dt,

ψ  (t) (ψ (t) − ψ (x))α−1

C

Dx0 − f (t) dt,

x0

(7.33)

∀ x ∈ [x0 , b] , and f (x) − f (x0 ) =

1  (α)



x0 x

α,ψ

(7.34)

∀ x ∈ [a, x0 ] . When 0 < α < 1, i.e. n = 1, then (7.33), (7.34) are valid without any initial condition. We have that f ψ[k] ∈ C ([a, b]), for all k = 0, 1, . . . , n.

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

142

Here α ∈ / N and we have C

α,ψ

Dx0 + f (x) =

1  (n − α)



x x0

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

(7.35)

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

(7.36)

∀ x ∈ [x0 , b] , and C

α,ψ Dx0 −

(−1)n f (x) =  (n − α)



x0 x

∀ x ∈ [a, x0 ] . Thus, we derive    C α,ψ  Dx0 + f (x) ≤

1  (n − α)



x x0

    ψ  (t) (ψ (x) − ψ (t))n−α−1  f ψ[n] (t) dt ≤

[n] fψ

[n]  x fψ ∞,[a,b] ∞,[a,b] (ψ (x) − ψ (x0 ))n−α . (ψ (x) − ψ (t))n−α−1 dψ (t) =  (n − α)  (n − α + 1) x0

(7.37)

We found that [n] fψ    C α,ψ ∞,[a,b] (ψ (x) − ψ (x0 ))n−α ,  Dx0 + f (x) ≤  (n − α + 1)

(7.38)

∀ x ∈ [x0 , b] . α,ψ Clearly, it holds C Dx0 + f (x0 ) = 0. Similarly, it holds   C α,ψ   Dx0 − f (x) ≤

1  (n − α)



x0 x

    ψ  (t) (ψ (t) − ψ (x))n−α−1  f ψ[n] (t) dt ≤

[n] fψ

[n]  x0 fψ ∞,[a,b] ∞,[a,b] (ψ (x0 ) − ψ (x))n−α . (ψ (t) − ψ (x))n−α−1 dψ (t) =  (n − α)  (n − α + 1) x

(7.39)

That is

[n] fψ    C α,ψ ∞,[a,b] (ψ (x0 ) − ψ (x))n−α ,  Dx0 − f (x) ≤  (n − α + 1)

∀ x ∈ [a, x0 ] . α,ψ Clearly, it holds C Dx0 − f (x0 ) = 0.

(7.40)

7.2 Main Results

143

Therefore, we can write  x

 1 α,ψ α,ψ ψ  (t) (ψ (x) − ψ (t))α−1 C Dx0 + f (t) −C Dx0 + f (x0 ) dt,  (α) x0

f (x) − f (x0 ) =

(7.41)

∀ x ∈ [x0 , b], and f (x) − f (x0 ) =

 x0

 1 α,ψ α,ψ ψ  (t) (ψ (t) − ψ (x))α−1 C Dx0 − f (t) −C Dx0 − f (x0 ) dt,  (α) x

(7.42)

∀ x ∈ [a, x0 ] . We can rewrite (7.41) and (7.42), as follows (by z := ψ (t)): 1  (α)

f (x) − f (x0 ) =



C

α,ψ





Dx0 + f ◦ ψ −1 (z) −



C



ψ(x) ψ(x0 )

(ψ (x) − z)α−1

   α,ψ Dx0 + f ◦ ψ −1 (ψ (x0 )) dz,

(7.43)

∀ x ∈ [x0 , b], and f (x) − f (x0 ) =



C

1  (α)



ψ(x0 )

ψ(x)

(z − ψ (x))α−1



    α,ψ α,ψ Dx0 − f ◦ ψ −1 (z) − C Dx0 − f ◦ ψ −1 (ψ (x0 )) dz,

(7.44)

∀ x ∈ [a, x0 ] . We have that (x0 ≤ x ≤ b) (7.43)

| f (x) − f (x0 )| =

1  (α)



ψ(x) ψ(x0 )

(ψ (x) − z)α−1



 

    C α,ψ  α,ψ Dx0 + f ◦ ψ −1 (z) − C Dx0 + f ◦ ψ −1 (ψ (x0 )) dz ≤  (δ1 > 0)    ψ(x)  1 δ1 |z − ψ (x0 )| α,ψ α−1 C −1 ω D f ◦ ψ , dz ≤ − z) (ψ (x) 1 x0 +  (α) ψ(x0 ) δ1 [ψ(x0 ),ψ(b)] ω1



C D α,ψ f x0 +



◦ ψ −1 , δ1  (α)

 [ψ(x0 ),ψ(b)]

  (z − ψ (x0 )) dz (ψ (x) − z)α−1 1 + δ1 ψ(x0 )

 ψ(x)

(7.45)

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

144

= 

ω1



C

  α,ψ Dx0 + f ◦ ψ −1 , δ1

1 (ψ (x) − ψ (x0 ))α + α δ1 ω1

[ψ(x0 ),ψ(b)]

 (α)



C



ψ(x)

ψ(x0 )

 (ψ (x) − z)α−1 (z − ψ (x0 ))2−1 dz =

  α,ψ Dx0 + f ◦ ψ −1 , δ1

[ψ(x0 ),ψ(b)]

 (α) 

 1  (α)  (2) (ψ (x) − ψ (x0 ))α α+1 = + (ψ (x) − ψ (x0 )) α δ1  (α + 2) ω1



C

  α,ψ Dx0 + f ◦ ψ −1 , δ1

[ψ(x0 ),ψ(b)]

 (α) 

 1 1 (ψ (x) − ψ (x0 ))α + (ψ (x) − ψ (x0 ))α+1 , α δ1 (α + 1) α

(7.46)

∀ x ∈ [x0 , b] . We have proved that

| f (x) − f (x0 )| ≤

ω1



C

  α,ψ Dx0 + f ◦ ψ −1 , δ1

[ψ(x0 ),ψ(b)]

 (α + 1)



 (ψ (x) − ψ (x0 ))α+1 , (ψ (x) − ψ (x0 )) + δ1 (α + 1) α

(7.47)

∀ x ∈ [x0 , b] , δ1 > 0. By (7.44) we obtain | f (x) − f (x0 )| ≤

1  (α)



ψ(x0 )

ψ(x)

(z − ψ (x))α−1



 

    C α,ψ  α,ψ Dx0 − f ◦ ψ −1 (z) − C Dx0 − f ◦ ψ −1 (ψ (x0 )) dz ≤  (δ2 > 0)    ψ(x0 )  1 δ2 (ψ (x0 ) − z) α,ψ α−1 C −1 ω1 Dx0 − f ◦ ψ , dz ≤ (z − ψ (x))  (α) ψ(x) δ2 [ψ(a),ψ(x0 )]

7.2 Main Results

ω1



C D α,ψ f x0 −



145

◦ ψ −1 , δ2



 ψ(x0 )

[ψ(a),ψ(x0 )]

 (α)

ψ(x)

= 

ω1



C

−1

f ◦ ψ , δ2

(7.48)

 [ψ(a),ψ(x0 )]

 (α)

1 (ψ (x0 ) − ψ (x))α + α δ2 ω1

α,ψ Dx0 −



  (ψ (x0 ) − z) dz (z − ψ (x))α−1 1 + δ2



C



 (ψ (x0 ) − z)2−1 (z − ψ (x))α−1 dz =

ψ(x0 ) ψ(x)

  α,ψ Dx0 − f ◦ ψ −1 , δ2

[ψ(a),ψ(x0 )]

 (α) 

 1  (2)  (α) (ψ (x0 ) − ψ (x))α + (ψ (x0 ) − ψ (x))α+1 = α δ2  (α + 2) ω1



C

  α,ψ Dx0 − f ◦ ψ −1 , δ2

[ψ(a),ψ(x0 )]

 (α) 

 1 1 (ψ (x0 ) − ψ (x))α α+1 = + (ψ (x0 ) − ψ (x)) α δ2 α (α + 1) ω1



C

(7.49)

  α,ψ Dx0 − f ◦ ψ −1 , δ2

[ψ(a),ψ(x0 )]

 (α + 1)  (ψ (x0 ) − ψ (x))α +

 1 (ψ (x0 ) − ψ (x))α+1 , δ2 (α + 1)

(7.50)

∀ x ∈ [a, x0 ] . In conclusion, we derive

| f (x) − f (x0 )| ≤

ω1



C

  α,ψ Dx0 − f ◦ ψ −1 , δ2

[ψ(a),ψ(x0 )]

 (α + 1)



 (ψ (x0 ) − ψ (x))α+1 , (ψ (x0 ) − ψ (x)) + δ2 (α + 1) α

∀ x ∈ [a, x0 ] , δ2 > 0. Choosing δ = δ1 = δ2 > 0, by (7.47) and (7.51), we get (7.30).

(7.51)



7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

146

Corollary 7.12 All as in Theorem 7.11. Then ω1

| f (·) − f (x0 )| ≤



C

  α,ψ Dx0 f ◦ ψ −1 , δ  (α + 1)



 |ψ (·) − ψ (x0 )|α+1 |ψ (·) − ψ (x0 )| + , δ (α + 1) α

(7.52)

true over [a, b], δ > 0. Our first result on the convergence of positive sublinear operators comes next. Theorem 7.13 Let α > 0, α ∈ / N, n ∈ N such that n = α, [a, b] ⊂ R, and f ∈ C+ ([a, b]), f, ψ ∈ C n ([a, b]) with ψ being strictly increasing. Let x0 ∈ [a, b] be fixed such that f ψ[k] (x0 ) = 0, k = 1, . . . , n − 1. Let L N : C+ ([a, b]) → C+ ([a, b]), ∀ N ∈ N, be positive sublinear operators, such that L N (1) = 1, ∀ N ∈ N. Then |L N ( f ) (x0 ) − f (x0 )| ≤ 

ω1



C

  α,ψ Dx0 f ◦ ψ −1 , δ  (α + 1)

   L N |ψ (·) − ψ (x0 )|α+1 (x0 ) L N (|ψ (·) − ψ (x0 )| ) (x0 ) + , δ (α + 1) α

(7.53)

where δ > 0, ∀ N ∈ N. In particular (7.53) is true for α > 1, α ∈ / N. Proof By (7.7) and (7.52) we have |L N ( f ) (x0 ) − f (x0 )| ≤ L N (| f (·) − f (x0 )|) (x0 ) ≤ 

ω1



C

  α,ψ Dx0 f ◦ ψ −1 , δ  (α + 1)

   L N |ψ (·) − ψ (x0 )|α+1 (x0 ) L N (|ψ (·) − ψ (x0 )| ) (x0 ) + , δ > 0, (7.54) δ (α + 1) α

proving the claim.



Corollary 7.14 (to Theorem 7.13) When 0 < α < 1, then (7.53) is valid for any x0 ∈ [a, b], and without any initial conditions.

7.2 Main Results

147

We give   Theorem 7.15 All as in Theorem 7.13, and assume that L N |ψ (·) − ψ (x0 )|α+1 (x0 ) > 0, ∀ N ∈ N. Then |L N ( f ) (x0 ) − f (x0 )| ≤ ω1



C

(α + 2)  (α + 2)

1       α+1 α+1 −1 |ψ − ψ Dxα,ψ f ◦ ψ , L (·) (x (x )| ) N 0 0 0



   α L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 ,

(7.55)

∀ N ∈ N. If 0 < α < 1, then (7.55) is valid without any initial conditions and true for any x0 ∈ [a, b]. Proof By Theorem 7.2, see (7.4), we get     α L N (|ψ (·) − ψ (x0 )|α ) (x0 ) ≤ L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 . Choose

   1  δ := L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 > 0,

i.e.

(7.56)

(7.57)

  δ α+1 = L N |ψ (·) − ψ (x0 )|α+1 (x0 ) .

By (7.53) we obtain |L N ( f ) (x0 ) − f (x0 )| ≤ ω1



C

     1  α,ψ Dx0 f ◦ ψ −1 , L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1  (α + 1)

 

ω1

   α L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 +



C

 δ α+1 = δ (α + 1)

     1  α,ψ Dx0 f ◦ ψ −1 , L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1  (α + 1)

 

    α  α    α+1 L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 α+1 + L N |ψ (·) − ψ (x0 )| = (x0 ) (α + 1) (7.58)

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

148

ω1



C

     1  α,ψ Dx0 f ◦ ψ −1 , L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1  (α + 1)



ω1



C

   α L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1

 1+

1 α+1

 =

     1  α,ψ Dx0 f ◦ ψ −1 , L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 (α + 2) (7.59)

 (α + 2) 

   α L N |ψ (·) − ψ (x0 )|α+1 (x0 ) α+1 , 

proving (7.55).

Remark of Theorem 7.15:  7.16  In the terms andassumptions  let L N |ψ (·) − ψ (x0 )|α+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → +∞. We need   Corollary 7.17 All as in Theorem 7.13, and assume that L N |· − x0 |α+1 (x0 ) > 0, ∀ N ∈ N. Then (α + 2) |L N ( f ) (x0 ) − f (x0 )| ≤  (α + 2) ω1



C

1       α+1 −1  α+1 | |· ψ L − x Dxα,ψ f ◦ ψ , (x ) N 0 0 0 ∞,[a,b]

 α ψ

∞,[a,b]



   α L N |· − x0 |α+1 (x0 ) α+1 ,

(7.60)

∀ N ∈ N. Proof Use of Theorem 7.15 (7.55), ω1 and L N properties, and  that |ψ (x) −     ψ (x0 )| ≤ ψ ∞,[a,b] |x − x0 |, along with |ψ (x) − ψ (x0 )| ≥ inf ψ  (x) x∈[a,b]

|x − x0 |, ∀ x, x0∈ [a, b], given that ψ  (x) = 0, ∀ x ∈ [a, b]. Clearly then (7.60)  is valid, and L N |ψ (·) − ψ (x0 )|α+1 (x0 ) > 0.

7.3 Applications (I) Case 0 < α < 1. Here we apply Corollary 7.17 to well known Max-product operators. We make

7.3 Applications

149

Remark 7.18 The Max-product Bernstein operators B N(M) ( f ) (x) are defined by (7.8), see also [5, p. 10]; here f : [0, 1] → R+ is a continuous function. We have B N(M) (1) = 1, and B N(M) (|· − x|) (x) ≤ √

6 N +1

, ∀ x ∈ [0, 1] , ∀ N ∈ N,

(7.61)

see [5, p. 31]. B N(M) are positive sublinear operators and thus they possess the monotonicity property, also since |· − x| ≤ 1, then |· − x|β ≤ 1, ∀ x ∈ [0, 1], ∀ β > 0. Therefore it holds   6 , ∀ x ∈ [0, 1] , ∀ N ∈ N, ∀ β > 0. B N(M) |· − x|1+β (x) ≤ √ N +1

(7.62)

Furthermore, clearly it holds that   B N(M) |· − x|1+β (x) > 0, ∀ N ∈ N, ∀ β ≥ 0 and any x ∈ (0, 1) .

(7.63)

The operator B N(M) maps C+ ([0, 1]) into itself. We present Theorem 7.19 Let 0 < α < 1, f ∈ C+ ([0, 1]), f, ψ ∈ C 1 ([0, 1]) with ψ being strictly increasing. Then   (α + 2)   (M) B N ( f ) (x) − f (x) ≤  (α + 2) ω1

 C

Dxα,ψ

 α ψ

f ◦ ψ , ψ  ∞,[0,1] 

∞,[0,1]

−1

 √

6 N +1

α  α+1

 √

6

1   α+1

N +1

, ∀ N ∈ N,

(7.64)

∀ x ∈ (0, 1) . As N → +∞, we get B N(M) ( f ) (x) → f (x), for any x ∈ (0, 1) . Proof By Corollary 7.17. We continue with Remark 7.20 The truncated Favard–Szász–Mirakjan operators are given by



7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

150

TN(M)

 

N

k k=0 s N ,k (x) f N N k=0 s N ,k (x)

( f ) (x) =

, x ∈ [0, 1] , N ∈ N, f ∈ C+ ([0, 1]) , (7.65)

s N ,k (x) = (Nk!x) , see also [5, p. 11]. By [5, pp. 178–179], we get that k

3 TN(M) (|· − x|) (x) ≤ √ , ∀ x ∈ [0, 1] , ∀ N ∈ N. N

(7.66)

Clearly it holds   3 TN(M) |· − x|1+β (x) ≤ √ , ∀ x ∈ [0, 1] , ∀ N ∈ N, ∀ β > 0. N

(7.67)

The operators TN(M) are positive sublinear operators mapping C+ ([0, 1]) into itself, with TN(M) (1) = 1. Furthermore it holds TN(M)

  |· − x|λ (x) =

N



(N x)k  k − k! N  N (N x)k k=0 k!

k=0

λ x

> 0, ∀ x ∈ (0, 1], ∀ λ ≥ 1, ∀ N ∈ N. (7.68)

We give Theorem 7.21 All as in Theorem 7.19. Then   (α + 2)  (M)  TN ( f ) (x) − f (x) ≤  (α + 2) ω1

 C

Dxα,ψ

 α ψ

f ◦ ψ , ψ  ∞,[0,1] 

∞,[0,1]

−1



3 √ N

α  α+1



3 √ N

1   α+1

, ∀ N ∈ N,

(7.69)

∀ x ∈ (0, 1]. As N → +∞, we get TN(M) ( f ) (x) → f (x), for any x ∈ (0, 1]. Proof By Corollary 7.17.



We continue with Remark 7.22 Next we study the truncated Max-product Baskakov operators (see [5, p. 11])

7.3 Applications

U N(M)

151

N ( f ) (x) =

k=0

b N ,k (x) f

N

k N

k=0 b N ,k (x)



where b N ,k (x) =

, x ∈ [0, 1] , f ∈ C+ ([0, 1]) , N ∈ N,

N +k−1 k

(7.70) 

xk . (1 + x) N +k

(7.71)

From [5, pp. 217–218], we get (x ∈ [0, 1])  √ √

 2 3 2+2 U N(M) (|· − x|) (x) ≤ , N ≥ 2, N ∈ N. √ N +1

(7.72)

Let λ ≥ 1, clearly then it holds  √ √

 2 3 2+2   U N(M) |· − x|λ (x) ≤ , ∀ N ≥ 2, N ∈ N. √ N +1

(7.73)

Also it holds U N(M) (1) = 1, and U N(M) are positive sublinear operators from C+ ([0, 1]) into itself. Furthermore it holds   U N(M) |· − x|λ (x) > 0, ∀ x ∈ (0, 1], ∀ λ ≥ 1, ∀ N ∈ N.

(7.74)

We present Theorem 7.23 All as in Theorem 7.19. Then   (α + 2)  (M)  U N ( f ) (x) − f (x) ≤  (α + 2) ⎛

1 ⎞  ⎞ α+1 ⎛ √ √ 2 3 2 + 2  ⎜ ⎠ ⎟ ω1 ⎝ C Dxα,ψ f ◦ ψ −1 , ψ  ∞,[0,1] ⎝ √ ⎠ N +1

α  ⎞ α+1 ⎛ √ √ 2 3 2 + 2  α ψ ⎝ ⎠ , ∀ N ∈ N − {1}, √ ∞,[0,1] N +1

(7.75)

∀ x ∈ (0, 1]. As N → +∞, we get U N(M) ( f ) (x) → f (x), for any x ∈ (0, 1]. Proof By Corollary 7.17.



7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

152

We give Remark 7.24 Here we study the Max-product Meyer-Köning and Zeller operators (see [5, p. 11]) defined by Z (M) N



∞

k k=0 s N ,k (x) f N +k ∞ k=0 s N ,k (x)

( f ) (x) =

 , ∀ N ∈ N, f ∈ C+ ([0, 1]) ,

(7.76)



 N +k x k , x ∈ [0, 1]. s N ,k (x) = k By [5, p. 253], we get that

√  8 1 + 5 √x (1 − x) Z (M) , ∀ x ∈ [0, 1] , ∀ N ≥ 4, N ∈ N. √ N (|· − x|) (x) ≤ 3 N (7.77) As before we get that (for λ ≥ 1)

√  8 1 + 5 √x (1 − x)   |· − x|λ (x) ≤ Z (M) := ρ (x) , √ N 3 N

(7.78)

∀ x ∈ [0, 1], N ≥ 4, N ∈ N. (M) are positive sublinear operators from Also it holds Z (M) N (1) = 1, and Z N C+ ([0, 1]) into itself. Also it holds   |· − x|λ (x) > 0, ∀ x ∈ (0, 1), ∀ λ ≥ 1, ∀ N ∈ N. Z (M) N

(7.79)

We give Theorem 7.25 All as in Theorem 7.19. Then   (α + 2)  (M)  Z N ( f ) (x) − f (x) ≤  (α + 2) ω1

 C

  1 Dxα,ψ f ◦ ψ −1 , ψ  ∞,[0,1] (ρ (x)) α+1

 α ψ

∞,[0,1]

α

(ρ (x)) α+1 , ∀ N ≥ 4, N ∈ N,

(7.80)

∀ x ∈ (0, 1). As N → +∞, we get Z (M) N ( f ) (x) → f (x), for any x ∈ (0, 1). Proof By Corollary 7.17.



7.3 Applications

153

Corollary 7.26 All as in Theorem 7.19, for α = 21 . Then    √ 3    10 6  (M)  √ √  B N ( f ) (x) − f (x) ≤ ψ  ∞,[0,1] 6 3 π N +1 ω1



C

1 2 ,ψ

Dx

 √ 3  36 f ◦ ψ , ψ ∞,[0,1] √ , 3 N +1 

−1

(7.81)

∀ x ∈ (0, 1), ∀ N ∈ N. 

Proof By Theorem 7.19. (II) Case of α > 1, α ∈ / N. We need Corollary 7.27 (to Theorem 7.13) Same assumptions as in Theorem 7.13. Then

|L N ( f ) (x0 ) − f (x0 )| ≤  α ψ



∞,[a,b]

α

L N (|· − x0 | ) (x0 ) +



ω1

C

  α,ψ Dx0 f ◦ ψ −1 , δ  (α + 1)

 ψ

∞,[a,b]

   L N |· − x0 |α+1 (x0 ) δ (α + 1)

,

(7.82)

where δ > 0, ∀ N ∈ N.

Proof By Theorem 7.13, and |ψ (x) − ψ (x0 )| ≤ ψ  ∞,[a,b] |x − x0 |, ∀ x, x0 ∈  [a, b] and the properties of L N . We apply Corollary 7.27 to well-known Max-product operators. We present

Theorem 7.28 Let α > 1, α ∈ / N, n ∈ N such that n = α, and f ∈ C+ ([0, 1]), f, ψ ∈ C n ([0, 1]) with ψ being strictly increasing. For a fixed x ∈ [0, 1] assume that f ψ[k] (x) = 0, k = 1, . . . , n − 1. Then  ω1    (M)  B N ( f ) (x) − f (x) ≤  α ψ

∞,[0,1]

 √

6 N +1



C



−1

f ◦ψ ,

√6 N +1

1   α+1

(7.83)

 (α + 1) +

 ψ

∀ N ∈ N. We get lim B N(M) ( f ) (x) = f (x) . N →+∞

α,ψ Dx

∞,[0,1]

(α + 1)

 √

6 N +1

α   α+1

,

7 Generalized ψ-Fractional Quantitative Approximation by Sublinear Operators

154

Proof By (7.82) we get   ω1  (M)  B N ( f ) (x) − f (x) ≤ 

 α ψ

∞,[0,1]

B N(M)



C

α,ψ

Dx

  f ◦ ψ −1 , δ

 (α + 1)

   (M)  ψ |· − x|α+1 (x) (7.62) B N ∞,[0,1] ≤ (|· − x|α ) (x) + δ (α + 1)

  f ◦ ψ −1 , δ α ψ  ∞,[0,1]  (α + 1)    ψ √6 6 ∞,[0,1] N +1 + =: (∗) . √ δ (α + 1) N +1 ω1

C

α,ψ

Dx

1  α+1

, then δ α+1 = Choose δ = √ N6+1 We get that

ω1



C

(∗) =

α,ψ Dx

√6 , N +1



−1

f ◦ψ ,

and apply it to (7.84).

√6 N +1

1   α+1

6 ⎢ + ⎣√ N +1

 ψ

∞,[0,1]

 α ψ

∞,[0,1]

 (α + 1) ⎡

(7.84)

√6 N +1

(α + 1)

α ⎤  α+1

⎥ ⎦,

∀ N ∈ N, proving the claim.

(7.85)



Next comes Theorem 7.29 All as in Theorem 7.28. Then  1  

 α+1 3 C α,ψ −1 √ ω D f ◦ ψ , x 1   N  (M)  TN ( f ) (x) − f (x) ≤  (α + 1)   α   α+1  ψ  α 3 3 ∞,[0,1] ψ , √ + √ ∞,[0,1] (α + 1) N N ∀ N ∈ N. We get lim TN(M) ( f ) (x) = f (x) . N →+∞

(7.86)

7.3 Applications

Proof Similar to Theorem 7.28.

155



Next comes Theorem 7.30 All as in Theorem 7.28. Then   1  

√ √ 2 3( 2+2) α+1 C α,ψ −1 √ ω D f ◦ ψ , x 1   N +1  (M)  U N ( f ) (x) − f (x) ≤  (α + 1) ⎡ √ √ α ⎤   ⎞ α+1 ⎛ √ √ ψ  2 3 2 3 2+2 2+2 ⎢ ∞,[0,1] ⎝ ⎠ ⎥ + √ √ ⎦, ∞,[0,1] ⎣ (α + 1) N +1 N +1

 α ψ

(7.87)

∀ N ∈ N − {1}. We get lim U N(M) ( f ) (x) = f (x) . N →+∞

Proof Similar to Theorem 7.28.



We finish with Theorem 7.31 All as in Theorem 7.28. Then

  1   ω1 C Dxα,ψ f ◦ ψ −1 , (ρ (x)) α+1  (M)  Z N ( f ) (x) − f (x) ≤  (α + 1)    ψ  α α ∞,[0,1] ψ ρ (x) + (ρ (x)) α+1 , ∞,[0,1] (α + 1)

(7.88)

∀ N ∈ N, N ≥ 4. We get lim Z (M) N ( f ) (x) = f (x) , where ρ (x) is as in (7.78). N →+∞

Proof Similar to Theorem 7.28.



References 1. Almeida, R.: A Caputo fractional derivative of a function with respect to another function. Commun. Nonlinear Sci. Numer. Simul. 44, 460–481 (2017) 2. Anastassiou, G.A.: Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Max-Product Operators. Springer, Heidelberg (2018) 3. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 4. Anastassiou, G.: Generalized ψ-fractional approximation by sublinear operators (2019). Submitted 5. Bede, B., Coroianu, L., Gal, S.: Approximation by Max-Product type Operators. Springer, Heidelberg (2016)

Chapter 8

Generalized g-Iterated Fractional Quantitative Approximation By Sublinear Operators

Here we study the approximation of functions by sublinear positive operators with applications to several Max-Product operators under generalized g -iterated fractional differentiability. Our work is based on our generalized g-iterated fractional results about positive sublinear operators. We produce Jackson type inequalities under iterated initial conditions. So our approach is quantitative by deriving inequalities with their right hand sides involving the modulus of continuity of generalized g-iterated fractional derivative of the function under approximation. See also [3].

8.1 Background Here we study the approximation properties of positive sublinear operators converging the unit operator related to the iterated generalized fractional smoothness of the involved function under approximation. This is a quantitative study using inequalities. We need some background: We mention Definition 8.1 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]) ,

(8.1)

(ii) if f, g ∈ C+ ([a, b]) : f ≤ g, then L N ( f ) ≤ L N (g) , ∀ N ∈ N, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_8

(8.2) 157

8 Generalized g-Iterated Fractional Quantitative Approximation …

158

(iii) L N ( f + g) ≤ L N ( f ) + L N (g) , ∀ f, g ∈ C+ ([a, b]) .

(8.3)

We call {L N } N ∈N positive sublinear operators. We need a Hölder’s type inequality, see next: Theorem 8.2 (see [2, 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 (·) g (·)) (s∗ ) ≤ L ( f (·)) p (s∗ ) p L (g (·))q (s∗ ) q .

(8.4)

We make Remark 8.3 By [5, p. 17], we get: let f, g ∈ C+ ([a, b]), then |L N ( f ) (x) − L N (g) (x)| ≤ L N (| f − g|) (x) , ∀ x ∈ [a, b] .

(8.5)

Furthermore, we also have that |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) + | f (x)| |L N (e0 ) (x) − 1| , (8.6) ∀ x ∈ [a, b]; e0 (t) = 1. From now on we assume that L N (1) = 1. Hence it holds |L N ( f ) (x) − f (x)| ≤ L N (| f (·) − f (x)|) (x) , ∀ x ∈ [a, b] .

(8.7)

In [5, p. 10], the authors introduced the basic Max-product Bernstein operators, B N(M)

N ( f ) (x) =

k=0

p N ,k (x) f

N

k=0

where



k

p N ,k (x) 

stands for maximum, and p N ,k (x) =

N k



N

, N ∈ N,

(8.8)

x k (1 − x) N −k and f : [0, 1] →

R+ = [0, ∞). These are nonlinear and piecewise rational operators, in particular they are positive sublinear operators. The authors in [5] studied similar such nonlinear operators such as: the Maxproduct Favard–Szász–Mirakjan operators and their truncated version, the Maxproduct Baskakov operators and their truncated version, also many other similar specific operators. These Max-product operators tend to converge faster to the on hand function.

8.1 Background

159

So we mention from [5, p. 30], that for f : [0, 1] → R+ continuous, we have the estimate     1  (M)  , for all N ∈ N, x ∈ [0, 1] , B N ( f ) (x) − f (x) ≤ 12ω1 f, √ N +1 (8.9) Also from [5, p. 36], we mention that for f : [0, 1] → R+ being concave function we get that     1  (M)  , for all x ∈ [0, 1] , B N ( f ) (x) − f (x) ≤ 2ω1 f, N

(8.10)

a much faster convergence. Above ω1 is the first modulus of continuity. Inequalities (8.9) and (8.10) motivate our work. We mention Definition 8.4 Let f ∈ C ([a, b]), we define ω1 ( f, δ) :=

| f (x) − f (y)| , where 0 < δ ≤ b − a.

sup x,y∈[a,b]: |x−y|≤δ

We need Definition 8.5 ([1]) Let α > 0, α = m, · the ceiling of the number. Here g ∈ AC ([a, b]) (absolutely continuous functions) and strictly increasing. We assume that  (m) f ◦ g −1 ◦ g ∈ L ∞ ([a, b]). We define the left generalized g-fractional derivative of f of order α as follows: 

 α f (x) := Da+;g

1  (m − α)



x

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

a

(8.11)

all x ∈ [a, b] . α f ∈ C ([a, b]). If α ∈ / N, by [4, pp. 360–361], we have that Da+;g We see that 

(m)  α  m−α Ia+;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , x ≥ a. We set m f (x) := Da+;g

When g = id, then



f ◦ g −1

(m)

◦ g (x) ,

(8.12)

(8.13)

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

(8.14)

α α α f = Da+;id f = D∗a f, Da+;g

(8.15)

8 Generalized g-Iterated Fractional Quantitative Approximation …

160

the usual left Caputo fractional derivative. We need Definition 8.6 ([1]) Here g ∈ AC ([a, b]) and strictly increasing. We assume that  (m) f ◦ g −1 ◦ g ∈ L ∞ ([a, b]), where N  m = α, α > 0. We define the right generalized g-fractional derivative of f of order α as follows: 

α Db−;g



(−1)m f (x) :=  (m − α)



b

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

x

(8.16)

all x ∈ [a, b] .

α f ∈ C ([a, b]). If α ∈ / N, by [4, p. 378], we get that Db−;g We see that

 (m)  α  m−α ◦ g (x) = Db−;g f (x) , a ≤ x ≤ b. Ib−;g (−1)m f ◦ g −1 We set m f (x) = (−1)m Db−;g



f ◦ g −1

(m)

◦ g (x) ,

(8.17)

(8.18)

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

When g = id, then

α α α f (x) = Db−;id f (x) = Db− f, Db−;g

(8.19)

the usual right Caputo fractional derivative. Denote by nα α α α := Db−;g Db−;g . . . Db−;g (n-times), n ∈ N. Db−;g

(8.20)

We mention the following g-right generalized modified Taylor’s formula: kα Theorem 8.7 ([1]) Suppose that Fk := Db−;g f , for k = 0, 1, . . . , n + 1, fulfill: Fk ◦   g −1 ∈ AC ([c, d]) , where c = g (a), d = g (b), and Fk ◦ g −1 ◦ g ∈ L ∞ ([a, b]), where 0 < α ≤ 1. Then

f (x) =

n  (g (b) − g (x))iα  i=0

1  ((n + 1) α)



b x

 (iα + 1)

 iα Db−;g f (b) +



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

(8.21)

8.1 Background

161

n  (g (b) − g (x))iα  i=0

 (iα + 1)



iα Db−;g f (b) +

(n+1)α Db−;g f (ψx )

 ((n + 1) α + 1)

(g (b) − g (x))(n+1)α , (8.22)

where ψx ∈ [x, b], any x ∈ [a, b] . Denote by nα α α α Da+;g := Da+;g Da+;g . . . Da+;g (n-times), n ∈ N.

(8.23)

We mention the following g-left generalized modified Taylor’s formula: kα Theorem 8.8 ([1]) Suppose that Fk∗ := Da+;g f , for k = 0, 1, . . . , n + 1, fulfill:   ∗ −1 Fk ◦ g ∈ AC ([c, d]), where c = g (a), d = g (b), and Fk∗ ◦ g −1 ◦ g∈L ∞ ([a, b]) , where 0 < α ≤ 1. Then

f (x) =

n  (g (x) − g (a))iα 

 (iα + 1)

i=0

1  ((n + 1) α)



x

a

n  (g (x) − g (a))iα  i=0

 (iα + 1)

 iα Da+;g f (a) +

(8.24)

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

iα Da+;g f (a) +

(n+1)α Da+;g f (ψx )

 ((n + 1) α + 1)

(g (x) − g (a))(n+1)α , (8.25)

where ψx ∈ [a, x], any x ∈ [a, b] . We make Remark 8.9 Here we restrict ourselves to 0 < α < 1. In this case  x  α    1 Da+;g f (x) = (g (x) − g (t))−α g (t) f ◦ g −1 (g (t)) dt,  (1 − α) a (8.26) and  b   α   −1 Db−;g f (x) = (g (t) − g (x))−α g (t) f ◦ g −1 (g (t)) dt,  (1 − α) x (8.27) ∀ x ∈ [a, b]. We observe that  x    α   1   −α −1 D ≤ f ◦ g f g − g (x) (g (x) (t)) (t) (g (t))   dt a+;g  (1 − α) a

8 Generalized g-Iterated Fractional Quantitative Approximation …

162

≤

  f ◦ g −1 ◦ g



∞,[a,b]

 (1 − α)

(by [6, p. 107]) =

=

x

(g (x) − g (t))−α g (t) dt

a

  f ◦ g −1 ◦ g  (1 − α)   f ◦ g −1 ◦ g

∞,[a,b]

∞,[a,b]

 (2 − α)

(g (x) − g (a))1−α 1−α

(g (x) − g (a))1−α .

(8.28)

That is  α  D

a+;g

  f (x) ≤

  f ◦ g −1 ◦ g

∞,[a,b]

 (2 − α)

(g (x) − g (a))1−α < +∞,

∀ x ∈ [a, b].

α f (a) = 0. Clearly it holds Da+;g I.e.   α Dx0 +;g f (x0 ) = 0, ∀ x0 ∈ [a, b] . We define



 Dxα0 +;g f (x) = 0, for a ≤ x < x0 .

(8.29)

(8.30)

(8.31)

Similarly we get (by [6, p. 107])   b−;g f (x) ≤

 α D

≤

1  (1 − α)



b

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

x

  f ◦ g −1 ◦ g

∞,[a,b]

=

 (1 − α)   f ◦ g −1 ◦ g  (2 − α)



b

(g (t) − g (x))−α g (t) dt

x

∞,[a,b]

(g (b) − g (x))1−α .

(8.32)

That is   −1 ◦ g  f ◦g ∞,[a,b]  (g (b) − g (x))1−α < +∞, b−;g f (x) ≤  (2 − α)

 α D

∀ x ∈ [a, b].

(8.33)

8.1 Background

163



α Clearly it holds Db−;g f (b) = 0. In particular we have  Dxα0 −;g f (x0 ) = 0, ∀ x0 ∈ [a, b] .

(8.34)

 Dxα0 −;g f (x) = 0, for x0 < x ≤ b.

(8.35)

 We define



From Theorem 8.7 we have Theorem 8.10 Let 0 < α < 1, g ∈ C 1 ([a, b]) and strictly increasing, so that g −1 ∈ C 1 ([g (a) , g (b)]) . Assume that Fk := Dxkα0 −;g f , for k = 0, 1, . . . , n + 1, n ∈ N fulfill: Fk ∈ C 1 ([a, x0 ]) , where x0 ∈ [a, b] is fixed. Then f (x) − f (x0 ) =

n  (g (x0 ) − g (x))iα  i=2

1  ((n + 1) α)



x0 x

 (iα + 1)

 Dxiα0 −;g f (x0 ) +

f (t) dt, (g (t) − g (x))(n+1)α−1 g (t) Dx(n+1)α −;g 0

(8.36)

∀ x ∈ [a, x0 ] . From Theorem 8.8 we have Theorem 8.11 Let 0 < α < 1, g ∈ C 1 ([a, b]) and strictly increasing, so that g −1 ∈ C 1 ([g (a) , g (b)]) . Assume that Fk∗ := Dxkα0 +;g f , for k = 0, 1, . . . , n + 1, n ∈ N fulfill: Fk∗ ∈ C 1 ([x0 , b]) , where x0 ∈ [a, b] is fixed. Then f (x) − f (x0 ) =

n  (g (x) − g (x0 ))iα  i=2

1  ((n + 1) α)



x x0

 (iα + 1)

 Dxiα0 +;g f (x0 ) +

f (t) dt, (g (x) − g (t))(n+1)α−1 g (t) Dx(n+1)α 0 +;g

(8.37)

∀ x ∈ [x0 , b] . We make

Remark 8.12 All as in Theorem 8.10 and Dxiα0 −;g f (x0 ) = 0, for i = 2, . . . , n ∈ N, then we get f (x) − f (x0 ) =

 x0

1 (n+1)α (g (t) − g (x))(n+1)α−1 g (t) Dx −;g f (t) dt, 0  ((n + 1) α) x

(8.38)

∀ x ∈ [a, x0 ] .

8 Generalized g-Iterated Fractional Quantitative Approximation …

164



Next, all as in Theorem 8.11 and Dxiα0 +;g f (x0 ) = 0, for i = 2, . . . , n ∈ N, then we get f (x) − f (x0 ) =

 x

1 (n+1)α (g (x) − g (t))(n+1)α−1 g (t) Dx +;g f (t) dt, 0  ((n + 1) α) x0

(8.39)

∀ x ∈ [x0 , b] . We need Definition 8.13 Let Dx(n+1)α f denote any of Dx(n+1)α f , Dx(n+1)α f , n ∈ N and δ > 0. 0 ;g 0 −;g 0 +;g We set



f ◦ g −1 , δ := ω1 Dx(n+1)α 0 ;g





−1 max ω1 Dx(n+1)α f ◦ g , δ 0− ;g

, ω1

[g(a),g(x0 )]





−1 Dx(n+1)α f ◦ g , δ 0 +;g

 [g(x0 ),g(b)]

,

(8.40) where x0 ∈ [a, b]. Here the moduli of continuity are considered over [g (a) , g (x0 )] and [g (x0 ) , g (b)], respectively.

8.2 Main Results We will use the following important result. Theorem 8.14 Let 0 < α < 1, f, g ∈ C 1 ([a, b]), g is strictly increasing and g −1 ∈ C 1 ([g (a) , g (b)]). Assume that Dxkα0 −;g f ∈ C 1 ([a, x0 ]) and Dxkα0 +;g f ∈ C 1 ([x0 , b]),

for k = 1, . . . , n + 1; where x0 ∈ [a, b] is fixed. Further assume that Dxiα0 ±;g f (x0 ) = 0, i = 2, . . . , n + 1. Then | f (x) − f (x0 )| ≤  |g (x) − g (x0 )|

(n+1)α

ω1





−1 Dx(n+1)α f ◦ g , δ ;g 0

 ((n + 1) α + 1)  |g (x) − g (x0 )|(n+1)α+1 + , δ ((n + 1) α + 1)

∀ x ∈ [a, b] , δ > 0.

Proof By Dxiα0 +;g f (x0 ) = 0, for i = 2, . . . , n + 1, we have f (x) − f (x0 ) =

1  ((n + 1) α)



x x0

(g (x) − g (t))(n+1)α−1 g (t)

(8.41)

8.2 Main Results

165







(n+1)α dt, Dx(n+1)α f f − D ) (t) (x 0 +;g +;g x 0 0

(8.42)

∀ x ∈ [x0 , b] . Hence (z := g (t)) 

1 f (x) − f (x0 ) =  ((n + 1) α) 

g(x) g(x0 )

(g (x) − z)(n+1)α−1





 Dx(n+1)α f ◦ g −1 (z) − Dx(n+1)α f ◦ g −1 (g (x0 )) dz, 0 +;g 0 +;g

(8.43)

∀ x ∈ [x 0 , b] .

By Dxiα0 −;g f (x0 ) = 0, for i = 2, . . . , n + 1 ∈ N, we have 1 f (x) − f (x0 ) =  ((n + 1) α) 



x0

(g (t) − g (x))(n+1)α−1 g (t)

x



 (n+1)α Dx(n+1)α f − D f dt, (t) (x ) 0 x0 −;g 0 −;g

(8.44)

∀ x ∈ [a, x0 ] . Hence (z := g (t)) f (x) − f (x0 ) = 



1  ((n + 1) α)



Dx(n+1)α f ◦ g −1 (z) − 0 −;g





g(x0 )

g(x)

(z − g (x))(n+1)α−1



 −1 Dx(n+1)α f ◦ g dz, (g (x )) 0 0 −;g

(8.45)

∀ x ∈ [a, x0 ] . We have that (x0 ≤ x ≤ b) | f (x) − f (x0 )| ≤

1  ((n + 1) α)



g(x)

g(x0 )

(g (x) − z)(n+1)α−1

 





  (n+1)α (n+1)α  Dx0 +;g f ◦ g −1 (z) − Dx0 +;g f ◦ g −1 (g (x0 )) dz ≤

(δ1 >0)

1  ((n + 1) α) ω1



Dx(n+1)α 0 +;g





g(x) g(x0 )

(g (x) − z)(n+1)α−1

δ1 |z − g (x0 )| f ◦g , δ1 −1

(8.46)

 [g(x0 ),g(b)]

dz ≤

8 Generalized g-Iterated Fractional Quantitative Approximation …

166

ω1





−1 Dx(n+1)α f ◦ g , δ 1 +;g 0

[g(x0 ),g(b)]

 ((n + 1) α) 

g(x)

g(x0 )

(n+1)α−1

(g (x) − z) ω1



  (z − g (x0 )) 1+ dz = δ1



−1 Dx(n+1)α f ◦ g , δ 1 +;g 0

[g(x0 ),g(b)]

 ((n + 1) α) 

1 (g (x) − g (x0 ))(n+1)α + δ1 (n + 1) α ω1





g(x) g(x0 )

 (g (x) − z)(n+1)α−1 (z − g (x0 ))2−1 dz = (8.47)



−1 Dx(n+1)α f ◦ g , δ 1 0 +;g

[g(x0 ),g(b)]

 ((n + 1) α) 

 1  ((n + 1) α)  (2) (g (x) − g (x0 ))(n+1)α (n+1)α+1 + = (g (x) − g (x0 )) δ1  ((n + 1) α + 2) (n + 1) α ω1





−1 Dx(n+1)α f ◦ g , δ 1 0 +;g

[g(x0 ),g(b)]

(8.48)

 ((n + 1) α) 

 1 (g (x) − g (x0 ))(n+1)α+1 (g (x) − g (x0 ))(n+1)α + . δ1 (n + 1) α ((n + 1) α + 1) (n + 1) α

We have proved that

| f (x) − f (x0 )| ≤

ω1





−1 Dx(n+1)α f ◦ g , δ 1 +;g 0

[g(x0 ),g(b)]

 ((n + 1) α + 1)



 (n+1)α+1 − g (g (x) (x )) 0 , (g (x) − g (x0 ))(n+1)α + δ1 ((n + 1) α + 1)

∀ x ∈ [x0 , b], δ1 > 0. We have that (a ≤ x ≤ x0 ) (8.44)

| f (x) − f (x0 )| ≤

1  ((n + 1) α)



g(x0 )

g(x)

(z − g (x))(n+1)α−1

(8.49)

8.2 Main Results

167

 





  (n+1)α (n+1)α  Dx0 −;g f ◦ g −1 (z) − Dx0 −;g f ◦ g −1 (g (x0 )) dz ≤

(δ2 >0)

1  ((n + 1) α) 

ω1

Dx(n+1)α 0 −;g ω1





g(x0 )

g(x)

(z − g (x))(n+1)α−1

δ2 |z − g (x0 )| f ◦g , δ2



−1

 [g(a),g(x0 )]

dz ≤

(8.50)



−1 Dx(n+1)α f ◦ g , δ 2 −;g 0

[g(a),g(x0 )]

 ((n + 1) α) 

g(x0 ) g(x)

  g (x0 ) − z dz = (z − g (x))(n+1)α−1 1 + δ2 ω1





−1 Dx(n+1)α f ◦ g , δ 2 −;g 0

[g(a),g(x0 )]

 ((n + 1) α) 

1 (g (x0 ) − g (x))(n+1)α + δ2 (n + 1) α ω1





g(x0 )

g(x)

 (g (x0 ) − z)2−1 (z − g (x))(n+1)α−1 dz = (8.51)



−1 Dx(n+1)α f ◦ g , δ 2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α) 

 1  (2)  ((n + 1) α) (g (x0 ) − g (x))(n+1)α (n+1)α+1 + = (g (x0 ) − g (x)) δ2  ((n + 1) α + 2) (n + 1) α ω1





Dx(n+1)α f ◦ g −1 , δ2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α) 

1 (g (x0 ) − g (x))(n+1)α+1 (g (x0 ) − g (x))(n+1)α + δ2 (n + 1) α ((n + 1) α + 1) (n + 1) α ω1



 =



Dx(n+1)α f ◦ g −1 , δ2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α + 1)  (g (x0 ) − g (x))

(n+1)α

 (g (x0 ) − g (x))(n+1)α+1 + . δ2 ((n + 1) α + 1)

(8.52)

8 Generalized g-Iterated Fractional Quantitative Approximation …

168

We have proved that ω1

| f (x) − f (x0 )| ≤





−1 Dx(n+1)α f ◦ g , δ 2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α + 1)



 (n+1)α+1 − g ) (x)) (g (x 0 , (g (x0 ) − g (x))(n+1)α + δ2 ((n + 1) α + 1)

(8.53)

∀ x ∈ [a, x0 ], δ2 > 0. By (8.49) and (8.53), setting δ = δ1 = δ2 > 0, we derive (8.41).



We state Corollary 8.15 (to Theorem 8.14) All as in Theorem 8.14. Then

| f (·) − f (x0 )| ≤

ω1

 (n+1)α

|g (·) − g (x0 )|





−1 Dx(n+1)α f ◦ g , δ 0 ;g

 ((n + 1) α + 1)  |g (·) − g (x0 )|(n+1)α+1 + , δ ((n + 1) α + 1)

(8.54)

δ > 0, true over [a, b] . Our first result on the approximation by positive sublinear operators follows: Theorem 8.16 Let 0 < α < 1, f ∈ C+ ([a, b]) , f, g ∈ C 1 ([a, b]), g is strictly increasing and g −1 ∈ C 1 ([g (a) , g (b)]). Assume that Dxkα0 −;g f ∈ C 1 ([a, x0 ]) and Dxkα0 +;g f ∈ C 1 ([x0 , b]), for k = 1, . . . , n + 1; where x0 ∈ [a, b] is fixed. Further

assume that Dxiα0 ±;g f (x0 ) = 0, i = 2, . . . , n + 1. Let L N : C+ ([a, b]) → C+ ([a, b]), ∀ N ∈ N, be positive sublinear operators, such that L N (1) = 1, ∀ N ∈ N. Then



−1 ω1 Dx(n+1)α f ◦ g , δ 0 ;g |L N ( f ) (x0 ) − f (x0 )| ≤  ((n + 1) α + 1) 



L N |g (·) − g (x0 )|

 (n+1)α

   L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) , (x0 ) + δ ((n + 1) α + 1)

∀ N ∈ N, where δ > 0. Proof By (8.7) and (8.54) we have |L N ( f ) (x0 ) − f (x)| ≤ L N (| f (·) − f (x0 )|) (x0 )

(8.55)

8.2 Main Results

169

≤

ω1





−1 Dx(n+1)α f ◦ g , δ ;g 0

 ((n + 1) α + 1)



     L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α L N |g (·) − g (x0 )| , δ > 0, (x0 ) + δ ((n + 1) α + 1) (8.56) proving the claim.  We give

  Theorem 8.17 All as in Theorem 8.16, and assume that L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) > 0, ∀ N ∈ N. Then |L N ( f ) (x0 ) − f (x)| ≤ ω1



((n + 1) α + 2)  ((n + 1) α + 2)



1     (n+1)α+1 −1 (n+1)α+1 |g Dx(n+1)α − g f ◦ g , L (·) (x (x )| ) N 0 0 0 ;g 

   (n+1)α L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 ,

(8.57)

∀ N ∈ N. Proof By Theorem 8.2, see (8.4), we get       (n+1)α L N |g (·) − g (x0 )|(n+1)α (x0 ) ≤ L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 . (8.58) Choose    1  (8.59) δ := L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 > 0, i.e.

  δ (n+1)α+1 = L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) .

By (8.55) we obtain |L N ( f ) (x0 ) − f (x0 )| ≤ ω1



    1

Dx(n+1)α f ◦ g −1 , L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 0 ;g  ((n + 1) α + 1)

 

ω1

   (n+1)α L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 +



 δ (n+1)α+1 = δ ((n + 1) α + 1)



1     (n+1)α+1 (n+1)α+1 −1 |g Dx(n+1)α − g f ◦ g , L (·) (x (x )| ) N 0 0 0 ;g  ((n + 1) α + 1)

(8.60)

8 Generalized g-Iterated Fractional Quantitative Approximation …

170

  

ω1



   (n+1)α L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 +

   (n+1)α ⎤ L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 ⎦= ((n + 1) α + 1)

(n+1)α Dx ;g f 0





◦ g −1 , L N





|g (·) − g (x0 )|(n+1)α+1 (x0 )



1 (n+1)α+1



 ((n + 1) α + 1)



 (n+1)α    (n+1)α+1 (n+1)α+1 1+ L N |g (·) − g (x0 )| (x0 )

 1 = (n + 1) α + 1

(8.61)

 





1 ((n + 1) α + 2) (n+1)α+1 (n+1)α −1 (n+1)α+1 Dx ;g ω1 f ◦ g , L N |g (·) − g (x0 )| (x0 ) 0  ((n + 1) α + 2)



   (n+1)α L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 , 

proving (8.57). We make

Remark  In the terms and assumptions   of Theorem 8.17:  8.18 let L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → +∞. We need

  Corollary 8.19 All as in Theorem 8.16, and assume that L N |· − x0 |(n+1)α+1 (x0 ) > 0, ∀ N ∈ N. Then |L N ( f ) (x0 ) − f (x0 )| ≤ ω1



((n + 1) α + 2)  ((n + 1) α + 2)



1     (n+1)α+1 −1 (n+1)α+1 | |· g Dx(n+1)α L − x f ◦ g , (x ) N 0 0 ∞,[a,b] 0 ;g (n+1)α     (n+1)α g L N |· − x0 |(n+1)α+1 (x0 ) (n+1)α+1 , ∞,[a,b]

(8.62)

∀ N ∈ N. |g (x) − g (x0 )| ≤ and that  Proof By Theorem 8.17 (8.57), ω1 and L N properties,    g |x − x0 |, along with |g (x) − g (x0 )| ≥ inf g (x) |x − x0 |, ∀ x, x0 ∞,[a,b] x∈[a,b]

∈ [a, b], ∀ x ∈ [a,  by g (x) > 0,(n+1)α+1  b]. Clearly then (8.62) is valid, and L N |g (·) − g (x0 )| (x0 ) > 0, ∀ N ∈ N.



8.3 Applications

171

8.3 Applications (I) Case 0 < (n + 1) α < 1. Here we apply Corollary 8.19 to well known Max-product operators. We make Remark 8.20 The Max-product Bernstein operators B N(M) ( f ) (x) are defined by (8.8), see also [5, p. 10]; here f : [0, 1] → R+ is a continuous function. We have B N(M) (1) = 1, and B N(M) (|· − x|) (x) ≤ √

6 N +1

, ∀ x ∈ [0, 1] , ∀ N ∈ N,

(8.63)

see [5, p. 31]. B N(M) are positive sublinear operators and thus they possess the monotonicity property, also since |· − x| ≤ 1 , then |· − x|β ≤ 1, ∀ x ∈ [0, 1], ∀ β > 0. Therefore it holds   6 , ∀ x ∈ [0, 1] , ∀ N ∈ N, ∀ β > 0. B N(M) |· − x|1+β (x) ≤ √ N +1

(8.64)

Furthermore, clearly it holds that   B N(M) |· − x|1+β (x) > 0, ∀ N ∈ N, ∀ β ≥ 0 and any x ∈ (0, 1) .

(8.65)

The operator B N(M) maps C+ ([0, 1]) into itself. We present Theorem 8.21 Let 0 < α < 1, f ∈ C+ ([0, 1]), f, g ∈ C 1 ([0, 1]) , g is strictly kα f ∈ C 1 ([0, x]) and increasing and g −1 ∈ C 1 ([g (0) , g (1)]). Assume that Dx−;g kα f ∈ C 1 ([x, 1]), for k = 1, . . . , n + 1; where x0 ∈ (0, 1) is fixed. Further Dx+;g

iα assume that Dx±;g f (x) = 0, i = 2, . . . , n + 1. Then   ((n + 1) α + 2)   (M)  B N ( f ) (x) − f (x) ≤  ((n + 1) α + 2)  ω1



(n+1)α Dx;g



f ◦ g , g ∞,[0,1]

(n+1)α g

∞,[0,1]

−1

 √

6



(n+1)α  (n+1)α+1

N +1



As N → +∞, we get B N(M) ( f ) (x) → f (x) .

√

6

 1  (n+1)α+1

N +1

, ∀ N ∈ N.

(8.66)

8 Generalized g-Iterated Fractional Quantitative Approximation …

172



Proof By Corollary 8.19. We continue with Remark 8.22 The truncated Favard–Szász–Mirakjan operators are given by TN(M)

 

N

k k=0 s N ,k (x) f N N k=0 s N ,k (x)

( f ) (x) =

, x ∈ [0, 1] , N ∈ N, f ∈ C+ ([0, 1]) , (8.67)

s N ,k (x) = (Nk!x) , see also [5, p. 11]. By [5, pp. 178–179], we get that k

3 TN(M) (|· − x|) (x) ≤ √ , ∀ x ∈ [0, 1] , ∀ N ∈ N. N

(8.68)

Clearly it holds   3 TN(M) |· − x|1+β (x) ≤ √ , ∀ x ∈ [0, 1] , ∀ N ∈ N, ∀ β > 0. N

(8.69)

The operators TN(M) are positive sublinear operators mapping C+ ([0, 1]) into itself, with TN(M) (1) = 1. Furthermore it holds TN(M)



|· − x|

λ



N (x) =



(N x)k  k − k! N  N (N x)k k=0 k!

k=0

λ x

> 0, ∀ x ∈ (0, 1], ∀ λ ≥ 1, ∀ N ∈ N. (8.70)

We give Theorem 8.23 All as in Theorem 8.21, but now x ∈ (0, 1]. Then   ((n + 1) α + 2)  (M)  TN ( f ) (x) − f (x) ≤  ((n + 1) α + 2) ω1



(n+1)α Dx;g



f ◦ g , g ∞,[0,1]

(n+1)α g

∞,[0,1]

−1



3 √ N

(n+1)α  (n+1)α+1



As N → +∞, we get TN(M) ( f ) (x) → f (x).



3 √ N

 1  (n+1)α+1

, ∀ N ∈ N.

(8.71)

8.3 Applications

173



Proof By Corollary 8.19. We continue with

Remark 8.24 Next we study the truncated Max-product Baskakov operators (see [5, p. 11]) U N(M)

N ( f ) (x) =

k=0

b N ,k (x) f

N

k

k=0 b N ,k (x)



where b N ,k (x) =

N

, x ∈ [0, 1] , f ∈ C+ ([0, 1]) , N ∈ N,

N +k−1 k

(8.72) 

xk . (1 + x) N +k

(8.73)

From [5, pp. 217–218], we get (x ∈ [0, 1])

√ √

2 3 2+2 U N(M) (|· − x|) (x) ≤ , N ≥ 2, N ∈ N. √ N +1

(8.74)

Let λ ≥ 1, clearly then it holds

√ √

2 3 2+2   U N(M) |· − x|λ (x) ≤ , ∀ N ≥ 2, N ∈ N. √ N +1

(8.75)

Also it holds U N(M) (1) = 1, and U N(M) are positive sublinear operators from C+ ([0, 1]) into itself. Furthermore it holds   U N(M) |· − x|λ (x) > 0, ∀ x ∈ (0, 1], ∀ λ ≥ 1, ∀ N ∈ N.

(8.76)

We present Theorem 8.25 Here all as in Theorem 8.23. Then   ((n + 1) α + 2)  (M)  U N ( f ) (x) − f (x) ≤  ((n + 1) α + 2) ⎛ ⎞ 1

⎞ (n+1)α+1 ⎛ √ √

2 3 2+2 ⎜ (n+1)α ⎟ ⎠ f ◦ g −1 , g ∞,[0,1] ⎝ ω1 ⎝ Dx;g √ ⎠ N +1 (n+1)α

⎞ (n+1)α+1 ⎛ √ √ 2 3 2 + 2 (n+1)α g ⎠ ⎝ , ∀ N ≥ 2, N ∈ N. √ ∞,[0,1] N +1

As N → +∞, we get U N(M) ( f ) (x) → f (x).

(8.77)

8 Generalized g-Iterated Fractional Quantitative Approximation …

174



Proof By Corollary 8.19. We give

Remark 8.26 Here we study the Max-product Meyer-Köning and Zeller operators (see [5, p. 11]) defined by Z (M) N

∞



k k=0 s N ,k (x) f N +k ∞ k=0 s N ,k (x)

( f ) (x) =

 , ∀ N ∈ N, f ∈ C+ ([0, 1]) ,

(8.78)



 N +k s N ,k (x) = x k , x ∈ [0, 1]. k By [5, p. 253], we get that √

8 1 + 5 √x (1 − x) Z (M) , ∀ x ∈ [0, 1] , ∀ N ≥ 4, N ∈ N. √ N (|· − x|) (x) ≤ 3 N (8.79) As before we get that (for λ ≥ 1) √

8 1 + 5 √x (1 − x)   |· − x|λ (x) ≤ Z (M) := ρ (x) , √ N 3 N

(8.80)

∀ x ∈ [0, 1], N ≥ 4, N ∈ N. (M) are positive sublinear operators from Also it holds Z (M) N (1) = 1, and Z N C+ ([0, 1]) into itself. Also it holds   |· − x|λ (x) > 0, ∀ x ∈ (0, 1), ∀ λ ≥ 1, ∀ N ∈ N. Z (M) N

(8.81)

We give Theorem 8.27 Here all as in Theorem 8.21. Then   ((n + 1) α + 2)  (M)  Z N ( f ) (x) − f (x) ≤  ((n + 1) α + 2) ω1





1 (n+1)α Dx;g f ◦ g −1 , g ∞,[0,1] (ρ (x)) (n+1)α+1

(n+1)α (n+1)α g (ρ (x)) (n+1)α+1 , ∀ N ≥ 4, N ∈ N. ∞,[0,1]

(8.82)

As N → +∞, we get Z (M) N ( f ) (x) → f (x). Proof By Corollary 8.19. (II) Case of (n + 1) α > 1. We need



8.3 Applications

175

Corollary 8.28 All as in Theorem 8.16. Then

|L N ( f ) (x0 ) − f (x0 )| ≤



ω1



−1 Dx(n+1)α f ◦ g , δ 0 ;g

 ((n + 1) α + 1)

   g L |· − x0 |(n+1)α+1 (x0 )   ∞,[a,b] N (n+1)α L N |· − x0 | , (x0 ) + ∞,[a,b] δ ((n + 1) α + 1) (8.83) ∀ N ∈ N, where δ > 0. Proof By Theorem 8.16, and |g (x) − g (x0 )| ≤ g ∞,[a,b] |x − x0 |, ∀ x, x0 ∈ [a, b]  and the properties of L N . (n+1)α g



Applications follow: We give Theorem 8.29 Let 0 < α < 1, and (n + 1) α > 1, n ∈ N, f ∈ C+ ([0, 1]), f, g ∈ C 1 ([0, 1]) , g is strictly increasing and g −1 ∈ C 1 ([g (0) , g (1)]). Assume that Dxkα0 −;g f ∈ C 1 ([0, x0 ]) and Dxkα0 +;g f ∈ C 1 ([x0 , 1]), for k = 1, . . . , n + 1; where

x0 ∈ [0, 1] is fixed. Further assume that Dxiα0 ±;g f (x0 ) = 0, i = 2, . . . , n + 1. Then     (M) B N ( f ) (x0 ) − f (x0 ) ≤

ω1



 1

(n+1)α+1

−1 √6 Dx(n+1)α f ◦ g , 0 ;g N +1

⎡ (n+1)α ⎢ g ⎢√ 6 + ∞,[0,1] ⎣ N +1

 (α + 1)

(8.84)

⎤ (n+1)α

(n+1)α+1 ⎥ ∞,[0,1] ⎥, ⎦ ((n + 1) α + 1)

g



√6 N +1

∀ N ∈ N. As N → +∞, we get B N(M) ( f ) (x0 ) → f (x0 ). Proof By (8.83) we have   ω1  (M)  B N ( f ) (x0 ) − f (x0 ) ≤ 





−1 Dx(n+1)α f ◦ g , δ (n+1)α 0 ;g g ∞,[0,1]  ((n + 1) α + 1)

   g B (M) |· − x0 |(n+1)α+1 (x0 ) (8.64)   ∞,[0,1] N (M) (n+1)α ≤ B N |· − x| (x0 ) + δ ((n + 1) α + 1) ω1





−1 Dx(n+1)α f ◦ g , δ (n+1)α 0 ;g g ∞,[0,1]  ((n + 1) α + 1)

8 Generalized g-Iterated Fractional Quantitative Approximation …

176

 √

6 N +1

+

g ∞,[0,1]

δ ((n + 1) α + 1)

1

(n+1)α+1 , then δ (n+1)α+1 = Choose δ = √ N6+1 We get that

ω1 (∗) =

ω1



√6 N +1

√6 , N +1

 =: (∗) .

and apply it to (8.85).

 1

(n+1)α+1

−1 √6 Dx(n+1)α f ◦ g , 0 ;g N +1

(n+1)α g

∞,[0,1]

 ((n + 1) α + 1)   g δ (n+1)α+1 6 ∞,[0,1] + = √ δ ((n + 1) α + 1) N +1 

Dx(n+1)α 0 ;g



−1

f ◦g ,



√6 N +1

⎢ ⎢√ 6 ⎣ N +1 +

(8.86)

 1

(n+1)α+1 (n+1)α g

∞,[0,1]

 ((n + 1) α + 1) ⎡

(8.85)

⎤ (n+1)α

(n+1)α+1 ⎥ ∞,[0,1] ⎥, ⎦ ((n + 1) α + 1)

g



√6 N +1

∀ N ∈ N, proving the claim.



Next comes Theorem 8.30 All as in Theorem 8.29. Then   1

(n+1)α+1 (n+1)α 3 −1 Dx0 ;g f ◦ g , √ N   ω1  (M)  TN ( f ) (x0 ) − f (x0 ) ≤  ((n + 1) α + 1) (n+1)α g

∞,[0,1]



(n+1)α   (n+1)α+1  g 3 3 ∞,[0,1] , √ + √ ((n + 1) α + 1) N N

∀ N ∈ N. We get lim TN(M) ( f ) (x0 ) = f (x0 ) . N →+∞

(8.87)

8.3 Applications

177



Proof Similar to Theorem 8.29. Next follows: Theorem 8.31 All as in Theorem 8.29. Then    (M)  U N ( f ) (x0 ) − f (x0 ) ≤ ω1



1 

√ √ 2 3( 2+2) (n+1)α+1 −1 g (n+1)α √ Dx(n+1)α f ◦ g , ∞,[0,1] 0 ;g N +1  ((n + 1) α + 1)

⎡

⎤ (n+1)α

⎞ (n+1)α+1 ⎛ √ √ √ √ g 2 3 2+2 2+2 ⎢2 3 ⎥ ∞,[0,1] ⎢ ⎥, ⎝ ⎠ + √ √ ⎣ ⎦ ((n + 1) α + 1) N +1 N +1

(8.88)

∀ N ≥ 2, N ∈ N. We get lim U N(M) ( f ) (x0 ) = f (x0 ) . N →+∞

Proof Similar to Theorem 8.29.



We continue with Theorem 8.32 All as in Theorem 8.29. Then



1 −1 g (n+1)α (n+1)α+1   ω1 Dx(n+1)α f ◦ g , (ρ (x)) ∞,[0,1] 0 ;g  (M)   Z N ( f ) (x0 ) − f (x0 ) ≤  ((n + 1) α + 1)   g (n+1)α ∞,[0,1] ρ (x) + (8.89) (ρ (x)) (n+1)α+1 , ((n + 1) α + 1) ∀ N ∈ N, N ≥ 4. We get lim Z (M) N ( f ) (x 0 ) = f (x 0 ) , where ρ (x) is as in (8.80). N →+∞

Proof Similar to Theorem 8.29.



8 Generalized g-Iterated Fractional Quantitative Approximation …

178

We finish with Corollary 8.33 (to Theorem 8.29) Let 0 < α < 1, and (n + 1) α > 1, n ∈ N, f ∈ C+ ([0, 1]) and f ∈ C 1 ([0, 1]). Assume that Dxkα0 −;ex f ∈ C 1 ([0, x0 ]) and Dxkα0 +;ex f ∈ 1 C ([x0 , 1]),  for k = 1, . . . , n + 1; where x0 ∈ [0, 1] is fixed. Further assume that  iα Dx0 ±;ex f (x0 ) = 0, i = 2, . . . , n + 1. Then   ω1  (M)   B N ( f ) (x0 ) − f (x0 ) ≤



Dx(n+1)α x 0 ;e



f ◦ ln t,



√6 N +1

 1

(n+1)α+1

 ((n + 1) α + 1) ⎤ (n+1)α

(n+1)α+1 6 e √ N +1 ⎢ ⎥ ⎢√ 6 ⎥ ⎣ N + 1 + ((n + 1) α + 1) ⎦ , ∀ N ∈ N. ⎡

e(n+1)α (8.90)

It holds lim B N(M) ( f ) (x0 ) = f (x0 ) . N →+∞

Proof By Theorem 8.29 and g (x) = e x .



References 1. Anastassiou, G.: Advanced fractional Taylor’s formulae. J. Comput. Anal. Appl. 21(7), 1185– 1204 (2016) 2. Anastassiou, G.A.: Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Max-Product Operators. Springer, Heidelberg (2018) 3. Anastassiou, G.: Generalized g-iterated fractional approximations by sublinear operators. Appl. Math. (2020). Accepted 4. Anastassiou, G., Argyros, I.: Intelligent Numerical Methods: Applications to Fractional Calculus. Springer, Heidelberg (2016) 5. Bede, B., Coroianu, L., Gal, S.: Approximation by Max-Product type Operators. Springer, Heidelberg (2016) 6. Royden, H.L.: Real Analysis, 2nd edn. Macmillan, New York (1968)

Chapter 9

Generalized g-Fractional Vector Representation Formula And Bochner Integral Type Inequalities for Banach Space Valued Functions

Here we give a very general fractional Bochner integral representation formula for Banach space valued functions. We derive generalized left and right fractional Opial type inequalities, fractional Ostrowski type inequalities and fractional Grüss type inequalities. All these inequalities are very general having in their background Bochner type integrals. See also [4].

9.1 Background We need Definition 9.1 ([2]) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the left Riemann–Liouville generalized fractional Bochner integral operator 

 ν f (x) := Ia+;g

1  (ν)



x

(g (x) − g (z))ν−1 g  (z) f (z) dz,

(9.1)

a

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈C ([a, b] , X ), then f ∈L ∞ ν 0 f ∈ C ([a, b] , X ). Above we set Ia+;g f := f ([a, b] , X ). By [2] we get that Ia+;g   ν and see that Ia+;g f (a) = 0. ν ν When g is the identity function id, we get that Ia+;id = Ia+ , the ordinary left Riemann–Liouville fractional integral

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_9

179

9 Generalized g-Fractional Vector Representation …

180



 ν f (x) = Ia+

1  (ν)



x

(x − t)ν−1 f (t) dt,

(9.2)

a

 ν  ∀ x ∈ [a, b], Ia+ f (a) = 0. We need Theorem 9.2 ([2]) Let μ, ν > 0 and f ∈ C ([a, b] , X ). Then μ

μ+ν

μ

ν ν f = Ia+;g f = Ia+;g Ia+;g f. Ia+;g Ia+;g

(9.3)

We need Definition 9.3 ([2]) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the right Riemann–Liouville generalized fractional Bochner integral operator 

 ν f (x) := Ib−;g

1  (ν)



b

(g (z) − g (x))ν−1 g  (z) f (z) dz,

(9.4)

x

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈C ([a, b] , X ), then f ∈L ∞ ν 0 f ∈ C ([a, b] , X ). Above we set Ib−;g f := f ([a, b] , X ). By [2] we get that Ib−;g   ν and see that Ib−;g f (b) = 0. ν ν When g is the identity function id, we get that Ib−;id = Ib− , the ordinary right Riemann–Liouville fractional integral 

ν Ib−



1 f (x) =  (ν)



b

(t − x)ν−1 f (t) dt,

(9.5)

x

 ν  ∀ x ∈ [a, b], with Ib− f (b) = 0. We need Theorem 9.4 ([2]) Let μ, ν > 0 and f ∈ C ([a, b] , X ). Then μ

μ+ν

μ

ν ν Ib−;g Ib−;g f = Ib−;g f = Ib−;g Ib−;g f.

(9.6)

We will use Definition 9.5 ([2]) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) .

9.1 Background

181

We define the left generalized g-fractional derivative X -valued of f of order α as follows:  x  α (n)   1 Da+;g f (x) := (g (x) − g (t))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) a (9.7) ∀ x ∈ [a, b]. The last integral is of Bochner type. α f ∈ C ([a, b] , X ). If α ∈ / N, by [2], we have that Da+;g We see that    (n)  α  n−α Ia+;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , ∀ x ∈ [a, b] . (9.8) We set n f (x) := Da+;g



f ◦ g −1

(n)

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

(9.9)

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

When g = id, then

α α α f = Da+;id f = D∗a f, Da+;g

(9.10)

the usual left X -valued Caputo fractional derivative, see [3]. We will use Definition 9.6 ([2]) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the right generalized g-fractional derivative X -valued of f of order α as follows:  b (n)    α (−1)n Db−;g f (x) := (g (t) − g (x))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) x (9.11) ∀ x ∈ [a, b]. The last integral is of Bochner type. α f ∈ C ([a, b] , X ). If α ∈ / N, by [2], we have that Db−;g We see that    (n)  α  n−α ◦ g (x) = Db−;g f (x) , a ≤ x ≤ b. (9.12) Ib−;g (−1)n f ◦ g −1 We set n f (x) := (−1)n Db−;g



f ◦ g −1

n

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

(9.13)

9 Generalized g-Fractional Vector Representation …

182

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

When g = id, then

α α α f (x) = Db−;id f (x) = Db− f, Db−;g

(9.14)

the usual right X -valued Caputo fractional derivative, see [3]. We make Remark 9.7 All as in Definition 9.5. We have (by Theorem 2.5, p. 7, [6])      α  Da+;g f (x) ≤

≤

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

  (n)   ◦ g  f ◦ g −1



∞,[a,b]

 (n − α)

g(x)

g(a)

  (n)   ◦ g  f ◦ g −1

(g (x) − g (t))n−α−1 dg (t) =

∞,[a,b]

(g (x) − g (a))n−α .

 (n − α + 1) That is  α  D

a+;g

  f (x) ≤

  (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (x) − g (a))n−α ,

(9.15)

(9.16)

∀ x ∈ [a, b] .   α f (a) = 0. If α ∈ / N, then Da+;g Similarly, by Definition 9.6 we derive      α  Db−;g f (x) ≤

≤

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

  (n)   ◦ g  f ◦ g −1

∞,[a,b]



 (n − α)   (n)   ◦ g  f ◦ g −1

g(x)

∞,[a,b]

 (n − α + 1) That is

g(b)

(g (t) − g (x))n−α−1 dg (t) =

(g (b) − g (x))n−α .

     −1 (n) f ◦ g ◦ g     ∞,[a,b]  (g (b) − g (x))n−α , b−;g f (x) ≤  (n − α + 1)

 α  D

∀ x ∈ [a, b] .

(9.17)

(9.18)

9.1 Background

183

  α If α ∈ / N, then Db−;g f (b) = 0. Notation 9.8 We denote by nα α α α := Da+;g Da+;g . . . Da+;g (n times), n ∈ N, Da+;g

(9.19)

nα α α α := Ia+;g Ia+;g . . . Ia+;g , Ia+;g

(9.20)

nα α α α := Db−;g Db−;g . . . Db−;g , Db−;g

(9.21)

nα α α α := Ib−;g Ib−;g . . . Ib−;g , Ib−;g

(9.22)

and (n times), n ∈ N.

We are motivated by the following generalized fractional Ostrowski type inequality: Theorem 9.9 ([2]) Let g ∈ C 1 ([a, b]) and strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]), and 0 < α < 1, n ∈ N, f ∈ C 1 ([a, b] , X ), where (X, ·) is a Banach space. Let x0 ∈ [a, b] be fixed. Assume that Fkx0 := Dxkα0 −;g f , for k =   1, . . . , n, fulfill Fkx0 ∈ C 1 ([a, x0 ] , X ) and Dxiα0 −;g f (x0 ) = 0, i = 2, . . . , n. Similarly, we assume that G kx0 := Dxkα0 +;g f , for k = 1, . . . , n, fulfill G kx0 ∈ C 1   ([x0 , b] , X ) and Dxiα0 +;g f (x0 ) = 0, i = 2, . . . , n. Then    b   1 1   · f d x − f (x) (x ) 0 ≤ b − a − a)  + 1) α + 1) (b ((n a      f (g (b) − g (x0 ))(n+1)α (b − x0 ) Dx(n+1)α  +;g 0 (n+1)α

(g (x0 ) − g (a))

    f (x0 − a) Dx(n+1)α  −;g 0

∞,[x0 ,b]

+

∞,[a,x0 ]

.

(9.23)

In this work we will present several generalized fractional Bochner integral inequalities. We mention the following g-left generalized X -valued Taylor’s formula: Theorem 9.10 ([2]) Let α > 0, n = α , and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]). Then f (x) = f (a) +

n−1

(g (x) − g (a))i  i=1

i!

f ◦ g −1

(i)

(g (a)) +

9 Generalized g-Fractional Vector Representation …

184



1  (α)

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

x

a

f (a) +

n−1

(g (x) − g (a))i 

i!

i=1

1  (α)



g(x)

(g (x) − z)α−1

g(a)



f ◦ g −1

(i)

(g (a)) +

(9.24)

  α Da+;g f ◦ g −1 (z) dz, ∀ x ∈ [a, b] .

We mention the following g-right generalized X -valued Taylor’s formula: Theorem 9.11 ([2]) Let α > 0, n = α , and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]). Then f (x) = f (b) +

n−1

(g (x) − g (b))i 

i!

i=1

1  (α)



b x

f (b) + 

g(b)

g(x)

(i)

(g (b)) +

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

n−1

(g (x) − g (b))i 

i!

i=1

1  (α)

f ◦ g −1

(z − g (x))α−1



f ◦ g −1

(i)

(g (b)) +

(9.25)

  α f ◦ g −1 (z) dz, ∀ x ∈ [a, b] . Db−;g

For the Bochner integral excellent resources are [5, 7, 8] and [1, pp. 422–428].

9.2 Main Results We give the following representation formula: Theorem 9.12 All as in Theorem 9.10. Then  b n−1

1 f (y) = f (x) d x − b−a a k=1



f ◦ g −1

(k)

(g (y))  b

k! (b − a)

a

(g (x) − g (y))k d x + R1 (y) ,

(9.26) for any y ∈ [a, b], where R1 (y) = −

1  (α) (b − a)

9.2 Main Results



185



b

χ[a,y) (x)

a



b

y

|g (x) − g (t)|

α−1



g (t)



x



x

χ[y,b] (x)

a

y

D αy−;g



f (t) dt d x

   |g (x) − g (t)|α−1 g  (t) D αy+;g f (t) dt d x .

(9.27)

here χ A stands for the characteristic function set A, where A is an arbitrary set. One may write also that  y  y

  α 1 α−1  R1 (y) = − g (t) D y−;g f (t) dt d x (g (t) − g (x))  (α) (b − a) a x (9.28)

  b  x   + (g (x) − g (t))α−1 g  (t) D αy+;g f (t) dt d x , y

y

for any y ∈ [a, b] . Putting things together, one has 1 f (y) = b−a −

1  (α) (b − a) 

b



(k)  n−1 

f ◦ g −1 (g (y)) b f (x) d x − (g (x) − g (y))k d x k! − a) (b a k=1

b

a



b



a

x



x

χ[y,b] (x)

a

y

χ[a,y) (x)

y

  |g (x) − g (t)|α−1 g  (t) D αy−;g f (t) dt d x

   |g (x) − g (t)|α−1 g  (t) D αy+;g f (t) dt d x .

(9.29)

In particular, one has

f (y) −

 b

n−1

1 f (x) d x + b−a a k=1



f ◦ g −1

(k)

(g (y))  b

k! (b − a)

a

(g (x) − g (y))k d x = R1 (y) ,

(9.30) for any y ∈ [a, b] . Proof Here x, y ∈ [a, b]. We keep y as fixed. By Theorem 9.10 we get: (k) n−1 

f ◦ g −1 (g (y)) f (x) = f (y) + (g (x) − g (y))k + k! k=1 1  (α)



x y

  (g (x) − g (t))α−1 g  (t) D αy+;g f (t) dt, for any x ≥ y.

(9.31)

9 Generalized g-Fractional Vector Representation …

186

By Theorem 9.11 we get: (k) n−1 

f ◦ g −1 (g (y)) f (x) = f (y) + (g (x) − g (y))k + k! k=1 1  (α)



(9.32)

  (g (t) − g (x))α−1 g  (t) D αy−;g f (t) dt, for any x ≤ y.

y x

By (9.31), (9.32) we notice that 



b

f (x) d x =

a



y

a

b y



b

f (x) d x +

a

f (x) d x =

(9.33)

y

(k)  n−1 

f ◦ g −1 (g (y)) y f (y) d x + (g (x) − g (y))k d x+ k! a k=1 

1  (α) 

y

y



a

y

(g (t) − g (x))

α−1



g (t)

x



D αy−;g



f (t) dt d x+

(k)  n−1 

f ◦ g −1 (g (y)) b f (y) d x + (g (x) − g (y))k d x+ k! y k=1 

1  (α)

b



y

x y

  α α−1  g f dt d x. − g D (g (x) (t)) (t) y+;g (t)

Hence it holds (k)  n−1 

f ◦ g −1 (g (y)) b f (x) d x = f (y) + (g (x) − g (y))k d x+ k! − a) (b a a k=1 (9.34)  y  y

  α 1 α−1  |g (x) − g (t)| g (t) D y−;g f (t) dt d x+  (α) (b − a) a x

1 b−a



b



b



y

x y

|g (x) − g (t)|

α−1



g (t)



D αy+;g





f (t) dt d x .

Therefore we obtain 1 f (y) = b−a

 a

b

(k)  n−1 

f ◦ g −1 (g (y)) b f (x) d x − (g (x) − g (y))k d x− k! − a) (b a k=1 (9.35)

9.2 Main Results

187

1  (α) (b − a)



y



a

b 



y

x

y x

  |g (x) − g (t)|α−1 g  (t) D αy−;g f (t) dt d x+

|g (x) − g (t)|

α−1



g (t)

y



D αy+;g





f (t) dt d x .

Hence the remainder 1  (α) (b − a)

R1 (y) := −



b

+



y

−

x

 a

y



y x

|g (x) − g (t)|

  |g (x) − g (t)|α−1 g  (t) D αy−;g f (t) dt d x

α−1



g (t)

y



D αy+;g





f (t) dt d x =

 y  b

  1 |g (x) − g (t)|α−1 g  (t) D αy−;g f (t) dt d x χ[a,y) (x)  (α) (b − a) a x  x

 (9.36)  b   α α−1  |g (x) − g (t)| χ[y,b] (x) g (t) D y+;g f (t) dt d x . + a

y



The theorem is proved. Next we present a left fractional Opial type inequality:

Theorem 9.13 All as in Theorem 9.10. Additionally assume that α ≥ 1, g ∈ C 1 (k)  ([a, b]), and f ◦ g −1 (g (a)) = 0, for k = 0, 1, . . . , n − 1. Let p, q > 1 : 1p + 1 = 1. Then q 

x

a

 x  w a

a

 α    f (w)  Da+;g f (w) g  (w) dw ≤

1 1

 (α) 2 q

·

(9.37)

1  x

2 q   p q   q    α f (w) dw , g (w)  Da+;g (g (w) − g (t)) p(α−1) dt dw a

∀ x ∈ [a, b] . Proof By Theorem 9.10, we have that f (x) =

1  (α)



x

a

  α f (t) dt, ∀ x ∈ [a, b] . (9.38) (g (x) − g (t))α−1 g  (t) Da+;g

Then, by Hölder’s inequality we obtain,  f (x) ≤

1  (α)

 a

x

1  (g (x) − g (t)) p(α−1) dt

p

a

x



 q q1  q   α  .  Da+;g f (t) dt

g  (t)

(9.39)

9 Generalized g-Fractional Vector Representation …

188



Call

  q  α  q g (t)  Da+;g f (t) dt,

(9.40)

 q  q  α f (x) ≥ 0, z  (x) = g  (x)  Da+;g

(9.41)

x

z (x) := a

z (a) = 0. Thus

and



z  (x)

 q1

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

(9.42)

  α   f (w) g  (w)  Da+;g f (w) ≤

(9.43)

Consequently, we get



1  (α)

w

(g (w) − g (t))

1p dt



z (w) z  (w)

 q1

, ∀ w ∈ [a, b] .

a



Then

 α    f (w)  Da+;g f (w) g  (w) dw ≤

x

a



1  (α) 1  (α)

p(α−1)



x

x



a

1  (α)

(g (w) − g (t))

p(α−1)

1p dt



1 z (w) z  (w) q dw ≤

a



a

w

(9.44)

w

1p 

(g (w) − g (t))

p(α−1)



z (w) z (w) dw

dt dw

a



x

a x



a

w

(g (w) − g (t)) p(α−1) dt dw

1p

a

1  (α)



x

a

 a

x



w

z (x) 2

(g (w) − g (t))

p(α−1)

2

1p

dt dw

q1

q1

=

(9.45) =

·

a

  q  α  q g (t)  Da+;g f (t) dt

q2

The theorem is proved.

· 2− q . 1

(9.46) 

We also give a right fractional Opial type inequality: Theorem 9.14 All as in Theorem 9.11. Additionally assume that α ≥ 1, g ∈ C 1 (k)  ([a, b]), and f ◦ g −1 (g (b)) = 0, k = 0, 1, . . . , n − 1. Let p, q > 1 : 1p + q1 = 1. Then

9.2 Main Results

189



b x

  b b x

w

 α    f (w)  Db−;g f (w) g  (w) dw ≤ 

(g (t) − g (w)) p(α−1) dt

1 1 q

2  (α)

·

(9.47)

 1  2 q p q   b q    α  g (w)  Db−;g f (w) dw dw , x

all a ≤ x ≤ b. Proof By Theorem 9.11, we have that 1  (α)

f (x) =



b x

  α f (t) dt, all a ≤ x ≤ b. (g (t) − g (x))α−1 g  (t) Db−;g (9.48)

Then, by Hölder’s inequality we obtain, 1  (α)

 f (x) ≤



b

(g (t) − g (x)) p(α−1) dt

x



(9.49) (9.50)

 q  q  α f (x) ≤ 0, z  (x) = − g  (x)  Db−;g

(9.51)

 q  q  α f (x) ≥ 0, − z  (x) = g  (x)  Db−;g

(9.52)

 α   1   f (x) ≥ 0, ∀ x ∈ [a, b] . −z (x) q = g  (x)  Db−;g

(9.53)



x

and

  q q1   q   α  g (t)  Db−;g f (t) dt .

 q q  α f (t) dt, g  (t)  Db−;g

b

z (x) :=

and

b x

Call

z (b) = 0. Hence

1p 

Consequently, we get   α   f (w) g  (w)  Db−;g f (w) ≤ 1  (α) Then



b w

(g (t) − g (w)) 

b x

1p p(α−1)

dt



  1 z (w) −z  (w) q , ∀ w ∈ [a, b] . (9.54)

 α    f (w)  Db−;g f (w) g  (w) dw ≤

(9.55)

9 Generalized g-Fractional Vector Representation …

190

1  (α) 1  (α)



b



b



w

x



b w

x

1  (α)

b

(g (t) − g (w)) p(α−1) dt

1p



−z (w) z  (w)

1p  − dt dw

(g (t) − g (w))

p(α−1)

 q1

b

dw ≤



z (w) z (w) dw x



b



b

(g (t) − g (w)) p(α−1) dt dw

w

x



1

b



1

2 q  (α)

w

x



b x

b

1p

z (x) 2 2

(g (t) − g (w))

p(α−1)

  q  α  q g (t)  Db−;g f (t) dt

dt dw

q2

1p

q1

q1

=

(9.56) =

·

.

(9.57) 

The theorem is proved.

Two extreme fractional Opial type inequalities follow (case p = 1, q = ∞).  (k) Theorem 9.15 All as in Theorem 9.10. Assume that f ◦ g −1 (g (a)) = 0, k = 0, 1, . . . , n − 1. Then  2  α 

 x Da+;g f   x  α  ∞    f (w) Da+;g f (w) dw ≤ (g (w) − g (a))α dw ,  (α + 1) a a (9.58) all a ≤ x ≤ b. Proof For any w ∈ [a, b], we have that f (x) = and

1  (α)

 f (x) ≤



w

a

1  (α)

  α f (t) dt, (g (w) − g (t))α−1 g  (t) Da+;g



=

a

w

(9.59)

  α f ∞ (g (w) − g (t))α−1 g  (t) dt  Da+;g

   α  Da+;g f 

∞

 (α + 1)

(g (w) − g (a))α .

(9.60)

Hence we obtain

 2  α   Da+;g f ∞  α  f (w)  Da+;g f (w) ≤ (g (w) − g (a))α .  (α + 1)

Integrating (9.61) over [a, x] we derive (9.58).

(9.61) 

9.2 Main Results

191

Theorem 9.16 All as in Theorem 9.11. Assume that k = 0, 1, . . . , n − 1. Then 

b x

 α   f (w)  Db−;g f (w) dw ≤

 2  α  Db−;g f  

b

∞

 (α + 1)



f ◦ g −1

(k)

(g (b)) = 0,

α

(g (b) − g (w)) dw ,

x

(9.62)

all a ≤ x ≤ b. Proof For any w ∈ [a, b], we have f (x) =

1  (α)



b

w

and  f (x) ≤

1  (α) =

  α f (t) dt, (g (t) − g (w))α−1 g  (t) Db−;g



b w

(9.63)

  α f ∞ (g (t) − g (w))α−1 g  (t) dt  Db−;g

   α  Db−;g f 

∞

 (α + 1)

(g (b) − g (w))α .

(9.64)

Hence we obtain  α   f (w)  Db−;g f (w) ≤

 2  α   Db−;g f 

∞

 (α + 1)

(g (b) − g (w))α .

(9.65) 

Integrating (9.65) over [x, b] we derive (9.62). Next we present three fractional Ostrowski type inequalities: Theorem 9.17 All as in Theorem 9.10. Then    (k)   −1  b  b n−1   f ◦ g (g (y))

1   k f (x) d x + (g (x) − g (y)) d x  ≤  f (y) −   b−a a k! (b − a) a   k=1

(9.66) 1 ·  (α + 1) (b − a) 

     (g (y) − g (a))α (y − a)  D αy−;g f ∞ + (g (b) − g (y))α (b − y)  D αy+;g f ∞ ,

∀ y ∈ [a, b] .

9 Generalized g-Fractional Vector Representation …

192



 D αy+;g f (t) = 0, for t < y,  and 

Proof Define

(9.67)

D αy−;g f (t) = 0, for t > y.

Notice for 0 < α ∈ / N by Remark 9.7 we have   α Da+;g f (a) = 0.

(9.68)

Similarly it holds (0 < α ∈ / N) by Remark 9.7 that   α Db−;g f (b) = 0.

(9.69)



Thus

   D αy+;g f (y) = 0, D αy−;g f (y) = 0,

(9.70)

0 1 :

1 p

+

1 q

= 1, α >

1 . q

Then

  (k)   b n−1   

f ◦ g −1 (g (y)) b 1   k f (x) d x + (g (x) − g (y)) d x   f (y) −   b−a a k! − a) (b a k=1 (9.80) 1 ≤ 1 ·  (α) (b − a) ( p (α − 1) + 1) p     1 (g (y) − g (a))α−1+ p (y − a)  D αy−;g f ◦ g −1 q,[g(a),g(y)]     1 + (g (b) − g (y))α−1+ p (b − y)  D αy+;g f ◦ g −1 q,[g(y),g(b)] , ∀ y ∈ [a, b] . Proof Here we use (9.75). We get that 1 R1 (y) ≤  (α) (b − a) 

g(y)

g(x)

 α  D

y−;g



f ◦g

−1





q (z) dz

y



g(y)

a

g(x)

q1



b

dx + y



g(x)

g(y)

 α  D

y+;g



f ◦g

 −1

(z − g (x)) 

g(x)

g(y)

q (z) dz

q1

1p p(α−1)

dz

(g (x) − z)  dx ≤

·

1p

p(α−1)

dz

·

9.2 Main Results

1  (α) (b − a)  +

y

a

195



y

(g (y) − g (x))(α−1)+ p 1

1 p



   d x  D αy−;g f ◦ g −1 q,[g(a),g(y)]

( p (α − 1) + 1)   (9.81) (α−1)+ 1p     (g (x) − g (y)) d x  D αy+;g f ◦ g −1 q,[g(y),g(b)] . 1 p ( p (α − 1) + 1) a

(here it is α − 1 + Hence it holds

> 0)

1 p

R1 (y) ≤

1 1

 (α) (b − a) ( p (α − 1) + 1) p

·

(9.82)

    1 (g (y) − g (a))α−1+ p (y − a)  D αy−;g f ◦ g −1 q,[g(a),g(y)] +     1 (g (b) − g (y))α−1+ p (b − y)  D αy+;g f ◦ g −1 q,[g(y),g(b)] . Clearly here  α  D

y−;g

     f ◦ g −1 q,[g(a),g(y)] ,  D αy+;g f ◦ g −1 q,[g(y),g(b)] < ∞. 

We have proved the theorem. Next we give some fractional Grüss type inequalities:

Theorem 9.20 Let f, h as in Theorem 9.10. Here R1 (y) will be renamed as R1 ( f, y), so we can consider R1 (h, y). Then (1)

n ( f, h) :=

1 b−a

 a

b

 b f (x) h (x) d x −

a

f (x) d x

 

b a

h (x) d x

(b − a)2

 +

 b  b  n−1

 (k) 1 1 h (y) f ◦ g −1 (g (y)) + 2 2 (b − a) k=1 k! a a    (k)  f (y) h ◦ g −1 (g (y)) (g (x) − g (y))k d x dy = 1 2 (b − a)

 a

b

 (h (y) R1 ( f, y) + f (y) R1 (h, y)) dy =: K n ( f, h) , (9.83)

9 Generalized g-Fractional Vector Representation …

196

(2) it holds         (g (b) − g (a))α  α   α  h∞ n ( f, h) ≤ sup D y−;g f  + D y+;g f  ∞ ∞ 2 (α + 1) y∈[a,b]  +  f ∞



    h sup D α y−;g 

∞

y∈[a,b]

     + D α h y+;g 

,

∞

(9.84)

(3) if α ≥ 1, we get: n ( f, h) ≤ 

 h1

1 (g (b) − g (a))α−1 · 2 (α) (b − a)

     sup  D αy−;g f ◦ g −1 

     +  D αy+;g f ◦ g −1 

     sup  D αy−;g h ◦ g −1 

     +  D αy+;g h ◦ g −1 

1,[g(a),g(b)]

y∈[a,b]

  f 1

1,[g(a),g(b)]

y∈[a,b]

(4) if p, q > 1 :

1 p

+

1 q

(9.85)



1,[g(a),g(b)]

+

 ,

1,[g(a),g(b)]

= 1, α > q1 , we get: (g (b) − g (a))α−1+ p 1

n ( f, h) ≤ 



h∞

sup y∈[a,b]

  f ∞

1

2 (α) ( p (α − 1) + 1) p

    α   D y−;g f ◦ g −1 

q,[g(a),g(b)]

     sup  D αy−;g h ◦ g −1 

q,[g(a),g(b)]

y∈[a,b]

·

(9.86)

     +  D αy+;g f ◦ g −1 



q,[g(a),g(b)]

     +  D αy+;g h ◦ g −1 

q,[g(a),g(b)]

+

 .

All right hand sides of (9.84)–(9.86) are finite. Proof By Theorem 9.10 we have h (y) h (y) f (y) = b−a



b

f (x) d x− a

(k)   n−1

h (y) f ◦ g −1 (g (y)) b (g (x) − g (y))k d x + h (y) R1 ( f, y) , k! − a) (b a k=1

(9.87)

9.2 Main Results

197

and

f (y) b−a

f (y) h (y) =



b

h (x) d x− a

(k)   n−1

f (y) h ◦ g −1 (g (y)) b (g (x) − g (y))k d x + f (y) R1 (h, y) , k! − a) (b a k=1

(9.88)

∀ y ∈ [a, b] . Then integrating (9.87) we find 

b

 b h (y) f (y) dy =

b−a

a n−1

k=1

1 k! (b − a)



b



a

b

 h (y) dy 

a

b

f (x) d x −

a

(k)  h (y) f ◦ g −1 (g (y)) (g (x) − g (y))k d xd y

a



b

+

h (y) R1 ( f, y) dy,

(9.89)

a

and integrating (9.88) we obtain 

b

 b a

f (y) h (y) dy =

f (y) dy

k=1

1 k! (b − a)



b



a

b

b a

h (x) d x

b−a

a n−1

 

 −

(k)  f (y) h ◦ g −1 (g (y)) (g (x) − g (y))k d xd y

a



b

+

f (y) R1 (h, y) dy.

(9.90)

a

Adding the last two equalities (9.89) and (9.90), we get:     b b  b 2 a f (x) d x h d x (x) a − 2 f (x) h (x) d x = b − a a n−1

k=1

1 k! (b − a)

 a

b



b



h (y) f ◦ g

 −1 (k)



(g (y)) + f (y) h ◦ g

 −1 (k)

 (g (y)) ·

a

 (g (x) − g (y))k d xd y +



b

(h (y) R1 ( f, y) + f (y) R1 (h, y)) dy.

a

Divide the last (9.91) by 2 (b − a) to obtain (9.83).

(9.91)

9 Generalized g-Fractional Vector Representation …

198

Then, we upper bound K n ( f, h) using Theorems 9.17–9.19, to obtain (9.84)– (9.86), respectively. We use also that a norm is a continuous function. The theorem is proved.  We make Remark 9.21 (in support of the proof of Theorem 9.20) Let α > 0, α ∈ / N, α = n. We have  x (n)   α  1 D y+;g f (x) = (g (x) − g (t))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) y (9.92) ∀ x ∈ [y, b] , and 

 D αy−;g f (x) =

(−1)n  (n − α)



y

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

x

(9.93) ∀ x ∈ [a, y] , both are Bochner type integrals. By change of variables for Bochner integrals, see [7, Lemma B. 4.10] and [8, p. 158], we get: 

 D αy+;g f (x) =

1  (n − α)





g(x) g(y)

 (n) (g (x) − z)n−α−1 f ◦ g −1 (z) dz =

  α f ◦ g −1 (g (x)) , ∀ x ∈ [y, b] , Dg(y)+

(9.94)

and 

 D αy−;g f (x) =

(−1)n  (n − α)





g(y) g(x)

 (n) (z − g (x))n−α−1 f ◦ g −1 (z) dz =

  α f ◦ g −1 (g (x)) , ∀ x ∈ [a, y] . Dg(y)−

(9.95)

α α , Dg(y)− are the left and right X -valued Caputo fractional differentiation Here Dg(y)+ operators. Fix w : w ≥ x0 ≥ y0 ; w, x0 , y0 ∈ [a, b], then g (w) ≥ g (x0 ) ≥ g (y0 ). Hence

 α  D

y0 +;g

 α  D

g(y0 )+



f ◦ g −1



    f (w) − Dxα0 +;g f (w) =

  α   f ◦ g −1 (g (w)) = (g (w)) − Dg(x 0 )+

 g(x0 )      1 n−α−1 −1 (n)  f ◦g (g (w) − z) (z) dz   ≤  (n − α) g(y0 )

(9.96)

9.2 Main Results

199



1  (n − α)

g(x0 )

g(y0 )

 (n)    (g (w) − z)n−α−1  f ◦ g −1 (z) dz ≤

 (n)     f ◦ g −1 

∞,[g(a),g(b)]

 (n − α)  (n)     f ◦ g −1 

∞,[g(a),g(b)]

 (n − α + 1)



g(x0 ) g(y0 )

(g (w) − z)n−α−1 dz =

  (g (y0 ) − z)n−α − (g (x0 ) − z)n−α → 0,

   α f ◦ g −1 (g (x)) as y0 → x0 , then g (y0 ) → g (x0 ), proving continuity of Dg(y)+   with respect to g (y), and of course continuity of D αy+;g f (x) in y ∈ [a, b] .   Similarly, it is proved that D αy−;g f (x) is continuous in y ∈ [a, b], the proof is omitted. Remark 9.22 Some examples for g follow: g (x) = e x , x ∈ [a, b] ⊂ R, g (x) = sin x, g (x) = tan x,   π π where x ∈ − + ε, − ε , where ε > 0 small. 2 2 Indeed, the above examples of g are strictly increasing and continuous functions. One can apply all of our results here for the above specific choices of g. We choose to omit this job.

References 1. Aliprantis, C.D., Border, K.C.: Infinite Dimensional Analysis. Springer, New York (2006) 2. Anastassiou, G.A.: Principles of general fractional analysis for Banach space valued functions. Bull. Allahabad Math. Soc. 32(1), 71–145 (2017) 3. Anastassiou, G.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 4. Anastassiou, G.: Generalized g-fractional vector representation formula and integral Inequalities for Banach space valued functions. J. Comput. Anal. Appl. 29(6), 1063–1081 (2021) 5. Bochner integral. Encyclopedia of Mathematics. http://www.encyclopediaofmath.org/index. php?title=Bochner_integral&oldid=38659 6. Kreuter, M.: Sobolev space of vector-valued functions. Ulm University, Master thesis in Mathematics, Ulm, Germany (2015) 7. Mikkola, K.: Appendix B integration and differentiation in Banach spaces. http://math.aalto.fi/ ~kmikkola/research/thesis/contents/thesisb.pdf 8. Mikusinski, J.: The Bochner Integral. Academic, New York (1978)

Chapter 10

Iterated g-Fractional Vector Bochner Integral Representation Formulae and Inequalities for Banach Space Valued Functions

Here we present very general iterated fractional Bochner integral representation formulae for Banach space valued functions. Based on these we derive generalized and iterated left and right: fractional Poincaré type inequalities, fractional Opial type inequalities and fractional Hilbert–Pachpatte inequalities. All these inequalities are very general having in their background Bochner type integrals. See also [4].

10.1 Background We need Definition 10.1 ([2]) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the left Riemann–Liouville generalized fractional Bochner integral operator 

 ν f (x) := Ia+;g

1  (ν)



x

(g (x) − g (z))ν−1 g  (z) f (z) dz,

(10.1)

a

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈ C ([a, b] , X ), then f ∈ L ∞ ν 0 f ∈ C ([a, b] , X ). Above we set Ia+;g f := f ([a, b] , X ). By [2] we get that Ia+;g   ν and see that Ia+;g f (a) = 0. ν ν When g is the identity function id, we get that Ia+;id = Ia+ , the ordinary left Riemann–Liouville fractional integral 

ν Ia+



1 f (x) =  (ν)



x

(x − t)ν−1 f (t) dt,

(10.2)

a

 ν  ∀ x ∈ [a, b], Ia+ f (a) = 0. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_10

201

10 Iterated g-Fractional Vector Bochner …

202

We need Theorem 10.2 ([2]) Let μ, ν > 0 and f ∈ C ([a, b] , X ). Then μ

μ+ν

μ

ν ν f = Ia+;g f = Ia+;g Ia+;g f. Ia+;g Ia+;g

(10.3)

We need Definition 10.3 ([2]) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the right Riemann–Liouville generalized fractional Bochner integral operator 

 ν f (x) := Ib−;g

1  (ν)



b

(g (z) − g (x))ν−1 g  (z) f (z) dz,

(10.4)

x

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈ C ([a, b] , X ), then f ∈ L ∞ ν 0 f ∈ C ([a, b] , X ). Above we set Ib−;g f := f ([a, b] , X ). By [2] we get that Ib−;g   ν and see that Ib−;g f (b) = 0. ν ν = Ib− , the ordinary right When g is the identity function id, we get that Ib−;id Riemann–Liouville fractional integral 

ν Ib−



1 f (x) =  (ν)



b

(t − x)ν−1 f (t) dt,

(10.5)

x

 ν  ∀ x ∈ [a, b], with Ib− f (b) = 0. We need Theorem 10.4 ([2]) Let μ, ν > 0 and f ∈ C ([a, b] , X ). Then μ

μ+ν

μ

ν ν f = Ib−;g f = Ib−;g Ib−;g f. Ib−;g Ib−;g

(10.6)

We will use Definition 10.5 ([2]) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the left generalized g-fractional derivative X -valued of f of order α as follows:  x (n)    α 1 Da+;g f (x) := (g (x) − g (t))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) a (10.7) ∀ x ∈ [a, b]. The last integral is of Bochner type.

10.1 Background

203

  α If α ∈ / N, by [2], we have that Da+;g f ∈ C ([a, b] , X ). We see that    (n)   α n−α Ia+;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , ∀ x ∈ [a, b] .

(10.8)

We set n f (x) := Da+;g



f ◦ g −1

(n)

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

(10.9)

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

When g = id, then

α α α f = Da+;id f = D∗a f, Da+;g

(10.10)

the usual left X -valued Caputo fractional derivative, see [3]. We will use Definition 10.6 ([2]) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the right generalized g-fractional derivative X -valued of f of order α as follows:  b  (n)   α (−1)n Db−;g f (x) := (g (t) − g (x))n−α−1 g  (t) f ◦ g −1 (g (t)) dt,  (n − α) x (10.11) ∀ x ∈ [a, b]. The last integral is of Bochner type. α f ∈ C ([a, b] , X ). If α ∈ / N, by [2], we have that Db−;g We see that    (n)   α n−α ◦ g (x) = Db−;g f (x) , a ≤ x ≤ b. (10.12) Ib−;g (−1)n f ◦ g −1 We set n f (x) := (−1)n Db−;g



f ◦ g −1

n

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

(10.13)

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

When g = id, then

α α α f (x) = Db−;id f (x) = Db− f, Db−;g

the usual right X -valued Caputo fractional derivative, see [3]. We make

(10.14)

10 Iterated g-Fractional Vector Bochner …

204

Remark 10.7 All as in Definition 10.5. We have (by [6, Theorem 2.5, p. 7])      α  Da+;g f (x) ≤

≤

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

  (n)   ◦ g  f ◦ g −1



∞,[a,b]

 (n − α)

g(x)

g(a)

  (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (x) − g (t))n−α−1 dg (t) =

(g (x) − g (a))n−α .

(10.15)

That is      −1 (n) ◦ g   f ◦g ∞,[a,b]  (g (x) − g (a))n−α , a+;g f (x) ≤  (n − α + 1)

 α  D



(10.16)

∀ x ∈ [a, b] .   α f (a) = 0. If α ∈ / N, then Da+;g Similarly, by Definition 10.6 we derive     α   Db−;g f (x) ≤

≤

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

  (n)   ◦ g  f ◦ g −1

∞,[a,b]



 (n − α)

g(b) g(x)

  (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (t) − g (x))n−α−1 dg (t) =

(g (b) − g (x))n−α .

(10.17)

That is  α  D

b−;g

  f (x) ≤

  (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (b) − g (x))n−α ,

(10.18)

∀ x ∈ [a, b] .   α f (b) = 0. If α ∈ / N, then Db−;g Notation 10.8 We denote by nα α α α Da+;g := Da+;g Da+;g . . . Da+;g (n times), n ∈ N,

(10.19)

10.1 Background

205

and

nα α α α Ia+;g := Ia+;g Ia+;g . . . Ia+;g ,

(10.20)

nα α α α := Db−;g Db−;g . . . Db−;g , Db−;g

(10.21)

nα α α α := Ib−;g Ib−;g . . . Ib−;g , Ib−;g

(10.22)

(n times), n ∈ N. We are motivated by the following generalized fractional Ostrowski type inequality: Theorem 10.9 ([2]) Let g ∈ C 1 ([a, b]) and strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]), and 0 < α < 1, n ∈ N, f ∈ C 1 ([a, b] , X ), where (X, ·) is a Banach space. Let x0 ∈ [a, b] be fixed. Assume that Fkx0 := Dxkα0 −;g f , for k =   1, . . . , n, fulfill Fkx0 ∈ C 1 ([a, x0 ] , X ) and Dxiα0 −;g f (x0 ) = 0, i = 2, . . . , n. Similarly, we assume that G kx0 := Dxkα0 +;g f , for k = 1, . . . , n, fulfill G kx0 ∈ C 1   ([x0 , b] , X ) and Dxiα0 +;g f (x0 ) = 0, i = 2, . . . , n. Then    b   1 1  f (x) d x − f (x0 )  ≤ (b − a)  ((n + 1) α + 1) · b − a a      f (g (b) − g (x0 ))(n+1)α (b − x0 ) Dx(n+1)α  0 +;g     f (g (x0 ) − g (a))(n+1)α (x0 − a)  Dx(n+1)α  0 −;g

∞,[x0 ,b]

+

∞,[a,x0 ]

.

(10.23)

In this work we will present several generalized fractional Bochner integral inequalities. We mention the following g-left generalized modified X -valued Taylor’s formula. Theorem 10.10 ([2]) Let 0 < α ≤ 1, n ∈ N, f ∈ C 1 ([a, b] , X ), g ∈ C 1 ([a, b]), kα f, k = strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]). Let Fk := Da+;g 1 1, . . . , n, that fulfill Fk ∈ C ([a, b] , X ) . Then n

 (g (x) − g (a))iα  iα Da+;g f (a) + f (x) =  + 1) (iα i=0 1  ((n + 1) α)

 a

x

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

∀ x ∈ [a, b] . (n+1)α f ∈ C ([a, b] , X ) . Clearly here Da+;g

(10.24)

10 Iterated g-Fractional Vector Bochner …

206

We also mention the following g-right generalized modified X -valued Taylor’s formula. Theorem 10.11 ([2]) Let f ∈ C 1 ([a, b] , X ), g ∈ C 1 ([a, b]), strictly increasing, kα f , k = 1, . . . , n, fulfill such that g −1 ∈ C 1 ([g (a) , g (b)]). Suppose that Fk := Db−;g 1 Fk ∈ C ([a, b] , X ) , where 0 < α ≤ 1, n ∈ N. Then n

 (g (b) − g (x))iα  iα Db−;g f (b) + f (x) =  (iα + 1) i=0 1  ((n + 1) α)



b x

  (n+1)α f (t) dt, (g (t) − g (x))(n+1)α−1 g  (t) Db−;g

(10.25)

∀ x ∈ [a, b] . (n+1)α Clearly here Db−;g f ∈ C ([a, b] , X ) . For differentiation of functions from real numbers to normed linear spaces the definition is the same as for the real valued functions, however the limit and convergence is in the norm of linear space (X, ·) . We state Corollary 10.12 (to Theorem 10.10) Let 0 < α < 1, n ∈ N , f ∈ C 1 ([a, b] , X ), −1 1 g ∈ C 1 ([a, b]), strictly increasing, such that  g ∈C ([g (a) , g (b)]). Assume kα iα Da+;g f ∈ C 1 ([a, b] , X ), k = 1, . . . , n, and Da+;g f (a) = 0, i = 0, 2, 3, . . . , n. Then  x   1 (n+1)α f (t) dt, f (x) = (g (x) − g (t))(n+1)α−1 g  (t) Da+;g  ((n + 1) α) a (10.26) ∀ x ∈ [a, b] . We state Corollary 10.13 (to Theorem 10.11) Let f ∈ C 1 ([a, b] , X ), g ∈ C 1 ([a, b]), kα f ∈ C1 strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]). Suppose that Db−;g ([a,  b] , X), for k = 1, . . . , n; where 0 < α < 1, n ∈ N. We further assume that iα Db−;g f (b) = 0, i = 0, 2, 3, . . . , n. Then  b   1 (n+1)α f (t) dt, f (x) = (g (t) − g (x))(n+1)α−1 g  (t) Db−;g  ((n + 1) α) x (10.27) ∀ x ∈ [a, b] . For the Bochner integral excellent resources are [5, 8] and [1, pp. 422–428].

10.2 Main Results

207

10.2 Main Results We give the following representation formula: Theorem 10.14 All as in Corollary 10.12. Let γ > 0 with γ = m < n + 1, such m that m < (n + 1) α (i.e. n+1 < α < 1). Then 

1  ((n + 1) α − γ) 

γ



x

a

∀ x ∈ [a, b] and Da+;g f



 γ Da+;g f (x) =

(10.28)

  (n+1)α f (t) dt, (g (x) − g (t))(n+1)α−γ−1 g  (t) Da+;g ∈ C ([a, b] , X ) .

Proof By Corollary 10.12 we have f (x) =

1  ((n + 1) α)



x

a

  (n+1)α f (t) dt, (g (x) − g (t))(n+1)α−1 g  (t) Da+;g (10.29)

∀ x ∈ [a, b] . We can write f (x) =

1  ((n + 1) α)

 a

x

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

(set z := g (t), a ≤ t ≤ b) 1  ((n + 1) α)



g(x)

g(a)

(g (x) − z)(n+1)α−1



  (n+1)α Da+;g f ◦ g −1 (z) dz,

(10.30)

∀ x ∈ [a, b] . Hence it holds (y = g (x)) 

f ◦g

−1



1 (y) =  ((n + 1) α)



y

g(a)

(y − z)(n+1)α−1



  (n+1)α Da+;g f ◦ g −1 (z) dz, (10.31)

∀ y ∈ [g (a) , g (b)] . By assuming (n + 1) α − 1 > 0 (equivalently, α > 

1 ) n+1

and by [2], we get:

      (n + 1) α − 1 y (n+1)α f ◦ g −1 (y) = (y − z)(n+1)α−2 Da+;g f ◦ g −1 (z) dz,  ((n + 1) α) g(a)

(10.32) ∀ y ∈ [g (a) , g (b)] .

10 Iterated g-Fractional Vector Bochner …

208

If (n + 1) α − 2 > 0 (equivalently, α > 

2 ), n+1

we get

  ((n + 1) α − 1) ((n + 1) α − 2) f ◦ g −1 (y) =  ((n + 1) α)



y

g(a)

(y − z)(n+1)α−3



  (n+1)α Da+;g f ◦ g −1 (z) dz,

∀ y ∈ [g (a) , g (b)] . In general, if (n + 1) α − m > 0 (equivalently, α > m 

 

y

g(a)

f ◦g

 −1 (m)

(y) =

(y − z)(n+1)α−m−1

m ) n+1

(10.33)

we get that there exists

((n + 1) α − j)

j=1



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

(10.34)

∀ y ∈ [g (a) , g (b)] . (m)  ∈ C ([g (a) , g (b)]) . By [2] we get f ◦ g −1 By (10.2) we have that m 



f ◦g

(n+1)α−m Ig(a)+



 −1 (m)

(y) =

((n + 1) α − j)  ((n + 1) α − m)

j=1

 ((n + 1) α)

     (n+1)α (n+1)α−m (n+1)α Da+;g Da+;g f ◦ g −1 (y) = Ig(a)+ f ◦ g −1 (y) . (10.35)

That is 

f ◦ g −1

(m)

(n+1)α−m (y) = Ig(a)+



  (n+1)α Da+;g f ◦ g −1 (y) ,

(10.36)

  (n+1)α Da+;g f ◦ g −1 (g (x)) ,

(10.37)

∀ y ∈ [g (a) , g (b)], and 

f ◦ g −1

(m)

(n+1)α−m (g (x)) = Ig(a)+



∀ x ∈ [a, b] . Clearly, it holds (see [7, Lemma B.4.10] and [8, p. 158])  ∀ x ∈ [a, b] .

f ◦ g −1

(m)

   (n+1)α−m (n+1)α Da+;g f (x) , (g (x)) = Ia+;g

(10.38)

10.2 Main Results

209

Let γ > 0 with γ = m < n + 1, such that m < (n + 1) α (equivalently, α >

m ). n+1

We have that (case of γ < m) 

    (m) (10.8) (10.38) γ m−γ  ◦ g (x) = Da+;g f (x) = Ia+;g f ◦ g −1

      m−γ (n+1)α−γ (n+1)α−m (n+1)α (n+1)α Da+;g Da+;g f (x) = Ia+ f (x) , Ia+;g Ia+;g

(10.39)

(by (10.3)). We have proved that 

    γ (n+1)α−γ (n+1)α Da+;g f (x) = Ia+ Da+;g f (x) ,

(10.40)

∀ x ∈ [a, b] , which is continuous, by [2].



We continue with Theorem 10.15 All as in Corollary 10.12 and let γ > 0 with γ = m. When γ+m < n+1 α < 1 , we get that   1 2γ Da+;g f (x) =  ((n + 1) α − 2γ) 

x

a

  (n+1)α f (t) dt, (g (x) − g (t))(n+1)α−2γ−1 g  (t) Da+;g

(10.41)

  2γ ∀ x ∈ [a, b] and Da+;g f ∈ C ([a, b] , X ) . Proof Call λ := (n + 1) α − γ − 1, i.e. λ + 1 = (n + 1) α − γ, δ := (n + 1) α. Then we can rewrite (10.28) as 

γ Da+;g



1 f (x) =  (λ + 1)



x

a

  δ f (t) dt, (g (x) − g (t))λ g  (t) Da+;g

and

call

(10.42)

∀ x ∈ [a, b] . That is (z = g (t), a ≤ t ≤ b) 

γ Da+;g



1 f (x) =  (λ + 1)



g(x) g(a)

(g (x) − z)λ



  δ f ◦ g −1 (z) dz, (10.43) Da+;g

∀ x ∈ [a, b] . Hence it holds (y = g (x)) 

  γ Da+;g f ◦ g −1 (y) =

1  (λ + 1)



y

g(a)

(y − z)λ



  δ f ◦ g −1 (z) dz, Da+;g (10.44)

10 Iterated g-Fractional Vector Bochner …

210

∀ y ∈ [g (a) , g (b)] . If λ > 0, then 

  γ Da+;g f ◦ g −1 (y) =

λ  (λ + 1)



y g(a)

(y − z)λ−1



  δ Da+;g f ◦ g −1 (z) dz, (10.45)

∀ y ∈ [g (a) , g (b)] . If λ − 1 > 0, then 

γ Da+;g



f ◦g

−1



λ (λ − 1) (y) =  (λ + 1)



y g(a)

(y − z)λ−2



  δ Da+;g f ◦ g −1 (z) dz, (10.46)

∀ y ∈ [g (a) , g (b)] . If λ − 2 > 0, then 

      λ (λ − 1) (λ − 2) y γ δ Da+;g f ◦ g −1 (y) = f ◦ g −1 (z) dz, (y − z)λ−3 Da+;g  (λ + 1) g(a)

(10.47) ∀ y ∈ [g (a) , g (b)] , etc. In general, if λ − m + 1 > 0, then 

(m)  λ (λ − 1) (λ − 2) . . . (λ − m + 1) γ Da+;g f ◦ g −1 (y) =  (λ + 1) 

y g(a)

(y − z)(λ−m+1)−1



  δ f ◦ g −1 (z) dz = Da+;g

(λ−m+1) λ (λ − 1) (λ − 2) . . . (λ − m + 1) Ig(α)+



  δ Da+;g f ◦ g −1 (y)

 (λ + 1) (λ−m+1) = Ig(a)+



  δ Da+;g f ◦ g −1 (y) ,

(10.48)

∀ y ∈ [g (a) , g (b)] . That is, if λ − m + 1 > 0, we have 

(m)    γ (λ−m+1)  δ Da+;g f ◦ g −1 Da+;g f ◦ g −1 (y) , (y) = Ig(a)+

(10.49)

∀ y ∈ [g (a) , g (b)] . Hence 

(m)     γ (λ−m+1) (n+1)α Da+;g f ◦ g −1 Da+;g f ◦ g −1 (g (x)) , (10.50) (g (x)) = Ig(a)+

∀ x ∈ [a, b] .

10.2 Main Results

211

Clearly, it holds by change of variables 

(m)     γ (λ−m+1) (n+1)α Da+;g f ◦ g −1 Da+;g f (x) , (g (x)) = Ia+

(10.51)

∀ x ∈ [a, b], under λ − m + 1 > 0. We see that      2γ γ γ Da+;g f (x) = Da+;g Da+;g f (x) = m−γ

Ia+;g



(m)      (10.51) γ m−γ (λ−m+1) (n+1)α Da+;g f Da+;g f ◦ g −1 ◦g Ia+;g Ia+;g (x) = (x)

      λ−γ+1 (n+1)α−γ−1−γ+1 (n+1)α (n+1)α Da+;g f (x) = Ia+;g Da+;g f (x) = Ia+;g

(10.52)

   (n+1)α−2γ (n+1)α Da+;g f (x) , = Ia+;g

∀ x ∈ [a, b] . We have derived that      2γ (n+1)α−2γ (n+1)α Da+;g f (x) = Ia+;g Da+;g f (x) , ∀ x ∈ [a, b], under the condition

γ+m n+1

< α < 1.

(10.53) 

Similarly to the last two theorems, it holds Theorem 10.16 All as in Corollary 10.12 and let γ > 0 with γ = m. When m+2γ < α < 1, we obtain that n+1   a

x

 3γ Da+;g f (x) =

1  ((n + 1) α − 3γ)

  (n+1)α f (t) dt, (g (x) − g (t))(n+1)α−3γ−1 g  (t) Da+;g

(10.54)

  3γ ∀ x ∈ [a, b] and Da+;g f ∈ C ([a, b] , X ) . In general, we derive the iterated left fractional derivative representation formula: Theorem 10.17 All as in Corollary 10.12 and let γ > 0 with γ = m. When m+(k−1)γ < α < 1, k ∈ N, we obtain that n+1 

 kγ Da+;g f (x) =

1  ((n + 1) α − kγ)

10 Iterated g-Fractional Vector Bochner …

212



x

a

  (n+1)α f (t) dt, (g (x) − g (t))(n+1)α−kγ−1 g  (t) Da+;g

(10.55)

  kγ ∀ x ∈ [a, b] and Da+;g f ∈ C ([a, b] , X ) . Similarly, it holds the iterated right fractional derivative representation formula: Theorem 10.18 All as in Corollary 10.13 and let γ > 0 with γ = m. Here we assume m+(k−1)γ < α < 1, k ∈ N. Then n+1  

b x

 kγ Db−;g f (x) =

1  ((n + 1) α − kγ)

  (n+1)α f (t) dt, (g (t) − g (x))(n+1)α−kγ−1 g  (t) Db−;g

(10.56)

  kγ ∀ x ∈ [a, b] and Db−;g f ∈ C ([a, b] , X ) . It follows a left generalized and iterated fractional Poincaré type inequality: Theorem 10.19 Here all as in Theorem 10.17. Let p, q > 1 : assume that (k ∈ N) 1 > α > max

m + (k − 1) γ kγq + 1 , n+1 (n + 1) q

1 p

+

1 q

= 1. We further

.

(10.57)

Then     kγ Da+;g f 

q,[a,b]

((n+1)α−kγ−1)+ 1p

(g (b) − g (a))

≤

1

(b − a) q

1

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1) p

    (n+1)α   Da+;g f ◦ g −1 

q,[g(a),g(b)]

.

(10.58)

Proof We use (10.55). We observe that     kγ   Da+;g f (x) ≤  a

x

1 ·  ((n + 1) α − kγ)

(10.59)

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

1  ((n + 1) α − kγ)



g(x) g(a)

    (n+1)α  −1  f g (z)  dz (g (x) − z)(n+1)α−kγ−1  Da+;g

10.2 Main Results

≤

213

1  ((n + 1) α − kγ)  ·

g(b) g(a)

(g (x) − g (a))

=



g(x) g(a)

(g (x) − z) p((n+1)α−kγ−1) dz

1p (10.60)

q1     (n+1)α  −1 q  Da+;g f g (z)  dz

p((n+1)α−kγ−1)+1 p

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) +

1 1) p

    (n+1)α   Da+;g f ◦ g −1 

.

q,[g(a),g(b)]

Thus we have   q   kγ  Da+;g f (x) ≤

q

(g (x) − g (a))( p((n+1)α−kγ−1)+1) p q

 ((n + 1) α − kγ)q ( p ((n + 1) α − kγ − 1) + 1) p (10.61)  q   (n+1)α −1  ·  Da+;g f ◦ g  . q,[g(a),g(b)]

Therefore it holds 

b

a

  q  kγ   Da+;g f (x) d x ≤

q

(g (x) − g (a))( p((n+1)α−kγ−1)+1) p (b − a) q

 ((n + 1) α − kγ)q ( p ((n + 1) α − kγ − 1) + 1) p (10.62)  q   (n+1)α  ·  Da+;g f ◦ g −1  , q,[g(a),g(b)]



proving the claim. It follows a right generalized and iterated fractional Poincaré type inequality: Theorem 10.20 Here all as in Theorem 10.18. Let p, q > 1 : assume that (k ∈ N)

m + (k − 1) γ kγq + 1 , 1 > α > max n+1 (n + 1) q     kγ Db−;g f 

Then

q,[a,b]

(g (b) − g (a))

((n+1)α−kγ−1)+ 1p

1

(b − a) q

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) +

1 1) p

1 p

+

1 q

= 1. We further

.

(10.63)

≤     (n+1)α   Db−;g f ◦ g −1 

q,[g(a),g(b)]

.

(10.64)

Proof As similar to Theorem 10.19 is omitted. Next comes a left generalized and iterated fractional Opial type inequality:



10 Iterated g-Fractional Vector Bochner …

214

Theorem 10.21 All here as in Theorem 10.17. Let p, q > 1 : assume that (k ∈ N) 1 > α > max 

Then

y

g(a)

m + (k − 1) γ kγq + 1 , n+1 (n + 1) q

+

1 p

1 q

= 1. We further

.

(10.65)

           kγ (n+1)α  Da+;g f ◦ g −1 (w)  Da+;g f ◦ g −1 (w) dw ≤ (y − g (a))

((n+1)α−kγ−1)+ 2p

1

1

2 q  ((n + 1) α − kγ) [( p ((n + 1) α − kγ − 1) + 1) ( p ((n + 1) α − kγ − 1) + 2)] p



y g(a)

q2 q      (n+1)α −1 (w) dw , ∀ y ∈ [g (a) , g (b)] .  Da+;g f ◦ g

·

(10.66)

Proof We use (10.55). We observe that     kγ   Da+;g f (x) ≤  a

=

x

1 ·  ((n + 1) α − kγ)

(10.67)

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

1  ((n + 1) α − kγ)

g(x)

g(a)

1 ≤  ((n + 1) α − kγ) 

g(x)

g(a)

    (n+1)α  −1  f g (z)  dz (g (x) − z)(n+1)α−kγ−1  Da+;g



g(x)

g(a)

(g (x) − z)

1p p((n+1)α−kγ−1)

dz

·

(10.68)

q1     (n+1)α  −1 q  Da+;g f g (z)  dz p((n+1)α−kγ−1)+1

p 1 (g (x) − g (a)) = ·  ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1) 1p



g(x)

g(a)

q1     (n+1)α  −1 q  Da+;g f g (z)  dz .

(10.69)

10.2 Main Results

215

That is     kγ   Da+;g f (x) ≤ 

(g (x) − g (a))

p((n+1)α−kγ−1)+1 p 1

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1) p g(x)

g(a)

q1     (n+1)α  −1 q  Da+;g f g (z)  dz ,

·

(10.70)

∀ x ∈ [a, b] . Call y = g (x) , then

x = g −1 (y) .

(10.71)

Hence it holds      kγ  Da+;g f g −1 (y)  ≤  ·

y g(a)

∀ y ∈ [g (a) , g (b)] . Call

(y − g (a))

1

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1) p

q1     (n+1)α  −1 q  Da+;g f g (z)  dz ,



η (y) :=

p((n+1)α−kγ−1)+1 p

y g(a)

    (n+1)α  −1 q  Da+;g f g (z)  dz,

(10.72)

(10.73)

and η (g (a)) = 0. Thus

and

    (n+1)α  −1 q f g (y)  ≥ 0, η  (y) =  Da+;g

(10.74)

  1     (n+1)α  −1  f g (y)  ≥ 0, η (y) q =  Da+;g

(10.75)

∀ y ∈ [g (a) , g (b)] . Consequently, we get        kγ   (n+1)α    Da+;g f g −1 (w)   Da+;g f g −1 (w)  ≤

10 Iterated g-Fractional Vector Bochner …

216

(w − g (a))

p((n+1)α−kγ−1)+1 p



 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1)

1 p

η (w) η  (w)

 q1

,

∀ w ∈ [g (a) , g (b)] . Then it holds  y        kγ   (n+1)α    Da+;g f g −1 (w)   Da+;g f g −1 (w)  dw ≤

(10.76)

(10.77)

g(a)

1 1

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1) p 

y g(a)

≤ 

y g(a)

(w − g (a))

p((n+1)α−kγ−1)+1 p

 1 η (w) η  (w) q dw

1 1

 ((n + 1) α − kγ) ( p ((n + 1) α − kγ − 1) + 1) p

(w − g (a))

( p((n+1)α−kγ−1)+1)

1p 

y

dw g(a)

(y − g (a))

·

·

(10.78)



η (w) η (w) dw

q1

=

p((n+1)α−kγ−1)+2 p 1

 ((n + 1) α − kγ) [( p ((n + 1) α − kγ − 1) + 1) ( p ((n + 1) α − kγ − 1) + 2)] p ·

η 2 (y) 2

q1

=

(10.79)

((n+1)α−kγ−1)+ 2p

(y − g (a)) 1

1

2 q  ((n + 1) α − kγ) [( p ((n + 1) α − kγ − 1) + 1) ( p ((n + 1) α − kγ − 1) + 2)] p

 ·

y

g(a)

q2 q      (n+1)α −1 (w) dw ,  Da+;g f ◦ g

(10.80)

∀ y ∈ [g (a) , g (b)] , proving the claim.



Also we give a right generalized and iterated fractional Opial type inequality: Theorem 10.22 All here as in Theorem 10.18. Let p, q > 1 : assume that (k ∈ N)

m + (k − 1) γ kγq + 1 1 > α > max , n+1 (n + 1) q

1 p

+

1 q

= 1. We further

.

(10.81)

10.2 Main Results

217

Then 

g(b) y

           kγ (n+1)α  Db−;g f ◦ g −1 (w)  Db−;g f ◦ g −1 (w) dw ≤ ((n+1)α−kγ−1)+ 2p

(g (b) − y) 1

1

2 q  ((n + 1) α − kγ) [( p ((n + 1) α − kγ − 1) + 1) ( p ((n + 1) α − kγ − 1) + 2)] p



g(b) y

q2  q     (n+1)α −1 (w) dw , ∀ y ∈ [g (a) , g (b)] .  Db−;g f ◦ g

·

(10.82) 

Proof It is omitted as similar to Theorem 10.21.

Next we present a left generalized and iterated fractional Hilbert–Pachpatte type inequality: Theorem 10.23 Here i = 1, 2. Let ai , bi ∈ R, ai < bi , 0 < αi < 1, as in (10.83), (10.84), and gi ∈ C 1 ([ai , bi ]) that are strictly increasing, f i ∈ C 1 ([ai , bi ] , X ) , αi f i ∈ C 1 ([ai , bi ] , X ), Assume that Dakii +;g gi−1 ∈ C 1 ([gi (ai ) , gi (bi )]) . i   αi f i (ai ) = 0, λi = 0, 2, 3, . . . , n i , where n i ∈ N. Let ki = 1, . . . , n i , and Daλii+;g i γi > 0 with γi = m i , ki ∈ N, p, q > 1 :

1 p

+

1 q

= 1. We further assume:

m 1 + (k1 − 1) γ1 k1 γ1 q + 1 , , 1 > α1 > max n1 + 1 (n 1 + 1) q

(10.83)

m 2 + (k2 − 1) γ2 k2 γ2 p + 1 , . 1 > α2 > max n2 + 1 (n 2 + 1) p

(10.84)





and

Then 

g1 (b1 )

g1 (a1 )



g2 (b2 ) g2 (a2 )



       k1 γ1   k γ2  −1 g f (y )  Da1 +;g1 f 1 g1−1 (y1 )   Da22 +;g 2 2  2 2 (y1 −g1 (a1 )) p((n 1 +1)α1 −k1 γ1 −1)+1 p( p((n 1 +1)α1 −k1 γ1 −1)+1)

+

(y2 −g2 (a2 ))q ((n 2 +1)α2 −k2 γ2 −1)+1 q(q((n 2 +1)α2 −k2 γ2 −1)+1)

(g1 (b1 ) − g1 (a1 )) (g2 (b2 ) − g2 (a2 )) · ≤  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 ) 

g1 (b1 )

g1 (a1 )



g2 (b2 ) g2 (a2 )

q1     (n 1 +1)α1 q −1  Da1 +;g1 f 1 g1 (z 1 )  dz 1 ·

1p   −1   (n 2 +1)α2 p . (Da2 +;g2 f 2 ) g2 (z 2 )  dz 2

 dy1 dy2 (10.85)

10 Iterated g-Fractional Vector Bochner …

218

Proof As in the proof of Theorem 10.21, we obtain:     k1 γ1   Da1 +;g1 f 1 g1−1 (y1 )  ≤ (y1 − g1 (a1 ))

p ((n 1 +1)α1 −k1 γ1 −1)+1

p 1

 ((n 1 + 1) α1 − k1 γ1 ) ( p ((n 1 + 1) α1 − k1 γ1 − 1) + 1) p 

y1

g1 (a1 )

·

(10.86)

q1     (n 1 +1)α1 q  Da1 +;g1 f 1 g1−1 (z 1 )  dz 1 ,

∀ y1 ∈ [g1 (a1 ) , g1 (b1 )] . Also we obtain:      k2 γ2  Da2 +;g2 f 2 g2−1 (y2 )  ≤ (y2 − g2 (a2 ))

q ((n 2 +1)α2 −k2 γ2 −1)+1 q 1

 ((n 2 + 1) α2 − k2 γ2 ) (q ((n 2 + 1) α2 − k2 γ2 − 1) + 1) q 

y2 g2 (a2 )

1p     (n 2 +1)α2 p −1 ,  Da2 +;g2 f 2 g2 (z 2 )  dz 2

·

(10.87)

∀ y2 ∈ [g2 (a2 ) , g2 (b2 )] . Multiplying (10.86) and (10.87), we get:         k γ2   k1 γ1 −1 g f (y ) ≤  Da1 +;g1 f 1 g1−1 (y1 )   Da22 +;g 2 2 2 2 1 ·  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 ) 

(y1 − g1 (a1 )) p((n 1 +1)α1 −k1 γ1 −1)+1 ( p ((n 1 + 1) α1 − k1 γ1 − 1) + 1) 

y1 g1 (a1 )



y2

g2 (a2 )

 1p 

  (n 1 +1)α1  Da1 +;g1

 q1 (y2 − g2 (a2 ))q((n 2 +1)α2 −k2 γ2 −1)+1 · (q ((n 2 + 1) α2 − k2 γ2 − 1) + 1) (10.88)

q1 q    f 1 g1−1 (z 1 )  dz 1 ·

1p     (n 2 +1)α2 p −1  Da2 +;g2 f 2 g2 (z 2 )  dz 2 1

1

(using Young’s inequality for a, b ≥ 0, a p b q ≤

a p

+ qb )

10.2 Main Results

219

1 ·  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 )

≤ 

(y1 − g1 (a1 )) p((n 1 +1)α1 −k1 γ1 −1)+1 p ( p ((n 1 + 1) α1 − k1 γ1 − 1) + 1)

 ·

y1

g1 (a1 )

 ·

y2 g2 (a2 )



 +

(y2 − g2 (a2 ))q((n 2 +1)α2 −k2 γ2 −1)+1 q (q ((n 2 + 1) α2 − k2 γ2 − 1) + 1)

q1     (n 1 +1)α1 q −1  Da1 +;g1 f 1 g1 (z 1 )  dz 1



(10.89)

1p     (n 2 +1)α2 p −1 .  Da2 +;g2 f 2 g2 (z 2 )  dz 2

So far we have         k γ2   k1 γ1 f 2 g2−1 (y2 )   Da1 +;g1 f 1 g1−1 (y1 )   Da22 +;g 2     ≤ (y1 −g1 (a1 )) p((n 1 +1)α1 −k1 γ1 −1)+1 (y2 −g2 (a2 ))q ((n 2 +1)α2 −k2 γ2 −1)+1 + p( p((n 1 +1)α1 −k1 γ1 −1)+1) q(q((n 2 +1)α2 −k2 γ2 −1)+1)

(10.90)

1 ·  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 ) 

y1 g1 (a1 )



y2 g2 (a2 )

q1     (n 1 +1)α1 q −1  Da1 +;g1 f 1 g1 (z 1 )  dz 1 ·

1p     (n 2 +1)α2 p .  Da2 +;g2 f 2 g2−1 (z 2 )  dz 2

The denominator in (10.90) can be zero only when y1 = g1 (a1 ) and y2 = g2 (a2 ). Therefore we obtain        k1 γ1   k γ2  −1  g1 (b1 )  g2 (b2 ) g f (y )  Da1 +;g1 f 1 g1−1 (y1 )   Da22 +;g  2 2 2 2     dy1 dy2 p ((n 1 +1)α1 −k1 γ1 −1)+1 q ((n 2 +1)α2 −k2 γ2 −1)+1 (y1 −g1 (a1 )) (y2 −g2 (a2 )) g1 (a1 ) g2 (a2 ) + p( p((n 1 +1)α1 −k1 γ1 −1)+1) q(q((n 2 +1)α2 −k2 γ2 −1)+1) (10.91) 1 ≤ ·  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 ) 

g1 (b1 ) g1 (a1 )



y1 g1 (a1 )



q1     (n 1 +1)α1 q −1  Da1 +;g1 f 1 g1 (z 1 )  dz 1 dy1 ·

10 Iterated g-Fractional Vector Bochner …

220



g2 (b2 )



g2 (a2 )

g2 (a2 )

g1 (b1 )



g1 (a1 )



g2 (b2 )

1p     (n 2 +1)α2 p dy2  Da2 +;g2 f 2 g2−1 (z 2 )  dz 2



1 ·  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 )

≤ 

y2

g1 (b1 )

g1 (a1 )



g2 (a2 )

g2 (b2 ) g2 (a2 )



q1     (n 1 +1)α1 q −1  Da1 +;g1 f 1 g1 (z 1 )  dz 1 dy1 ·

1p     (n 2 +1)α2 p −1 dy2  Da2 +;g2 f 2 g2 (z 2 )  dz 2

 (10.92)

(g1 (b1 ) − g1 (a1 )) (g2 (b2 ) − g2 (a2 )) ·  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 )

=



g1 (b1 ) g1 (a1 )



g2 (b2 )

g2 (a2 )

q1     (n 1 +1)α1 q −1  Da1 +;g1 f 1 g1 (z 1 )  dz 1 ·

(10.93)

1p     (n 2 +1)α2 p −1 .  Da2 +;g2 f 2 g2 (z 2 )  dz 2 

The theo is proved.

Finally we present a right generalized and iterated fractional Hilbert–Pachpatte type inequality: Theorem 10.24 Here i = 1, 2. Let ai , bi ∈ R, ai < bi , 0 < αi < 1, as in (10.94), (10.95), and gi ∈ C 1 ([ai , bi ]) that are strictly increasing, f i ∈ C 1 ([ai , bi ] , X ) , αi f i ∈ C 1 ([ai , bi ] , X ), gi−1 ∈ C 1 ([gi (ai ) , gi (bi )]) . Assume that Dbkii −;g i   αi ki = 1, . . . , n i , and Dbλii−;g f i (bi ) = 0, λi = 0, 2, 3, . . . , n i , where n i ∈ N. Let i γi > 0 with γi = m i , ki ∈ N, p, q > 1 : 1 > α1 > max

and 1 > α2 > max Then

1 p

+

1 q

= 1. We further assume:

m 1 + (k1 − 1) γ1 k1 γ1 q + 1 , . n1 + 1 (n 1 + 1) q

(10.94)

m 2 + (k2 − 1) γ2 k2 γ2 p + 1 , . n2 + 1 (n 2 + 1) p

(10.95)

10.2 Main Results



g1 (b1 )

g1 (a1 )



g2 (b2 ) g2 (a2 )

221



       k1 γ1   k γ2  −1 g f (y )  Db1 −;g1 f 1 g1−1 (y1 )   Db22 −;g  2 2 2 2 (g1 (b1 )−y1 ) p((n 1 +1)α1 −k1 γ1 −1)+1 p( p((n 1 +1)α1 −k1 γ1 −1)+1)

+

(g2 (b2 )−y2 )q ((n 2 +1)α2 −k2 γ2 −1)+1 q(q((n 2 +1)α2 −k2 γ2 −1)+1)

(g1 (b1 ) − g1 (a1 )) (g2 (b2 ) − g2 (a2 )) · ≤  ((n 1 + 1) α1 − k1 γ1 )  ((n 2 + 1) α2 − k2 γ2 ) 

g1 (b1 )

g1 (a1 )



g2 (b2 )

g2 (a2 )

 dy1 dy2 (10.96)

q1     (n 1 +1)α1 q −1  Db1 −;g1 f 1 g1 (z 1 )  dz 1 ·

1p     (n 2 +1)α2 p −1 .  Db2 −;g2 f 2 g2 (z 2 )  dz 2

Proof It is similar to Theorem 10.23, thus it is omitted.



Remark 10.25 Some examples for g follow: g (x) = e x , x ∈ [a, b] ⊂ R, g (x) = sin x, g (x) = tan x,   π π where x ∈ − + ε, − ε , where ε > 0 small. 2 2 Indeed, the above examples of g are strictly increasing and continuous functions. One can apply all of our results here for the above specific choices of g. We choose to omit this job.

References 1. Aliprantis, C.D., Border, K.C.: Infinite Dimensional Analysis. Springer, New York (2006) 2. Anastassiou, G.A.: Principles of general fractional analysis for Banach space valued functions. Bull. Allahabad Math. Soc. 32(1), 71–145 (2017) 3. Anastassiou, G.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 4. Anastassiou, G.: Iterated g-fractional vector representation formulae and inequalities for Banach space valued functions. Issues Anal. 9(27)(1), 3–26 (2020) 5. Bochner integral. Encyclopedia of Mathematics. http://www.encyclopediaofmath.org/index. php?title=Bochner_integral&oldid=38659 6. Kreuter, M.: Sobolev space of vector-valued functions. Ulm University, Master thesis in Mathematics, Ulm, Germany (2015) 7. Mikkola, K.: Appendix B integration and differentiation in Banach spaces. http://math.aalto.fi/ ~kmikkola/research/thesis/contents/thesisb.pdf 8. Mikusinski, J.: The Bochner Integral. Academic, New York (1978)

Chapter 11

Vectorial Generalized g-Fractional Direct and Iterated Quantitative Approximation by Linear Operators

In this work we consider quantitatively with rates the convergence of sequences of linear operators applied on Banach space valued functions. The results are pointwise estimates with rates. To prove our main results we use an elegant and natural boundedness property of our linear operators by their companion positive linear operators. Our inequalities are generalized g-direct and iterated fractional involving the right and left vector Caputo type generalized g-direct and iterated fractional derivatives, built in vector moduli of continuity. We treat wide and general classes of Banach space valued functions. We give applications to vectorial Bernstein operators. See also [6].

11.1 Motivation Let (X, ·) be a Banach space, N ∈ N. Consider g ∈ C ([0, 1]) and the classic Bernstein polynomials 

N    g B N g (t) =



k=0

k N



N k



t k (1 − t) N −k , ∀ t ∈ [0, 1] .

(11.1)

Let al.so f ∈ C ([0, 1] , X ) and define the vector valued in X Bernstein linear operators (B N f ) (t) =

N  k=0

 f

k N



N k



t k (1 − t) N −k , ∀ t ∈ [0, 1] .

(11.2)

That is (B N f ) (t) ∈ X. Clearly here  f  ∈ C ([0, 1]). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_11

223

11 Vectorial Generalized g-Fractional Direct and Iterated …

224

We notice that    N         f k  N t k (1 − t) N −k =  (B N f ) (t) ≤ B N ( f ) (t) ,   k N k=0

(11.3)

∀ t ∈ [0, 1] . The property   (B N f ) (t) ≤  B N ( f ) (t) , ∀t ∈ [0, 1] ,

(11.4)

is shared by almost all summation/integration similar operators and motivates our work here. If f (x) = c ∈ X the constant function, then (B N c) = c.

(11.5)

If g ∈ C ([0, 1]) and c ∈ X , then cg ∈ C ([0, 1] , X ) and B N (g) . (B N (cg)) = c 

(11.6)

Again (11.5), (11.6) are fulfilled by many summation/integration operators. In fact here (11.6) implies (11.5), when g ≡ 1. The above can be generalized from [0, 1] to any interval [a, b] ⊂ R. All this discussion motivates us to consider the following situation. Let L N : C ([a, b] , X ) → C ([a, b] , X ), (X, ·) a Banach space, L N is a linear L N : C ([a, b]) → C ([a, b]), a sequence operator, ∀ N ∈ N, x0 ∈ [a, b]. Let also  of positive linear operators, ∀ N ∈ N. We assume that   (L N ( f )) (x0 ) ≤  L N ( f ) (x0 ) , (11.7) ∀ N ∈ N, ∀ x0 ∈ [a, b], ∀ f ∈ C ([a, b] , X ) . When g ∈ C ([a, b]), c ∈ X , we assume that

The special case of

L N (g) . (L N (cg)) = c

(11.8)

 L N (1) = 1,

(11.9)

L N (c) = c, ∀ c ∈ X.

(11.10)

implies We call  L N the companion operator of L N .

11.1 Motivation

225

Based on the above fundamental properties we study the fractional approximation properties of the sequence of linear operators {L N } N ∈N , i.e. their fractional convergence to the unit operator. No kind of positivity property of {L N } N ∈N is assumed.

11.2 Background We need Definition 11.1 ([3]) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the left Riemann–Liouville generalized fractional Bochner integral operator 

ν Ia+;g



1 f (x) :=  (ν)



x

(g (x) − g (z))ν−1 g (z) f (z) dz,

(11.11)

a

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈ C ([a, b] , X ), then f ∈ ν 0 L ∞ ([a, b] , X ). By [3] we get that Ia+;g f ∈ C ([a, b] , X ). Above we set Ia+;g f :=

 ν f and see that Ia+;g f (a) = 0. ν ν When g is the identity function id, we get that Ia+;id = Ia+ , the ordinary left Riemann–Liouville fractional integral 

ν Ia+



1 f (x) =  (ν)



x

(x − t)ν−1 f (t) dt,

(11.12)

a

 ν  ∀ x ∈ [a, b], Ia+ f (a) = 0. We mention Theorem 11.2 ([3]) Let μ, ν > 0 and f ∈ C ([a, b] , X ). Then μ

μ+ν

μ

ν ν Ia+;g Ia+;g f = Ia+;g f = Ia+;g Ia+;g f.

(11.13)

We need Definition 11.3 ([3]) Let [a, b] ⊂ R, (X, ·) a Banach space, g ∈ C 1 ([a, b]) and increasing, f ∈ C ([a, b] , X ), ν > 0. We define the right Riemann–Liouville generalized fractional Bochner integral operator

11 Vectorial Generalized g-Fractional Direct and Iterated …

226



ν Ib−;g



1 f (x) :=  (ν)



b

(g (z) − g (x))ν−1 g (z) f (z) dz,

(11.14)

x

∀ x ∈ [a, b], where  is the gamma function. The last integral is of Bochner type. Since f ∈ C ([a, b] , X ), then ν f ∈ L ∞ ([a, b] , X ). By [3] we get that Ib−;g f ∈ C ([a, b] , X ). Above we set

 0 ν Ib−;g f := f and see that Ib−;g f (b) = 0. ν ν When g is the identity function id, we get that Ib−;id = Ib− , the ordinary right Riemann–Liouville fractional integral 

 ν f (x) = Ib−

1  (ν)



b

(t − x)ν−1 f (t) dt,

(11.15)

x

 ν  ∀ x ∈ [a, b], with Ib− f (b) = 0. We mention Theorem 11.4 ([3]) Let μ, ν > 0 and f ∈ C ([a, b] , X ). Then μ

μ+ν

μ

ν ν f = Ib−;g f = Ib−;g Ib−;g f. Ib−;g Ib−;g

(11.16)

We will use Definition 11.5 ([3]) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]) , strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]) . We define the left generalized g-fractional derivative X -valued of f of order α as follows: x (n)   α  1 Da+;g f (x) := (g (x) − g (t))n−α−1 g (t) f ◦ g −1 (g (t)) dt,  (n − α) a (11.17) ∀ x ∈ [a, b]. The last integral is of Bochner type. α f ∈ C ([a, b] , X ). If α ∈ / N, by [3], we have that Da+;g We see that

  (n)  α  n−α Ia+;g f ◦ g −1 ◦ g (x) = Da+;g f (x) , ∀ x ∈ [a, b] . (11.18) We set n f (x) := Da+;g



f ◦ g −1

(n)

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ Nm,

0 f (x) = f (x) , ∀x ∈ [a, b] . Da+;g

When g = id, then

(11.19)

11.2 Background

227 α α α Da+;g f = Da+;id f = D∗a f,

(11.20)

the usual left X -valued Caputo fractional derivative, see [5]. We will use Definition 11.6 ([3]) Let α > 0, α = n, · the ceiling of the number. Let f ∈ C n ([a, b] , X ), where [a, b] ⊂ R, and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]). We define the right generalized g-fractional derivative X -valued of f of order α as follows: b (n)    α (−1)n Db−;g f (x) := (g (t) − g (x))n−α−1 g (t) f ◦ g −1 (g (t)) dt,  (n − α) x (11.21) ∀ x ∈ [a, b]. The last integral is of Bochner type. α f ∈ C ([a, b] , X ). If α ∈ / N, by [3], we have that Db−;g We see that

  (n)  α  n−α ◦ g (x) = Db−;g f (x) , a ≤ x ≤ b. (11.22) Ib−;g (−1)n f ◦ g −1 We set n f (x) := (−1)n Db−;g



f ◦ g −1

 ◦ g (x) ∈ C ([a, b] , X ) , n ∈ N,

n

(11.23)

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

When g = id, then

α α α f (x) = Db−;id f (x) = Db− f, Db−;g

(11.24)

the usual right X -valued Caputo fractional derivative, see [5]. We make Remark 11.7 All as in Definition 11.5. We have (by [8, Theorem 2.5, p. 7] and [10]) x 

    (n) 1  α    (g (x) − g (t))n−α−1 g (t)  f ◦ g −1 (g (t)) dt  Da+;g f (x) ≤  (n − α) a   (n)   g(x) ◦ g  f ◦ g −1 ∞,[a,b] ≤ (g (x) − g (t))n−α−1 dg (t) =  (n − α) g(a)   (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (x) − g (a))n−α .

(11.25)

11 Vectorial Generalized g-Fractional Direct and Iterated …

228

That is  α  D

a+;g

  f (x) ≤

  (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (x) − g (a))n−α ,

(11.26)

∀ x ∈ [a, b] . 

α f (a) = 0. If α ∈ / N, then Da+;g Similarly, by Definition 11.6 we derive 

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

 (n)    f ◦ g −1  ◦ g g(b)   ∞,[a,b] ≤ (g (t) − g (x))n−α−1 dg (t) =  (n − α) g(x)



    α  Db−;g f (x) ≤

 

(n)     f ◦ g −1 ◦ g  

∞,[a,b]

 (n − α + 1)

(g (b) − g (x))n−α .

(11.27)

That is  α  D

b−;g

  f (x) ≤

  (n)   ◦ g  f ◦ g −1

∞,[a,b]

 (n − α + 1)

(g (b) − g (x))n−α ,

(11.28)

∀ x ∈ [a, b] . 

α f (b) = 0. If α ∈ / N, then Db−;g Notation 11.8 We denote by

and

nα α α α := Da+;g Da+;g . . . Da+;g (n times), n ∈ N, Da+;g

(11.29)

nα α α α := Ia+;g Ia+;g . . . Ia+;g , Ia+;g

(11.30)

nα α α := Db−;g Db−;g . . .αb−;g , Db−;g

(11.31)

nα α α α := Ib−;g Ib−;g . . . Ib−;g , Ib−;g

(11.32)

(n times), n ∈ N. We mention the following g-left generalized modified X -valued Taylor’s formula.

11.2 Background

229

Theorem 11.9 ([3]) Let 0 < α ≤ 1, n ∈ N, f ∈ C 1 ([a, b] , X ), g ∈ C 1 ([a, b]), kα f , k = 1, strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]). Let Fk := Da+;g 1 . . . , n, that fulfill Fk ∈ C ([a, b] , X ) . Then n   (g (x) − g (a))iα  iα Da+;g f (a) + f (x) =  (iα + 1) i=0 1  ((n + 1) α)



x

a

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

(11.33)

∀ x ∈ [a, b] . (n+1)α f ∈ C ([a, b] , X ) . Clearly here Da+;g We also mention the following g-right generalized modified X -valued Taylor’s formula. Theorem 11.10 ([3]) Let f ∈ C 1 ([a, b] , X ), g ∈ C 1 ([a, b]), strictly increasing, kα f , k = 1, . . . , n, fulfill such that g −1 ∈ C 1 ([g (a) , g (b)]). Suppose that Fk := Db−;g 1 Fk ∈ C ([a, b] , X ) , where 0 < α ≤ 1, n ∈ N. Then n   (g (b) − g (x))iα  iα Db−;g f (b) + f (x) =  (iα + 1) i=0 1  ((n + 1) α)



b x



(n+1)α f (t) dt, (g (t) − g (x))(n+1)α−1 g (t) Db−;g

(11.34)

∀ x ∈ [a, b] . (n+1)α Clearly here Db−;g f ∈ C ([a, b] , X ) . For differentiation of functions from real numbers to normed linear spaces the definition is the same as for the real valued functions, however the limit and convergence is in the norm of linear space (X, ·) . We state Corollary 11.11 (to Theorem 11.9) Let 0 < α < 1, n ∈ N, f ∈ C 1 ([a, b] , X ), g ∈ kα f ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]). Assume Da+;g 

iα f (a) = 0, i = 0, 2, 3, . . . , n. C 1 ([a, b] , X ), k = 1, . . . , n, and Da+;g Then x 

1 (n+1)α f (t) dt, f (x) = (g (x) − g (t))(n+1)α−1 g (t) Da+;g  ((n + 1) α) a (11.35) ∀ x ∈ [a, b] . We state

11 Vectorial Generalized g-Fractional Direct and Iterated …

230

Corollary 11.12 (to Theorem 11.10) Let f ∈ C 1 ([a, b] , X ), g ∈ C 1 ([a, b]), kα f ∈ strictly increasing, such that g −1 ∈ C 1 ([g (a) , g (b)]). Suppose that Db−;g 1 C

([a, b], X ), for k = 1, . . . , n; where 0 < α < 1, n ∈ N. We further assume that iα Db−;g f (b) = 0, i = 0, 2, 3, . . . , n. Then b 

1 (n+1)α f (t) dt, f (x) = (g (t) − g (x))(n+1)α−1 g (t) Db−;g  ((n + 1) α) x (11.36) ∀ x ∈ [a, b] . We mention the following g-left generalized X -valued Taylor’s formula: Theorem 11.13 ([3]) Let α > 0, n = α , and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]). Then f (x) = f (a) +

n−1  (g (x) − g (a))i 

i!

i=1

1  (α)

a

n−1  (g (x) − g (a))i 

i!

i=1

1  (α)



g(x) g(a)

(i)

(g (a)) +

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

x

f (a) +

f ◦ g −1

(g (x) − z)α−1



f ◦ g −1

(i)

(g (a)) +

(11.37)

  α Da+;g f ◦ g −1 (z) dz, ∀ x ∈ [a, b] .

We also mention the following g-right generalized X -valued Taylor’s formula: Theorem 11.14 ([3]) Let α > 0, n = α , and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing, such that g −1 ∈ C n ([g (a) , g (b)]). Then f (x) = f (b) +

n−1  (g (x) − g (b))i 

i!

i=1

1  (α)



b x

f (b) +

f ◦ g −1

(i)

(g (b)) +

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

n−1  (g (x) − g (b))i  i=1

i!

f ◦ g −1

(i)

(g (b)) +

(11.38)

11.2 Background

1  (α)



231 g(b) g(x)

(z − g (x))α−1



  α Db−;g f ◦ g −1 (z) dz, ∀ x ∈ [a, b] .

Integrals in (11.33)–(11.38) are of Bochner type. For the Bochner integral excellent resources are [7, 9, 10] and [1, pp. 422–428]. We need Definition 11.15 ([4]) Let f ∈ C ([a, b] , X ), [a, b] ⊂ R, (X, ·) a Banach space. We define the first modulus of continuity of f as ω1 ( f, δ) :=

sup

 f (x) − f (y) , 0 < δ ≤ b − a.

(11.39)

x,y∈[a,b]: |x−y|≤δ

If δ > b − a, then ω1 ( f, δ) = ω1 ( f, b − a). Notice ω1 ( f, δ) is increasing in δ > 0. Clearly f is uniformly continuous and ω1 ( f, δ) < ∞. For f ∈ B ([a, b] , X ) (bounded functions) ω1 ( f, δ) is defined the same way. Lemma 11.16 ([4]) We have ω1 ( f, δ) → 0 as δ ↓ 0 iff f ∈ C ([a, b] , X ). We need Theorem 11.17 ([4]) Let f : [a, b]2 → X be jointly continuous, X is a Banach space. Consider (11.40) G (x) = ω1 ( f (·, x) , δ, [x, b]) , δ > 0, x ∈ [a, b] . Then G is continuous on [a, b] . Theorem 11.18 ([4]) Let f : [a, b]2 → X be jointly continuous, X is a Banach space. Then (11.41) H (x) = ω1 ( f (·, x) , δ, [a, x]) , x ∈ [a, b], is continuous in x ∈ [a, b], δ > 0. We need Lemma 11.19 ([2, p. 208, Lemma 7.1.1]) Let f ∈ B ([a, b] , X ), (X, ·) is a Banach space. Then  f (x) − f (x0 ) ≤ ω1 ( f, h) ∀ x, x0 ∈ [a, b], h > 0. We make

|x − x0 | h

  |x − x0 | ≤ ω1 ( f, h) 1 + , (11.42) h

11 Vectorial Generalized g-Fractional Direct and Iterated …

232

Remark 11.20 All here as in (11.7)–(11.10). Let f, g ∈ C ([a, b] , X ). We have that (11.7)

L N ( f ) (x0 ) − L N (g) (x0 ) = (L N ( f − g)) (x0 ) ≤



  L N ( f − g) (x0 ) , (11.43)

∀ N ∈ N, ∀ x0 ∈ [a, b] . In this work, for simplicity, from now on we will assume that  L N (1) = 1, ∀ N ∈ N.

(11.44)

We observe that (11.10)

(L N ( f )) (x0 ) − f (x0 ) = (L N ( f )) (x0 ) − (L N ( f (x0 ))) (x0 ) = (11.45) (L N ( f − f (x0 ))) (x0 )

(11.43.)

≤

 L N ( f (·) − f (x0 )) (x0 ) ,

∀ x0 ∈ [a, b], ∀ N ∈ N. We have proved that (L N ( f )) (x0 ) − f (x0 ) ≤  L N ( f (·) − f (x0 )) (x0 ) ,

(11.46)

∀ x0 ∈ [a, b], ∀ N ∈ N.

11.3 Main Results We need Definition 11.21 Let Dx(n+1)α f denote any of Dx(n+1)α f , Dx(n+1)α f , n ∈ N and 0 ;g 0 −;g 0 +;g δ > 0. We set

  f ◦ g −1 , δ := ω1 Dx(n+1)α 0 ;g 

  −1 max ω1 Dx(n+1)α f ◦ g , δ 0− ;g

[g(a),g(x0 )]

, ω1



Dx(n+1)α 0 +;g



−1

f ◦ g ,δ



 [g(x0 ),g(b)]

,

(11.47) where x0 ∈ [a, b]. Here the moduli of continuity are considered over [g (a) , g (x0 )] and [g (x0 ) , g (b)], respectively. We will use the following: Theorem 11.22 Let 0 < α < 1, f ∈ C 1 ([a, b] , X ) , g ∈ C 1 ([a, b]), g is strictly increasing and g −1 ∈ C 1 ([g (a) , g (b)]). Assume that Dxkα0 −;g f ∈ C 1 ([a, x0 ] , X ) and Dxkα0 +;g f ∈ C 1 ([x0 , b] , X ), for k = 1, . . . , n; where x0 ∈ [a, b] is fixed. Fur

 ther assume that Dxiα0 ±;g f (x0 ) = 0, i = 2, . . . , n + 1, n ∈ N. Then

11.3 Main Results

233

 f (x) − f (x0 ) ≤  |g (x) − g (x0 )|

(n+1)α

ω1



  −1 Dx(n+1)α f ◦ g , δ ;g 0

 ((n + 1) α + 1)

(11.48)

 |g (x) − g (x0 )|(n+1)α+1 + , δ ((n + 1) α + 1)

∀ x ∈ [a, b] , δ > 0.

 Proof By Dxiα0 +;g f (x0 ) = 0, for i = 2, . . . , n + 1, and (11.33) we have f (x) − f (x0 ) = 

1  ((n + 1) α)



x

(g (x) − g (t))(n+1)α−1 g (t)

x0



  (n+1)α Dx(n+1)α f − D f dt, (t) (x ) 0 x0 +;g 0 +;g

(11.49)

∀ x ∈ [x0 , b] . Here we apply the change of variables method for Bochner integrals, see [8, Theorem 2.5, p. 7] and [10]. Hence (z := g (t)) f (x) − f (x0 ) = 

1  ((n + 1) α)



g(x) g(x0 )

(g (x) − z)(n+1)α−1



    (n+1)α −1 −1 Dx(n+1)α − D f ◦ g f ◦ g dz, (z) (g (x )) 0 x +;g +;g 0 0

(11.50)

∀ x ∈ [x

0 , b] .  By Dxiα0 −;g f (x0 ) = 0, for i = 2, . . . , n + 1 ∈ N, and (11.34) we have f (x) − f (x0 ) = 

1  ((n + 1) α)



x0

(g (t) − g (x))(n+1)α−1 g (t)

x

  

Dx(n+1)α f (t) − Dx(n+1)α f (x0 ) dt, 0 −;g 0 −;g

(11.51)

∀ x ∈ [a, x0 ] . Hence (z := g (t)) f (x) − f (x0 ) = 





1  ((n + 1) α)

Dx(n+1)α f ◦ g −1 (z) − 0 −;g





g(x0 )

g(x)

(z − g (x))(n+1)α−1

   −1 dz, f ◦ g Dx(n+1)α )) (g (x 0 0 −;g

(11.52)

11 Vectorial Generalized g-Fractional Direct and Iterated …

234

∀ x ∈ [a, x0 ] . We have that (x0 ≤ x ≤ b) 1  f (x) − f (x0 ) ≤  ((n + 1) α)



g(x) g(x0 )

(g (x) − z)(n+1)α−1



    

  (n+1)α (n+1)α  Dx0 +;g f ◦ g −1 (z) − Dx0 +;g f ◦ g −1 (g (x0 )) dz ≤

(δ1 >0)

1  ((n + 1) α) ω1



Dx(n+1)α 0 +;g ω1





g(x) g(x0 )

(g (x) − z)(n+1)α−1 

δ1 |z − g (x0 )| f ◦g , δ1



(11.53)

−1

dz [g(x0 ),g(b)]

(11.42.)

≤

  Dx(n+1)α f ◦ g −1 , δ1 0 +;g

[g(x0 ),g(b)]

 ((n + 1) α)

g(x)

g(x0 )

(n+1)α−1

(g (x) − z) ω1



  (z − g (x0 )) 1+ dz = δ1

  −1 Dx(n+1)α f ◦ g , δ 1 +;g 0

[g(x0 ),g(b)]

 ((n + 1) α) 

1 (g (x) − g (x0 ))(n+1)α + δ1 (n + 1) α ω1





g(x) g(x0 )

 (g (x) − z)

(n+1)α−1

(z − g (x0 ))

2−1

  −1 Dx(n+1)α f ◦ g , δ 1 0 +;g

dz = (11.54)

[g(x0 ),g(b)]

 ((n + 1) α) 

 1  ((n + 1) α)  (2) (g (x) − g (x0 ))(n+1)α (n+1)α+1 + = (g (x) − g (x0 )) δ1  ((n + 1) α + 2) (n + 1) α ω1



  −1 Dx(n+1)α f ◦ g , δ 1 0 +;g

[g(x0 ),g(b)]

 ((n + 1) α) 

 1 (g (x) − g (x0 ))(n+1)α+1 (g (x) − g (x0 ))(n+1)α + . δ1 (n + 1) α ((n + 1) α + 1) (n + 1) α

(11.55)

11.3 Main Results

235

We have proved that ω1

 f (x) − f (x0 ) ≤



  −1 Dx(n+1)α f ◦ g , δ 1 0 +;g

[g(x0 ),g(b)]

(11.56)

 ((n + 1) α + 1)



 (n+1)α+1 − g )) (g (x) (x 0 , (g (x) − g (x0 ))(n+1)α + δ1 ((n + 1) α + 1)

∀ x ∈ [x0 , b], δ1 > 0. We have that (a ≤ x ≤ x0 ) (11.52)

 f (x) − f (x0 ) ≤

1  ((n + 1) α)



g(x0 ) g(x)

(z − g (x))(n+1)α−1



 

     (n+1)α (n+1)α  Dx0 −;g f ◦ g −1 (z) − Dx0 −;g f ◦ g −1 (g (x0 )) dz ≤

(δ2 >0)

1  ((n + 1) α) ω1





g(x0 )

g(x)

(z − g (x))(n+1)α−1

  −1 δ2 |z − g (x 0 )| Dx(n+1)α f ◦ g , dz 0 −;g δ2 [g(a),g(x0 )] ω1



(11.42.)

≤

(11.57)

  −1 Dx(n+1)α f ◦ g , δ 2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α)

g(x0 ) g(x)

(n+1)α−1

(z − g (x)) ω1



  g (x0 ) − z 1+ dz = δ2

  −1 Dx(n+1)α f ◦ g , δ 2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α) 

1 (g (x0 ) − g (x))(n+1)α + δ2 (n + 1) α ω1





g(x0 )

g(x)

Dx(n+1)α 0 −;g

 (g (x0 ) − z)

2−1



−1

f ◦ g , δ2

 ((n + 1) α)

(n+1)α−1

(z − g (x))

 [g(a),g(x0 )]

dz = (11.58)

11 Vectorial Generalized g-Fractional Direct and Iterated …

236



 1  (2)  ((n + 1) α) (g (x0 ) − g (x))(n+1)α (n+1)α+1 + = (g (x0 ) − g (x)) δ2  ((n + 1) α + 2) (n + 1) α ω1



  Dx(n+1)α f ◦ g −1 , δ2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α) 

1 (g (x0 ) − g (x))(n+1)α+1 (g (x0 ) − g (x))(n+1)α + δ2 (n + 1) α ((n + 1) α + 1) (n + 1) α ω1



 =

(11.59)

  Dx(n+1)α f ◦ g −1 , δ2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α + 1)  (g (x0 ) − g (x))

(n+1)α

 (g (x0 ) − g (x))(n+1)α+1 + . δ2 ((n + 1) α + 1)

We have proved that

 f (x) − f (x0 ) ≤

ω1



 (g (x0 ) − g (x))

(n+1)α

  −1 Dx(n+1)α f ◦ g , δ 2 0 −;g

[g(a),g(x0 )]

 ((n + 1) α + 1)  (g (x0 ) − g (x))(n+1)α+1 + , δ2 ((n + 1) α + 1)

∀ x ∈ [a, x0 ], δ2 > 0. By (11.56) and (11.60), setting δ = δ1 = δ2 > 0, we derive (11.48).

(11.60)



We state Corollary 11.23 (to Theorem 11.22) All as in Theorem 11.22. Then

 f (·) − f (x0 ) ≤  (n+1)α

|g (·) − g (x0 )|

ω1



  Dx(n+1)α f ◦ g −1 , δ 0 ;g

 ((n + 1) α + 1)  |g (·) − g (x0 )|(n+1)α+1 + , δ ((n + 1) α + 1)

δ > 0, true over [a, b] . We need Definition 11.24 Let Dxα0 ;g f denote any of Dxα0 ±;g f and δ > 0. We set

(11.61)

11.3 Main Results

237

ω1



  Dxα0 ;g f ◦ g −1 , δ :=

    α f ◦ g −1 , δ [g(x max ω1 Dx+;g

0 ),g(b)]

, ω1



(11.62)

   Dxα0 −;g f ◦ g −1 , δ [g(a),g(x )] , 0

where x0 ∈ [a, b]. Here the moduli of continuity are considered over [g (x0 ) , g (b)] and [g (a) , g (x0 )], respectively. We will use: Theorem 11.25 Let α > 0, n = α , α ∈ / N, and f ∈ C n ([a, b] , X ) , where [a, b] ⊂ R and (X, ·) is a Banach space. Let g ∈ C 1 ([a, b]), strictly increasing such that  (i) g −1 ∈ C n ([g (a) , g (b)]). Let x0 ∈ [a, b], be fixed such that f ◦ g −1 (g (x0 )) = 0, for, i = 1, . . . , n − 1. Then  f (x) − f (x0 ) ≤

ω1



  Dxα0 ;g f ◦ g −1 , δ  (α + 1)

(11.63)

  |g (x) − g (x0 )|α+1 |g (x) − g (x0 )|α + , δ (α + 1) ∀ δ > 0, ∀ x ∈ [a, b]. If 0 < α < 1, then we do not need any initial conditions.

 Proof By Remark 11.7 we get that Dxα0 ±;g f (x0 ) = 0, for α ∈ / N. By (11.37) we obtain x   1 f (x) − f (x0 ) = (g (x) − g (t))α−1 g (t) Dxα0 +;g f (t) dt, (11.64)  (α) x0 ∀ x ∈ [x0 , b] . And, by (11.38) we get f (x) − f (x0 ) =

1  (α)



x0 x

  (g (t) − g (x))α−1 g (t) Dxα0 −;g f (t) dt,

(11.65)

∀ x ∈ [a, x0 ] . When 0 < α < 1, i.e. n = 1, then (11.64), (11.65) are valid without any initial conditions. We can write

f (x) − f (x0 ) =

1  (α)



x x0

(g (x) − g (t))α−1 g (t)



 

 Dxα0 +;g f (t) − Dxα0 +;g f (x0 ) dt,

(11.66)

11 Vectorial Generalized g-Fractional Direct and Iterated …

238

∀ x ∈ [x0 , b] , f (x) − f (x0 ) =

1  (α)



x0

(g (t) − g (x))α−1 g (t)



x

 

 Dxα0 −;g f (t) − Dxα0 −;g f (x0 ) dt,

(11.67)

∀ x ∈ [a, x0 ] . We can rewrite (11.66) and (11.67) as follows (by z := g (t)):

1 f (x) − f (x0 ) =  (α) 

g(x)

(g (x) − z)α−1

g(x0 )

      Dxα0 +;g f ◦ g −1 (z) − Dxα0 +;g f ◦ g −1 (g (x0 )) dz,

(11.68)

∀ x ∈ [x0 , b] , and f (x) − f (x0 ) = 

1  (α)



g(x0 )

g(x)

(z − g (x))α−1

      Dxα0 −;g f ◦ g −1 (z) − Dxα0 −;g f ◦ g −1 (g (x0 )) dz,

(11.69)

∀ x ∈ [a, x0 ] . We have that (x0 ≤ x ≤ b)  f (x) − f (x0 ) =

g(x) 1 (g (x) − z)α−1  (α) g(x0 )



 

      Dxα0 +;g f ◦ g −1 (z) − Dxα0 +;g f ◦ g −1 (g (x0 )) dz ≤ (δ1 > 0) 

 g(x)  (11.42.) 1 δ1 |z − g (x0 )| Dxα0 +;g f ◦ g −1 , dz ≤ (g (x) − z)α−1 ω1  (α) g(x0 ) δ1 [g(x0 ),g(b)] ω1



  Dxα +;g f ◦ g −1 , δ1 0

[g(x0 ),g(b)]

 (α)

g(x) g(x0 )

ω1



(g (x) − z)α−1



Dxα +;g f ◦ g −1 , δ1 0

  (z − g (x0 )) 1+ dz = δ1

(11.70)

 [g(x0 ),g(b)]

 (α) 

 1 g(x) (g (x) − g (x0 ))α α−1 2−1 + dz = (g (x) − z) (z − g (x0 )) α δ1 g(x0 )

11.3 Main Results

239

ω1



  Dxα +;g f ◦ g −1 , δ1 0

[g(x0 ),g(b)]

 (α) 

 1  (α)  (2) (g (x) − g (x0 ))α α+1 + = (g (x) − g (x0 )) α δ1  (α + 2) ω1



  Dxα +;g f ◦ g −1 , δ1 0

[g(x0 ),g(b)]

 (α) 

 1 1 (g (x) − g (x0 ))α + (g (x) − g (x0 ))α+1 , α δ1 α (α + 1)

(11.71)

∀ x ∈ [x0 , b] . We have proved that

 f (x) − f (x0 ) ≤

ω1



  Dxα0 +;g f ◦ g −1 , δ1

[g(x0 ),g(b)]

 (α + 1)

  (g (x) − g (x0 ))α+1 α , (g (x) − g (x0 )) + δ1 (α + 1)

(11.72)

∀ x ∈ [x0 , b], δ1 > 0. By (11.69) we obtain  f (x) − f (x0 ) ≤

1  (α)



g(x 0 )

(z − g (x))α−1

g(x)



 

      Dxα0 −;g f ◦ g −1 (z) − Dxα0 −;g f ◦ g −1 (g (x0 )) dz ≤ (δ2 > 0) 1  (α) ω1





g(x 0 ) g(x)

(z − g (x))α−1 ω1



  Dxα0 −;g f ◦ g −1 , δ2

[g(a),g(x 0 )]

  δ2 (g (x0 ) − z) dz Dxα0 −;g f ◦ g −1 , δ2 [g(a),g(x 0 )]

 (α)

g(x 0 ) g(x)

ω1



  Dxα0 −;g f ◦ g −1 , δ2

1 (g (x0 ) − g (x))α + α δ2



g(x 0 ) g(x)

≤

  g (x0 ) − z dz = (11.73) (z − g (x))α−1 1 + δ2

[g(a),g(x 0 )]

 (α) 

( 11.42.)

 (g (x0 ) − z)

2−1

(z − g (x))

α−1

dz =

11 Vectorial Generalized g-Fractional Direct and Iterated …

240 ω1



  Dxα0 −;g f ◦ g −1 , δ2

[g(a),g(x 0 )]

 (α) 

 1  (2)  (α) (g (x0 ) − g (x))α + (g (x0 ) − g (x))α+1 = α δ2  (α + 2) ω1



  Dxα0 −;g f ◦ g −1 , δ2

[g(a),g(x 0 )]

 (α) 

 1 1 (g (x0 ) − g (x))α + (g (x0 ) − g (x))α+1 = α δ2 α (α + 1) ω1



  Dxα0 −;g f ◦ g −1 , δ2

(11.74)

[g(a),g(x 0 )]

 (α + 1) 

(g (x0 ) − g (x))α +

 1 (g (x0 ) − g (x))α+1 , δ2 (α + 1)

(11.75)

∀ x ∈ [a, x0 ] . In conclusion we derive

 f (x) − f (x0 ) ≤

ω1



  Dxα0 −;g f ◦ g −1 , δ2

[g(a),g(x0 )]

 (α + 1)

  (g (x0 ) − g (x))α+1 , (g (x0 ) − g (x))α + δ2 (α + 1) ∀ x ∈ [a, x0 ], δ2 > 0. Choosing δ = δ1 = δ2 > 0 by (11.72) and (11.76), we get (11.63).

(11.76)



We state Corollary 11.26 All as in Theorem 11.25. Then

 f (·) − f (x0 ) ≤

ω1



  Dxα0 ;g f ◦ g −1 , δ  (α + 1)

  |g (·) − g (x0 )|α+1 α |g (·) − g (x0 )| + , δ (α + 1) true over [a, b], δ > 0. If 0 < α < 1, no initial conditions are needed. We make

(11.77)

11.3 Main Results

241

Assumption 11.27 Let L N : C ([a, b] , X ) → C ([a, b] , X ), (X, ·) a Banach L N : C ([a, b]) → space, L N is a linear operator, ∀ N ∈ N, x0 ∈ [a, b]. Let also  C ([a, b]), a sequence of positive linear operators, ∀ N ∈ N. We assume that   (L N ( f )) (x0 ) ≤  L N ( f ) (x0 ) , (11.78) ∀ N ∈ N, ∀ x0 ∈ [a, b], ∀ f ∈ C ([a, b] , X ) . When g ∈ C ([a, b]), c ∈ X , we assume that L N (g) , ∀ N ∈ N. (L N (cg)) = c

(11.79)

We treat the special case of  L N (1) = 1, ∀ N ∈ N,

(11.80)

L N (c) = c, ∀ c ∈ X , ∀ N ∈ N.

(11.81)

which implies We call  L N the companion operator of L N . Operators  B N see (1), and B N see (11.2), fulfill Assumption 11.27. We make Remark 11.28 Let  L N : C ([a, b]) → C ([a, b]), N ∈ N, be a sequence of positive linear operators. By Riesz representation theorem (see [11, p. 304]) we have  L N ( f, x0 ) =

[a,b]

f (t) dμ Nx0 (t) ,

(11.82)

∀ x0 ∈ [a, b], where μ Nx0 is a unique positive finite measure on σ-Borel algebra of [a, b]. Call  (11.83) L N (1, x0 ) = μ Nx0 ([a, b]) = M Nx0 . In our case of  L N (1, x0 ) = 1, we have that M Nx0 = 1, and μ Nx0 is a probability measure. By Hölder’s inequality we obtain  L N (|g (·) − g (x0 )|α ) (x0 ) =

[a,b]

 [a,b]

|g (t) − g (x0 )|α+1 dμ Nx0 (t)

α  α+1

|g (t) − g (x0 )|α dμ Nx0 (t) ≤

    α =  L N |g (·) − g (x0 )|α+1 (x0 ) α+1 . (11.84)

11 Vectorial Generalized g-Fractional Direct and Iterated …

242

Similarly, it holds       (n+1)α  L N |g (·) − g (x0 )|(n+1)α (x0 ) ≤  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 . (11.85) We present Theorem 11.29 Under Assumption 11.27 and all as in Theorem 11.22, we derive

(L N ( f )) (x0 ) − f (x0 ) ≤ 

ω1



  −1 Dx(n+1)α f ◦ g , δ 0 ;g

 ((n + 1) α + 1)

(11.86)

   (n+1)α+1    |g − g L (·) (x (x )| ) N 0 0  L N |g (·) − g (x0 )|(n+1)α (x0 ) + , δ ((n + 1) α + 1)

∀ N ∈ N, δ > 0. Proof By (11.61) we get  L N ( f (·) − f (x0 )) (x0 ) ≤ 

ω1



  −1 Dx(n+1)α f ◦ g , δ ;g 0

 ((n + 1) α + 1)

(11.87)

      L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α  L N |g (·) − g (x0 )| , (x0 ) + δ ((n + 1) α + 1) 

along with (11.46) proving the claim. We also give

Theorem 11.30 Under Assumption 11.27 and all as in Theorem 11.25, we derive

(L N ( f )) (x0 ) − f (x0 ) ≤ 

ω1



  Dxα0 ;g f ◦ g −1 , δ  (α + 1)

   α+1  |g − g L (·) (x (x )| ) N 0 0  L N (|g (·) − g (x0 )|α ) (x0 ) + , δ (α + 1)

∀ N ∈ N, δ > 0.

(11.88)

11.3 Main Results

243

Proof By (11.77) we get  L N ( f (·) − f (x0 )) (x0 ) ≤ 

ω1



  Dxα0 ;g f ◦ g −1 , δ  (α + 1)

    L N |g (·) − g (x0 )|α+1 (x0 )  L N (|g (·) − g (x0 )| ) (x0 ) + , δ (α + 1) α

(11.89) 

along with (11.46) proving the claim. We give

Theorem 11.31 Under Assumption 11.27 and all as in Theorem 11.22, we obtain (L N ( f )) (x0 ) − f (x0 ) ≤ ω1



((n + 1) α + 2)  ((n + 1) α + 2)

  1     (n+1)α+1 −1  (n+1)α+1 |g Dx(n+1)α − g f ◦ g , L (·) (x (x )| ) N 0 0 ;g 0 

(11.90)

   (n+1)α  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 , ∀ N ∈ N.

  If  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → ∞. Proof By (11.87) and (11.85) we obtain that Right hand side (11.87) ≤ 

ω1



  −1 , δ f ◦ g Dx(n+1)α ;g 0

 ((n + 1) α + 1)

  

  (n+1)α  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 (n+1)α+1  L N |g (·) − g (x0 )| + =: (∗) . (x0 ) δ ((n + 1) α + 1)

(11.91)   Momentarily we assume that  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) > 0, and choose    1  δ :=  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 > 0.

(11.92)

Thus, we have  ω1



(n+1)α Dx ;g f 0





◦ g −1 ,  LN

(∗) = 



 LN





|g (·) − g (x0 )|(n+1)α+1 (x0 )



1 (n+1)α+1



 ((n + 1) α + 1)



|g (·) − g (x0 )|(n+1)α+1 (x0 )



(n+1)α (n+1)α+1

δ (n+1)α+1 + δ ((n + 1) α + 1)

 =

(11.93)

11 Vectorial Generalized g-Fractional Direct and Iterated …

244



ω1

(n+1)α Dx ;g f 0





◦ g −1 ,  LN





|g (·) − g (x0 )|(n+1)α+1 (x0 )



1 (n+1)α+1



 ((n + 1) α + 1)

 (n+1)α 

 (n+1)α+1  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) 1+

 1 = (n + 1) α + 1







  1  ((n + 1) α + 2) (n+1)α+1 −1  (n+1)α+1 |g f ◦ g , L − g Dx(n+1)α ω1 (x )| ) (·) (x N 0 0 0 ;g  ((n + 1) α + 2)

(11.94)



 (n+1)α

 (n+1)α+1  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) .

So far we have proved that ((n + 1) α + 2)  L N ( f (·) − f (x0 )) (x0 ) ≤  ((n + 1) α + 2) ω1



  1     (n+1)α+1 −1  (n+1)α+1 |g Dx(n+1)α − g f ◦ g , L (·) (x (x )| ) N 0 0 0 ;g 

(11.95)

   (n+1)α  L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) (n+1)α+1 ,

when δ > 0. If δ = 0, we get R.H.S.(11.95)= 0, and    L N |g (·) − g (x0 )|(n+1)α+1 (x0 ) =

[a,b]

|g (t) − g (x0 )|(n+1)α+1 dμ Nx0 (t) = 0.

(11.96) Hence |g (t) − g (x0 )| = 0, a.e. on [a, b]. That is g (t) = g (x0 ), a.e. on [a, b] and t = x0 , a.e. on [a, b], because g is strictly increasing. More precisely μ Nx0 {t ∈ [a, b] : t = x0 } = 0. Therefore μ Nx0 concentrates on x0 , that is μ Nx0 = δx0 , the unit Dirac measure. Consequently, it holds  L N ( f (·) − f (x0 )) (x0 ) =

 f (t) − f (x0 ) dδx0 (t) [a,b]

=  f (x0 ) − f (x0 ) = 0.

(11.97)

Therefore (11.95) is true always, in both cases, completing the proof of the theorem, by the use of (11.46).  We also give

11.3 Main Results

245

Theorem 11.32 Under Assumption 11.27 and all as in Theorem 11.25, we obtain (L N ( f )) (x0 ) − f (x0 ) ≤ ω1



(α + 2)  (α + 2)

     1  Dxα0 ;g f ◦ g −1 ,  L N |g (·) − g (x0 )|α+1 (x0 ) α+1 

(11.98)

   α  L N |g (·) − g (x0 )|α+1 (x0 ) α+1 , ∀ N ∈ N.

  If  L N |g (·) − g (x0 )|α+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → ∞. Proof Similar to the proof of Theorem 11.31. It is based on (11.46), (11.84) and (11.89). We omit the details.  We give Corollary 11.33 (to Theorem 11.31) All as in Theorem 11.31. Then (L N ( f )) (x0 ) − f (x0 ) ≤ ω1



((n + 1) α + 2)  ((n + 1) α + 2)

  1       (n+1)α+1 −1   (n+1)α+1  | |· g Dx(n+1)α L − x f ◦ g , (x ) N 0 0 ∞,[a,b] 0 ;g (n+1)α 

  g 

∞,[a,b]

   (n+1)α  L N |· − x0 |(n+1)α+1 (x0 ) (n+1)α+1 , ∀ N ∈ N.

(11.99)

  If  L N |· − x0 |(n+1)α+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → ∞. Proof By Theorem 11.31 and   |g (x) − g (x0 )| ≤ g ∞,[a,b] |x − x0 | , ∀ x, x0 ∈ [a, b] .

(11.100) 

We give Corollary 11.34 (to Theorem 11.32) All as in Theorem 11.32. Then (L N ( f )) (x0 ) − f (x0 ) ≤ ω1



  g 

(α + 2)  (α + 2)

       1  Dxα0 ;g f ◦ g −1 , g ∞,[a,b]  L N |· − x0 |α+1 (x0 ) α+1

∞,[a,b]

α 

   α  L N |· − x0 |α+1 (x0 ) α+1 , ∀ N ∈ N.

  If  L N |· − x0 |α+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → ∞.

(11.101)

11 Vectorial Generalized g-Fractional Direct and Iterated …

246

Proof By Theorem 11.32 and   |g (x) − g (x0 )| ≤ g ∞,[a,b] |x − x0 | , ∀ x, x0 ∈ [a, b] .

(11.102) 

We make Remark 11.35 Theorems 11.31, 11.32 and Corollaries 11.33, 11.34 have direct B N Bernstein operators over [0, 1]. We obtain there B N ( f ) (x0 ) applications to B N ,  → f (x0 ), as N → ∞. We notice that |· − x0 |α+1 = |· − x0 |α |· − x0 | ≤ |· − x0 | , over [0, 1] . Hence

   B N |· − x0 |α+1 (x0 ) ≤  B N (|· − x0 |) (x0 ) , x0 ∈ [0, 1] .

Similarly, it holds |· − x0 |(n+1)α+1 = |· − x0 |(n+1)α |· − x0 | ≤ |· − x0 | , over [0, 1] . Hence

   B N |· − x0 |(n+1)α+1 (x0 ) ≤  B N (|· − x0 |) (x0 ) , x0 ∈ [0, 1] .

We have that

   1  B N (· − x0 )2 (x0 ) 2 B N (|· − x0 |) (x0 ) ≤   =

1 x0 (1 − x0 ) ≤ √ , ∀ N ∈ N. N 2 N

(11.103)

We give Corollary 11.36 Here [a, b] = [0, 1] and all as in Theorem 11.22. Then (B N ( f )) (x0 ) − f (x0 ) ≤

ω1



Dx(n+1)α 0 ;g



((n + 1) α + 2)  ((n + 1) α + 2)

  f ◦ g , g ∞,[a,b] −1



1 √

 1  (n+1)α+1

2 N

(n+1)α (n+1)α  1  (n+1)α+1 , ∀ N ∈ N, x0 ∈ [0, 1] . √ ∞,[a,b] 2 N

  g 

As N → ∞, we get B N ( f ) (x0 ) → f (x0 ).

(11.104)

11.3 Main Results

247



Proof By Corollary 11.33 and (11.103). We give Corollary 11.37 Here [a, b] = [0, 1] and all as in Theorem 11.25. Then

ω1

(B N ( f )) (x0 ) − f (x0 ) ≤

(α + 2)  (α + 2)

 



  Dxα0 ;g f ◦ g −1 , g ∞,[a,b] 

1 √

1   α+1

2 N

α α  1  α+1 , ∀ N ∈ N. √ ∞,[a,b] 2 N

  g 

(11.105)

As N → ∞, we get B N ( f ) (x0 ) → f (x0 ), x0 ∈ [0, 1] . 

Proof By Corollary 11.34 and (11.103). We continue with the following

Corollary 11.38 (to Corollary 11.37) Here [a, b] = [0, 1] , all as in Theorem 11.25, and g (x) = e x , x ∈ [0, 1] . Then (B N ( f )) (x0 ) − f (x0 ) ≤

ω1

 

Dxα0 ;ex





f ◦ ln x, e

1 √

1   α+1

2 N

eα



(α + 2)  (α + 2)

1 √

α  α+1

2 N

, ∀ N ∈ N.

(11.106)

As N → ∞, we get B N ( f ) (x0 ) → f (x0 ), x0 ∈ [0, 1] . 

Proof By Corollary 11.37. We finish with

Corollary 11.39 (to Corollary 11.38) Here [a, b] = [0, 1] , all as in Theorem 11.25, g (x) = e x , x ∈ [0, 1] and α = 21 (no initial conditions needed). Then 10 (B N ( f )) (x0 ) − f (x0 ) ≤ 3 ω1



1 2



Dx0 ;ex f ◦ ln x, e



1 √

2 N

 23  

1 √

2 N

 13



e π

∀ N ∈ N, ∀ x0 ∈ [0, 1] . (11.107)

As N → ∞, we get B N ( f ) (x0 ) → f (x0 ), any x0 ∈ [0, 1] . Proof By Corollary 11.38.



248

11 Vectorial Generalized g-Fractional Direct and Iterated …

References 1. Aliprantis, C.D., Border, K.C.: Infinite Dimensional Analysis. Springer, New York (2006) 2. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman Research Notes in Mathematics, vol. 287. Longman Sci. & Tech., Harlow (1993) 3. Anastassiou, G.A.: Principles of General Fractional Analysis for Banach space valued functions. Bull. Allahabad Math. Soc. 32(1), 71–145 (2017) 4. Anastassiou, G.A.: Vector fractional Korovkin type Approximation. Dyn. Syst. Appl. 26, 81– 104 (2017) 5. Anastassiou, G.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg, New York (2018) 6. Anastassiou, G.: Vectorial generalized g-fractional direct and iterated approximations by linear operators. Fasciculi Mathematici (2020). Accepted 7. Bochner integral. Encyclopedia of Mathematics. http://www.encyclopediaofmath.org/index. php?title=Bochner_integral&oldid=38659 8. Kreuter, M.: Sobolev space of vector-valued functions. Ulm University, Master thesis in Mathematics, Ulm, Germany (2015) 9. Mikkola, K.: Appendix B integration and differentiation in Banach spaces. http://math.aalto. fi/~kmikkola/research/thesis/contents/thesisb.pdf 10. Mikusinski, J.: The Bochner Integral. Academic, New York (1978) 11. Royden, H.L.: Real Analysis, 2nd edn. Macmillan, New York (1968)

Chapter 12

Quantitative Multivariate Complex Korovkin Approximation Theory

Let K be a compact convex subset of Ck , k ≥ 2, and C (K , C) be the space of continuous functions from K into C. We consider bounded linear operators from C (K , C) into itself. We assume that these are bounded by companion positive linear operators from C (K , R) into itself. We study quantitatively the rate of convergence of the approximation and high order approximation of these multivariate complex operators to the unit operators. Our results are inequalities of Korovkin type involving the multivariate complex modulus of continuity of the engaged function or its partial derivatives and basic test functions. On the way to prove our main results we establish a complex multivariate Taylor’s formula and results on complex multivariate modulus of continuity. We consider related approximations under convexity and we finish with important applications. See also [5].

12.1 Introduction The study of the convergence of positive linear operators became more intensive and attractive when Korovkin (1953) proved his famous theorem (see [7, 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 Mamedov (1959) (see [10]) was the first to put Korovkin’s theorem in a quantitative scheme. in Mamedov’s Theorem. Let {L n }n∈N be a sequence of positive linear  operators  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 © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_12

249

250

12 Quantitative Multivariate Complex Korovkin …

   L n ( f, x) − f (x)∞ ≤ 3ω1 f, dn ,

(12.1)

where ω1 is the first modulus of continuity and dn = βn (x) − 2xαn (x)∞ . An improvement of the last result was the following. Shisha and Mond’s Theorem. (1968, see [13]). 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)∞

(12.2)

(12.3)

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. 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 ShishaMond 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 Ck , k ≥ 2, and L be a linear operator from C (K , C) into itself, and let  L be a positive linear operator from L (| f |), ∀ f ∈ C (K , C), where λ > 0. C (K , R) into itself, such that |L ( f )| ≤ λ Clearly then L is a bounded linear operator. Here we create a complete quantitative Korovkin type theory over the last described setting.

12.2 Background

251

12.2 Background We need   Theorem 12.1 Let K ⊆ Ck , | |·| | , k ∈ N − {1}, and f a function from K into C. Consider the first multivariate complex modulus of continuity ω1 ( f, δ) :=

sup | f (x) − f (y)| , δ > 0.

(12.4)

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,

(12.5)

that is the subadditivity property is true. Also it holds ω1 ( f, nδ) ≤ nω1 ( f, δ)

(12.6)

ω1 ( f, λδ) ≤ λ ω1 ( f, δ) ≤ (λ + 1) ω1 ( f, δ) ,

(12.7)

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 ω1 ( f + g, δ) ≤ ω1 ( f, δ) + ω1 (g, δ) ,

(12.8)

for δ > 0, any f, g : K → C, K ⊂ Ck 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

252

12 Quantitative Multivariate Complex Korovkin …

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. (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.



12.2 Background

253

We need Theorem 12.2 ([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.

(12.9)

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.

(12.10)

(12.11)

  Corollary 12.3 Let K ⊆ Ck , · be a compact convex subset, and f ∈ C (K , C). Then  x − x0  | f (x) − f (x0 )| ≤ ω1 ( f, δ) , δ > 0, (12.12) δ ∀ x, x0 ∈ K . We make

  Remark 12.4 Let K ⊆ Ck , · 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, [11, p. 304], there exists a unique Borel measure μ on K such that

g (t) dμ (t) ,

I (g) =

(12.13)

K

∀ g ∈ C (K , R), where μ (K ) < ∞. Next the vector derivatives and vector Riemann integral are as the numerical ones, see [12, pp. 83–95]. We have the following vector Taylor type formula: Theorem 12.5 Let n ∈ N and f ∈ C n ([a, b] , X ), where [a, b] ⊂ R and (X, ·) is a Banach space. Then    tn−1

b  t1 (b − a)i (i) (n) f (b) = f (a) + ... f f (a) + (tn ) dtn . . . dt1 . i! a a a i=1 n−1

(12.14)

254

12 Quantitative Multivariate Complex Korovkin …

We also have that f (b) = f (a) +

n

(b − a)i

i!

i=1

a

b





t1

...

a



tn−1

f (i) (a) +

   f (n) (tn ) − f (n) (a) dtn . . . dt1 .

(12.15)

a

Proof By [12, p. 92] we have that (n−1)

f

(tn−1 ) = f

(n−1)

tn−1

f (n) (tn ) dtn ,

(12.16)

f (n−1) (tn−1 ) dtn−1 ,

(12.17)

(a) + a

and f (n−2) (tn−2 ) = f (n−2) (a) +

tn−2

a

where a ≤ tn−1 ≤ tn−2 ≤ b. Hence it holds f (n−2) (tn−2 ) = f (n−2) (a) +

tn−2



f (n−1) (a) +

a

tn−1

 f (n) (tn ) dtn dtn−1

a

= f (n−2) (a) + f (n−1) (a) (tn−2 − a) +

tn−2 

a

tn−1



(12.18)

f (n) (tn ) dtn dtn−1 .

a

Similarly, we have f (n−3) (tn−3 ) = f (n−3) (a) +

tn−3

f (n−2) (tn−2 ) dtn−2 ,

(12.19)

a

where a ≤ tn−1 ≤ tn−2 ≤ tn−3 ≤ b. Consequently we get f (n−3) (tn−3 ) = f (n−3) (a) +

tn−3



f (n−2) (a) + f (n−1) (a) (tn−2 − a)

a

tn−2

+



a

tn−1

  f (n) (tn ) dtn dtn−1 dtn−2 =

a

(tn−3 − a)2 + 2   f (n) (tn ) dtn dtn−1 dtn−2 .

f (n−3) (a) + f (n−2) (a) (tn−3 − a) + f (n−1) (a)

a

tn−3

 a

tn−2

 a

tn−1

(12.20)

12.2 Background

255

Similarly, we have (n−4)

f

(tn−4 ) = f

(n−4)

(a) +

tn−4

f (n−3) (tn−3 ) dtn−3 ,

(12.21)

a

where a ≤ tn−1 ≤ tn−2 ≤ tn−3 ≤ tn−4 ≤ b. Thus, we find f (n−4) (tn−4 ) = f (n−4) (a) +

tn−4



f (n−3) (a) +

a

f (n−1) (a) (tn−3 − a)2 + 2    f (n) (tn ) dtn dtn−1 dtn−2 dtn−3

f (n−2) (a) (tn−3 − a) +

tn−3



a

tn−2



a

tn−1

(12.22)

a

= f (n−4) (a) + f (n−3) (a) (tn−4 − a) + f (n−2) (a)

(tn−4 − a)2 2

f (n−1) (a) (tn−4 − a)3 + 3!   

tn−4  tn−3  tn−2  tn−1 f (n) (tn ) dtn dtn−1 dtn−2 dtn−3 . +

a

a

a

a

We have proved that f (n−4) (tn−4 ) = f (n−4) (a) + f (n−3) (a) (tn−4 − a) +

(12.23)

f (n−2) (a) f (n−1) (a) (tn−4 − a)2 + (tn−4 − a)3 + 2 3!   

tn−4  tn−3  tn−2  tn−1 (n) f (tn ) dtn dtn−1 dtn−2 dtn−3 , a

a

a

a

where a ≤ tn−1 ≤ tn−2 ≤ tn−3 ≤ tn−4 ≤ b. Etc. Working inductively we have proved the claim.



We make Remark 12.6 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 [6, 8] f is jointly holomorphic and jointly continuous on Q. Let n ∈ N. Each nth order complex partial

256

12 Quantitative Multivariate Complex Korovkin …

derivative is denoted by f α := k  and |α| := αi = n.

∂α f ∂xα

, where α := (α1 , . . . , αk ), αi ∈ Z+ , i = 1, . . . , k

i=1

Consider gz (t) := f (x0 + t (z − x0 )) , 0 ≤ t ≤ 1.

(12.24)

Clearly it holds that x0 + t (z − x0 ) ∈ Q and gz (t) ∈ C, ∀ t ∈ [0, 1] . Then we derive

( j) gz (t)

⎡ j ⎤ k

∂ =⎣ f ⎦ (x01 + t (z 1 − x01 ) , . . . , x0k + t (z k − x0k )) , (z i − x0i ) ∂ xi i=1

(12.25) for all j = 0, 1, . . . , n. Notice here that any mixed partials commute. We remind that (C, |·|) is a Banach space. We also have for j = 0, 1, . . . , n : ⎞

⎛

gz( j) (0) =

+

α:=(α1 ,...,αk ),α j ∈Z k  i=1,...,k;|α|:= αi = j

⎜ j! ⎟ ⎟ ⎜ ⎟ ⎜ k ⎠ ⎝ αi !





k

(z i − x0i )αi

f α (x0 ) .

(12.26)

i=1

i=1

i=1

Furthermore it holds ⎞

⎛ gz(n) (θ ) =

+

α:=(α1 ,...,αk ),α j ∈Z k  i=1,...,k;|α|:= αi =n

⎜ n! ⎟ ⎟ ⎜ ⎟ ⎜ k ⎠ ⎝ αi !





k

(z i − x0i )αi

f α (x0 + θ (z − x0 )) ,

i=1

i=1

i=1

(12.27) 0 ≤ θ ≤ 1. Notice that gz (0) = f (x0 ) . ( j) By (12.26), if all f α (x0 ) = 0, then gz (0) = 0, where j ∈ {0, 1, . . . , n} . We give Example 12.7 Let n = k = 2. Then gz (t) = f (x01 + t (z 1 − x01 ) , x02 + t (z 2 − x02 )) , t ∈ [0, 1] , and

12.2 Background

257

∂f ∂f (x0 + t (z − x0 )) + (z 2 − x02 ) (x0 + t (z − x0 )) . ∂ x1 ∂ x2 (12.28)

gz (t) = (z 1 − x01 ) In addition,

gz (t) = (z 1 − x01 )



∂f (x0 + t (z − x0 )) ∂ x1



 + (z 2 − x02 )

∂f (x0 + t (z − x0 )) ∂ x2

" ∂f 2 ∂f 2 = (z 1 − x01 ) (z 1 − x01 ) 2 (∗) + (z 2 − x02 ) (∗) + ∂ x2 ∂ x1 ∂ x1



!

(12.29)

!

" ∂f 2 ∂f 2 (z 2 − x02 ) (z 1 − x01 ) (∗) + (z 2 − x02 ) 2 (∗) . ∂ x1 ∂ x2 ∂ x2

Hence gz (t) = (z 1 − x01 )2

∂f 2 ∂f 2 − x − x + (∗) (z ) (z ) (∗) 1 01 2 02 ∂ x2 ∂ x1 ∂ x12

+ (z 1 − x01 ) (z 2 − x02 )

∂f 2 ∂f 2 (∗) + (z 2 − x02 )2 2 (∗) , ∂ x1 ∂ x2 ∂ x2

(12.30)

where ∗ := x0 + t (z − x0 ) . Notice that gz (t) , gz (t) , gz (t) ∈ C. Etc. We give the following Taylor’s formula on Ck , k ≥ 2. Theorem 12.8 It holds f (z 1 , . . . , z k ) = gz (1) =

n ( j)

gz (0) + Rn (z, 0) , j! j=0

(12.31)

where

1

Rn (z, 0) := 0



t1 0



tn−1

... 0



   gz(n) (tn ) − gz(n) (0) dtn . . . dt1 , ∀ z ∈ Q.

Proof By Theorem 12.5 (12.15), it is applied to gz (t), t ∈ [0, 1] . Next we will estimate Rn (z, 0) .

(12.32) 

258

12 Quantitative Multivariate Complex Korovkin …

We need Remark 12.9 See [1, p. 210]. Let h > 0 be fixed, x ∈ R. Denote by

φn (x) =

|x|

0

Clearly φn (0) = 0. It holds φn (x) ≤

x1

 ...

0

0

xn−1

#x $ n

h

 d xn . . . d x1 .

|x|n |x|n+1 h |x|n−1 + , + 2n! 8 (n − 1)! (n + 1)!h

(12.33)

(12.34)

with equality only at x = 0.   Theorem 12.10 Let Q ⊆ Ck , ·1 be a compact convex subset, k ≥ 2, where ·1 is the l1 -norm. Here f : Q → C is a coordinate-wise holomorphic function and Rn (z, 0) as in (12.32). Then   |Rn (z, 0)| ≤ maxω1 ( f α , h) φn (z − x0 1 ) , |α|=n

(12.35)

∀ z, x0 ∈ Q, n ∈ N. Proof By Corollary 12.3 we have that  | f α (x0 + t (z − x0 )) − f α (x0 )| ≤ ω1 ( f α , h)

t z − x0 1 , h

(12.36)

∀ t ∈ [0, 1], h > 0, all α : |α| = n. For z − x0 1 = 0, it follows from (12.27), using (12.33), that

|Rn (z, 0)| ≤

1

⎛

⎝

0

t1

⎛

...⎝

0

⎛ tn−1

|α|=n

n! |z 1 − x01 |α1 . . . |z k − x0k |αk α1 ! . . . αk !

|α|=n

 

  tn z − x0 1 dtn . . . dt1 = h

maxω1 ( f α , h)





0

 |α|=n

⎝

k 

|z i − x0i |αi   n! i=1 maxω1 ( f α , h) φn (z − x0 1 ) = |α|=n α1 ! . . . αk ! z − x0 1 

 maxω1 ( f α , h) φn (z − x0 1 ) ,

|α|=n

(12.37)

12.2 Background

by z − x0 1 =

259 k 

|z i − x0i |. Therefore

i=1





|Rn (z, 0)| ≤ maxω1 ( f α , h) φn (z − x0 1 ) , |α|=n

for all z ∈ Q. If z = x0 , then z − x0 1 = 0, and (12.38) is trivially true.

(12.38)



We state the following main result:   Theorem 12.11 Let Q ⊆ Ck , ·1 be a compact convex subset, k ≥ 2, where ·1 is the l1 -norm. Here f : Q → C is a coordinate-wise holomorphic function and z, x0 ∈ Q, n ∈ N. Assume that f α (x0 ) = 0, for all α : |α| = j, j = 1, . . . , n. Then (1) f (z) = f (x0 ) + Rn (z, 0) ,

(12.39)

where Rn (z, 0) as in (12.32), and (2)





| f (z) − f (x0 )| ≤ maxω1 ( f α , h) φn (z − x0 1 ) ≤ |α|=n

%

 maxω1 ( f α , h)

|α|=n

& z − x0 n1 z − x0 n+1 h z − x0 n−1 1 1 + + , (12.40) 2n! 8 (n − 1)! (n + 1)!h

where h > 0. Proof By (12.32), (12.34) and (12.35).



We give Corollary 12.12 All as in Theorem 12.11. Then & % · − x0 n1 · − x0 n+1 h · − x0 n−1 1 1 | f (·) − f (x0 )| ≤ max ω1 ( f α , h) + + , 2n! 8 (n − 1)! (n + 1)!h |α|=n 

(12.41) h > 0, true over Q. Above ω1 is defined with respect to ·1 .

260

12 Quantitative Multivariate Complex Korovkin …

12.3 Main Results Let K be a compact and convex subset of Ck , k ≥ 2. Consider L : C (K , C) → C (K , C) a linear operator and  L : C (K , R) → C (K , R) a positive linear operator L ( f1) ≥  L ( f 2 )) both over the field (i.e. for f 1 . f 2 ∈ C (K , R) with f 1 ≥ f 2 we get  of R. We assume that |L ( f )| ≤ λ L (| f |) , ∀ f ∈ C (K , C) , where λ > 0,

(12.42)

L (| f |) (z), ∀ z ∈ K ). (i.e. |L ( f ) (z)| ≤ λ We call  L the companion operator of L . Let z 0 ∈ K . Clearly, then L (·) (z 0 ) is a linear functional from C (K , C) into C, and  L (·) (z 0 ) 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 (· − z 0 r ) (z 0 ), r > 0, is a continuous function in z 0 ∈ K . Here · is an arbitrary norm. We have the following approximation result with rates of Korovkin type.   Theorem 12.13 Here K is a convex and compact subset of Ck , · , k ≥ 2, 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 )| ≤ λ L n (| f |) , λ > 0, ∀ f ∈ C (K , C) , ∀ n ∈ N (12.43)   L n (| f |) (z 0 ), ∀ z 0 ∈ K ). (i.e. |L n ( f ) (z 0 )| ≤ λ  Additionally, we assume that L n (g) , ∀ g ∈ C (K , R) , ∀ c ∈ C L n (cg) = c

(12.44)

  L n (g) (z 0 ) , ∀ z 0 ∈ K ). (i.e. (L n (cg)) (z 0 ) = c  Then, for any f ∈ C (K , C), we have |(L n ( f )) (z 0 ) − f (z 0 )| ≤ | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 +     λ  L n (1 (·)) (z 0 ) + 1 ω1 f,  L n (· − z 0 ) (z 0 ) ,

(12.45)

∀ z 0 ∈ K , ∀ n ∈ N. If  L n (1 (·)) (z 0 ) = 1, ∀ z 0 ∈ K , then   |(L n ( f )) (z 0 ) − f (z 0 )| ≤ 2λω1 f,  L n (· − z 0 ) (z 0 ) ,

(12.46)

12.3 Main Results

261

∀ z 0 ∈ K , ∀ n ∈ N. If  L n (1 (·)) (z 0 ) → 1, and  L n (· − z 0 ) (z 0 ) → 0, as n → ∞, then L n ( f ) (z 0 ) L n (1 (·)) (z 0 ) is bounded. → f (z 0 ), ∀ f ∈ C (K , C) . Here  Proof We notice that |(L n ( f )) (z 0 ) − f (z 0 )| = (12.44)

|(L n ( f )) (z 0 ) − L n ( f (z 0 ) (·)) (z 0 ) + L n ( f (z 0 ) (·)) (z 0 ) − f (z 0 )| =

(L n ( f )) (z 0 ) − L n ( f (z 0 ) (·)) (z 0 ) + f (z 0 )  L n (1 (·)) (z 0 ) − f (z 0 ) ≤ |(L n ( f )) (z 0 ) − L n ( f (z 0 ) (·)) (z 0 )| + | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 =

(12.47)

(12.43) |L n ( f (·) − f (z 0 )) (z 0 )| + | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 ≤   (12.4.) | f (z 0 )|  L n (| f (·) − f (z 0 )|) (z 0 ) ≤ L n (1 (·)) (z 0 ) − 1 + λ      (12.7.) δ · − z 0  | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 + λ  L n ω1 f, (z 0 ) ≤ δ     1 | f (z 0 )|  L n ω1 ( f, δ) 1 (·) + · − z 0  L n (1 (·)) (z 0 ) − 1 + λ  (z 0 ) = δ '  1  ) | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 + λω1 ( f, δ) L n (1 (·)) (z 0 ) + L n (· − z 0 (z 0 ) = δ (12.48)   ( | f (z 0 )|  L n (· − z 0 ) (z 0 )  L n (1 (·)) (z 0 ) − 1 + λω1 f,  L n (1 (·)) (z 0 ) + 1 , by choosing

δ :=  L n (· − z 0 ) (z 0 ) ,

(12.49)

if  L n (· − z 0 ) (z 0 ) > 0. Next we consider the case of  L n (· − z 0 ) (z 0 ) = 0.

(12.50)

By Riesz representation theorem [11, p. 304] there exists a positive finite measure μz0 such that  L n (g) (z 0 ) =

g (t) dμz0 (t) , ∀ g ∈ C (K , R) . K

That is

(12.51)

262

12 Quantitative Multivariate Complex Korovkin …

t − z 0  dμz0 (t) = 0 , K

which implies t − z 0  = 0, a.e., hence t − z 0 = 0, a.e., and t = z 0 , a.e. on K . Consequently μz0 ({t ∈ K : t = z 0 }) = 0. L n(1 (·)) (z 0 )). Hence, in That is μz0 = δz0 M (where 0 < M := μz0 (K ) =   that L n (· − z 0 ) (z 0 ) = 0, case  L n (g) (z 0 ) = g (z 0 ) M. Consequently, it holds ω1 f,  and the right hand side of (12.45) equals | f (z 0 )| |M − 1| . Also, it is  L n (| f (·) − f (z 0 ) (·)|) (z 0 ) = | f (z 0 ) − f (z 0 )| M = 0. And by (12.43) we obtain |(L n ( f (·) − f (z 0 ) (·))) (z 0 )| = 0, that is |L n ( f ) (z 0 ) − L n ( f (z 0 ) (·)) (z 0 )| = 0. The last says that (12.44)

L n ( f ) (z 0 ) = L n ( f (z 0 ) (·)) (z 0 ) =

f (z 0 )  L n (1 (·)) (z 0 ) = M f (z 0 ) .

Consequently the left hand side of (12.45) becomes |L n ( f ) (z 0 ) − f (z 0 )| = |M f (z 0 ) − f (z 0 )| = | f (z 0 )| |M − 1| . So that (12.45) becomes an equality, and both sides equal | f (z 0 )| |M − 1| in the extreme case of  L n (· − z 0 ) (z 0 ) = 0. Thus inequality (12.45) is proved completely in both cases.  A similar result follows: Theorem 12.14 Here all as in Theorem 12.13. Then, for any f ∈ C (K , C), we have |(L n ( f )) (z 0 ) − f (z 0 )| ≤ | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 +

(12.52)

      1  λ  L n (1 (·)) (z 0 ) + 1 ω1 f,  L n · − z 0 2 (z 0 ) 2 , ∀ z 0 ∈ K , ∀ n ∈ N. If  L n (1 (·)) (z 0 ) = 1, ∀ z 0 ∈ K , then     1  |(L n ( f )) (z 0 ) − f (z 0 )| ≤ 2λω1 f,  L n · − z 0 2 (z 0 ) 2 , ∀ z 0 ∈ K , ∀ n ∈ N. Proof Let t, z 0 ∈ K and δ > 0. If t − z 0  > δ, then

(12.53)

12.3 Main Results

263

  | f (t) − f (z 0 )| ≤ ω1 ( f, t − z 0 ) = ω1 f, t − z 0  δ −1 δ ≤

(12.54)

    t − z 0 2 t − z 0  ω1 ( f, δ) . 1+ ω1 ( f, δ) ≤ 1 + δ δ2 The estimate

  t − z 0 2 | f (t) − f (z 0 )| ≤ 1 + ω1 ( f, δ) δ2

(12.55)

also holds trivially when t − z 0  ≤ δ. So (12.55) is true always, ∀ t ∈ K , for any z 0 ∈ K . As in the proof of Theorem 12.13 we have |(L n ( f )) (z 0 ) − f (z 0 )| ≤ · · · ≤ | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 (12.55)   +λ  L n (| f (·) − f (z 0 )|) (z 0 ) ≤ | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 +  λ  Ln



· − z 0 2 1 (·) + δ2





ω1 ( f, δ)

(z 0 ) =

'    1 | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 + λω1 ( f, δ)  L n (1 (·)) (z 0 ) + 2  L n · − z 0 2 (z 0 ) = δ | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 + λω1









1

f,  L n · − z 0 2 (z 0 )

2





(  L n (1 (·)) (z 0 ) + 1 ,

(12.56) by choosing

   1 δ :=  L n · − z 0 2 (z 0 ) 2 ,

(12.57)

  if  L n · − z 0 2 (z 0 ) > 0. Next we consider the case of    L n · − z 0 2 (z 0 ) = 0. By Riesz representation theorem there exists a positive finite measure μz0 such that  L n (g) (z 0 ) =

g (t) dμz0 (t) , ∀ g ∈ C (K , R) . K

That is

t − z 0 2 dμz0 (t) = 0, K

(12.58)

264

12 Quantitative Multivariate Complex Korovkin …

which implies t − z 0 2 = 0, a.e., hence t − z 0  = 0, a.e., thus t − z 0 = 0, a.e., and t = z 0 , a.e. on K . Consequently μz0 ({t ∈ K : t = z 0 }) = 0. L n (1 (·)) (z 0 )). Hence, in that That is μz0 = δz0 M (where 0 < M := μz0 (K ) =  case  L n (g) (z 0 ) = g (z 0 ) M.     1  L n · − z 0 2 (z 0 ) 2 = 0, and the right hand Consequently, it holds ω1 f,  side of (12.52) equals | f (z 0 )| |M − 1| . Also, it is  L n (| f (·) − f (z 0 ) (·)|) (z 0 ) = | f (z 0 ) − f (z 0 )| M = 0. And by (12.43) we obtain |(L n ( f (·) − f (z 0 ) (·))) (z 0 )| = 0, that is |L n ( f ) (z 0 ) − L n ( f (z 0 ) (·)) (z 0 )| = 0. The last says that (12.44)

L n ( f ) (z 0 ) = L n ( f (z 0 ) (·)) (z 0 ) =

f (z 0 )  L n (1 (·)) (z 0 ) = M f (z 0 ) .

Consequently the left hand side of (12.52) becomes |L n ( f ) (z 0 ) − f (z 0 )| = |M f (z 0 ) − f (z 0 )| = | f (z 0 )| |M − 1| . So that (12.52) becomes an equality, and both sides equal | f (z 0 )| |M − 1| in the   extreme case of  L n · − z 0 2 (z 0 ) = 0. Thus inequality (12.52) is proved completely in both cases.  We give Corollary 12.15 All as in Theorem 12.13, z 0 ∈ K . Then   L n ( f ) − f ∞ ≤  f ∞  L n (1 (·)) − 1∞ +

(12.59)

      L n (· − z 0 ) (z 0 )∞,z0 , λ  L n (1 (·)) + 1∞ ω1 f,  ∀ n ∈ N. If  L n (1 (·)) = 1, then     L n ( f ) − f ∞ ≤ 2λω1 f,  L n (· − z 0 ) (z 0 )∞,z0 ,

(12.60)

∀ n ∈ N. u u L n (· − z 0 ) (z 0 ) → 0 (u is uniformly), as n → ∞, then As  L n (1) → 1, and  u L n ( f ) → f , ∀ f ∈ C (K , C). Notice  L n (1) is bounded, and all suprema in (12.59) are finite.

12.3 Main Results

265

Corollary 12.16 All as in Theorem 12.14, z 0 ∈ K . Then   L n ( f ) − f ∞ ≤  f ∞  L n (1 (·)) − 1∞ +

(12.61)

      21   L n · − z 0 2 (z 0 )∞,z L n (1 (·)) + 1∞ ω1 f,  , λ  0 ∀ n ∈ N. If  L n (1 (·)) = 1, then     21   L n ( f ) − f ∞ ≤ 2λω1 f,  L n · − z 0 2 (z 0 )∞,z , 0

(12.62)

∀ n ∈ N.   u u u L n · − z 0 2 (z 0 ) → 0, then L n ( f ) → f , as n → ∞, ∀ As  L n (1) → 1, and  f ∈ C (K , C). We need   Theorem 12.17 Let a convex subset K ⊆ Ck , · , k ≥ 2, where · is an arbitrary norm, 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 ball B (x0 , δ) ⊂ K . Then | f (t) − f (x0 )| ≤

ω1 ( f, δ) t − x0  , ∀ t ∈ K . δ

(12.63)

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

(12.64)

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) .

(12.65)

That is ω1 (g, δ) ≤ ω1 ( f, δ) . The last (12.66) and (12.64) imply (12.63).

(12.66) 

We present a convex Korovkin type result: Theorem 12.18 Here all as in Theorem 12.13. Let a fixed z 0 ∈ K 0 and assume that | f (t) − f (z 0 )| is convex in t ∈ K . Assume the closed ball B z 0 ,  L n (· − z 0 ) (z 0 ) ⊂ K . Then

266

12 Quantitative Multivariate Complex Korovkin …

|(L n ( f )) (z 0 ) − f (z 0 )| ≤ | f (z 0 )|  L n (1 (·)) (z 0 ) − 1   + λω1 f,  L n (· − z 0 ) (z 0 ) , ∀ n ∈ N.

(12.67)

As  L n (1 (·)) (z 0 ) → 1, and  L n (· − z 0 ) (z 0 ) → 0, then (L n ( f )) (z 0 ) → f (z 0 ), as n → ∞. Proof As in the proof of Theorem 12.13 we have |(L n ( f )) (z 0 ) − f (z 0 )| ≤   (12.63) | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 + λ  L n (| f (·) − f (z 0 )|) (z 0 ) ≤ (δ > 0 : B (z 0 , δ) ⊂ K ) λω1 ( f, δ)  | f (z 0 )|  L n (1 (·)) (z 0 ) − 1 + L n (· − z 0 ) (z 0 ) = δ   | f (z 0 )|  L n (· − z 0 ) (z 0 ) , L n (1 (·)) (z 0 ) − 1 + λω1 f,  by choosing

(12.68)

δ :=  L n (· − z 0 ) (z 0 ) ,

if  L n (· − z 0 ) (z 0 ) > 0. The case  L n (· − z 0 ) (z 0 ) = 0 is treated similarly as in the proof of Theorem 12.13. The theorem is proved.  We make   Assumption 12.19 Let Q ⊆ Ck , ·1 be a compact convex subset, k ≥ 2, where ·1 is the l1 -norm. Here we consider f : Q → C that are coordinate-wise holomorphic functions and x0 ∈ Q, n ∈ N. Assume that f α (x0 ) = 0 for all α : |α| = j, j = 1, . . . , n. Let {L N } N ∈N be a sequence of linear operators from) C (Q, * C) into itself. There is a sequence of companion positive linear operators  L N N ∈N from C (Q, R) into itself, such that  L N (1) = 1, ∀ N ∈ N, and |L N (F)| ≤ λ L N (|F|) , λ > 0,

(12.69)

∀ F ∈ C (Q, C), ∀ N ∈ N, and L N (g) , ∀ g ∈ C (Q, R) , ∀ c ∈ C, ∀ N ∈ N. L N (cg) = c We give

(12.70)

12.3 Main Results

267

Theorem 12.20 All as in Assumption 12.19. Then   |(L N ( f )) (x0 ) − f (x0 )| ≤ λ max ω1 ( f α , h)

(12.71)

|α|=n

⎡ ⎣

   L N · − x0 n+1 (x0 ) 1 (n + 1)!h

⎤      L N · − x0 n1 (x0 ) h n−1  L N · − x0 1 + + (x0 )⎦ , 2n! 8 (n − 1)!

∀ N ∈ N, where h > 0. Proof We have that (12.70) |(L N ( f )) (x0 ) − f (x0 )| = (L N ( f )) (x0 ) − f (x0 )  L N (1) = |L N ( f ) (x0 ) − L N ( f (x0 )) (x0 )| = |L N ( f (·) − f (x0 )) (x0 )| λ L N (| f (·) − f (x0 )|) (x0 ) ⎡ ⎣

   L N · − x0 n+1 (x0 ) 1 (n + 1)!h

(12.41)

≤

(12.69)

≤

(12.72)

  λ max ω1 ( f α , h) |α|=n

⎤      L N · − x0 n1 (x0 ) h  L N · − x0 n−1 + + (x0 )⎦ , 1 2n! 8 (n − 1)!

h > 0, ∀ N ∈ N.



We present Theorem 12.21 All as in Assumption 12.19. Then |(L N ( f )) (x0 ) − f (x0 )| ≤  maxω1

|α|=n



'  λ 1 1 n + + n! (n + 1) 2 8

 1        n+1  n n+1   f α , L N · − x0 1 L N · − x0 n+1 (x0 ) (x0 ) n+1 , 1 

(12.73) ∀ N ∈ N, n ∈ N.  If  L N · − x0 n+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → +∞. 1   Proof (i) Assume that  L N · − x0 n+1 (x0 ) = 0. Then the right hand side of 1 (12.73) equals zero. By Riesz representation theorem there exists a probability measure μx0 such that  L n (g) (z 0 ) =

g (t) dμx0 (t) , ∀ g ∈ C (Q, R) . Q

That is

268

12 Quantitative Multivariate Complex Korovkin …

Q

t − x0 n+1 dμx0 (t) = 0, 1

(12.74)

= 0, a.e., hence t = x0 , a.e. on Q. Consequently which implies t − x0 n+1 1 μx0 ({t ∈ Q : t = x0 }) = 0. That is μx0 = δx0 (the Dirac measure). L N (| f (·) − Thus  L n (g) (x0 ) = g (x0 ), ∀ g ∈ C (Q, R) and therefore  f (x0 )|) (x0 ) = 0. By (12.69) we have |L N ( f (·) − f (x0 )) (x0 )| = 0, which implies |(L N ( f )) (x0 ) − f (x0 )| = 0.Therefore inequality (12.73) is valid trivially. (ii) Assume that  L N · − x0 n+1 (x0 ) > 0. Choose 1    1  h :=  L N · − x0 n+1 (x0 ) n+1 > 0, 1 then

(12.75)

  L N · − x0 n+1 h n+1 =  (x0 ) . 1

By Riesz representation theorem and Hölder’s inequality we obtain       n  L N · − x0 n+1 L N · − x0 n1 (x0 ) ≤  (x0 ) n+1 = h n , 1

(12.76)

      n−1  L N · − x0 n+1 L N · − x0 n−1 (x0 ) ≤  (x0 ) n+1 = h n−1 , 1 1

(12.77)

and

for n ∈ N. The right hand side of (12.71) is less equal to      1   n+1 λ max ω1 f α ,  L N · − x0 n+1 (x ) 0 1 |α|=n

'

 hn hn hn = + + 2n! 8 (n − 1)! (n + 1)! '

λ 

 max ω1

|α|=n

1 1 1 + + 2n! 8 (n − 1)! (n + 1)!

     1    n  n+1 n+1  L N · − x0 n+1 L N · − x0 n+1 fα ,  (x (x0 ) ) 0 1 1 =



 max ω1

|α|=n



'  λ 1 1 n + + n! (n + 1) 2 8

(12.78)

     1    n  n+1 n+1  fα ,  L N · − x0 n+1 L N · − x0 n+1 , (x (x0 ) ) 0 1 1

12.3 Main Results

269

∀ N ∈ N, proving the claim.



We give Corollary 12.22 As in Assumption 12.19 with n = 1. Then |(L N ( f )) (x0 ) − f (x0 )| ≤ 

 max ω1

i∈{1,...,k}

 1 ∂ f   , L N · − x0 21 (x0 ) 2 ∂ xi





9λ 8

  1  L N · − x0 21 (x0 ) 2 ,

(12.79)

∀ N ∈ N.  If  L N · − x0 21 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → +∞. 

Proof By Theorem 12.21. To go further we need

  Theorem 12.23 Let Q ⊆ Ck , ·1 be a compact convex subset, k ≥ 2, where ·1 is the l1 -norm. Here f : Q → C is a coordinate-wise holomorphic function and x0 ∈ Q 0 , n ∈ N. Assume that f α (x0 ) = 0 for all α : |α| = j, j = 1, . . . , n. For all α : |α| = n, suppose that | f α (z) − f α (x0 )| is convex in z ∈ Q. For h > 0 assume that the ball B (x0 , h) ⊂ Q. Then | f (z) − f (x0 )| ≤

maxω1 ( f α , h)

|α|=n

h

That is | f (·) − f (x0 )| ≤

z − x0 n+1 1 , ∀ z ∈ Q. (n + 1)!

maxω1 ( f α , h)

|α|=n

h

· − x0 n+1 1 , (n + 1)!

(12.80)

(12.81)

valid over Q. Proof We apply Theorem 12.8. Thus, we must estimate

1

Rn (z, 0) =



0

t1 0

 ... 0

tn−1



   gz(n) (tn ) − gz(n) (0) dtn . . . dt1 , z ∈ Q. (12.82)

Here when |α| = n, by assumption | f α (z) − f α (x0 )| is convex in z ∈ Q. Furthermore for h > 0, the ball B (x0 , h) ⊂ Q, where x0 ∈ Q 0 . By Theorem 12.17 we get | f α (z) − f α (x0 )| ≤ ∀ z ∈ Q.

ω1 ( f α , h) z − x0 1 ≤ h

maxω1 ( f α , h)

|α|=n

h

z − x0 1 , (12.83)

270

12 Quantitative Multivariate Complex Korovkin …

In particular, we have that | f α (x0 + t (z − x0 )) − f α (x0 )| ≤

maxω1 ( f α , h)

|α|=n

h

t z − x0 1 , all t ∈ [0, 1] . (12.84)

It follows, from (12.27) and (12.84), that ⎛

|Rn (z, 0)| ≤ 0

1

⎛

⎛

k  αi ⎜ t1 ⎜ tn−1 ⎜ n! |z i − x0i | ⎜ ⎜ ⎜ i=1 ...⎜ ⎜ ⎜ ⎝ 0 ⎝ 0 ⎝|α|=n α1 ! . . . αk !

maxω1 ( f α , h)

|α|=n

h

⎞

⎞

(12.85)

⎞

z − x0 1 tn ⎠ dtn ⎠ . . .⎠ dt1 =

maxω1 ( f α , h)

|α|=n

h

z − x0 n+1 1 . (n + 1)!

Therefore, it holds |Rn (z, 0)| ≤

maxω1 ( f α , h)

|α|=n

h

z − x0 n+1 1 , ∀ z ∈ Q, (n + 1)!

(12.86) 

proving the claim. We make

  Assumption 12.24 Let Q ⊆ Ck , ·1 be a compact convex subset, k ≥ 2, where ·1 is the l1 -norm. Here we consider f : Q → C to be coordinate-wise holomorphic functions and x0 ∈ Q 0 , n ∈ N. Assume that f α (x0 ) = 0 for all α : |α| = j, j = 1, . . . , n. For all α : |α| = n, suppose that | f α (z) − f α (x0 )| is convex in z ∈ Q. For h > 0 assume that the ball B (x0 , h) ⊂ Q. Let {L N } N ∈N be a sequence of linear operators from ) C*(Q, C) into itself. There is a sequence of companion positive linear operators  L N N ∈N from C (Q, R) into  itself, such that L N (1) = 1, ∀ N ∈ N, and |L N (F)| ≤ λ L N (|F|) , λ > 0,

(12.87)

∀ F ∈ C (Q, C), ∀ N ∈ N, and L N (g) , ∀ g ∈ C (Q, R) , ∀ c ∈ C, ∀ N ∈ N. L N (cg) = c

(12.88)

12.3 Main Results

271

We give Theorem 12.25 All as in Assumption 12.24. Then 

 λ maxω1 ( f α , h) |(L N ( f )) (x0 ) − f (x0 )| ≤

|α|=n

h (n + 1)!

   L N · − x0 n+1 (x0 ) , (12.89) 1

∀ N ∈ N. Proof We have that (12.88) |(L N ( f )) (x0 ) − f (x0 )| = (L N ( f )) (x0 ) − f (x0 )  L N (1) = (12.87)

|L N ( f ) (x0 ) − L N ( f (x0 )) (x0 )| = |L N ( f (·) − f (x0 )) (x0 )| ≤

(12.81)

λ L N (| f (·) − f (x0 )|) (x0 ) ≤

  λ maxω1 ( f α , h)

∀ N ∈ N.

|α|=n

h (n + 1)!

   L N · − x0 n+1 (x0 ) , 1 (12.90) 

We also give

  Theorem 12.26 All as in Assumption 12.24, with h =  L N · − x0 n+1 (x0 ), n ∈ 1 N. Then       λ |(L N ( f )) (x0 ) − f (x0 )| ≤ maxω1 f α ,  L N · − x0 n+1 , (x ) 0 1 (n + 1)! |α|=n (12.91) ∀ N ∈ N.  If  L N · − x0 n+1 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → +∞. 1   Proof If  L N · − x0 n+1 (x0 ) > 0, then (12.91) comes directly from (12.89). If 1    L N · − x0 n+1 = 0, then (12.91) is trivially true, by acting as in the proof of (x ) 0 1 Theorem 12.21 . It comes

  Corollary 12.27 All as in Assumption 12.24 with n = 1, h =  L N · − x0 21 (x0 ). Then     λ ∂f   2 |(L N ( f )) (x0 ) − f (x0 )| ≤ , max ω1 , L N · − x0 1 (x0 ) 2 i∈{1,...,k} ∂ xi (12.92) ∀ N ∈ N.  If  L N · − x0 21 (x0 ) → 0, then L N ( f ) (x0 ) → f (x0 ), as N → +∞. Proof By Theorem 12.26, for n = 1.



272

12 Quantitative Multivariate Complex Korovkin …

12.4 Applications √ Example 12.28 Let z j = x j + i y j ; x j , y j ∈ R, i = −1, j = 1, . . . , k ∈ N − {1}. Here z = (z 1 , . . . , z k ) ∈ K , which is a compact convex subset of Ck . Let f ∈ C (K , C), which is f (z) = u (x, y) + iv (x, y) = u + iv, where x = (x1 , . . . , xk ), y = (y1 , . . . , yk ). Consider  L : C (K , R) → C (K , R) a positive linear operator. Also consider L : C (K , C) → C (K , C) the linear operator such that:     L ( f ) (z) :=  L (u) (x, y) + i  L (v) (x, y) ,

(12.93)

as f ∈ C (K , C) iff u, v ∈ C (K , R) . Here u = u (x, y), v = v (x, y) ∈ R, and from |u| ≤ |u| ⇔ − |u| ≤ u ≤ |u| ⇔ L (|u|). − L (|u|) ≤  L (u) ≤  L (|u|) ⇔  L (u) ≤  We have that     |L ( f ) (z)| ≤  L (u) (x, y) +  L (v) (x, y) ≤ 

      L (|u|) (x, y) +  L (|v|) (x, y) =  L (|u| + |v|) (x, y) ≤

(12.94)

 √   √  √  2  L u 2 + v 2 (x, y) = 2 L (| f (z)|) = 2  L (| f |) (z) . We have proved that (1) |L ( f )| (z) ≤

√

2 L (| f |) (z) , ∀ z ∈ K .

(12.95)

Next, let g ∈ C (K , R), and c ∈ C, i.e. c = a + bi; a, b ∈ R. Then cg = ag + ibg. Clearly, L (cg) =  L (ag) + i  L (bg) = a  L (g) + ib L (g) = c L (g) . Thus, it is true (2)

L (cg) = c L (g) , ∀ c ∈ C and ∀ g ∈ C (K , R) .

(12.96)

So our main assumptions in this work are natural and possible.   Example 12.29 (a realization of Example 12.28) Let F ∈ C [0, 1]k , k ∈ N − {1}, and n 1 , . . . , n k ∈ N. The multivariate Bernstein polynomials ((x1 , . . . , xk ) ∈ [0, 1]k ) 

   nk  n1



 n1 nk Bn 1 ,...,n k (F) (x1 , . . . , xk ) := ... ··· ν1 νk ν1 =0

νk =0

12.4 Applications

273

 f

ν1 νk ,..., n1 nk



x ν1 (1 − x)n 1 −ν1 . . . xkνk (1 − xk )n k −νk

(12.97)

converge to F, uniformly, as n 1 , . . . , n k → ∞, see [9, p. 51]. These polynomials are positive linear operators. Notice that Bn 1 ,...,n k (1) = 1. Here z j = x j + i y j ; x j , y j ∈ [0, 1], j = 1, . . . , k ∈ N − {1}; where z = (z 1 , . . . , z k ) .   We take K = [0, 1]2k , which is a compact and convex subset of Ck , · , where · is an arbitrary norm.   Thus, for f ∈ C [0, 1]2k , C : f (z) = u (x, y) + iv (x, y) = u + iv, we define 

     k BnC1 ,...,n 2k ( f ) (z) := Bn 1 ,...,n 2k (u) (x, y) + i Bn 1 ,...,n 2k (v) (x, y) ,

(12.98)

the multivariate complex Bernstein polynomial operators; where x = (x1 , . . . , xk ) and y = (y1 , . . . , yk ) . By (12.95) we get that √ k C Bn 1 ,...,n 2k ( f ) ≤ 2Bn 1 ,...,n 2k (| f |) ,

(12.99)

true over [0, 1]2k , and by (12.96) we have that BnC1 ,...,n 2k (cg) = cBn 1 ,...,n 2k (g) , k

(12.100)

  ∀ c ∈ C and ∀ g ∈ C [0, 1]2k , R . So the main assumptions of our work are fulfilled. Notice that  k  C Bn 1 ,...,n 2k ( f ) (z) − f (z) =  Bn  Bn

1 ,...,n 2k

1 ,...,n 2k

   (u) (x, y) + i Bn 1 ,...,n 2k (v) (x, y) − u (x, y) − iv (x, y) =      (u) (x, y) − u (x, y) + i Bn 1 ,...,n 2k (v) (x, y) − v (x, y) =

+

 2   2 Bn 1 ,...,n 2k (u) (x, y) − u (x, y) + Bn 1 ,...,n 2k (v) (x, y) − v (x, y) =: (∗) . (12.101)   < ε1 , ∀ x, y ∈ [0, 1]k , ∀ n 1 , . . . , n 2k ≥ We have that Bn 1 ,...,n 2k (u) (x, y) − u (x, y)   N1 , and Bn 1 ,...,n 2k (v) (x, y) − v (x, y) < ε2 , ∀ x, y ∈ [0, 1]k , ∀ n 1 , . . . , n 2k ≥ N2 ; N1 , N2 ∈ N, where ε1 , ε2 > 0. Thus, it holds + (∗) ≤

ε12 + ε22 =: ε,

(12.102)

274

12 Quantitative Multivariate Complex Korovkin …

∀ x, y ∈ [0, 1]k , ∀ n 1 , . . . , n 2k ≥ max (N1 , N2 ) =: N ∗ , ε > 0. Hence  k  C Bn 1 ,...,n 2k ( f ) (z) − f (z) ≤ ε, ∀ z ∈ [0, 1]2k ,

(12.103)

∀ n 1 , . . . , n 2k ≥ N ∗ ∈ N, where ε > 0. k Therefore BnC1 ,...,n 2k ( f ) → f , converges uniformly as n 1 , . . . , n 2k → ∞.   Example 12.30 (continuation of Example 12.29) For f ∈ C [0, 1]2k , C , by (12.46), we have that  k  √     C Bn 1 ,...,n 2k ( f ) (z 0 ) − f (z 0 ) ≤ 2 2ω1 f, Bn 1 ,...,n 2k · − z 0 l1 (z 0 ) , (12.104) ∀ z 0 ∈ [0, 1]2k , ∀ n 1 , . . . , n 2k ∈ N, see also (12.99). Here z 0 = x0 + i y0 , where x0 = (x01 , . . . , x0k ), y0 = (y01 , . . . , y0k ). It is z 0 = (x0 , y0 ). We notice that: 

⎞⎞ ⎛ ⎛ k

  · − z 0 j ⎠⎠ (z 0 ) = Bn 1 ,...,n 2k · − z 0 l1 (z 0 ) = ⎝ Bn 1 ,...,n 2k ⎝ j=1

k



k    21

 2   Bn 1 ,...,n 2k · − z 0 j Bn 1 ,...,n 2k · − z 0 j (z 0 ) ≤ (z 0 ) =

j=1

j=1 k 

  21 2  2  Bn 1 ,...,n 2k · − x0 j + · − y0 j (x0 , y0 ) =

(12.105)

j=1 k ,

Bn 1 ,...,n 2k

- 21    2  2  · − x0 j (x0 , y0 ) + Bn 1 ,...,n 2k · − y0 j (x0 , y0 ) =

j=1 k ,

Bn j

  2     2   - 21 · − x0 j x0 j + Bn j+k · − y0 j y0 j

j=1

(where Bn j is the basic univariate real Bernstein polynomial on [0, 1]) %    &1 k

x0 j 1 − x0 j y0 j 1 − y0 j 2 = + ≤ nj n j+k j=1 2 k ' k

1 1 1

+ = 4n j 4n j+k 2 j=1 j=1 1

.

1 1 + nj n j+k

 .

(12.106)

12.4 Applications

275

We have found that 

k   1

Bn 1 ,...,n 2k · − z 0 l1 (z 0 ) ≤ 2 j=1

.

1 1 + nj n j+k

 .

(12.107)

By (12.104) and (12.107) we derive that ⎛ ⎞⎞ ⎛ . k  k 

√ 1 1 1 ⎠⎠ C , + Bn 1 ,...,n 2k ( f ) (z 0 ) − f (z 0 ) ≤ 2 2ω1 ⎝ f, ⎝ 2 j=1 n j n j+k   ∀ z 0 ∈ [0, 1]2k , ∀ n 1 , . . . , n 2k ∈ N, and ∀ f ∈ C [0, 1]2k , C . Clearly, it holds  k   C  Bn 1 ,...,n 2k ( f ) − f 

∞

⎛ ⎞⎞ ⎛ . k

√ 1 1 1 ⎠⎠ , ≤ 2 2ω1 ⎝ f, ⎝ + 2 j=1 n j n j+k

(12.108)

(12.109)

∀ n 1 , . . . , n 2k ∈ N. k We get BnC1 ,...,n 2k ( f ) → f , uniformly, as n 1 , . . . , n 2k → ∞.

References 1. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman Research Notes in Math, vol. 287. Longman Sci. & Tech., Harlow (1993) 2. Anastassiou, G.A.: Lattice homomorphism - Korovkin type inequalities for vector valued functions. Hokkaido Math. J. 26, 337–364 (1997) 3. Anastassiou, G.A.: Quantitative Approximations. Chapman & Hall/CRC, Boca Raton (2001) 4. Anastassiou, G.A.: Intelligent Computations: Abstract Fractional Calculus, Inequalities, Approximations. Springer, Heidelberg (2018) 5. Anastassiou, G.: Quantitative multivariate complex Korovkin theory. RACSAM 114 (2020). Article 59 6. Caratheodory, C.: Theory of Functions of a complex variable, vol. 2. Chelsea publishing Company, New York (1954) 7. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 8. Krantz, S.G.: Function Theory of Several Complex Variables, 2nd edn. AMS Chelsea publishing, Providence (2001) 9. Lorentz, G.G.: Bernstein Polynomials. Chelsea Publishing Company, New York (1986) 10. Mamedov, R.G.: On the order of the approximation of functions by linear positive operators. Dokl. Akad. Nauk USSR 128, 674–676 (1959) 11. Royden, H.L.: Real Analysis, 2nd edn. Macmillan, New York (1968) 12. Shilov, G.E.: Elementary Functional Analysis. Dover Publications Inc., New York (1996) 13. Shisha, O., Mond, B.: The degree of convergence of sequences of linear positive operators. Natl. Acad. Sci. 60, 1196–1200 (1968)

Chapter 13

M-Fractional Integral Type Inequalities

Here we present M-fractional integral inequalities of Ostrowski and Polya types. See also [5].

13.1 Introduction We are inspired by the following results: Theorem 13.1 ([2, p. 498], [1, 6] Ostrowski inequality) Let f ∈ C 1 ([a, b]), x ∈ [a, b]. Then    1  b − a

a

b

     (x − a)2 + (b − x)2    f  . f (z) dz − f (x) ≤ ∞ 2 (b − a)

(13.1)

Inequality (13.1) is sharp. In particular the optimal function is f ∗ (z) := |z − x|α (b − a) , α > 1.

(13.2)

Theorem 13.2 ([7], [8, p. 62], [9], [10, p. 83] Polya integral inequality) 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.

(13.3)

a

In this short work we present inequalities of types (13.1) and (13.3) involving the left and right fractional local general M-derivatives, see [3, 4].

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_13

277

278

13 M-Fractional Integral Type Inequalities

13.2 Background We need Definition 13.3 ([3]) Let f : [a, ∞) → R and t > a, a ∈ R. For 0 < α ≤ 1 we define the left local general M-derivative of order α of function f , denoted by α,β D M,a f (t), by α,β D M,a

   f tEβ ε (t − a)−α − f (t) f (t) := lim , ε→0 ε

∀ t > a, where Eβ (t) =

∞ k=0

tk , (βk+1)

(13.4)

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+

(13.5)

Theorem 13.4 ([3]) If a function f : [a, ∞) → R has the left local general Mderivative of order α ∈ (0, 1], β > 0, at t0 > a, then f is continuous at t0 . We need Theorem 13.5 ([3] Mean value theorem) Let f : [γ, δ] → R with γ > a, 0 ∈ / [γ, 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

(13.6)

We need Definition 13.6 ([4]) Let f : (−∞, b] → R and t < b, b ∈ R. For 0 < α ≤ 1 we define the right local general M-derivative of order α of function f , denoted as α,β M,b D f (t), by α,β M,b D f

∀ t < b.

(t) := −lim

ε→0

   f tEβ ε (b − t)−α − f (t) , ε

(13.7)

13.2 Background

279

α,β

α,β Df t→b− M,b

If M,b D f (t) exists over (γ, b), γ ∈ R and lim α,β M,b D f

(b) = lim

α,β

t→b− M,b

(t) exists, then

D f (t) .

(13.8)

Theorem 13.7 ([4]) If a function f : (−∞, b] → R has the right local general M-derivative of order α ∈ (0, 1], β > 0, at t0 < b, then f is continuous at t0 . We also need Theorem 13.8 ([4] Mean value theorem) Let f : [γ, δ] → R with δ < b, 0 ∈ / [γ, 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 α,β

(13.9)

α,β

Fractional derivatives D M,a and M,b D possess all basic properties of the ordinary derivatives and beyond, see [3, 4].

13.3 Main Results We present the following M-fractional Ostrowski type inequality: Theorem 13.9 Let a < γ < δ < b, 0 ∈ / [γ, δ], f : [a, b] → R, which is continuous α,β α,β over [γ, δ]. We assume that D M,a , M,b D exist and are continuous over [γ, x0 ] and [x0 , δ], respectively, where x0 ∈ [γ, δ], for some α ∈ (0, 1]. Then

⎡    α,β D M,a f (x)  ⎢   ⎣  x  

   1  δ − γ

∞,[γ,x0 ]

δ γ

   (β + 1) f (x) d x − f (x0 ) ≤ 2 (δ − γ)

   α,β   M,b D f (x)  α 2   (x0 − a) (x0 − γ) +   x  

⎤

⎥ (b − x0 )α (δ − x0 )2 ⎦ .

∞,[x0 ,δ ]

(13.10) Proof Let x ∈ [γ, x0 ], the by Theorem 13.5, there exists c1 ∈ (x, x0 ), such that  f (x0 ) − f (x) =

 α,β D M,a f (c1 )  (β + 1) (c1 − a)α (x0 − x) . c1

(13.11)

280

13 M-Fractional Integral Type Inequalities

Thus

 α,β  D   M,a f (c1 )  | f (x) − f (x0 )| =    (β + 1) (c1 − a)α |x − x0 | ≤   c1  α,β  D   M,a f (x)      x

 (β + 1) (x0 − a)α |x − x0 | ,

(13.12)

∞,[γ,x0 ]

∀ x ∈ [γ, x0 ] . Let now x ∈ [x0 , δ], then by Theorem 13.8, there exists c2 ∈ (x0 , x), such that  α,β

M,b D f

f (x) − f (x0 ) = − Thus

(c2 )

  (β + 1) (b − c2 )α (x − x0 ) .

c2

(13.13)

 α,β     M,b D f (c2 )  | f (x) − f (x0 )| =    (β + 1) (b − x0 )α |x − x0 | ≤   c2  α,β     M,b D f (x)      x

 (β + 1) (b − x0 )α |x − x0 | ,

(13.14)

∞,[x0 ,δ]

∀ x ∈ [x0 , δ] . We have that    δ  1  1   f d x − f (x) (x ) 0 = δ − γ δ − γ γ 1 δ−γ 1 δ−γ

 γ

x0



δ γ

   

δ

γ

  ( f (x) − f (x0 )) d x  ≤

| f (x) − f (x0 )| d x = 

| f (x) − f (x0 )| d x +

δ

(13.15) 

| f (x) − f (x0 )| d x

(by (13.12), (13.14))

≤

x0

⎡  α,β  1 ⎣  D M,a f (x)     δ−γ  x

 (β + 1) (x0 − a)

α

∞,[γ,x0 ]

  α,β    M,b D f (x)  +    x

∞,[x0 ,δ]

 (β + 1) (b − x0 )

α



δ x0

 γ

x0

(x0 − x) d x 

(x − x0 ) d x =

13.3 Main Results

281

⎡  α,β   (β + 1) ⎣  D M,a f (x)     2 (δ − γ)  x

(x0 − a)α (x0 − γ)2 +

(13.16)

∞,[γ,x0 ]

  α,β    M,b D f (x)      x

 α

(b − x0 ) (δ − x0 )

2

.

∞,[x0 ,δ]



The theorem is proved. Next we give two M-fractional Polya type inequalities: Theorem 13.10 All as in Theorem 13.9 and f (x0 ) = 0. Then    

γ

δ

   f (x) d x  ≤

δ

γ

⎡   α,β  D M,a f (x)  ⎢  ⎣   x  

| f (x)| d x ≤ 

 (β + 1) 2 ⎤



 α,β   M,b D f (x)  ⎥  (x0 − a)α (x0 − γ)2 +  (b − x0 )α (δ − x0 )2 ⎦ .   x   ∞,[γ,x0 ] ∞,[x0 ,δ ]

(13.17)

Proof Same as in the proof of Theorem 13.9, by setting f (x0 ) = 0.



) All as in Theorem 13.9 and Corollary 13.11 (to Theorem 13.10, case of x0 = γ+δ 2 

γ+δ = 0. Then f 2  δ  (β + 1) (δ − γ)2 | f (x)| d x ≤ 8 γ ⎡

   α,β  ⎢ D M,a f (x)   ⎢  ⎣ x  

  ∞, γ, γ+δ 2



γ+δ 2

 −a

α

   α,β   M,b D f (x)   +   x  

  ∞, γ+δ 2 ,δ



 b−

γ+δ 2

α

⎤ ⎥ ⎥. ⎦

(13.18) Proof Apply (13.17) for x0 =

γ+δ . 2



References 1. Anastassiou, G.A.: Ostrowski type inequalities. Proc. AMS 123, 3775–3781 (1995) 2. Anastassiou, G.A.: Quantitative Approximations. Chapmann & Hall / CRC, Boca Raton (2001) 3. Anastassiou, G.: On the left fractional local general M-derivative (2019). Submitted for publication 4. Anastassiou, G.: About the right fractional local general M-derivative. Analele Univ. Oradea, Fasc. Mate., Tom XXVII(1), 87–94 (2020)

282

13 M-Fractional Integral Type Inequalities

5. Anastassiou, G.: M-fractional integral inequalities. J. Comput. Anal. Appl. 29(6), 1153–1158 (2021) 6. Ostrowski, A.: Über die Absolutabweichung einer differtentiebaren Funktion von ihrem Integralmittelwert. Comment. Math. Helv. 10, 226–227 (1938) 7. Polya, G.: Ein mittelwertsatz für Funktionen mehrerer Veränderlichen. Tohoku Math. J. 19, 1–3 (1921) 8. Polya, G., Szegö, G.: Aufgaben und Lehrsätze aus der Analysis, vol. I. Springer, Berlin (1925). (German) 9. Polya, G., Szegö, G.: Problems and Theorems in Analysis. Classics in Mathematics, vol. I. Springer, Berlin (1972) 10. Polya, G., Szegö, G.: Problems and Theorems in Analysis, vol. I, Chinese edn. (1984)

Chapter 14

Principles of Stochastic Caputo Fractional Calculus with Fractional Approximation of Stochastic Processes

Here we consider and study very general stochastic positive linear operators induced by general positive linear operators that are acting on continuous functions. These are acting on the space of real fractionally differentiable stochastic processes. Under some very mild, general and natural assumptions on the stochastic processes we produce related fractional stochastic Shisha-Mond type inequalities of L q -type 1 ≤ q < ∞ and corresponding fractional stochastic Korovkin type theorems. These are regarding the stochastic q-mean fractional convergence of a sequence of stochastic positive linear operators to the stochastic unit operator for various cases. All convergences are produced with rates and are given via the fractional stochastic inequalities involving the stochastic modulus of continuity of the αth fractional derivatives of the engaged stochastic process, α > 0, α ∈ / N. The impressive fact is that the basic real Korovkin test functions assumptions are enough for the conclusions of our fractional stochastic Korovkin theory. We give applications to stochastic Bernstein operators. See also [10].

14.1 Introduction Motivation for this work comes from [2, 3, 15, 16]. This work continues our earlier work [5], now at the stochastic fractional level. First we lay down the foundations of Stochastic fractional calculus in the direct analytical sense, see Sect. 14.2, this is in the Caputo fractional direction. In the Sect. 14.3, about background, we talk about the q-mean (1 ≤ q < ∞) first modulus of continuity of a stochastic process and its upper bounds. There we describe completely our setting by introducing our stochastic positive linear operator M, see (14.15), which is based on the positive linear operator  L from C ([a, b]) into itself. The operator M is acting on a wide space of Caputo fractional differentiable real valued stochastic processes X . See there Assumptions 14.10, 14.13, 14.14. We first give the main pointwise fractional © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_14

283

284

14 Principles of Stochastic Caputo Fractional Calculus …

stochastic Shisha-Mond type inequalities [14], see Theorems 14.16, 14.17, and their several corollaries covering important special cases. We continue with fractional q-mean uniform Shisha-Mond type inequalities, see Theorems 14.23, 14.24, and their interesting corollaries. All this theory is regarding the fractional stochastic convergence of operators M to I (stochastic unit operator) given quantitatively with rates. An extensive application about the stochastic Bernstein operators follows in full details. Based on our Shisha-Mond type inequalities of our main Theorems 14.16, 14.17, 14.23, 14.24 we derive pointwise and uniform Stochastic Korovkin theorems [12] on stochastic processes, see Theorems 14.33– 14.36. The amazing fact here is, that basic conditions on operator  L regarding two simple real valued functions that are not stochastic, are able to enforce fractional stochastic convergence on all stochastic processes we are dealing with; see Concepts 14.9 and Assumptions 14.10, 14.13, 14.14.

14.2 Foundation of Stochastic Fractional Calculus Let t ∈ [a, b] ⊂ R, ω ∈ , where (, F, P) is a probability space. Here X (t, ω) stands for a stochastic process. Case of X (·, ω) being continuous on [a, b], ∀ ω ∈ . Then by Caratheodory Theorem 20.15, p.156, [1], we get that X (t, ω) is jointly measurable. Next we define the left and right respectively, Riemann–Liouville stochastic fractional integrals, where α > 0 is not an integer: α X (x, ω) = Ia+

1  (α)

α Ib− X (x, ω) =

1  (α)

and



x

(x − t)α−1 X (t, ω) dt,

(14.1)

(t − x)α−1 X (t, ω) dt,

(14.2)

a



b x

∀ x ∈ [a, b], ∀ ω ∈ , where  is the gamma function. α α In the following important cases we prove that Ia+ X , Ib− X are stochastic processes: (i) Assume that (, F, P) is a complete probability space, and that (x − t)α−1 X (t, ω) is an integrable function on [a, x] × , ∀ x ∈ [a, b], then by Fubini’s α X (x, ·) is an integrable function on , ∀ x ∈ [a, b]. theorem, [13], p. 269, Ia+ α−1 X (t, ω) is an integrable function on [x, b] × , ∀ x ∈ Similarly, if (t − x) α X (x, ·) is an integrable function on [a, b], then again by Fubini’s theorem Ib− α α X (x, ω) are stochastic processes. , ∀ x ∈ [a, b] . That is Ia+ X (x, ω) and Ib− (ii) Assume a general probability space (, F, P) and the Lebesgue measure spaces on [a, x], [x, b], ∀ x ∈ [a, b]. These are clearly σ-finite measure spaces. We assume that the jointly measurable stochastic process X (t, ω) ≥ 0 on [a, b] × ,

14.2 Foundation of Stochastic Fractional Calculus

285

hence (x − t)α−1 X (t, ω) ≥ 0 on [a, x] × , and (t − x)α−1 X (t, ω) ≥ 0 on [x, b] × , ∀ x ∈ [a, b], and both are jointly measurable. Then by Tonelli’s theα α X (x, ·) , Ib− X (x, ·) are measurable functions orem, [13, p. 270], we get that Ia+ α α on , ∀ x ∈ [a, b]. That is Ia+ X, Ib− X are stochastic processes. The above facts provide the foundation of stochastic fractional calculus in the direct analytical α α X, Ib− X are stochastic processes. sense. So it is not unusual to consider that Ia+ α (iii) Given that X (·, ω) is in L 1 ([a, b]) then Ia+ X (·, ω) ∈ L 1 ([a, b]), ∀ ω ∈ , see α X (·, ω) ∈ L 1 ([a, b]), ∀ ω ∈ , see [8, p. 334]. [11, p. 13], and Ib− α X (·, ω) ∈ C ([a, b]), when 0 < And given that X (·, ω) ∈ L ∞ ([a, b]), then Ia+ α α < 1, and Ia+ X (·, ω) ∈ AC ([a, b]) (absolutely continuous functions), when α ≥ 1, ∀ ω ∈ , see [6, p. 388]. Similarly, if X (·, ω) ∈ L ∞ ([a, b]), then α α X (·, ω) ∈ C ([a, b]), when 0 < α < 1, and Ib− X (·, ω) ∈ AC ([a, b]), when Ib− α ≥ 1, ∀ ω ∈ , see [9]. We need Definition 14.1 Let non-integer α > 0, n = α (· is the ceiling of the number), t ∈ [a, b] ⊂ R, ω ∈ , where (, F, P) is a general probability space. Here X (t, ω) stands for a stochastic process. Assume that X (·, ω) ∈ AC n ([a, b]) (spaces of functions X (·, ω) with X (n−1) (·, ω) ∈ AC ([a, b])), ∀ ω ∈ . We call stochastic left Caputo fractional derivative α D∗a X



1 (x, ω) =  (n − α)

x

(x − t)n−α−1 X (n) (t, ω) dt,

(14.3)

a

∀ x ∈ [a, b], ∀ ω ∈ . And, we call stochastic right Caputo fractional derivative α Db− X (x, ω) =

(−1)n  (n − α)



b

(z − x)n−α−1 X (n) (z, ω) dz,

x

∀ x ∈ [a, b], ∀ ω ∈ . Remark 14.2 (to Definition 14.1) We further assume here that  (n)   X (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Then, by (14.3), we have   α  D X (x, ω) ≤ ∗a ≤

M∗  (n − α)

1  (n − α)  a

x



x

  (x − t)n−α−1  X (n) (t, ω) dt

a

(x − t)n−α−1 dt =

M ∗ (x − a)n−α .  (n − α + 1)

(14.4)

286

14 Principles of Stochastic Caputo Fractional Calculus …

That is

n−α ∗   α  D X (x, ω) ≤ M (x − a) , ∀x ∈ [a, b] . ∗a  (n − α + 1)

(14.5)

Also, from (14.4) we get  α   D X (x, ω) ≤ b− M∗  (n − α) That is

1  (n − α) 

b



b

  (z − x)n−α−1  X (n) (z, ω) dz ≤

x

(z − x)n−α−1 dz =

x

M ∗ (b − x)n−α .  (n − α + 1)

n−α ∗  α   D X (x, ω) ≤ M (b − x) , ∀x ∈ [a, b] . b−  (n − α + 1)

(14.6)

α α By (14.1)–(14.4), it is not strange to assume that D∗a X , Db− X are stochastic processes.

14.3 Background We need Definition 14.3 We define the relative q-mean first modulus of continuity of stochastic process X (t, ω) by 1 (X, δ) L q ,[c,d] :=  sup





1 |X (x, ω) − X (y, ω)|q P (dω)

q

: x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ ,

(14.7) δ > 0, 1 ≤ q < ∞. Definition 14.4 Let 1 ≤ q < ∞. Let X (x, ω) be a stochastic process. We call X a q-mean uniformly continuous stochastic process over [a, b], iff ∀ ε > 0 ∃ δ > 0 : whenever |x − y| ≤ δ; x, y ∈ [a, b] implies that   U

|X (x, s) − X (y, s)|q P (ds) ≤ ε.

We denote it as X ∈ CR q ([a, b]) .

(14.8)

14.3 Background

287

It holds U

Proposition 14.5 ([5]) Let X ∈ CR q ([a, b]), then 1 (X, δ) L q ,[a,b] < ∞, any δ > 0. Also it holds Proposition 14.6 ([5]) Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R. Then following are true ([c, d] ⊆ [a, b]): (i) 1 (X, δ) L q ,[c,d] is nonnegative and nondecreasing in δ > 0, U (ii) lim1 (X, δ) L q ,[c,d] = 1 (X, 0) L q ,[c,d] = 0, iff X ∈ CR q ([c, d]) , δ↓0

(iii) 1 (X, δ1 + δ2 ) L q ,[c,d] ≤ 1 (X, δ1 ) L q ,[c,d] + 1 (X, δ2 ) L q ,[c,d] , δ1 , δ2 > 0, (iv) 1 (X, mδ) L q ,[c,d] ≤ m1 (X, δ) L q ,[c,d] , δ > 0, m ∈ N, (v) 1 (X, λδ) L q ,[c,d] ≤ λ 1 (X, δ) L q ,[c,d] ≤ (λ + 1) 1 (X, δ) L q ,[c,d] , λ > 0, δ > 0, (vi) 1 (X + Y, δ) L q ,[c,d] ≤ 1 (X, δ) L q ,[c,d] + 1 (Y, δ) L q ,[c,d] , δ > 0, U (vii) 1 (X, ·) L q ,[c,d] is continuous on R+ for X ∈ CR q ([c, d]) . We give Remark 14.7 (to Proposition 14.6) By Proposition 14.6(v) we get 1 (X, |x − y|) L q ,[c,d] ≤

|x − y| 1 (X, δ) L q ,[c,d] , δ

(14.9)

∀ x, y ∈ [c, d], any δ > 0. We give   Remark 14.8 (continuation of Remark 14.2) We assume again that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Let δ > 0, 1 ≤ q < ∞. Then 



α 1 D∗c X, δ L q ,[c,d]

 = sup



  α   D X (x, ω) − D α X (y, ω)q P (dω) ∗c ∗c

 q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤  sup 

    α  D X (x, ω) +  D α X (y, ω) q P (dω) ∗c ∗c

 q1

:

([5])

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ M∗ sup  (n − α + 1)

 

 q (x − c)n−α + (y − c)n−α P (dω)

(14.10)  q1

:

288

14 Principles of Stochastic Caputo Fractional Calculus …

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} =

 M∗ sup (x − c)n−α + (y − c)n−α :  (n − α + 1) x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ That is

 α X, δ L q ,[c,d] ≤ 1 D∗c

2M ∗ (d − c)n−α .  (n − α + 1)

2M ∗ (d − c)n−α ,  (n − α + 1)

(14.11)

2M ∗ (d − c)n−α ,  (n − α + 1)

(14.12)

where a ≤ c < d ≤ b. Similarly, it holds  α X, δ L q ,[c,d] ≤ 1 Dd− where a ≤ c < d ≤ b. Next let x0 ∈ [a, b], then  α sup 1 D∗x X, δ L q ,[x 0

x0 ∈[a,b]

=

0 ,b]

≤

2M ∗ sup (b − x0 )n−α  (n − α + 1) x0 ∈[a,b]

2M ∗ (b − a)n−α .  (n − α + 1)

(14.13)

Similarly, it holds  sup 1 Dxα0 − X, δ L q ,[a,x ] ≤ 0

x0 ∈[a,b]

=

2M ∗ sup (x0 − a)n−α  (n − α + 1) x0 ∈[a,b]

2M ∗ (b − a)n−α .  (n − α + 1)

(14.14)

We need Concepts 14.9 Let  L be a positive linear operator from C ([a, b]) into itself. Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R, where (, F, P) is a probability space. Here we assume that for non-integer α > 0, α = n, X (·, ω) ∈ AC n ([a, b]) with X (n) (·, ω) ∈ L ∞ ([a, b]), ∀ ω ∈ . Also we assume for each t ∈ [a, b] that X (k) (t, ·) is measurable for all k = α X (z, ω) is a stochastic process for 1, . . . , n − 1. Further we assume that D∗t α z ∈ [t, b], ω ∈ , and Dt− X (z, ω) is a stochastic process for z ∈ [a, t] , ω ∈ ; ∀ t ∈ [a, b]. Define

14.3 Background

289

M (X ) (t, ω) :=  L (X (·, ω)) (t) , ∀ω ∈ , ∀ t ∈ [a, b] ,

(14.15)

and assume that it is a random variable in ω. Clearly M is a positive linear operator on stochastic processes. We make Assumption 14.10 Let non-integer α > 0. α (i) For any t ∈ [a, b] we assume that D∗t X (z, ω) is continuous in z ∈ [t, b], uniformly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ > 0 : whenever |z 1 − z 2 | ≤ δ; α α z 1 , z 2 ∈ [t, b], then  D∗t X (z 1 , ω) − D∗t X (z 2 , ω) ≤ ε, ∀ ω ∈ . U α We denote this by D∗t X ∈ CR ([t, b]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [t, b] . α X (z, ω) is continuous in z ∈ [a, t], uni(ii) For any t ∈ [a, b] we assume that Dt− formly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ >0 : whenever |z 1 − z 2 | ≤ δ; α α z 1 , z 2 ∈ [a, t], then  Dt− X (z 1 , ω) − Dt− X (z 2 , ω) ≤ ε, ∀ ω ∈ . U α We denote this by Dt− X ∈ CR ([a, t]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [a, t] .

Remark 14.11 Assumption 14.10 implies: α α X (·, ω) ∈ C ([t, b]), ∀ ω ∈ , and D∗t X is q-mean uniformly continuous in (i) D∗t Uq α z ∈ [t, b], that is D∗t X ∈ CR ([t, b]), for any 1 ≤ q < ∞. α α X (·, ω) ∈ C ([a, t]), ∀ ω ∈ , and Dt− X is q-mean uniformly continuous (ii) Dt− U q α in z ∈ [a, t], that is Dt− X ∈ CR ([a, t]) , for any 1 ≤ q < ∞.

We need Definition 14.12 Denote by  (E X ) (t) :=



X (t, ω) P (dω) , ∀ t ∈ [a, b] ,

(14.16)

the expectation operator. We make Assumption 14.13 We assume that   q  E  X (k)  (t) < ∞, ∀ t ∈ [a, b] , q > 1, for all k = 0, 1, . . . , n − 1, n = α ; α > 0 non-integer. We make

(14.17)

290

14 Principles of Stochastic Caputo Fractional Calculus …

Assumption 14.14 We assume that   (k)  E  X  (t) < ∞, ∀ t ∈ [a, b] ,

(14.18)

for all k = 0, 1, . . . , n − 1, n = α ; α > 0 non-integer. We give Remark 14.15 By the Riesz representation theorem [13] we have that there exists μt unique, completed Borel measure on [a, b] with L (1) (t) > 0, m t := μt ([a, b]) = 

(14.19)

f (x) dμt (x) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(14.20)

such that  L ( f ) (t) =

 [a,b]

Consequently we have that  X (x, ω) dμt (x) , ∀ (t, ω) ∈ [a, b] × ,

M (X ) (t, ω) =

(14.21)

[a,b]

and X as in Concepts 14.9. Here χ[γ,δ] (s) stands for the characteristic function on [γ, δ] ⊆ [a, b] . Notice that (r > 0)    χ[t,b] (s) |s − t|r μt (ds) =  L |· − t|r χ[t,b] (·) (t) , (s − t)r μt (ds) = [t,b]

[a,b]

(14.22) and 

 (t − s)r μt (ds) =

[a,t]

 χ[a,t] (s) |s − t|r μt (ds) =  L |· − t|r χ[a,t] (·) (t) .

[a,b]

(14.23) Let now n = α , α ∈ / N, α > 0, k = 1, . . . , n − 1. Then by Hölder’s inequality we obtain      k  |x − t|k dμt (x) ≤ (x − t) dμt (x) ≤  [a,b]



[a,b]

|x − t|α+1 dμt (x)

k ( α+1 )

(μt ([a, b]))(

α+1−k α+1

).

[a,b]

Therefore it holds        L (· − t)k (t)∞,[a,b] ≤  L |· − t|k (t)∞,[a,b] ≤

(14.24)

14.3 Background

291 k   ( α+1−k ) ( α+1 )  α+1   , L |· − t|α+1 (t)∞,[a,b] L (1)∞,[a,b]

(14.25)

all k = 1, . . . , n − 1. Also, we observe that

and

C ([a, b])  |· − t|α+1 χ[a,t] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] ,

(14.26)

C ([a, b])  |· − t|α+1 χ[t,b] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] .

(14.27)

By positivity of  L we obtain        L |· − t|α+1 (t)∞,[a,b] < ∞, L |· − t|α+1 χ[a,t] (·) (t)∞,[a,b] ≤ 

(14.28)

 by  L |· − t|α+1 (t) being continuous in t ∈ [a, b], see p. 388 of [4], and        L |· − t|α+1 (t)∞,[a,b] . L |· − t|α+1 χ[t,b] (·) (t)∞,[a,b] ≤ 

(14.29)

Take a ≤ s ≤ t, then (t − s)α+1 ≤ (t − s)α+1 · 1 + 0, and for t ≤ s ≤ b, we get (s − t)α+1 ≤ 0 + (s − t)α+1 · 1. So we have |· − t|α+1 ≤ |· − t|α+1 χ[a,t] (·) + |· − t|α+1 χ[t,b] (·) , ∀ t ∈ [a, b] .

(14.30)

Thus, by positivity of  L we obtain        L |· − t|α+1 χ[a,t] (·) (t)∞,[a,b] L |· − t|α+1 (t)∞,[a,b] ≤     +  L |· − t|α+1 χ[t,b] (·) (t)∞,[a,b] .

(14.31)

 L (1) (t), and by In this work we denote by  L χ[a,t] (·) (t) := μt ([a, t]) ≤    L χ[t,b] (·) (t) := μt ([t, b]) ≤  L (1) (t) .

14.4 Main Results Next we present our first main result on the quantitative stochastic fractional approximation regarding stochastic processes: Theorem 14.16 Suppose Concepts 14.9, Assumptions 14.10 and 14.13. Here the non-integer α > 0 is such that α > q1 , where p, q > 1 such that 1p + q1 = 1. Then, ∀ t ∈ [a, b], we have:

292

14 Principles of Stochastic Caputo Fractional Calculus …

   1  1  L (1) (t) − 1 + E |M (X ) − X |q (t) q ≤ E |X |q (t) q  n−1  k=1

  (k) q q1 1  E  X  (t)   2p k   L (· − t) (t) + 1 α k!  (α) ( p (α − 1) + 1) p (q + 1) q(α+1) (14.32)   α 1p   q(α+1) q(α+1)   L χ[t,b] (·) (t) L |· − t| χ[t,b] (·) (t)    q1 1 α  L χ[t,b] (·) (t) α+1 (q + 1) (α+1) + 1  1

 1  q(α+1)  1 α  X, D∗t L |· − t|q(α+1) χ[t,b] (·) (t) (q + 1)





+ L q ,[t,b]

  1   α  L |· − t|q(α+1) χ[a,t] (·) (t) q(α+1) L χ[a,t] (·) (t) p     q1 1 α  L χ[a,t] (·) (t) α+1 (q + 1) (α+1) + 1  1

α X, Dt−



 1  q(α+1) 1  q(α+1) L |· − t| χ[a,t] (·) (t) (q + 1)

 . L q ,[a,t]

Proof Here non-integer α > 0, α = n, α > q1 , p, q > 1 : 1p + q1 = 1, X (·, ω) ∈ AC n ([a, b]), with X (n) (·, ω) ∈ L ∞ ([a, b]), ∀ ω ∈ . We get by left Caputo fractional Taylor’s formula that [11, p. 54] X (s, ω) = 1  (α)



s

t

n−1  X (k) (t, ω) (s − t)k + k! k=0

 α α X (z, ω) − D∗t X (t, ω) dz, (s − z)α−1 D∗t

(14.33)

for all t ≤ s ≤ b, ∀ ω ∈ . Also from [8, p. 341], using the right Caputo fractional Taylor formula we get X (s, ω) = 1  (α)



t s

n−1  X (k) (t, ω) (s − t)k + k! k=0

 α α X (z, ω) − Dt− X (t, ω) dz, (z − s)α−1 Dt−

(14.34)

14.4 Main Results

293

for all a ≤ s ≤ t, ∀ ω ∈ . Above notice that α α X (t, ω) = Dt− X (t, ω) = 0, D∗t

(14.35)

∀ ω ∈ , see [8, pp.358–359]. We assume that α X (s, ω) = 0, for s < t, D∗t and

α X (s, ω) = 0, for s > t, Dt−

∀ ω ∈ . Then working on the remainder related to (14.33) (t ≤ s ≤ b), we have 

s t



s

  α α X (z, ω) − D∗t X (t, ω) dz ≤ (s − z)α−1  D∗t

  α  D X (z, ω) − D α X (t, ω)q dz ∗t

t

∗t



s

t

s

 q1

∗t

(s − t)

∗t

 1p

∗t



qα−1 q 1

( p (α − 1) + 1) p

  α α X (z, ω) − D∗t X (t, ω) dz (s − z)α−1  D∗t

  α  D X (z, ω) − D α X (t, ω)q dz

t

 p (s − z)α−1 dz

=

(14.36)

t

∗t

Hence we have 

s

s

  α  D X (z, ω) − D α X (t, ω)q dz

t



 q1 

.

q ≤

(s − t)(qα−1) , ( p (α − 1) + 1)(q−1)

(14.37)

t ≤ s ≤ b. Applying again Hölder’s inequality we obtain: 1 :=   

t

b



s t

α−1

(s − z)

1  (α)

 q1  q   α  D X (z, ω) − D α X (t, ω) dz μt (ds) P (dω) ∗t ∗t 1

(μt ([t, b])) p ≤  (α)

294

14 Principles of Stochastic Caputo Fractional Calculus …

  

b



t

s t

  α α X (z, ω) − D∗t X (t, ω) dz (s − z)α−1  D∗t

q

 q1  μt (ds) P (dω) (14.38)

1

(14.37)

≤

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

 1      q b s   D α X (z, ω) − D α X (t, ω)q dz (s − t)(qα−1) μt (ds) P (dω) ∗t ∗t 

t

t

=: (∗) , q  α α X (z, ω) − D∗t X (t, ω) ≥ 0, is a real valued random variwhere ϕ (z, ω) :=  D∗t α X (z, ω), t ≤ z ≤ b, is continuous in z able for each z ∈ [t, b] (we assumed that D∗t and measurable in ω ). Hence ϕ in z and jointly measurable, see (z, ω) is continuous s [7, p. 353]. Hence λ (s, ω) := t ϕ (z, ω) dz (s − t)qα−1 is a real valued random variable in ω and continuous in s ∈ [t, b]. Again by [7, p. 353], λ (s, ω) is jointly measurable in (s, ω) . Therefore by applying twice Tonelli’s theorem, [13, p. 270], we get 1

(∗) = 

b  

t

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

1   s  q q  α α (qα−1)  D X (z, ω) − D X (t, ω) dz P (dω) (s − t) μt (ds) = ∗t ∗t t

1

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p 1      q b s   α  D X (z, ω) − D α X (t, ω)q P (dω) dz (s − t)(qα−1) μt (ds) . ∗t ∗t t



t

(14.39) We have proved that 1 :=   

t

b



s t

1  (α)

 q1  q   α α X (z, ω) − D∗t X (t, ω) dz μt (ds) P (dω) (s − z)α−1  D∗t (14.40) 1

≤

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

14.4 Main Results



b

   q1   α  D X (z, ω) − D α X (t, ω)q P (dω) dz (s − t)(qα−1) μt (ds) ∗t ∗t

 s 

t



t

295

(continuing estimation) 1

≤ 

b



t

s

q 1

t



α X, z D∗t

(let h 1 > 0)

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

−t



L q ,[t,b]

(qα−1)

dz (s − z)

μt (ds)

 q1

(by (14.9))

≤

 α 1 X, h 1 L q ,[t,b] (μt ([t, b])) p 1 D∗t 1

 (α) ( p (α − 1) + 1) p 

b

 s

t

t

z−t h1

q

 dz (s − t)

(qα−1)

μt (ds)

 q1

≤

(14.41)

 α 1 1 D∗t X, h 1 L q ,[t,b] (μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p 

b t

 q1  s    z−t q (qα−1) 1+ dz (s − t) μt (ds) ≤ h1 t



b

τ t

 q1  s    (z − t)q (qα−1) 1+ dz (s − t) μt (ds) = q h1 t

(where τ= )

(14.42)

 α 1 1 2 p 1 D∗t X, h 1 L q ,[t,b] (μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

 q1   s   1 qα q (qα−1) μt (ds) = (z − t) dz (s − t) (s − t) + q h1 t [t,b]

 τ

 q1   1 (s − t)q+1 qα (qα−1) μt (ds) = (s − t) (s − t) + q h 1 (q + 1) [t,b]

 τ

 τ

 (s − t)

qα

[t,b]

 q1  1 (s − t)q(α+1) + q = μt (ds) h 1 (q + 1)

(14.43)

296

14 Principles of Stochastic Caputo Fractional Calculus …

 τ

(s − t)qα μt (ds) + [t,b]

 τ



1 q h 1 (q + 1)

(s − t)q(α+1) μt (ds)

(s − t)q(α+1) μt (ds)

 q1

≤

[t,b] α  α+1

1

(μt ([t, b])) α+1 +

[t,b]

1 q h 1 (q + 1)

 (s − t)

q(α+1)

μt (ds)

 q1

=: (∗∗) .

(14.44)

[t,b]

We set and assume  h 1 :=

1 (q + 1)

I.e. q(α+1)

h1

=

 (s − t)q(α+1) μt (ds)

1  q(α+1)

> 0.

(14.45)

[t,b]

1 (q + 1)

 (s − t)q(α+1) μt (ds) > 0.

(14.46)

[t,b]

Therefore we get 1  1 α qα qα q = (∗∗) = τ (μt ([t, b])) α+1 h 1 (q + 1) α+1 + h 1   q1 1 α τ h α1 (μt ([t, b])) α+1 (q + 1) α+1 + 1 .

(14.47)

  q1 1 α 1 ≤ τ h α1 (μt ([t, b])) α+1 (q + 1) α+1 + 1 .

(14.48)

We have proved that

The last means that 1 :=   

t

b



s t

 q1  q   α α X (z, ω) − D∗t X (t, ω) dz μt (ds) P (dω) (s − z)α−1  D∗t 

1 p

≤ 2 1 

1  (α)

α X, D∗t



1 (q + 1)



1

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

 (s − t)

q(α+1)

L q ,[t,b]

[t,b]

1 (q + 1)

μt (ds)

 1  q(α+1)

 (s − t)q(α+1) μt (ds) [t,b]

α  q(α+1)

(14.49)

14.4 Main Results

297

  q1 1 α (μt ([t, b])) α+1 (q + 1) α+1 + 1 . Next we work on the remainder related to (14.34) (a ≤ s ≤ t), we have 

t

s



t s



  α α X (z, ω) − Dt− X (t, ω) dz ≤ (z − s)α−1  Dt−

  α  D X (z, ω) − D α X (t, ω)q dz t− t− t

s

 q1 

  α  D X (z, ω) − D α X (t, ω)q dz t− t−

t

(z − s) p(α−1) dz

 1p

=

s

 q1

(t − s)

qα−1 q 1

.

(14.50)

1

,

(14.51)

( p (α − 1) + 1) p

Hence we have  s



t s

t

  α α X (z, ω) − Dt− X (t, ω) dz ≤ (z − s)α−1  Dt−

  α  D X (z, ω) − D α X (t, ω)q dz t− t−

 q1

(t − s)

(qα−1) q

( p (α − 1) + 1) p

a ≤ s ≤ t. Applying again Hölder’s inequality we obtain: 2 :=   t  

a

t

α−1

(z − s)

s

1  (α)

  q1 q   α α  D X (z, ω) − D X (t, ω) dz μt (ds) P (dω) t− t− 1

≤   t  

a

t s

α−1

(z − s)

(μt ([a, t])) p  (α)

  α  D X (z, ω) − D α X (t, ω) dz t− t−

q

 μt (ds) P (dω)

 q1

(14.52) 1

(14.51)

≤

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

1    t  t   q   α  D X (z, ω) − D α X (t, ω)q dz (t − s)(qα−1) μt (ds) P (dω) t− t− 

a

s

298

14 Principles of Stochastic Caputo Fractional Calculus …

(as before) 1

=  t   

a

t s

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

  q1    α  D X (z, ω) − D α X (t, ω)q dz P (dω) (t − s)(qα−1) μt (ds) t− t− 1

=

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

  1  t  t  q   α  D X (z, ω) − D α X (t, ω)q P (dω) dz (t − s)(qα−1) μt (ds) t− t− a



s

(14.53) (μt ([a, t]))

≤

1 p 1

 (α) ( p (α − 1) + 1) p  q1  t  t  q α 1 Dt− X, t − z L q ,[a,t] dz (t − s)(qα−1) μt (ds) ≤ a

s

(let h 2 > 0)



1



α X, h 2 L q ,[a,t] 1 Dt−

 t  t a

s

t −z h2

q

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

 q1  dz (t − s)(qα−1) μt (ds) ≤

 α X, h 2 L q ,[a,t] 1 Dt−

1

(μt ([a, t])) p

(14.54)

1

 (α) ( p (α − 1) + 1) p

 t  t   q1   t−z q (qα−1) 1+ dz (t − s) μt (ds) ≤ h2 a s ρ

 q1    t  t  (t − z)q (qα−1) 1+ dz μ = − s) (ds) (t t q h2 a s

(where ρ := 2 )

1 p





α 1 Dt− X, h 2 L q ,[a,t]

1

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

14.4 Main Results

299



 ρ

(t − s)qα + [a,t]

1 q h2



t

 q1   (t − z)q dz (t − s)(qα−1) μt (ds) =

s

 q1   1 (t − s)q+1 qα (qα−1) μt (ds) = (t − s) (t − s) + q h 2 (q + 1) [a,t]

 ρ





ρ [a,t]

 ρ

(t − s)

qα

[a,t]

 q1  q(α+1) 1 − s) (t = μt (ds) (t − s)qα + q h 2 (q + 1)

1 μt (ds) + q h 2 (q + 1)

 ρ

(t − s)

q(α+1)

μt (ds)

 μt (ds)

(t − s)

q(α+1)

(14.55)  q1

≤

[a,t] α  α+1

1

(μt ([a, t])) α+1 +

[a,t]

1 q h 2 (q + 1)

 (t − s)

q(α+1)

μt (ds)

 q1

=: (∗ ∗ ∗) .

(14.56)

[a,t]

We set and assume  h 2 :=

1 (q + 1)

I.e. q(α+1) h2

 (t − s)

q(α+1)

μt (ds)

1  q(α+1)

> 0.

(14.57)

[a,t]

1 = + (q 1)

 (t − s)q(α+1) μt (ds) > 0.

(14.58)

[a,t]

Therefore we get 1  1 α qα qα q = (∗ ∗ ∗) = ρ (μt ([a, t])) α+1 h 2 (q + 1) α+1 + h 2

That is

  q1 1 α ρh α2 (μt ([a, t])) α+1 (q + 1) α+1 + 1 .

(14.59)

  q1 1 α 2 ≤ ρh α2 (μt ([a, t])) α+1 (q + 1) α+1 + 1 .

(14.60)

The last means that 2 :=

1  (α)

300

14 Principles of Stochastic Caputo Fractional Calculus …

  t  

a

t

s

  q1 q   α α X (z, ω) − Dt− X (t, ω) dz μt (ds) P (dω) (z − s)α−1  Dt− 

1 p

α X, Dt−

≤ 2 1 



1 (q + 1) 

1

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

 (t − s)

q(α+1)

μt (ds)

 1  q(α+1)

[a,t]

1 (q + 1)

L q ,[a,t]

 (t − s)

q(α+1)

μt (ds)

α  q(α+1)

(14.61)

[a,t]

  q1 1 α (μt ([a, t])) α+1 (q + 1) α+1 + 1 . Next we connect facts together. We have that M (X ) (t, ω) − X (t, ω)  L (1) (t) =  X (s, ω) μt (ds) − X (t, ω)  L (1) (t) = [a,b]



 X (s, ω) μt (ds) +

(t,b]

[a,t]



1  (α)



t

α−1

(z − s)

s



1  (α)



s

t

X (s, ω) μt (ds) − X (t, ω)  L (1) (t)

(by (14.33), (14.34))

=

 n−1  X (k) (t, ω) (s − t)k + k! [a,t] k=0 

α X Dt−

(z, ω) −

α Dt− X





(t, ω) dz μt (ds) +

(14.62)

 n−1  X (k) (t, ω) (s − t)k + k! (t,b] k=0

  α α X (z, ω) − D∗t X (t, ω) dz μt (ds) (s − z)α−1 D∗t −X (t, ω)  L (1) (t) =  n−1  X (k) (t, ω) (s − t)k μt (ds) + k! [a,t] k=0

1  (α)



 [a,t]

s

t

  α α X (z, ω) − Dt− X (t, ω) dz μt (ds) + (z − s)α−1 Dt−

14.4 Main Results

301

 n−1  X (k) (t, ω) (s − t)k μt (ds) + k! (t,b] k=0 1  (α)



 (t,b]

  α α X (z, ω) − D∗t X (t, ω) dz μt (ds) (s − z)α−1 D∗t

s

t

− X (t, ω)  L (1) (t) =

(14.63)

n−1  X (k) (t, ω)   L (· − t)k (t) + k! k=1





1  (α)

 (t,b]

α−1

(z − s)

s

[a,t]



t

s

t



α Dt− X

(z, ω) −

α Dt− X





(t, ω) dz μt (ds) +

   α α X (z, ω) − D∗t X (t, ω) dz μt (ds) . (s − z)α−1 D∗t

Furthermore we have    M (X ) (t, ω) − X (t, ω)  L (1) (t) ≤   n−1  (k)   X (t, ω)    L (· − t)k (t) + k! k=1 1  (α) 





s

[a,t]



[t,b]

t

s

   α α X (z, ω) − Dt− X (t, ω) dz μt (ds) + (z − s)α−1  Dt− α−1

(s − z)

t

  (14.64)   α α  D X (z, ω) − D X (t, ω) dz μt (ds) . ∗t ∗t

We have eventually that |M (X ) (t, ω) − X (t, ω)| ≤      M (X ) (t, ω) − X (t, ω)  L (1) (t) − 1 ≤ L (1) (t) + |X (t, ω)| 

(14.65)

 n−1  (k)     X (t, ω)    |X (t, ω)|  L (1) (t) − 1 + L (· − t)k (t) + k! k=1 1  (α)

 [a,t]



t s

   α α X (z, ω) − Dt− X (t, ω) dz μt (ds) + (z − s)α−1  Dt−

302

14 Principles of Stochastic Caputo Fractional Calculus …



 [t,b]

s t

    α α X (z, ω) − D∗t X (t, ω) dz μt (ds) , (s − z)α−1  D∗t

∀ t ∈ [a, b] . Hence it holds    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n−1  k=1

⎧ ⎨ 

1  (α) ⎩



 [a,t]

t

  (k) q q1  E  X  (t)    L (· − t)k (t) + k!

(z − s)



α−1 

s

α Dt− X

(z, ω) −

α Dt− X

q   q1   + (t, ω) dz μt (ds) P (dω)

(14.66)   

 s

[t,b]





α X (z, ω) − D α X (t, ω) dz (s − z)α−1  D∗t ∗t

t



⎫ 1 ⎬ q q μt (ds) P (dω) . ⎭

By (14.49) and (14.61) we obtain    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n−1  k=1



 1 p

2 1 

  (k) q q1  E  X  (t)    L (· − t)k (t) + k! 

1 α X, D∗t (q + 1) 

1

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

 (s − t)q(α+1) μt (ds)

 1  q(α+1)

[t,b]

L q ,[t,b]

1 (q + 1)

 (s − t)

q(α+1)

μt (ds)

α  q(α+1)

(14.67)

[t,b]

  q1  1 α α+1 α+1 + (μt ([t, b])) (q + 1) + 1 

 1 p

2 1 

α X, Dt−



1 (q + 1)





1

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

(t − s)

q(α+1)

μt (ds)

 1  q(α+1)

[a,t]

1 (q + 1)

L q ,[a,t]

 (t − s)

q(α+1)

[a,t]

μt (ds)

α  q(α+1)

14.4 Main Results

303

  q1  1 α . (μt ([a, t])) α+1 (q + 1) α+1 + 1 Clearly (14.67) is trivially valid when t = a or t = b, from what follows, see (14.69) and (14.73), just replace [t, b] by [a, b] or [a, t] by [a, b], respectively. Next assume that  (14.68) (s − t)q(α+1) μt (ds) = 0, [t,b]

then (s − t)q(α+1) = 0, a.e. on [t, b], thus s = t, a.e. on [t, b], i.e. μt {s ∈ [t, b] : s = t} = 0, hence μt (t, b] = 0. Therefore μt concentrates on [a, t]. In that case inequality (14.67) becomes    1  1  L (1) (t) − 1 + E |M (X ) − X |q (t) q ≤ E |X |q (t) q  n−1  k=1

 1 p

α X, Dt−

2 1 



  (k) q q1  E  X  (t)    L (· − t)k (t) + k!

1 (q + 1)

 (t − s)

q(α+1)

μt (ds)

 1  q(α+1)

[a,t]



1

(μt ([a, t])) p 1

 (α) ( p (α − 1) + 1) p

(14.69) L q ,[a,t]

1 (q + 1)

 (t − s)q(α+1) μt (ds)

α  q(α+1)

[a,t]

  q1 1 α (μt ([a, t])) α+1 (q + 1) α+1 + 1 . So far inequality (14.69) is valid when  (t − s)q(α+1) μt (ds) > 0. [a,t]

 In the case of [a,t] (t − s)q(α+1) μt (ds) = 0, we have (t − s)q(α+1) = 0, a.e. on [a, t], thus s = t, a.e. on [a, t], i.e. μt {s ∈ [a, t] : s = t} = 0, hence μt [a, t) = 0. Therefore μt = δt μt ([a, b]) = δt m t , where δt is the unit Dirac measure at t. Clearly then  L (· − t)k (t) = 0, all k = 1, . . . , n − 1. Then (14.69) collapses to    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 ,

(14.70)

equivalently, q    E |M (X ) − X |q (t) ≤ E |X |q (t)  L (1) (t) − 1 .

(14.71)

304

14 Principles of Stochastic Caputo Fractional Calculus …

 By (14.21) we have that  M (X ) (t, ω) = X (t, ω) L (1) (t). Therefore M (X ) (t, ω) − X (t, ω) = X (t, ω)  L (1) (t) − 1 . Hence, obviously, (14.71) holds as equality. Thus inequality (14.69) is valid trivially. Finally, let us go the other way around. Let us assume that  (t − s)q(α+1) μt (ds) = 0,

(14.72)

[a,t]

then (t − s)q(α+1) = 0, a.e. on [a, t], thus s = t, a.e. on [a, t], i.e. μt {s ∈ [a, t] : s = t} = 0, hence μt [a, t) = 0. Therefore μt concentrates on [t, b] . In that case inequality (14.67) becomes    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n−1  k=1

 1 p

2 1 

α X, D∗t

  (k) q q1  E  X  (t)    L (· − t)k (t) + k!



1 (q + 1) 

1

(μt ([t, b])) p 1

 (α) ( p (α − 1) + 1) p

 (s − t)

q(α+1)

μt (ds)

 1  q(α+1)

[t,b]

1 (q + 1)

L q ,[t,b]

 (s − t)q(α+1) μt (ds)

α  q(α+1)

(14.73)

[t,b]

  q1 1 α (μt ([t, b])) α+1 (q + 1) α+1 + 1 . So inequality (14.73) is valid when  (s − t)q(α+1) μt (ds) > 0. [t,b]

 In the case of [t,b] (s − t)q(α+1) μt (ds) = 0, we have (s − t)q(α+1) = 0, a.e. on [t, b], hence s = t, a.e. on [t, b]. Therefore μt {s ∈ [t, b] : s = t} = 0, thus μ t (t, b] = 0. Consequently μt = δt μt ([a, b]) = δt m t . Clearly then  L (· − t)k (t) = 0, all k = 1, . . . , n − 1. Therefore (14.73) collapses to    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 .

(14.74)

The last is an equality as earlier, see (14.71), etc. Thus inequality (14.73) is valid trivially. Inequality (14.67) has been proved in all possible cases. Inequality (14.32) is the equivalent of (14.67) via (14.20).

14.4 Main Results

305



Theorem 14.16 now is proved completely.

It follows our second main results, the L 1 -quantitative stochastic fractional approximation of stochastic processes, it is the q = 1 analog of Theorem 14.16. Inequality (14.75) is much simpler than (14.32). Theorem 14.17 Suppose Concepts 14.9, Assumptions 14.10 and 14.14. Here α > 0, α∈ / N, n = α . Then, ∀ t ∈ [a, b], we have   E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) − 1 +   n−1   (k)    E X (t)   1  L (· − t)k (t) + k!  + 1) (α k=1 

  1  L χ[a,t] (·) (t) α+1 +

   α 1  L |· − t|α+1 χ[a,t] (·) (t) α+1 (α + 1)

   1  α 1 Dt− X,  L |· − t|α+1 χ[a,t] (·) (t) α+1    1  L χ[t,b] (·) (t) α+1 +

L 1 ,[a,t]

+

   α 1  L |· − t|α+1 χ[t,b] (·) (t) α+1 (α + 1)

   1  α 1 D∗t X,  L |· − t|α+1 χ[t,b] (·) (t) α+1

(14.75)

 L 1 ,[t,b]

.

Proof We have that   E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 + n−1   (k)    E X (t)    L (· − t)k (t) + k! k=1

1  (α)

  t  

  

a

b  t

t

s

t

    α  α X (z, ω) − Dt− X (t, ω) dz μt (ds) P (dω) + (z − s)α−1  Dt−

(14.76)  s   α−1  α α  D∗t X (z, ω) − D∗t X (t, ω) dz μt (ds) P (dω) = (s − z) 



(by applying Tonelli’s theorem twice)   n−1   (k)      E X (t)      L (· − t)k (t) + (E |X |) (t) L (1) (t) − 1 + k! k=1

306

14 Principles of Stochastic Caputo Fractional Calculus …

1  (α)



b t

 t  t  a



s

     α  D X (z, ω) − D α X (t, ω) P (dω) (s − z)α−1 dz μt (ds) ≤ ∗t ∗t

 s  

t

     α  D X (z, ω) − D α X (t, ω) P (dω) (z − s)α−1 dz μt (ds) + t− t−

n−1   (k)      E X (t)    L (1) (t) − 1 + L (· − t)k (t) + (E |X |) (t)  k! k=1

1  (α) 

 t  a

b  t

t

s

t

s

(14.77)

   α 1 Dt− X, t − z L 1 ,[a,t] (z − s)α−1 dz μt (ds) +



α 1 D∗t X, z

−t

L 1 ,[t,b]

α−1

(s − z)

 dz μt (ds)

 ≤

(h 1 ,h 2 >0)

n−1   (k)      E X (t)    L (· − t)k (t) + L (1) (t) − 1 + (E |X |) (t)  k! k=1

  t  t      α 1 t−z 1+ 1 Dt− X, h 1 L 1 ,[a,t] (z − s)α−1 dz μt (ds) +  (α) h1 a s (14.78)  b  s     α z−t α−1 1 D∗t X, h 2 L 1 ,[t,b] 1+ dz μt (ds) = (s − z) h2 t t   n−1  % ( E | X (k) |)(t)   (set λ := (E |X |) (t)  L (1) (t) − 1 + L (· − t)k (t)) k! k=1

λ+  t  a

1   α 1 Dt− X, h 1 L 1 ,[a,t]  (α)

1 (t − s)α + α h1

 α 1 D∗t X, h 2 L 1 ,[t,b]



b



t

=λ+  t  a



t

  (t − z)2−1 (z − s)α−1 dz μt (ds) +

s

1 (s − t)α + α h2



s

  (s − z)α−1 (z − t)2−1 dz μt (ds)

t

1   α 1 Dt− X, h 1 L 1 ,[a,t]  (α)

  1  (2)  (α) (t − s)α α+1 μt (ds) + + (t − s) α h 1  (α + 2)

(14.79)

14.4 Main Results





307

α 1 D∗t X, h 2 L 1 ,[t,b]

λ+





b t

  1  (α)  (2) (s − t)α α+1 μt (ds) = + (s − t) α h 2  (α + 2)

  t     α 1 (t − s)α+1 1 Dt− μt (ds) + X, h 1 L 1 ,[a,t] (t − s)α +  (α + 1) h 1 (α + 1) a (14.80)   b  α+1   α − t) (s 1 D∗t X, h 2 L 1 ,[t,b] μt (ds) = (s − t)α + h 2 (α + 1) t λ+ 

t

  1 α 1 Dt− X, h 1 L 1 ,[a,t]  (α + 1) 1 h 1 (α + 1)

(t − s)α μt (ds) +

a

 α 1 D∗t X, h 2 L 1 ,[t,b]



b

(s − t)α μt (ds) +

t

(Choose



t

h1 =



t

 (t − s)α+1 μt (ds) +

a

1 h 2 (α + 1)

(t − s)α+1 μt (ds)



b

 (s − t)α+1 μt (ds) =: (ξ) .

t

1  α+1

(14.81) > 0,

a

and



b

h2 =

(s − t)

α+1

μt (ds)

1  α+1

> 0.

t



Then h α+1 1

t

=

(t − s)α+1 μt (ds) > 0,

a

and h α+1 = 2



b

(s − t)α+1 μt (ds) > 0.

t

) Hence it holds   1   α+1  t 1 α α+1 μt (ds) 1 Dt− X, (t − s) (ξ) = λ +  (α + 1) a  (μt ([a, t]))

 1 α+1

a

t

(t − s)

α+1

μt (ds)

α  α+1

L 1 ,[a,t]

 h α1 + + (α + 1)

308

14 Principles of Stochastic Caputo Fractional Calculus …

 α X, D∗t

1 



α+1

μt (ds)

(s − t)

1   α+1

(14.82)

t

 (μt ([t, b]))

b

b

1 α+1

L 1 ,[t,b]

(s − t)α+1 μt (ds)

α  α+1

h α2 + (α + 1)

t

 =

  1   α+1  t 1 α α+1 μt (ds) λ+ 1 Dt− X, (t − s)  (α + 1) a 

t

(t − s)α+1 μt (ds)

α   α+1 1 (μt ([a, t])) α+1 +

 1 + (α + 1)

a

 1 

b

α X, D∗t

α+1

(s − t)



b

α+1

μt (ds)

(s − t)

L 1 ,[a,t]

1   α+1

(14.83)

t

L 1 ,[t,b]

α   α+1 1 μt (ds) (μt ([t, b])) α+1 +

t

1 (α + 1)

 .

We have proved that   E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 + n−1   (k)    E X (t)    L (· − t)k (t) + k! k=1

1  (α + 1) 

t

α+1

(t − s)

μt (ds)

α  α+1

 1 (μt ([a, t])) α+1 +  α X, Dt−

1

a



b t

α+1

(s − t)

μt (ds)

α  α+1

 1

α X, D∗t

(t − s)

α+1



μt (ds)

1   α+1

a

 1 (μt ([t, b])) α+1 + 

t

1 (α + 1)



1 (α + 1) b

t

+ L 1 ,[a,t]

(14.84)



α+1

(s − t)

μt (ds)

1   α+1

 . L 1 ,[t,b]

The trivial cases of h 1 = 0 or h 2 = 0 or h 1 = h 2 = 0 are treated similarly as in the proof of Theorem 14.16.

14.4 Main Results

309



Theorem 14.17 is now proved. Next we give a series of corollaries:

Corollary 14.18 All as in Theorem 14.16. Further assume  L (1) = 1 and X (k) (t0 , ω) = 0 , ∀ ω ∈ , all k = 1, . . . , n − 1, for a fixed t0 ∈ [a, b]. Then   1 E |M (X ) − X |q (t0 ) q ≤

1

2p α

1

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1)

  1   α  L χ[t0 ,b] (·) (t0 ) p  L |· − t0 |q(α+1) χ[t0 ,b] (·) (t0 ) q(α+1)

(14.85)

   q1 1 α  L χ[t0 ,b] (·) (t0 ) α+1 (q + 1) (α+1) + 1  α X, D∗t 0

1



 1  q(α+1) 1  q(α+1) L |· − t0 | χ[t0 ,b] (·) (t0 ) (q + 1)

 + L q ,[t0 ,b]

  1   α  L |· − t0 |q(α+1) χ[a,t0 ] (·) (t0 ) q(α+1) L χ[a,t0 ] (·) (t0 ) p     q1 1 α  L χ[a,t0 ] (·) (t0 ) α+1 (q + 1) (α+1) + 1  1

Dtα0 − X,



1  L |· − t0 |q(α+1) χ[a,t0 ] (·) (t0 ) (q + 1)

 1  q(α+1)

 . L q ,[a,t0 ]

Corollary 14.19 All as in Theorem 14.16. Further assume that  L (1) = 1 and α < 1 (i.e n = 1). Then, ∀ t ∈ [a, b] , we have 1   E |M (X ) − X |q (t) q ≤

1

(14.86)

α

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1)

   q1 1 α  L χ[t,b] (·) (t) α+1 (q + 1) (α+1) + 1 1

α X, D∗t




2. Due to lack of space we omit this statement. Corollary 14.20 All as in Theorem 14.17. Further assume  L (1) = 1 and X (k) (t0 , ω) = 0 , ∀ ω ∈ , all k = 1, . . . , n − 1, for a fixed t0 ∈ [a, b]. Then E (|M (X ) − X |) (t0 ) ≤    1  L χ[a,t0 ] (·) (t0 ) α+1 +

1  (α + 1)

   α 1  L |· − t0 |α+1 χ[a,t0 ] (·) (t0 ) α+1 (α + 1)

   1  1 Dtα0 − X,  L |· − t0 |α+1 χ[a,t0 ] (·) (t0 ) α+1    1  L χ[t0 ,b] (·) (t0 ) α+1 + 1



α D∗t X, 0

L 1 ,[a,t0 ]

+

(14.87)

   α 1  L |· − t0 |α+1 χ[t0 ,b] (·) (t0 ) α+1 (α + 1)

  1   L |· − t0 |α+1 χ[t0 ,b] (·) (t0 ) α+1

 L 1 ,[t0 ,b]

.

Corollary 14.21 All as in Theorem 14.17. Further assume that  L (1) = 1, 0 < α < 1. Then, ∀ t ∈ [a, b] , we have E (|M (X ) − X |) (t) ≤ 

  1  L χ[a,t] (·) (t) α+1 +

1  (α + 1)

   α 1  L |· − t|α+1 χ[a,t] (·) (t) α+1 (α + 1)

   1  α 1 Dt− X,  L |· − t|α+1 χ[a,t] (·) (t) α+1    1  L χ[t,b] (·) (t) α+1 +

L 1 ,[a,t]

+

   α 1  L |· − t|α+1 χ[t,b] (·) (t) α+1 (α + 1)

(14.88)

14.4 Main Results

311

   1  α 1 D∗t X,  L |· − t|α+1 χ[t,b] (·) (t) α+1 The case α =

1 2

 .

L 1 ,[t,b]

follows:

Corollary 14.22 All as in Theorem 14.17. Further assume  L (1) = 1. Then, ∀ t ∈ [a, b] , we have 2 E (|M (X ) − X |) (t) ≤ √ π     13   2 2   3  L |· − t| 2 χ[a,t] (·) (t) L χ[a,t] (·) (t) 3 + 3  1

    23  3  2 Dt− X, L |· − t| χ[a,t] (·) (t) 1 2

L 1 ,[a,t]

+

    13   2 2   3  L χ[t,b] (·) (t) 3 + L |· − t| 2 χ[t,b] (·) (t) 3  1

    23  3 D∗t X,  L |· − t| 2 χ[t,b] (·) (t)

(14.89)



1 2

L 1 ,[t,b]

.

We give a uniform norm result (denote ·∞,[a,b] :=  f ∞ ). It is based on Theorem 14.16. Theorem 14.23 Suppose Concepts 14.9, Assumptions 14.10 and 14.13. Here the non-integer α > 0 is such that α > q1 , where p, q > 1 : 1p + q1 = 1. Additionally   assume that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Then       1  1   E |M (X ) − X |q  q ≤  E |X |q  q  L (1) − 1∞ + ∞ ∞ 1   1 n−1   (k) q  q     E X 2p ∞  L (· − t)k (t)∞ + 1 α k!  (α) ( p (α − 1) + 1) p (q + 1) q(α+1) k=1

  q1  1 α α+1   (α+1) L (1) ∞ (q + 1) +1  α  1    q(α+1)  L (1)∞ p  L |· − t|q(α+1) χ[t,b] (·) (t)∞  sup 1 t∈[a,b]



1    q(α+1) 1 α  D∗t L |· − t|q(α+1) χ[t,b] (·) (t)∞ X, (q + 1)





+ L q ,[t,b]

312

14 Principles of Stochastic Caputo Fractional Calculus …

 α  1    q(α+1)  L (1)∞ p  L |· − t|q(α+1) χ[a,t] (·) (t)∞  sup 1 t∈[a,b]

α Dt− X,



1    q(α+1) 1  L |· − t|q(α+1) χ[a,t] (·) (t)∞ (q + 1)





. L q ,[a,t]

(14.90) Based on Theorem 14.17 we give the following uniform estimate: Theorem 14.24 Suppose Concepts 14.9, Assumptions   14.10 and 14.14. Here α > 0, α∈ / N, n = α . Additionally assume that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Then   E (|M (X ) − X |)∞ ≤ E (|X |)∞  L (1) − 1∞ +   n−1   (k)     E X

∞

k!

k=1

   1 1 α+1  L (1)∞ + α+1     L (· − t)k (t)∞ +  (α + 1)

  α  α+1  L |· − t|α+1 χ[a,t] (·) (t)∞  1     α+1 α sup 1 Dt− X,  L |· − t|α+1 χ[a,t] (·) (t)∞

t∈[a,b]

L 1 ,[a,t]

+

α    α+1  L |· − t|α+1 χ[t,b] (·) (t)∞

sup 1 t∈[a,b]



1     α+1 |· − t|α+1 χ[t,b] (·) (t)∞

α D∗t X,  L

 L 1 ,[t,b]

.

(14.91)

Corollary 14.25 (to Theorem 14.24) All as in Theorem 14.24. Case of 0 < α < 1 and  L (1) = 1. Then E (|M (X ) − X |)∞ ≤

α+2  (α + 2)

  α  α+1  L |· − t|α+1 χ[a,t] (·) (t)∞  1     α+1 α sup 1 Dt− X,  L |· − t|α+1 χ[a,t] (·) (t)∞

t∈[a,b]

α    α+1  L |· − t|α+1 χ[t,b] (·) (t)∞

L 1 ,[a,t]

+

14.4 Main Results

sup 1

313



t∈[a,b]

 

α D∗t X,  L

α+1

|· − t|

1   α+1 χ[t,b] (·) (t)∞

 L 1 ,[t,b]

.

(14.92)

Corollary 14.26 (to Theorem 14.24, α = 21 ) All as in Theorem 14.24. Assume  L (1) = 1. Then 10 E (|M (X ) − X |)∞ ≤ √ 3 π     13 3    L |· − t| 2 χ[a,t] (·) (t)

∞

 sup 1 t∈[a,b]

    23  3  2 |· − t| χ[a,t] (·) (t)

 α L Dt− X, 

∞

L 1 ,[a,t]

+

    13 3    L |· − t| 2 χ[t,b] (·) (t)

(14.93)

∞

 sup 1 t∈[a,b]

    23  3   α 2 D∗t X,  L |· − t| χ[t,b] (·) (t) ∞

 L 1 ,[t,b]

.

14.5 Application Let f ∈ C ([0, 1]) and the Bernstein polynomials B N ( f ) (t) :=

N  k=0

 f

k N



N k



t k (1 − t) N −k ,

∀ t ∈ [0, 1], ∀ N ∈ N. We have that B N 1 = 1 and B N is a positive linear operator. We have that  t (1 − t) , ∀ t ∈ [0, 1] , B N (· − t)2 (t) = N and

  1 1  B N (· − t)2 (t) 2 ≤ √ , ∀ N ∈ N. ∞ 2 N

L in Concepts 14.9. So B N is like  Define the corresponding application of M by

(14.94)

(14.95)

(14.96)

314

14 Principles of Stochastic Caputo Fractional Calculus …

 B N (X ) (t, ω) := B N (X (·, ω)) (t) =

N 

 X

k=0

k ,ω N



N k



t k (1 − t) N −k , (14.97)

∀ t ∈ [0, 1], ∀ ω ∈ , N ∈ N, where X is a stochastic process. Clearly  B N is a stochastic process. We give Proposition 14.27 Let X (t, ω) be a stochastic process from [0, 1] × (, F, P) into R, where (, F, P) is a probability space. Here 0 < α < 1 (i.e. n = 1) and X (·, ω) ∈ AC ([0, 1]) with X (1) (·, ω) ∈ L ∞ ([0, 1]), ∀ ω ∈ . Further we assume α α X (z, ω) is a stochastic process for z ∈ [t, 1], ω ∈ , and Dt− X (z, ω) is a that D∗t stochastic process for z ∈ [0, t], ω ∈ ; ∀ t ∈ [0, 1] . For any t ∈ [0, 1] we assume α X (z, ω) is continuous in z ∈ [t, 1], uniformly with respect to ω ∈ . And that D∗t α X (z, ω) is continuous in z ∈ [0, t], uniformly for any t ∈ [0, 1] we assume that Dt− with respect to ω ∈ . Finally, we assume that (E |X |) (t) < ∞, ∀ t ∈ [0, 1]. Then, for any t ∈ [0, 1] , we have:   E  B N (X ) − X  (t) ≤  

 1 B N χ[0,t] (·) (t) α+1 +

1  (α + 1)

   α 1 B N |· − t|α+1 χ[0,t] (·) (t) α+1 (α + 1)

   1  α 1 Dt− X, B N |· − t|α+1 χ[0,t] (·) (t) α+1  

 1 B N χ[t,1] (·) (t) α+1 +

L 1 ,[0,t]

+

   α 1 B N |· − t|α+1 χ[t,1] (·) (t) α+1 (α + 1)

   1  α 1 D∗t X, B N |· − t|α+1 χ[t,1] (·) (t) α+1

(14.98)

 L 1 ,[t,1]

. 

Proof By Corollary 14.21. We give

Proposition 14.28 All as in Proposition 14.27 with α = 21 . Then, for any t ∈ [0, 1], we have:   2 E  B N (X ) − X  (t) ≤ √ π  

   13  23 2   3 B N |· − t| 2 χ[0,t] (·) (t) B N χ[0,t] (·) (t) + 3    23    1 3 2 1 Dt− X, B N |· − t| 2 χ[0,t] (·) (t)

L 1 ,[0,t]

+

(14.99)

14.5 Application

 

315

   13  2 2   3 B N χ[t,1] (·) (t) 3 + B N |· − t| 2 χ[t,1] (·) (t) 3  1

   23  3 D∗t X, B N |· − t| 2 χ[t,1] (·) (t)





1 2

.

L 1 ,[t,1]



Proof By Corollary 14.22. We give Proposition 14.29 All as in Proposition 14.27, α = 21 . Then   10 E  B N (X ) − X  (t) ≤ √ 3 π 



    13   23    1 3 3 2 B N |· − t| 2 (t) 1 Dt− X, B N |· − t| 2 (t) 



B N |· − t|

3 2



(t)

 13

 1

L 1 ,[0,t]

   23  3 D∗t X, B N |· − t| 2 (t) 1 2

+

(14.100) 



L 1 ,[t,1]

,

∀ t ∈ [0, 1], ∀ N ∈ N. Proof By (14.99) and the positivity of B N .



We further have Proposition 14.30 All as in Proposition 14.27, α = 21 . Then   10 E  B N (X ) − X  (t) ≤ √ √ 3 2π 4 N  1



1 Dt− X, √ 2 N 1 2



 L 1 ,[0,t]

+ 1

1 D∗t X, √ 2 N 1 2



 L 1 ,[t,1]

,

(14.101)

∀ t ∈ [0, 1], ∀ N ∈ N. Proof By discrete Hölder’s inequality we get that     3 3 B N |· − t| 2 (t) ≤ B N (· − t)2 (t) 4 

 43 1 1 t (1 − t) ≤ 3 , N (4N ) 4

(14.95)

=

(14.102)

316

14 Principles of Stochastic Caputo Fractional Calculus …

∀ t ∈ [0, 1], ∀ N ∈ N. Hence, the right hand side of (14.100) ≤ 

10 √ 3 π



1 1

(4N ) 4

10 √ √ 3 2π 4 N

1 Dt− X,





1



1

1 2

+

1

(4N ) 2

1 Dt− X, √ 2 N 1 2

L 1 ,[0,t]

1 D∗t X,

1

 + 1

1 D∗t X, √ 2 N 1 2



1

1 2

(4N ) 4

 L 1 ,[0,t]



1

 =

1

(4N ) 2

L 1 ,[t,1]



 L 1 ,[t,1]

, (14.103)

∀ t ∈ [0, 1], ∀ N ∈ N.



Consequently we obtain 1 Proposition  14.31∗ All as in Proposition 14.27, α ∗= 2 . Additionally assume that (1)  X (t, ω) ≤ M , ∀ (t, ω) ∈ [0, 1] × , where M > 0. Then

     E  B N (X ) − X  

∞



 sup 1

t∈[0,1]

1 Dt− X, √ 2 N 1 2

10 ≤ √ √ 3 2π 4 N



 L 1 ,[0,t]

+ sup 1 t∈[0,1]

1 D∗t X, √ 2 N 1 2



 L 1 ,[t,1]

< ∞, (14.104)

∀ N ∈ N. As N → +∞, we obtain      E  B N (X ) − X  

∞

That is  BN

“1-mean”

→

→ 0.

(14.105)

I (unit operator) with rates.

Proof By (14.101) and (14.13) with (14.14).



Remark 14.32 (to Proposition 14.31) Assume that  1

1 Dt− X, √ 2 N 1 2

 L 1 ,[0,t]

K1 ≤ √ , ∀ t ∈ [0, 1] , ∀ N ∈ N, 2 N

where K 1 > 0, and assume that   1 1 K2 ≤ √ , ∀ t ∈ [0, 1] , ∀ N ∈ N, 1 D∗t2 X, √ 2 N L 1 ,[t,1] 2 N where K 2 > 0.

(14.106)

(14.107)

14.5 Application

317

Conditions (14.106), (14.107) are of Lipschitz type of order 1. Then, by (14.104), (14.106) and (14.107), we get:      E  B N (X ) − X  

∞

=

5 1 1 ≤ √ (K 1 + K 2 ) √ √ 4 N N 3 2π

5 (K 1 + K 2 ) 1 √ 3 , ∀ N ∈ N. 3 2π N4

(14.108)

Thus, at smoothness of order 21 we achieve speed of convergence

1 3

N4

, N ∈ N. Without

any smoothness, in [5], we proved that the speed of convergence was √1N , N ∈ N. In the presence of the ordinary first derivative, see [5], in deterministic approximation the rate of convergence was N1 , N ∈ N. Naturally, we have as expected: 1 1 1 < 3 < √ , ∀ N ∈ N − {1}. N N N4

(14.109)

14.6 Stochastic Korovkin Results We give at first pointwise results. Theorem 14.33 Assume all as in Theorem 14.16. Here L, M are meant as sequences q(α+1)   |· of operators. Assume further that L → 1 and L − t| (1) (t) (t) → 0, point  wise in t ∈ [a, b]. Then E |M (X ) − X |q (t) → 0, pointwise in t ∈ [a, b], that is M → I (stochastic unit operator) in q-mean-pointwise with rates, quantitatively. Proof From now on R.H.S denotes the “right hand side”. By Hölder’s inequality we get that   1  R.H.S. (14.32) ≤ E |X |q (t) q  L (1) (t) − 1 + n−1  k=1

  (k) q q1 k  q(α+1)−k E  X  (t)    L |· − t|q(α+1) (t) q(α+1)  L (1) (t) q(α+1) k! 1

+

2p 1

 

α

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1)  q1 1 α  L (1) (t) α+1 (q + 1) (α+1) + 1

1   α  L (1) (t) p  L |· − t|q(α+1) (t) q(α+1)

(14.110)

318

14 Principles of Stochastic Caputo Fractional Calculus …



 α X, D∗t

1  1

α X, Dt−





 1  q(α+1) 1  q(α+1) L |· − t| (t) (q + 1)

 1  q(α+1) 1  q(α+1) L |· − t| (t) (q + 1)

+ L q ,[t,b]

 . L q ,[a,t]

Because  L (1) (t) converges  pointwise in t to 1 it is bounded. Assuming also that  L |· − t|q(α+1) (t) converges pointwise in t to zero, we get that R.H.S. (14.110) converges to zero pointwise in t, any t ∈ [a, b] .   1 Therefore E |M (X ) − X |q (t) q (by (14.32)) converges to zero pointwise in t, any t ∈ [a, b] .  We continue with Theorem 14.34 Assume all as in Theorem 14.17. Here  L, M are meant as sequences of operators. Assume further that  L (1) (t) → 1 and  L |· − t|(α+1) (t) → 0, pointwise in t ∈ [a, b]. Then E (|M (X ) − X |) (t) → 0, pointwise in t ∈ [a, b], that is M → I in 1-mean-pointwise with rates, quantitatively. Proof We observe that   (14.24) R.H.S. (14.75) ≤ (E |X |) (t)  L (1) (t) − 1 + n−1   (k)   k  α+1−k E X (t)    L |· − t|α+1 (t) α+1  L (1) (t) α+1 k! k=1

(14.111)

  1  α+1 1 1  L (1) (t) + +  (α + 1) α+1   α α+1 1      α+1 α  1 Dt− X,  L |· − t|α+1 (t) α+1 L |· − t| (t)  1    α 1 D∗t X,  L |· − t|α+1 (t) α+1

L 1 ,[a,t]

+

 L 1 ,[t,b]

.

Because  L (1) (t) converges in t to 1 it is bounded.  pointwise Assuming also that  L |· − t|α+1 (t) converges pointwise in t to zero, we get that R.H.S. (14.111) converges to zero pointwise in t, any t ∈ [a, b] . Therefore E (|M (X ) − X |) (t) (by (14.75)) converges to zero pointwise in t, any t ∈ [a, b] .  We continue with uniform convergence results.

14.6 Stochastic Korovkin Results

319

Theorem 14.35 All as in Theorem 14.23. Here  L, M are as sequences  meant q(α+1)  of L |· − t| operators. Assume further that  L (1) → 1 uniformly, and  (t)∞ →    “q-mean” q 0. Then  E |M (X ) − X | ∞ → 0 over [a, b] , that is M → I, the stochastic unit operator with rates, quantitatively. Proof We have that     q1   R.H.S. (14.90) ≤  E |X |q ∞ L (1) − 1∞ + 1   n−1   (k) q  q    k   q(α+1)−k  q(α+1) E X ∞   L (1)∞q(α+1) + L |· − t|q(α+1) (t)∞ k! k=1

1

2p 1 p

 (α) ( p (α − 1) + 1) (q + 1)

α q(α+1)

  q1  1 α α+1  L (1)∞ (q + 1) (α+1) + 1

(14.112)

α  1    q(α+1) p   L |· − t|q(α+1) (t)∞ L (1)∞



  α X, sup 1 D∗t

t∈[a,b]

 sup 1 t∈[a,b]

α Dt− X,



1    q(α+1) 1  L |· − t|q(α+1) (t)∞ (q + 1)

1    q(α+1) 1  L |· − t|q(α+1) (t)∞ (q + 1)

 + L q ,[t,b]





. L q ,[a,t]

 Because L (1) →  1 uniformly, it is bounded. Assuming also that   q(α+1)  L |· − t| (t)∞ → 0, we get that R.H.S. (14.112) converges uniformly to zero over [a, b] .    q1 Therefore  E |M (X ) − X |q ∞ (by (14.90)) converges to zero over [a, b] .  We also present: Theorem 14.36 All as in Theorem 14.24. Here  L, M are meant  as sequences  of operators. Assume further that  L (1) → 1 uniformly, and  L |· − t|(α+1) (t)∞ → 0. Then E (|M (X ) − X |)∞ → 0 over [a, b] , that is M titatively.

“1 -mean”

→

I with rates quan-

320

14 Principles of Stochastic Caputo Fractional Calculus …

Proof We have that   L (1) − 1∞ + R.H.S. (14.91) ≤ E (|X |)∞    n−1   (k)     E X k!

k=1

∞

k   α+1−k    α+1   L (1)∞α+1 + L |· − t|α+1 (t)∞

  α   1    α+1 1 1 α+1   + L (1)∞ L |· − t|α+1 (t)∞  (α + 1) α+1 

 1     α+1 α X,  L |· − t|α+1 (t)∞ sup 1 Dt−

t∈[a,b]

sup 1 t∈[a,b]



1     α+1 α D∗t X,  L |· − t|α+1 (t)∞

L 1 ,[a,t]

(14.113)

+

 L 1 ,[t,b]

.

   Because  L (1) → 1 uniformly, it is bounded. Assuming also  L |· − t|α+1 (t)∞ → 0, we get that R.H.S. (14.113) converges uniformly to zero over [a, b] . Therefore E (|M (X ) − X |)∞ (by (14.91)) converges to zero over [a, b] .  We finish with Remark 14.37 An amazing fact observation follows: In all convergence results here, see Theorems 14.33–14.36, the forcing conditions for convergences are based only on  L and basic real valued continuous functions on [a, b] and are not related to stochastic processes, but they are giving convergence results on stochastic processes!

References 1. Aliprantis, C., Burkinshaw, O.: Principles of Real Analysis, 3rd edn. Academic, San Diego (1998) 2. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 3. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 4. Anastassiou, G.: Quantitative Approximations. Chapman & Hall/CRC, Boca Raton (2001) 5. Anastassiou, G.: Stochastic Korovkin theory given quantitatively. Facta Univ. (Nis) Ser. Math. Inform. 22(1), 43–60 (2007) 6. Anastassiou, G.: Fractional Differentiation Inequalities. Springer, Heildelberg (2009) 7. Anastassiou, G.A.: Fuzzy Mathematics: Approximation Theory. Springer, Heildelberg (2010) 8. Anastassiou, G.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 9. Anastassiou, G.: Fractional representation formulae and right fractional inequalities. Math. Comput. Model. 54(11–12), 3098–3115 (2011)

References

321

10. Anastassiou, G.: Foundation of stochastic fractional calculus with fractional approximation of stochastic processes. RACSAM 114 (2020). Article 89 11. Diethelm, K.: The Analysis of Fractional Differential Equations. Springer, New York (2010) 12. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 13. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 14. Shisha, O., Mond, B.: The degree of convergence of sequences of linear positive operators. Natl. Acad. Sci. U.S. 60, 1196–1200 (1968) 15. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192, 73–80 (1986) 16. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61, 74–87 (1990)

Chapter 15

Trigonometric Caputo Fractional Approximation of Stochastic Processes

Here we encounter and study very general stochastic positive linear operators induced by general positive linear operators that are acting on continuous functions in the trigonometric sense. These are acting on the space of real fractionally differentiable stochastic processes. Under some very mild, general and natural assumptions on the stochastic processes we produce related trigonometric fractional stochastic ShishaMond type inequalities of L q -type 1 ≤ q < ∞ and corresponding trigonometric fractional stochastic Korovkin type theorems. These are regarding the trigonometric stochastic q-mean fractional convergence of a sequence of stochastic positive linear operators to the stochastic unit operator for various cases. All convergences are produced with rates and are given via the trigonometric fractional stochastic inequalities involving the stochastic modulus of continuity of the αth fractional derivatives of the engaged stochastic process, α > 0, α ∈ / N. The impressive fact is that only two basic real Korovkin test functions assumptions, one of them trigonometric, are enough for the conclusions of our trigonometric fractional stochastic Korovkin theory. We give applications to stochastic Bernstein operators in the trigonometric sense. See also [11].

15.1 Introduction Inspiration for this work comes from [2, 3, 15–17]. This work continues our earlier work [5], now at the stochastic fractional level. First we mention the foundations of Stochastic fractional calculus in the direct analytical sense, see [10], in Sect. 15.2, this is in the Caputo fractional direction. In the Sect. 15.3, about background, we talk about the q-mean (1 ≤ q < ∞) first modulus of continuity of a stochastic process and its upper bounds. There we describe completely our setting by introducing our stochastic positive linear operator M, see (15.15), which is based on the positive linear operator  L from C ([−π, π]) into itself. The operator M is acting on a wide space of Caputo fractional differentiable real valued stochastic processes X . See there © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_15

323

324

15 Trigonometric Caputo Fractional Approximation …

Assumptions 15.10, 15.13, 15.14. We first give the main trigonometric pointwise fractional stochastic Shisha-Mond type inequalities [15], see Theorems 15.21, 15.22, and their several corollaries covering important trigonometric special cases. We continue with trigonometric fractional q-mean uniform Shisha-Mond type inequalities, see Theorems 15.23, 15.24, and their interesting corollaries. All this theory is regarding the trigonometric fractional stochastic convergence of operators M to I (stochastic unit operator) given quantitatively with rates. An extensive trigonometric application about the stochastic Bernstein operators follows in full details. Based on our Shisha-Mond type inequalities of our main Theorems 15.21– 15.24 we derive trigonometric pointwise and uniform Stochastic Korovkin theorems [13] on stochastic processes, see Theorems 15.40–15.43. The amazing fact here is, that basic conditions on operator  L regarding two simple real valued functions, one of them trigonometric, that are not stochastic, are able to enforce fractional stochastic convergence on all stochastic processes we are dealing with; see Concepts 15.9 and Assumptions 15.10, 15.13, 15.14 on [−π, π].

15.2 Foundation of Stochastic Fractional Calculus ([10]) Let t ∈ [a, b] ⊂ R, ω ∈ , where (, F, P) is a probability space. Here X (t, ω) stands for a stochastic process. Case of X (·, ω) being continuous on [a, b], ∀ ω ∈ . Then by Caratheodory [1, Theorem 20.15, p.156], we get that X (t, ω) is jointly measurable. Next we define the left and right respectively, Riemann-Liouville stochastic fractional integrals [10], where α > 0 is not an integer: α X (x, ω) = Ia+

1  (α)

α Ib− X (x, ω) =

1  (α)

and



x

(x − t)α−1 X (t, ω) dt,

(15.1)

(t − x)α−1 X (t, ω) dt,

(15.2)

a



b x

∀ x ∈ [a, b], ∀ ω ∈ , where  is the gamma function. α α In the following important cases we prove that Ia+ X , Ib− X are stochastic processes: (i) Assume that (, F, P) is a complete probability space, and that (x − t)α−1 X (t, ω) is an integrable function on [a, x] × , ∀ x ∈ [a, b], then by Fubini’s α X (x, ·) is an integrable function on , ∀ x ∈ [a, b]. theorem, [14], p. 269, Ia+ α−1 X (t, ω) is an integrable function on [x, b] × , ∀ x ∈ Similarly, if (t − x) α X (x, ·) is an integrable function on [a, b], then again by Fubini’s theorem Ib− α α X (x, ω) are stochastic processes. , ∀ x ∈ [a, b] . That is Ia+ X (x, ω) and Ib−

15.2 Foundation of Stochastic Fractional Calculus ([10])

325

(ii) Assume a general probability space (, F, P) and the Lebesgue measure spaces on [a, x], [x, b] , ∀ x ∈ [a, b]. These are clearly σ-finite measure spaces. We assume that the jointly measurable stochastic process X (t, ω) ≥ 0 on [a, b] × , hence (x − t)α−1 X (t, ω) ≥ 0 on [a, x] × , and (t − x)α−1 X (t, ω) ≥ 0 on [x, b] × , ∀ x ∈ [a, b], and both are jointly measurable. Then by Tonelli’s theα α X (x, ·) , Ib− X (x, ·) are measurable functions orem, [14, p. 270], we get that Ia+ α α on , ∀ x ∈ [a, b]. That is Ia+ X, Ib− X are stochastic processes. The above facts provide the foundation of stochastic fractional calculus in the direct analytical α α X, Ib− X are stochastic processes. sense. So it is not unusual to consider that Ia+ α (iii) Given that X (·, ω) is in L 1 ([a, b]) then Ia+ X (·, ω) ∈ L 1 ([a, b]), ∀ ω ∈ , see α X (·, ω) ∈ L 1 ([a, b]), ∀ ω ∈ , see [8, p. 334]. [12, p. 13], and Ib− α X (·, ω) ∈ C ([a, b]), when 0 < And given that X (·, ω) ∈ L ∞ ([a, b]), then Ia+ α α < 1, and Ia+ X (·, ω) ∈ AC ([a, b]) (absolutely continuous functions), when α ≥ α X (·, ω) ∈ 1, ∀ ω ∈ , see [6, p. 388]. Similarly, if X (·, ω) ∈ L ∞ ([a, b]), then Ib− α C ([a, b]), when 0 < α < 1, and Ib− X (·, ω) ∈ AC ([a, b]), when α ≥ 1, ∀ ω ∈ , see [9]. We need

Definition 15.1 ([10]) Let non-integer α > 0, n = α (· is the ceiling of the number), t ∈ [a, b] ⊂ R, ω ∈ , where (, F, P) is a general probability space. Here X (t, ω) stands for a stochastic process. Assume that X (·, ω) ∈ AC n ([a, b]) (spaces of functions X (·, ω) with X (n−1) (·, ω) ∈ AC ([a, b])), ∀ ω ∈ . We call stochastic left Caputo fractional derivative α D∗a X (x, ω) =

1  (n − α)



x

(x − t)n−α−1 X (n) (t, ω) dt,

(15.3)

a

∀ x ∈ [a, b], ∀ ω ∈ . And, we call stochastic right Caputo fractional derivative α Db− X (x, ω) =

(−1)n  (n − α)



b

(z − x)n−α−1 X (n) (z, ω) dz,

x

∀ x ∈ [a, b], ∀ ω ∈ . Remark 15.2 (to Definition 15.1) We further assume here that  (n)   X (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Then, by (15.3), we have

(15.4)

326

15 Trigonometric Caputo Fractional Approximation …

  α  D X (x, ω) ≤ ∗a ≤ That is

M∗  (n − α)

1  (n − α) 

x



x

  (x − t)n−α−1  X (n) (t, ω) dt

a

(x − t)n−α−1 dt =

a

M ∗ (x − a)n−α .  (n − α + 1)

n−α ∗  α   D X (x, ω) ≤ M (x − a) , ∀ x ∈ [a, b] . ∗a  (n − α + 1)

(15.5)

Also, from (15.4) we get  α   D X (x, ω) ≤ b− M∗  (n − α) That is

1  (n − α) 

b



b

  (z − x)n−α−1  X (n) (z, ω) dz ≤

x

(z − x)n−α−1 dz =

x

M ∗ (b − x)n−α .  (n − α + 1)

n−α ∗  α   D X (x, ω) ≤ M (b − x) , ∀ x ∈ [a, b] . b−  (n − α + 1)

(15.6)

α α By (15.1)–(15.4), it is not strange to assume that D∗a X , Db− X are stochastic processes.

15.3 Background (See Also [10]) We need Definition 15.3 We define the relative q-mean first modulus of continuity of stochastic process X (t, ω) by 1 (X, δ) L q ,[c,d] :=  sup





1 |X (x, ω) − X (y, ω)|q P (dω)

q

: x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ ,

(15.7) δ > 0, 1 ≤ q < ∞. Definition 15.4 Let 1 ≤ q < ∞. Let X (x, ω) be a stochastic process. We call X a q-mean uniformly continuous stochastic process over [a, b], iff ∀ ε > 0 ∃ δ > 0 : whenever |x − y| ≤ δ; x, y ∈ [a, b] implies that  

|X (x, s) − X (y, s)|q P (ds) ≤ ε.

(15.8)

15.3 Background (See Also [10])

327

U

We denote it as X ∈ CR q ([a, b]) . It holds U

Proposition 15.5 ([5]) Let X ∈ CR q ([a, b]), then 1 (X, δ) L q ,[a,b] < ∞, any δ > 0. Also it holds Proposition 15.6 ([5]) Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R. Then following are true ([c, d] ⊆ [a, b]): (i) 1 (X, δ) L q ,[c,d] is nonnegative and nondecreasing in δ > 0, U (ii) lim1 (X, δ) L q ,[c,d] = 1 (X, 0) L q ,[c,d] = 0, iff X ∈ CR q ([c, d]) , δ↓0

(iii) 1 (X, δ1 + δ2 ) L q ,[c,d] ≤ 1 (X, δ1 ) L q ,[c,d] + 1 (X, δ2 ) L q ,[c,d] , δ1 , δ2 > 0, (iv) 1 (X, mδ) L q ,[c,d] ≤ m1 (X, δ) L q ,[c,d] , δ > 0, m ∈ N, (v) 1 (X, λδ) L q ,[c,d] ≤ λ 1 (X, δ) L q ,[c,d] ≤ (λ + 1) 1 (X, δ) L q ,[c,d] , λ > 0, δ > 0, (vi) 1 (X + Y, δ) L q ,[c,d] ≤ 1 (X, δ) L q ,[c,d] + 1 (Y, δ) L q ,[c,d] , δ > 0, U (vii) 1 (X, ·) L q ,[c,d] is continuous on R+ for X ∈ CR q ([c, d]) . We give Remark 15.7 (to Proposition 15.6) By Proposition 15.6(v) we get 1 (X, |x − y|) L q ,[c,d] ≤

|x − y| 1 (X, δ) L q ,[c,d] , δ

(15.9)

∀ x, y ∈ [c, d], any δ > 0. We give   Remark 15.8 (continuation of Remark 15.2) We assume again that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Let δ > 0, 1 ≤ q < ∞. Then 



α 1 D∗c X, δ L q ,[c,d]

 = sup



  α   D X (x, ω) − D α X (y, ω)q P (dω) ∗c ∗c

 q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤  sup 

    α  D X (x, ω) +  D α X (y, ω) q P (dω) ∗c ∗c ([5])

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤

 q1

:

(15.10)

328

15 Trigonometric Caputo Fractional Approximation …

M∗ sup  (n − α + 1)

 

 q (x − c)n−α + (y − c)n−α P (dω)

 q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} =

 M∗ sup (x − c)n−α + (y − c)n−α :  (n − α + 1) x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ That is

 α X, δ L q ,[c,d] ≤ 1 D∗c

2M ∗ (d − c)n−α .  (n − α + 1)

2M ∗ (d − c)n−α ,  (n − α + 1)

(15.11)

2M ∗ (d − c)n−α ,  (n − α + 1)

(15.12)

where a ≤ c < d ≤ b. Similarly, it holds  α X, δ L q ,[c,d] ≤ 1 Dd− where a ≤ c < d ≤ b. Next let x0 ∈ [a, b], then  α X, δ L q ,[x sup 1 D∗x 0

x0 ∈[a,b]

=

0 ,b]

≤

2M ∗ sup (b − x0 )n−α  (n − α + 1) x0 ∈[a,b]

2M ∗ (b − a)n−α .  (n − α + 1)

(15.13)

Similarly, it holds  sup 1 Dxα0 − X, δ L q ,[a,x ] ≤ 0

x0 ∈[a,b]

=

2M ∗ sup (x0 − a)n−α  (n − α + 1) x0 ∈[a,b]

2M ∗ (b − a)n−α .  (n − α + 1)

(15.14)

We need Concepts 15.9 (see also [10]) Let  L be a positive linear operator from C ([a, b]) into itself. Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R, where (, F, P) is a probability space. Here we assume that for non-integer α > 0, α = n, X (·, ω) ∈ AC n ([a, b]) with X (n) (·, ω) ∈ L ∞ ([a, b]), ∀ ω ∈ . Also we assume for each t ∈ [a, b] that X (k) (t, ·) is measurable for all k = α X (z, ω) is a stochastic process for 1, . . . , n − 1. Further we assume that D∗t

15.3 Background (See Also [10])

329

α z ∈ [t, b], ω ∈ , and Dt− X (z, ω) is a stochastic process for z ∈ [a, t] , ω ∈ ; ∀ t ∈ [a, b]. Define

M (X ) (t, ω) :=  L (X (·, ω)) (t) , ∀ ω ∈ , ∀ t ∈ [a, b] ,

(15.15)

and assume that it is a random variable in ω. Clearly M is a positive linear operator on stochastic processes. We mention Assumption 15.10 (as in [10]) Let non-integer α > 0. α (i) For any t ∈ [a, b] we assume that D∗t X (z, ω) is continuous in z ∈ [t, b], uniformly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ > 0 : whenever |z 1 − z 2 | ≤ δ; α α z 1 , z 2 ∈ [t, b], then  D∗t X (z 1 , ω) − D∗t X (z 2 , ω) ≤ ε, ∀ ω ∈ . U α We denote this by D∗t X ∈ CR ([t, b]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [t, b] . α X (z, ω) is continuous in z ∈ [a, t], uni(ii) For any t ∈ [a, b] we assume that Dt− formly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ >0 : whenever |z 1 − z 2 | ≤ δ; α α z 1 , z 2 ∈ [a, t], then  Dt− X (z 1 , ω) − Dt− X (z 2 , ω) ≤ ε, ∀ ω ∈ . U α We denote this by Dt− X ∈ CR ([a, t]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [a, t] .

Remark 15.11 Assumption 15.10 implies: α α (i) D∗t X (·, ω) ∈ C ([t, b]), ∀ ω ∈ , and D∗t X is q-mean uniformly continuous in U q α z ∈ [t, b], that is D∗t X ∈ CR ([t, b]), for any 1 ≤ q < ∞. α α X (·, ω) ∈ C ([a, t]), ∀ ω ∈ , and Dt− X is q-mean uniformly continuous (ii) Dt− U q α in z ∈ [a, t], that is Dt− X ∈ CR ([a, t]) , for any 1 ≤ q < ∞.

We need Definition 15.12 Denote by  (E X ) (t) :=



X (t, ω) P (dω) , ∀ t ∈ [a, b] ,

(15.16)

the expectation operator. We make Assumption 15.13 (as in [10]) We assume that   q  E  X (k)  (t) < ∞, ∀ t ∈ [a, b] , q > 1, for all k = 0, 1, . . . , n − 1, n = α ; α > 0 non-integer.

(15.17)

330

15 Trigonometric Caputo Fractional Approximation …

We make Assumption 15.14 (as in [10]) We assume that   (k)  E  X  (t) < ∞, ∀ t ∈ [a, b] ,

(15.18)

for all k = 0, 1, . . . , n − 1, n = α ; α > 0 non-integer. We give Remark 15.15 By the Riesz representation theorem [14] we have that there exists μt unique, completed Borel measure on [a, b] with L (1) (t) > 0, m t := μt ([a, b]) = 

(15.19)

f (x) dμt (x) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(15.20)

such that  L ( f ) (t) =

 [a,b]

Consequently we have that  X (x, ω) dμt (x) , ∀ (t, ω) ∈ [a, b] × ,

M (X ) (t, ω) =

(15.21)

[a,b]

and X as in Concepts 15.9. Here χ[γ,δ] (s) stands for the characteristic function on [γ, δ] ⊆ [a, b] . Notice that (r > 0)    χ[t,b] (s) |s − t|r μt (ds) =  L |· − t|r χ[t,b] (·) (t) , (s − t)r μt (ds) = [t,b]

[a,b]

(15.22) and 



 χ[a,t] (s) |s − t|r μt (ds) =  L |· − t|r χ[a,t] (·) (t) .

(t − s) μt (ds) = r

[a,t]

[a,b]

(15.23) Let now n = α , α ∈ / N, α > 0, k = 1, . . . , n − 1. Then by Hölder’s inequality we obtain      k  |x − t|k dμt (x) ≤ (x − t) dμt (x) ≤  [a,b]



[a,b]

α+1

|x − t| [a,b]

Therefore it holds

dμt (x)

k ( α+1 )

(μt ([a, b]))(

α+1−k α+1

).

(15.24)

15.3 Background (See Also [10])

331

       L |· − t|k (t)∞,[a,b] ≤ L (· − t)k (t)∞,[a,b] ≤  k   ( α+1−k ) ( α+1 )  α+1   , L |· − t|α+1 (t)∞,[a,b] L (1)∞,[a,b]

(15.25)

all k = 1, . . . , n − 1. Also, we observe that

and

C ([a, b])  |· − t|α+1 χ[a,t] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] ,

(15.26)

C ([a, b])  |· − t|α+1 χ[t,b] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] .

(15.27)

By positivity of  L we obtain        L |· − t|α+1 (t)∞,[a,b] < ∞, L |· − t|α+1 χ[a,t] (·) (t)∞,[a,b] ≤ 

(15.28)

 by  L |· − t|α+1 (t) being continuous in t ∈ [a, b], see p. 388 of [4], and        L |· − t|α+1 (t)∞,[a,b] . L |· − t|α+1 χ[t,b] (·) (t)∞,[a,b] ≤ 

(15.29)

Above (15.22)–(15.25) and (15.28), (15.29) can be used to derive convergence from (15.30), (15.32)–(15.34) next.  L (1) (t), and by In this work we denote by  L χ[a,t] (·) (t) := μt ([a, t]) ≤    L χ[t,b] (·) (t) := μt ([t, b]) ≤  L (1) (t) . Next we mention the first main result from [10], on the quantitative stochastic fractional approximation regarding stochastic processes: Theorem 15.16 ([10]) Suppose Concepts 15.9, Assumptions 15.10 and 15.13. Here the non-integer α > 0 is such that α > q1 , where p, q > 1 such that 1p + q1 = 1. Then, ∀ t ∈ [a, b], we have:    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n−1  k=1

  (k) q q1 1  E  X  (t)   2p  L (· − t)k (t) + 1 α k!  (α) ( p (α − 1) + 1) p (q + 1) q(α+1) (15.30)   1   α  L χ[t,b] (·) (t) p  L |· − t|q(α+1) χ[t,b] (·) (t) q(α+1)    q1 1 α  L χ[t,b] (·) (t) α+1 (q + 1) (α+1) + 1

332

15 Trigonometric Caputo Fractional Approximation …

 α X, D∗t

1



 1  q(α+1) 1  q(α+1) L |· − t| χ[t,b] (·) (t) (q + 1)

 + L q ,[t,b]

  1   α  L |· − t|q(α+1) χ[a,t] (·) (t) q(α+1) L χ[a,t] (·) (t) p     q1 1 α  L χ[a,t] (·) (t) α+1 (q + 1) (α+1) + 1  1

α X, Dt−



Here we have that

 1  q(α+1) 1  q(α+1) L |· − t| χ[a,t] (·) (t) (q + 1)

 . L q ,[a,t]

α α D∗t X (t, ω) = Dt− X (t, ω) = 0,

(15.31)

∀ ω ∈ , see [8, pp. 358–359]. We assume also α X (s, ω) = 0, for s < t, D∗t and

α X (s, ω) = 0, for s > t, Dt−

∀ ω ∈ . We mention the second main result from [10], the L 1 -quantitative stochastic fractional approximation of stochastic processes, it is the q = 1 analog of Theorem 15.16. Inequality (15.32) is much simpler than (15.30). Theorem 15.17 ([10]) Suppose Concepts 15.9, Assumptions 15.10 and 15.14. Here α > 0, α ∈ / N, n = α . Then, ∀ t ∈ [a, b], we have   E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 + n−1   (k)    E X (t)   1  L (· − t)k (t) + k!  (α + 1) k=1



  1  L χ[a,t] (·) (t) α+1 +

   α 1  L |· − t|α+1 χ[a,t] (·) (t) α+1 (α + 1)

   1  α 1 Dt− X,  L |· − t|α+1 χ[a,t] (·) (t) α+1 

  1  L χ[t,b] (·) (t) α+1 +

L 1 ,[a,t]

+

   α 1  L |· − t|α+1 χ[t,b] (·) (t) α+1 (α + 1)

(15.32)

15.3 Background (See Also [10])

333

   1  α 1 D∗t X,  L |· − t|α+1 χ[t,b] (·) (t) α+1

 L 1 ,[t,b]

.

We also mention a uniform norm result (denote ·∞,[a,b] :=  f ∞ ). It is based on Theorem 15.16. Theorem 15.18 ([10]) Suppose Concepts 15.9, Assumptions 15.10 and 15.13. Here the non-integer α > 0 is such that α > q1 , where p, q > 1 : 1p + q1 = 1. Additionally   assume that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Then       1  1   E |M (X ) − X |q  q ≤  E |X |q  q  L (1) − 1∞ + ∞ ∞ 1   1 n−1   (k) q  q     E X 2p k ∞   L (· − t) (t) ∞ + 1 α k!  (α) ( p (α − 1) + 1) p (q + 1) q(α+1) k=1

  q1  1  1 α α+1 p  L (1)∞ L (1)∞ (q + 1) (α+1) + 1    α  q(α+1)  L |· − t|q(α+1) χ[t,b] (·) (t)∞  sup 1 t∈[a,b]



(15.33)

   1 α  L |· − t|q(α+1) χ[t,b] (·) (t)∞ X, D∗t (q + 1)

⎫ ⎬

 1  q(α+1) L q ,[t,b]

⎭

  α  q(α+1) +  L |· − t|q(α+1) χ[a,t] (·) (t)∞  sup 1 t∈[a,b]



   1 α  L |· − t|q(α+1) χ[a,t] (·) (t)∞ X, Dt− (q + 1)

⎫⎫ ⎬⎬

 1  q(α+1) L q ,[a,t]

⎭⎭

.

Based on Theorem 15.17 we mention the following uniform estimate: Theorem 15.19 ([10]) Suppose Concepts 15.9, Assumptions 15.10   and 15.14. Here α > 0, α ∈ / N, n = α . Additionally assume that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Then   E (|M (X ) − X |)∞ ≤ E (|X |)∞  L (1) − 1∞ +   n−1   (k)     E X k=1

k!

∞

   1   α+1 + 1 L (1)    α+1 ∞  L (· − t)k (t)∞ +  (α + 1)

334

15 Trigonometric Caputo Fractional Approximation …

  α  α+1  L |· − t|α+1 χ[a,t] (·) (t)∞  1     α+1 α sup 1 Dt− X,  L |· − t|α+1 χ[a,t] (·) (t)∞

t∈[a,b]

L 1 ,[a,t]

+

(15.34)

α    α+1  L |· − t|α+1 χ[t,b] (·) (t)∞

sup 1



t∈[a,b]

1     α+1 α D∗t X,  L |· − t|α+1 χ[t,b] (·) (t)∞

 L 1 ,[t,b]

.

We make Remark 15.20 Next we specify [a, b] as [−π, π]. Clearly then  L : C ([−π, π]) → C ([−π, π]) is the positive linear operator on hand. Here n = α , α ∈ / N, α > 0, k = 1, . . . , n − 1. Next we use Hölder’s inequality. We notice that          |· − t| k |x − t| k  sin L sin dμt (x) ≤ (t) = 4 4 [−π,π]  [−π,π]





|x − t| sin 4

α+1 

k  α+1

dμt (x)

(μt ([−π, π]))

α+1−k α+1

=

(15.35)

k       α+1  α+1−k |· − t| α+1   sin L (1) (t) α+1 . L (t) 4

That is

    |· − t| k  sin L (t) ≤ 4

(15.36)

k       α+1  α+1−k |· − t| α+1   sin L (1) (t) α+1 , L (t) 4

for k = 1, . . . , n − 1; true also for q (α + 1) instead of (α + 1), for any 1 < q < ∞. Next ·∞ denotes ·∞,[−π,π] . Consequently, it holds         |· − t| k   sin (t) L   4

∞

≤

(15.37)

15.3 Background (See Also [10])

335

k        α+1    α+1−k |· − t| α+1    sin L (1) (t)∞α+1 , (t) L   4

∞

for k = 1, . . . , n − 1; true also for q (α + 1) instead of (α + 1), for any 1 < q < ∞. In this work we use a lot the following well known inequality:  |z| ≤ π sin

 |z| , ∀ z ∈ [−π, π] . 2

(15.38)

Notice that, for any t ∈ [−π, π], we have C ([−π, π])  |· − t| χ[−π,t] (·) ≤ |· − t| ∈ C ([−π, π]), therefore       |· − t| χ[−π,t] (·) α+1 |· − t| α+1 C ([−π, π])  sin ≤ sin ∈ C ([−π, π]) . 4 4

(15.39) Consequently, by positivity of  L we obtain         |· − t| χ[−π,t] (·) α+1   sin (t) L   4

∞

        |· − t| α+1   sin ≤ L (t) .   4 ∞ (15.40)

Similarly, for any t ∈ [−π, π], we have C ([−π, π])  |· − t| χ[t,π] (·) ≤ |· − t| ∈ C ([−π, π]), thus 



|· − t| χ[t,π] (·) C ([−π, π])  sin 4

α+1

Hence         |· − t| χ[t,π] (·) α+1   sin (t) L   4

∞

   |· − t| α+1 ≤ sin ∈ C ([−π, π]) . 4 (15.41)         |· − t| α+1   sin ≤ L (t) .   4 ∞ (15.42)

So, if the right hand side of (15.40), (15.42) goes to zero, so do their left hand sides. Above in (15.39)–(15.42), one can use q (α + 1) instead of (α + 1), 1 < q < ∞. A further detailed analysis reveals: We have that (1 ≤ q < ∞)  (15.20)  L |· − t|q(α+1) χ[t,π] (·) (t) =

 (x − t)q(α+1) dμt (x) = [t,π]

 2

q(α+1) [t,π]



x −t 2

q(α+1) dμt (x)

( 15.38)

≤

336

15 Trigonometric Caputo Fractional Approximation …

  q(α+1) x −t sin dμt (x) = 4 [t,π]

 (2π)

q(α+1)



 (2π)

q(α+1) [t,π]

 (2π)

q(α+1)



|x − t| sin 4

(15.43)

q(α+1) dμt (x) =

   |x − t| q(α+1) sin χ[t,π] (x) dμt (x) = 4 [−π,π]

q(α+1)   |x − t| (15.20) sin χ[t,π] (x) dμt (x) = 4 [−π,π]

 (2π)q(α+1)

L (2π)q(α+1) 





|· − t| sin χ[t,π] (·) 4

q(α+1)  (t) .

That is, we have obtained   L |· − t|q(α+1) χ[t,π] (·) (t) ≤ (2π)q(α+1)  L

  q(α+1)  |· − t| sin χ[t,π] (·) (t) . 4 (15.44)

Similarly, it holds   q(α+1)    |· − t|  sin χ[−π,t] (·) L |· − t|q(α+1) χ[−π,t] (·) (t) ≤ (2π)q(α+1)  L (t) . 4

(15.45) Above inequalities (15.44), (15.45) are valid for any 1 ≤ q < ∞. Furthermore, we observe that         k k    |x − t|k dμt (x) = L (· − t) (t) =  (x − t) dμt (x) ≤ [−π,π]



 2

k [−π,π]

That is

|x − t| 2

k

[−π,π]







|x − t| sin dμt (x) ≤ (2π) 4 [−π,π]     |· − t| k sin L = (2π)k  (t) . 4 (15.38)

    L (· − t)k (t) ≤ (2π)k  L

∀ t ∈ [−π, π], all k = 1, . . . , n − 1.

k





|· − t| sin 4

k dμt (x) (15.46)

k  (t) ,

(15.47)

15.4 Main Results

337

15.4 Main Results Next we present our first main result on the trigonometric quantitative stochastic fractional approximation of stochastic processes. It is a pointwise result. Theorem 15.21 Here [a, b] = [−π, π]. Suppose Concepts 15.9, Assumptions 15.10 and 15.13. Let α > 0, α ∈ / N, such that α > q1 , where p, q > 1 and 1p + q1 = 1. Then, ∀ t ∈ [−π, π], we have    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n−1 

  (k) q q1 E  X  (t) k!

k=1

L (2π) 



k



|· − t| sin 4

k  (t) +

2α+ p π α 1

1

(15.48)

α

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1)

⎧⎧ α    q(α+1)   q(α+1) ⎨⎨  1p |· − t|   L χ[t,π] (·) (t) L sin χ[t,π] (·) (t) ⎩⎩ 4    q1 1 α  L χ[t,π] (·) (t) α+1 (q + 1) (α+1) + 1 ⎛



α X, 2π 1 ⎝ D∗t

1  L (q + 1)





|· − t| sin χ[t,π] (·) 4

q(α+1) 



⎫ ⎬

⎞ 1 q(α+1)

(t)

⎠ L q ,[t,π]

⎭

+

⎧ α    q(α+1)   q(α+1) ⎨  1p |· − t|   sin χ[−π,t] (·) L χ[−π,t] (·) (t) L (t) ⎩ 4    q1 1 α+1 α  (α+1) L χ[−π,t] (·) (t) +1 (q + 1) ⎛ α 1 ⎝ Dt− X, 2π



1  L (q + 1)

⎞ 1   q(α+1)   q(α+1) |· − t| ⎠ sin χ[−π,t] (·) (t) 4

⎫⎫ ⎬⎬ L q ,[−π,t]

⎭⎭

.

338

15 Trigonometric Caputo Fractional Approximation …

Proof By Theorem 15.16 on [−π, π] we get (∀ t ∈ [−π, π]) that    1  1  E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n−1  k=1

  (k) q q1 1  E  X  (t)   2p k   L (· − t) (t) + 1 α k!  (α) ( p (α − 1) + 1) p (q + 1) q(α+1)   1   α  L χ[t,π] (·) (t) p  L |· − t|q(α+1) χ[t,π] (·) (t) q(α+1)    q1 1 α  L χ[t,π] (·) (t) α+1 (q + 1) (α+1) + 1  1

 1  q(α+1)  1 α  L |· − t|q(α+1) χ[t,π] (·) (t) X, D∗t (q + 1)





+ L q ,[t,π]

  1   α  L |· − t|q(α+1) χ[−π,t] (·) (t) q(α+1) L χ[−π,t] (·) (t) p 

(15.49)

 q1   1 α  L χ[−π,t] (·) (t) α+1 (q + 1) (α+1) + 1  1

α X, Dt−



1  L |· − t|q(α+1) χ[−π,t] (·) (t) (q + 1)

 1  q(α+1)

 =: (∗) . L q ,[−π,t]

Using (15.44), (15.45) and (15.47) we obtain: (∗) ≤ n−1  k=1



 1  E |X |q (t) q  L (1) (t) − 1 +

    (k) q q1 k  E  X  (t) |· − t| sin L (t) + (2π)k  k! 4 1

2p 1

α

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1) ⎧⎧ α    q(α+1)   q(α+1) ⎨⎨  1p |· − t| α   sin χ[t,π] (·) L χ[t,π] (·) (t) (2π) L (t) ⎩⎩ 4    q1 1 α  L χ[t,π] (·) (t) α+1 (q + 1) (α+1) + 1

(15.50)

15.4 Main Results

⎛

339



α X, 2π 1 ⎝ D∗t

1  L (q + 1)





|· − t| sin χ[t,π] (·) 4

q(α+1) 



⎫ ⎬

⎞ 1 q(α+1) ⎠

(t)

L q ,[t,π]

⎭

+

⎧ α    q(α+1)   q(α+1) ⎨  1p |· − t| α  L sin L χ[−π,t] (·) (t) (2π)  χ[−π,t] (·) (t) ⎩ 4    q1 1 α  L χ[−π,t] (·) (t) α+1 (q + 1) (α+1) + 1 ⎛



α 1 ⎝ Dt− X, 2π

1  L (q + 1)

⎞ 1   q(α+1)   q(α+1) |· − t| ⎠ sin χ[−π,t] (·) (t) 4

⎫⎫ ⎬⎬ L q ,[−π,t]

⎭⎭

,



proving the claim.

We continue with the trigonometric L 1 -quantitative stochastic fractional pointwise approximation of stochastic processes. Theorem 15.22 Here [a, b] = [−π, π]. Suppose Concepts 15.9, Assumptions 15.10 and 15.14. Here α > 0, α ∈ / N, n = α . Then, ∀ t ∈ [−π, π], we have   L (1) (t) − 1 + E (|M (X ) − X |) (t) ≤ (E |X |) (t)      n−1   (k)   E X (t) |· − t| k (2π)α k sin (t) + (2π) L k! 4  (α + 1) k=1 ⎧ α α+1   α+1     ⎨  1 |· − t| 1   L χ[−π,t] (·) (t) α+1 + χ[−π,t] (·) sin L (t) ⎩ 4 (α + 1)

⎛

1 ⎞    α+1   α+1 |· − t| α ⎠ sin χ[−π,t] (·) 1 ⎝ Dt− X, 2π  L (t) 4

   1  L χ[t,π] (·) (t) α+1 +

1 (α + 1)

+ L 1 ,[−π,t]

α α+1   α+1     |· − t|  χ[t,π] (·) sin L (t) 4

1 ⎞    α+1   α+1 |· − t| α ⎠ sin χ[t,π] (·) 1 ⎝ D∗t X, 2π  L (t) 4

⎫ ⎬

⎛

L 1 ,[t,π]

⎭

. (15.51)

340

15 Trigonometric Caputo Fractional Approximation …

Proof By Theorem 15.17 for [a, b] = [−π, π], and ∀ t ∈ [−π, π], we get   E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 + n−1   (k)    E X (t)   1  L (· − t)k (t) + k!  (α + 1) k=1

   1  L χ[−π,t] (·) (t) α+1 +

   α 1  L |· − t|α+1 χ[−π,t] (·) (t) α+1 (α + 1)

   1  α 1 Dt− X,  L |· − t|α+1 χ[−π,t] (·) (t) α+1    1  L χ[t,π] (·) (t) α+1 + 1



α D∗t X,

L 1 ,[−π,t]

+

(15.52)

   α 1  L |· − t|α+1 χ[t,π] (·) (t) α+1 (α + 1)

  1   L |· − t|α+1 χ[t,π] (·) (t) α+1

 L 1 ,[t,π]

=: (∗∗) .

Using (15.44), (15.45) and (15.47) we obtain:   L (1) (t) − 1 + (∗∗) ≤ (E |X |) (t)    k  n−1   (k)   E X (t) |· − t| (2π)α sin L (t) + (2π)k  k! 4  (α + 1) k=1 ⎧ α α+1   α+1     ⎨  1 |· − t| 1 α+1   L χ[−π,t] (·) (t) χ[−π,t] (·) sin + L (t) ⎩ 4 (α + 1)

⎛

1 ⎞    α+1   α+1 |· − t| α ⎠ sin χ[−π,t] (·) 1 ⎝ Dt− X, 2π  L (t) 4

   1  L χ[t,π] (·) (t) α+1 +

1 (α + 1)

+ L 1 ,[−π,t]

α α+1   α+1     |· − t|  χ[t,π] (·) sin L (t) 4

1 ⎞    α+1   α+1 |· − t| α ⎠ sin χ[t,π] (·) 1 ⎝ D∗t X, 2π  L (t) 4

⎫ ⎬

⎛

The claim is proved.

L 1 ,[t,π]

⎭

. (15.53) 

15.4 Main Results

341

We continue with trigonometric fractional uniform estimates (·∞,[−π,π] := ·∞ ) in L q -mean (1 ≤ q < ∞). Theorem 15.23 Here [a, b] = [−π, π]. Suppose Concepts 15.9, Assumptions 15.10 and 15.13. Let α > 0, α ∈ / N, n = α , such that α > q1 , where p, q > 1 and 1p +   1 = 1. Additionally assume that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [−π, π] × , where q M ∗ > 0. Then       1  1   E |M (X ) − X |q  q ≤  E |X |q  q  L (1) − 1∞ + ∞ ∞ 1   n−1   (k) q  q  E X

∞

k!

k=1

   k     |· − t|   sin L (t) + (2π)k    4 ∞

  q1   1 1 1 α α+1 p L (1)∞ π α 2α+ p  L (1)∞ (q + 1) (α+1) + 1  1

(15.54)

α

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1) ⎧⎧  α   q(α+1)    q(α+1) ⎨⎨ |· − t|   sin χ[t,π] (·) (t) L  ⎩⎩ 4 ∞

 sup 1 t∈[−π,π]

α X, 2π D∗t

 1 (q+1)

    1  q(α+1)   q(α+1)   |·−t|   (t)  L sin 4 χ[t,π] (·) ∞

 L q ,[t,π]

⎧  α   q(α+1)    q(α+1) ⎨ |· − t|   sin χ[−π,t] (·) + L (t)  ⎩ 4 ∞

 α X, 2π sup 1 Dt−

t∈[−π,π]

 1 (q+1)

    1  q(α+1)   q(α+1)   |·−t|   sin L χ (t) (·) [−π,t]   4 ∞

 . L q ,[−π,t]

Proof Based on Theorem 15.21. We take into account (15.13), (15.14) and the positivity of  L.  We continue with the L 1 -mean uniform stochastic fractional result. Theorem 15.24 Here [a, b] = [−π, π]. Suppose Concepts 15.9,Assumptions  15.10 and 15.14. Let α > 0, α ∈ / N, n = α . Additionally assume that  X (n) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [−π, π] × , where M ∗ > 0. Then   E (|M (X ) − X |)∞ ≤ E (|X |)∞  L (1) − 1∞ +   n−1   (k)     E X k=1

k!

∞

   k     |· − t|   sin L (t) + (2π)k    4 ∞

342

15 Trigonometric Caputo Fractional Approximation …

    1 1 (2π)α α+1  + L (1)∞  (α + 1) α+1 ⎧  α   α+1    α+1 ⎨ |· − t|   sin χ[−π,t] (·) (t) L  ⎩ 4 ∞

⎛

1 ⎞    α+1   α+1   |· − t|   α L sin χ[−π,t] (·) sup 1 ⎝ Dt− X, 2π  (t) ⎠   4 t∈[−π,π]

∞

L 1 ,[−π,t]

α    α+1    α+1  |· − t|   sin + L χ[t,π] (·) (t)   4

∞

1 ⎞    α+1   α+1   |· − t|   α sin L χ[t,π] (·) sup 1 ⎝ D∗t X, 2π  (t) ⎠   4 t∈[−π,π]

⎫ ⎬

⎛

∞

L 1 ,[t,π]

⎭

.

(15.55)

Proof Based on Theorem 15.22. We take into account (15.13), (15.14) and the positivity of  L.  We continue with interesting corollaries. Corollary 15.25 All as in Theorem 15.21. Further assume  L (1) = 1 and X (k) (t0 , ω) = 0 , ∀ ω ∈ , all k = 1, . . . , n − 1, for a fixed t0 ∈ [−π, π]. Then   1 E |M (X ) − X |q (t0 ) q ≤

2α+ p π α 1

1

α

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1)

(15.56)

⎧⎧ α     q(α+1) q(α+1)  ⎨⎨  1p |· − t0 |   sin χ[t0 ,π] (·) L χ[t0 ,π] (·) (t0 ) L (t0 ) ⎩⎩ 4    q1 1 α  L χ[t0 ,π] (·) (t0 ) α+1 (q + 1) (α+1) + 1  1

 α X, 2π D∗t 0

1  L (q+1)



 sin

|·−t0 | χ[t0 ,π] 4

1  q(α+1) q(α+1)  (·) (t0 )



 L q ,[t0 ,π]

⎧ α     q(α+1) q(α+1)  ⎨  1p |· − t0 |  sin +  L χ[−π,t0 ] (·) (t0 ) L χ[−π,t0 ] (·) (t0 ) ⎩ 4

15.4 Main Results

343

   q1 1 α  L χ[−π,t0 ] (·) (t0 ) α+1 (q + 1) (α+1) + 1 

 Dtα0 − X, 2π

1

1  L (q+1)

 1    q(α+1) q(α+1)  |·−t0 | sin 4 χ[−π,t0 ] (·) (t0 )

 . L q ,[−π,t0 ]

Corollary 15.26 All as in Theorem 15.21. Further assume that  L (1) = 1 and α < 1 (i.e. n = 1). Then, ∀ t ∈ [−π, π], we have 1   E |M (X ) − X |q (t) q ≤

2α+ p π α

1 q


2. Due to lack of space we omit this statement. Corollary 15.27 All as in Theorem 15.22. Further assume that  L (1) = 1 and X (k) (t0 , ω) = 0, ∀ ω ∈ , all k = 1, . . . , n − 1 , for a fixed t0 ∈ [−π, π]. Then E (|M (X ) − X |) (t0 ) ≤ 

  1  L χ[−π,t0 ] (·) (t0 ) α+1 +

1 (α+1)

(2π)α  (α + 1)

α  α+1     α+1  |·−t0 |  L sin 4 χ[−π,t0 ] (·) (t0 )

344

15 Trigonometric Caputo Fractional Approximation …

⎛

1 ⎞     α+1 α+1  |· | − t 0 ⎠ sin χ[−π,t0 ] (·) 1 ⎝ Dtα0 − X, 2π  L (t0 ) 4

  1  L χ[t0 ,π] (·) (t0 ) α+1 +

1 (α+1)

+ L 1 ,[−π,t0 ]

α  α+1     α+1  |·−t0 |  L sin 4 χ[t0 ,π] (·) (t0 )

1 ⎞     α+1 α+1  |· − t0 | α ⎠ sin χ[t0 ,π] (·) 1 ⎝ D∗t X, 2π  L (t0 ) 0 4

⎫ ⎬

⎛

L 1 ,[t0 ,π]

⎭

.

(15.58)

Corollary 15.28 All as in Theorem 15.22. Further assume that  L (1) = 1 and 0 < α < 1. Then, ∀ t ∈ [−π, π], we have E (|M (X ) − X |) (t) ≤

(2π)α  (α + 1)

⎧ α α+1   α+1     ⎨  1 α+1 |· − t| 1   L χ[−π,t] (·) (t) χ[−π,t] (·) sin + L (t) ⎩ 4 (α + 1) ⎛

1 ⎞    α+1   α+1 |· − t| α ⎠ sin χ[−π,t] (·) 1 ⎝ Dt− X, 2π  L (t) 4

   1  L χ[t,π] (·) (t) α+1 +

1 (α + 1)

+ L 1 ,[−π,t]

α α+1   α+1     |· − t|  χ[t,π] (·) sin L (t) 4

1 ⎞    α+1   α+1 |· − t| α ⎠ sin χ[t,π] (·) 1 ⎝ D∗t X, 2π  L (t) 4

⎫ ⎬

⎛

L 1 ,[t,π]

⎭

. (15.59)

Corollary 15.29 All as in Theorem 15.22. Further assume  L (1) = 1. Then ∀ t ∈ [−π, π], we have √ E (|M (X ) − X |) (t) ≤ 2 2 ⎧ ⎨

  2 2  L χ[−π,t] (·) (t) 3 + ⎩ 3

 23   13     |· − t|  sin χ[−π,t] (·) L (t) 4

15.4 Main Results

345

⎛

⎞     23   23 |· − t| sin χ[−π,t] (·) 1 ⎝ Dt− X, 2π  L (t) ⎠ 4 1 2

+ L 1 ,[−π,t]

  23   13       23 |· − t| 2   L χ[t,π] (·) (t) + χ[t,π] (·) sin L (t) 3 4 ⎞     23   23 |· − t| sin χ[t,π] (·) 1 ⎝ D∗t X, 2π  L (t) ⎠ 4

⎫ ⎬

⎛

1 2

L 1 ,[t,π]

⎭

.

(15.60)

Next we give uniform norm results. They are based on Theorems 15.23, 15.24. Corollary 15.30 All as in Theorem 15.23. Assume further 1. Then    1  E |M (X ) − X |q  q ≤ ∞

1 q

< α < 1 and  L (1) =

  q1 1 α π α 2α+ p (q + 1) (α+1) + 1 (15.61)

α

1

 (α) ( p (α − 1) + 1) p (q + 1) q(α+1)

⎧⎧  α   q(α+1)    q(α+1) ⎨⎨ |· − t|   sin χ[t,π] (·) (t) L  ⎩⎩ 4 ∞

 sup 1 t∈[−π,π]

α X, 2π D∗t

 1 (q+1)

    1  q(α+1)   q(α+1)   |·−t|   (t)  L sin 4 χ[t,π] (·) ∞

 L q ,[t,π]

⎧  α   q(α+1)    q(α+1) ⎨ |· − t|   sin χ[−π,t] (·) + L (t)  ⎩ 4 ∞

 α X, 2π sup 1 Dt−

t∈[−π,π]

 1 (q+1)

    1  q(α+1)   q(α+1)   |·−t|   sin χ L (t) (·) [−π,t]   4 ∞

 . L q ,[−π,t]

Corollary 15.31 All as in Theorem 15.24. Assume further 0 < α < 1 and  L (1) = 1. Then (2π)α (α + 2) E (|M (X ) − X |)∞ ≤  (α + 2) ⎧  α   α+1    α+1 ⎨ |· − t|   sin χ[−π,t] (·) (t) L  ⎩ 4 ∞

346

15 Trigonometric Caputo Fractional Approximation …

⎛

1 ⎞    α+1    α+1  |· − t|   α sin L χ[−π,t] (·) sup 1 ⎝ Dt− X, 2π  (t) ⎠   4 t∈[−π,π]

∞

L 1 ,[−π,t]

α    α+1    α+1  |· − t|   sin χ[t,π] (·) + L (t)   4

∞

1 ⎞    α+1    α+1  |· − t|   α sin L χ[t,π] (·) sup 1 ⎝ D∗t X, 2π  (t) ⎠   4 t∈[−π,π]

⎫ ⎬

⎛

∞

L 1 ,[t,π]

⎭

.

(15.62)

Corollary 15.32 All as in Theorem 15.24. Here α = E (|M (X ) − X |)∞

1 2

and  L (1) = 1. Then

√ 10 2 ≤ 3

⎧  1    23   3 ⎨ |· − t|   sin χ[−π,t] (·) (t) L  ⎩ 4

(15.63)

∞

⎛

2 ⎞     23   3   |· − t|   sin L sup 1 ⎝ Dt− X, 2π  χ[−π,t] (·) (t) ⎠   4 t∈[−π,π] 1 2

∞

L 1 ,[−π,t]

1     23   3  |· − t|   sin χ[t,π] (·) + L (t)   4

∞

2 ⎞     23   3  |· − t|   sin L sup 1 ⎝ D∗t X, 2π  χ[t,π] (·) (t) ⎠   4 t∈[−π,π]

⎫ ⎬

⎛

1 2

∞

L 1 ,[t,π]

⎭

.

15.5 Application Consider the Bernstein polynomials on [−π, π] for f ∈ C ([−π, π]) : B N ( f ) (x) =

 N   N k=0

k

 f

−π +

2πk N



x +π 2π

k 

π−x 2π

 N −k

,

(15.64)

15.5 Application

347

N ∈ N, any x ∈ [−π, π]. There are positive linear operators from C ([−π, π]) into itself. Setting g (t) = f (2πt − π), t ∈ [0, 1], we have g (0) = f (−π), g (1) = f (π), and    N   k N g t k (1 − t) N −k = (B N f ) (x) , x ∈ [−π, π] . (B N g) (t) = k N k=0 (15.65) Here x = ϕ (t) = 2πt − π is an 1 − 1 and onto map from [0, 1] onto [−π, π]. Clearly here g ∈ C ([0, 1]). Notice also that  %  & (2π)2 t (1 − t) B N (· − x)2 (x) = B N (· − t)2 (t) (2π)2 = N    1 π−x π2 (2π)2 x + π = = , ∀ x ∈ [−π, π] . (x + π) (π − x) ≤ N 2π 2π N N 

I.e.



 π2 , ∀ x ∈ [−π, π] . B N (· − x)2 (x) ≤ N

(15.66)

(B N 1) (x) = 1, ∀ x ∈ [−π, π] .

(15.67)

In particular

Define the corresponding application of M by  B N (X ) (t, ω) := B N (X (·, ω)) (t) =  N   N k=0

k

     t + π k π − t N −k 2πk ,ω , X −π + N 2π 2π

(15.68)

∀ N ∈ N, ∀ t ∈ [−π, π], ∀ ω ∈ , where X is a stochastic process. Clearly  B N is a stochastic process. We give Proposition 15.33 Let X (t, ω) be a stochastic process from [−π, π] × (, F, P) into R, where (, F, P) is a probability space. Here 0 < α < 1 (i.e. n = 1) and X (·, ω) ∈ AC ([−π, π]) with X (1) (·, ω) ∈ L ∞ ([−π, π]), ∀ ω ∈ . Further α X (z, ω) is a stochastic process for z ∈ [t, π], ω ∈ , and we assume that D∗t α Dt− X (z, ω) is a stochastic process for z ∈ [−π, t] , ω ∈ ; ∀ t ∈ [−π, π] . For α X (z, ω) is continuous in z ∈ [t, π], uniformly any t ∈ [−π, π] we assume that D∗t α X (z, ω) is conwith respect to ω ∈ . And for any t ∈ [−π, π] we assume that Dt− tinuous in z ∈ [−π, t], uniformly with respect to ω ∈ . Finally, we assume that (E |X |) (t) < ∞, ∀ t ∈ [−π, π]. Then, for any t ∈ [−π, π], we have:

348

15 Trigonometric Caputo Fractional Approximation …

  E  B N (X ) − X  (t) ≤ ⎧ ⎨ ⎩

 1 B N χ[−π,t] (·) (t) α+1 +

⎛

1 (α + 1)

(2π)α  (α + 1)



 BN

α α+1   α+1 |· − t| χ[−π,t] (·) sin (t) 4



1 ⎞   α+1   α+1 |· − t| α ⎠ sin χ[−π,t] (·) 1 ⎝ Dt− X, 2π B N (t) 4

 



 1 B N χ[t,π] (·) (t) α+1 + ⎛

1 (α + 1)



+ L 1 ,[−π,t]

α   α+1   α+1 |· − t| sin χ[t,π] (·) BN (t) 4

1 ⎞   α+1   α+1 |· − t| α ⎠ 1 ⎝ D∗t X, 2π B N sin χ[t,π] (·) (t) 4

⎫ ⎬



L 1 ,[t,π]

⎭

, (15.69)

∀ N ∈ N. 

Proof By Corollary 15.28. We give

Proposition 15.34 All as in Proposition 15.33 with α = 21 . Then, for any t ∈ [−π, π], we have: √   E  B N (X ) − X  (t) ≤ 2 2 ⎧ ⎨ ⎩

 2 2 B N χ[−π,t] (·) (t) 3 + 3 ⎛



   23   13 |· − t| sin χ[−π,t] (·) BN (t) 4

⎞    23   23 |· − t| sin χ[−π,t] (·) 1 ⎝ Dt− X, 2π B N (t) ⎠ 4 

1 2

 

 2 2 B N χ[t,π] (·) (t) 3 + 3

⎛

 1 2

1 ⎝ D∗t X, 2π B N ∀ N ∈ N.

 sin







BN

+ L 1 ,[−π,t]

 23   13 |· − t| sin χ[t,π] (·) (t) 4 

|· − t| χ[t,π] (·) 4

 23 

⎫ ⎬

 23 ⎞ (t) ⎠ L 1 ,[t,π]

⎭

,

(15.70)

15.5 Application

349



Proof By Proposition 15.33. We continue with

Proposition 15.35 All as in Proposition 15.33 with α = 21 . Then, for any t ∈ [−π, π] , we have: √     23   13   |· − t| 10 2 sin E  B N (X ) − X  (t) ≤ BN (t) 3 4 ⎡

⎛

⎞    23   23 |· − t| ⎣1 ⎝ Dt− X, 2π B N sin (t) ⎠ 4 

1 2

⎛

 1 2

1 ⎝ D∗t X, 2π B N





|· − t| sin 4

 23 

+ L 1 ,[−π,t]

 23 ⎞ (t)

⎤

⎠

⎦,

(15.71)

L 1 ,[t,π]

∀ N ∈ N. 

Proof By (15.70) and the positivity of B N , see also (15.39) and (15.41). We make

Remark 15.36 By |sin x| < |x|, ∀ x ∈ R − {0}, in particular sin x ≤ x, for x ≥ 0, we get    23  3 |· − t| |· − t| 2 1 3 sin ≤ = |· − t| 2 . 4 4 8 Hence     23     |· − t|   sin (t)  BN   4

∞

≤

   1 3   B N |· − t| 2 (t) . ∞ 8

(15.72)

We observe that 3       N      2πk  2 N t + π k π − t N −k 3  t + π − B N |· − t| 2 (t) =  k N  2π 2π k=0 (by discrete Hölder’s inequality)

350

15 Trigonometric Caputo Fractional Approximation …

+ N  , 43        2πk 2 N t + π k π − t N −k t +π− ≤ k N 2π 2π k=0  3 = B N (· − t)2 (t) 4 

Consequently it holds

3

π2

(15.66)

≤

3

N4

    3    B N |· − t| 2 (t)

∞

    23     |· − t|   sin (t)  BN   4

and

, ∀ t ∈ [−π, π] .

(15.73)

3

≤

π2

3

N4

,

(15.74)

3

≤

∞

π2

3

8N 4

, ∀ N ∈ N.

(15.75)

We further have Proposition 15.37 All as in Proposition 15.33 with α = 21 . Then, for any t ∈ [−π, π] we have: √   5 2π E  B N (X ) − X  (t) ≤ √ 4 N , +     1 1 π2 π2 2 2 + 1 D∗t X, √ , (15.76) 1 Dt− X, √ 2 N L 1 ,[−π,t] 2 N L 1 ,[t,π] ∀ N ∈ N.   As N → +∞, we get E  B N (X ) − X  (t) → 0. 

Proof By (15.71), (15.75) and positivity of B N . See also Proposition 15.5. Consequently we obtain

1 Proposition  15.38∗ All as in Proposition 15.33 with ∗α = 2 . Assume further that  (1)  X (t, ω) ≤ M , ∀ (t, ω) ∈ [−π, π] × , where M > 0. Then

     E  B N (X ) − X  

∞

+

 sup 1 t∈[−π,π]

π2 Dt− X, √ 2 N 1 2

√ 5 2π ≤ √ 4 N



 L 1 ,[−π,t]

+ sup 1 t∈[−π,π]

π2 D∗t X, √ 2 N 1 2

,

 L 1 ,[t,π]

,

(15.77) ∀ N ∈ N.     B N (X ) − X  ∞ → 0, i.e.  B N → I (stochastic unit As N → +∞, then  E  operator) in 1-mean.

15.5 Application

351



Proof By (15.76) and Remark 15.8, see (15.13), (15.14). Remark 15.39 (to Proposition 15.38) Assume that  1

π2 Dt− X, √ 2 N 1 2

 L 1 ,[−π,t]

K 1 π2 ≤ √ , ∀ t ∈ [−π, π] , ∀ N ∈ N, 2 N

(15.78)

where K 1 > 0. And assume that   1 π2 K 2 π2 2 ≤ √ , ∀ t ∈ [−π, π] , ∀ N ∈ N, 1 D∗t X, √ 2 N L 1 ,[t,π] 2 N

(15.79)

where K 2 > 0. Conditions (15.78), (15.79) are of Lipschitz type of order 1. By (15.77) and (15.78), (15.79), we easily derive that      E  B N (X ) − X  

∞

∀ N ∈ N, where θ > 0. Thus, at smoothness of order

1 2

≤

θ 3

N4

,

we achieve speed of convergence

(15.80)

1 3

N4

, N ∈ N.

Without any smoothness, in [5], we proved that the speed of convergence was √1N , N ∈ N. In the presence of the ordinary first derivative, see [5], in deterministic approximation the rate of convergence was N1 , N ∈ N. Naturally, we have as expected: 1 1 1 < 3 < √ , ∀ N ∈ N − {1}. N 4 N N

(15.81)

15.6 Trigonometric Stochastic Korovkin Results In this section  L, M are meant as sequences of operators. We give first pointwise results: Theorem Here all as in Theorem 15.21. Assume further that  L (1) (t) → 1  15.40 q(α+1)   |·−t| and  L sin 4 (t) → 0, pointwise in t ∈ [−π, π].  Then E |M (X ) − X |q (t) → 0, pointwise in t ∈ [−π, π], that is M → I (stochastic unit operator) in q-mean-pointwise with rates, quantitatively.

352

15 Trigonometric Caputo Fractional Approximation …

Proof We use (15.48), we take into account  L (1) (t) → 1, (15.36); and    L χ[t,π] (·) (t),  L χ[−π,t] (·) (t) ≤  L (1) (t), which  L (1) (t) is bounded, and by  ( 15.39), (15.41) and positivity of L we get that     q(α+1)  q(α+1)  |·−t| |·−t|   L sin 4 χ[t,π] (·) (t), L sin 4 χ[−π,t] (·) (t) ≤   q(α+1)   L sin |·−t| (t) → 0. 4 Finally, we use Proposition 15.6(ii) for the 1 (·, ·)’s to go to zero.



We continue with Theorem Here all as in Theorem 15.22. Assume further that  L (1) (t) → 1  15.41  (α+1) |·−t| and  L sin 4 (t) → 0, pointwise in t ∈ [−π, π]. Then E (|M (X ) − X |) (t) → 0, pointwise in t ∈ [−π, π], that is M → I in 1mean-pointwise with rates, quantitatively. Proof Based on (15.51), similar to the proof of Theorem 15.40, just take q = 1 there.  Next we give uniform results: Theorem 15.42 in Theorem Assume further that  L (1) → 1, uni Hereall as   15.23. q(α+1)   |·−t|  formly, and  (t)  L sin 4  → 0. ∞    Then  E |M (X ) − X |q ∞ → 0 over [−π, π] , that is M → I in the q-mean, quantitatively with rates.   Proof We use (15.54), we take into account  L (1) → 1, uniformly, (15.37);  L (1)∞ is bounded, use of (15.40), (15.42) and Remark 15.8, see there (15.13), (15.14).  Next we give the L 1 -mean uniform result  Theorem 15.43 in Theorem  Hereall as  15.24. Assume further that L (1) → 1, uni(α+1)   |·−t|  formly, and  (t)  L sin 4  → 0. ∞ Then E (|M (X ) − X |)∞ → 0 over [−π, π] , that is M → I in the 1-mean, quantitatively with rates. Proof Use of (15.55), similar to the proof of Theorem 15.42, just take q = 1 there.  We finish with Remark 15.44 An amazing fact/observation follows: In all trigonometric convergence results here, see Theorems 15.40–15.43, the forcing conditions for convergences are based only on  L and basic real valued continuous functions on [−π, π] and are not related to stochastic processes, but they are giving trigonometric convergence results on stochastic processes!

References

353

References 1. Aliprantis, C., Burkinshaw, O.: Principles of Real Analysis, 3rd edn. Academic, San Diego (1998) 2. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 3. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 4. Anastassiou, G.: Quantitative Approximations. Chapman & Hall/CRC, Boca Raton (2001) 5. Anastassiou, G.: Stochastic Korovkin theory given quantitatively, Facta Universitatis (Nis). Ser. Math. Inform. 22(1), 43–60 (2007) 6. Anastassiou, G.: Fractional Differentiation Inequalities. Springer, Heildelberg (2009) 7. Anastassiou, G.A.: Fuzzy Mathematics: Approximation Theory. Springer, Heildelberg (2010) 8. Anastassiou, G.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 9. Anastassiou, G.: Fractional representation formulae and right fractional inequalities. Math. Comput. Model. 54(11–12), 3098–3115 (2011) 10. Anastassiou, G.A.: Foundation of stochastic fractional calculus with fractional approximation of Stochastic processes. RACSAM 114 (2020). Article 89 11. Anastassiou, G.A.: Trigonometric fractional approximation of stochastic processes (2020). Submitted 12. Diethelm, K.: The Analysis of Fractional Differential Equations. Springer, New York (2010) 13. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 14. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 15. Shisha, O., Mond, B.: The degree of convergence of sequences of linear positive operators. Natl. Acad. Sci. U.S. 60, 1196–1200 (1968) 16. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192, 73–80 (1986) 17. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61, 74–87 (1990)

Chapter 16

Conformable Fractional Quantitative Approximation of Stochastic Processes

Here we consider and study very general stochastic positive linear operators induced by general positive linear operators that are acting on continuous functions. These are acting on the space of real conformable fractionally differentiable stochastic processes. Under some very mild, general and natural assumptions on the stochastic processes we produce related conformable fractional stochastic Shisha-Mond type inequalities of L q -type 1 ≤ q < ∞ and corresponding conformable fractional stochastic Korovkin type theorems. These are regarding the stochastic q -mean conformable fractional convergence of a sequence of stochastic positive linear operators to the stochastic unit operator for various cases. All convergences are produced with rates and are given via the conformable fractional stochastic inequalities involving the stochastic modulus of continuity of the αth conformable fractional derivatives of the engaged stochastic process, α > 0, α ∈ / N. The impressive fact is that the basic real Korovkin test functions assumptions are enough for the conclusions of our conformable fractional stochastic Korovkin theory. We give conformable fractional applications to stochastic Bernstein operators. See also [9].

16.1 Introduction Inspiration for this work comes from [2, 3, 14, 15]. This work continues our earlier work [5], now at the stochastic conformable fractional level. First we mention the basics of conformable fractional calculus. In the Sect. 16.3, about background, we talk about the q-mean (1 ≤ q < ∞) first modulus of continuity of a stochastic process and its upper bounds. There we describe completely our setting by introducing our stochastic positive linear operator M, see (16.31), which is based on the positive linear operator  L from C ([a, b]) into itself. The operator M is acting on a wide space of conformable fractional differentiable real valued stochastic processes X . See there Assumptions 16.16, 16.19, 16.20, 16.22. We first give the main © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_16

355

356

16 Conformable Fractional Quantitative Approximation …

pointwise conformable fractional stochastic Shisha-Mond type inequalities [13], see Theorems 16.23, 16.24, and their several corollaries covering important special cases. We continue with conformable fractional q-mean uniform Shisha-Mond type inequalities, see Theorems 16.30, 16.31, and their interesting corollaries. All this theory is regarding the conformable fractional stochastic convergence of operators M to I (stochastic unit operator) given quantitatively with rates. An extensive fractional conformable application about the stochastic Bernstein operators follows in full details. Based on our Shisha-Mond type inequalities of our main Theorems 16.23, 16.24, 16.30, 16.31 we derive pointwise and uniform Stochastic Korovkin theorems [11] on stochastic processes, see Theorems 16.41–16.44. The amazing fact here is, that basic conditions on operator  L regarding two simple real valued functions that are not stochastic, are able to enforce conformable fractional stochastic convergence on all stochastic processes we are dealing with; see Concepts 16.15 and Assumptions 16.16, 16.19, 16.20, 16.22.

16.2 Background—I Here we follow [1], see also [10]. We need Definition 16.1 ([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 

(16.1)

  If Tαa f (t) exists on (a, b), then 

   Tαa f (a) = lim Tαa f (t) . t→a+

(16.2)

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



(16.3)

 f (t) exists on (a, b), then b

αT

   f (b) = lim bα T f (t) .

Note that if f is differentiable then

t→b−

(16.4)

16.2 Background—I

357

 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,

(16.5)

(16.6)

(16.7)

a





Iα f (t) =

b

(b − x)α−1 f (x) d x,

(16.8)

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 16.2 ([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) .

(16.9)

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) .

(16.10)

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 (16.9), (16.10) collapse to {(16.1), (16.2)}, {(16.3), (16.4)} respectively. Lemma 16.3 ([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) .

(16.11)

We need Definition 16.4 (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.

(16.12)

358

16 Conformable Fractional Quantitative Approximation …

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.

(16.13)

t

We need Proposition 16.5 ([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

(16.14)

We also have Proposition 16.6 ([8, p. 154]) 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

(16.15)

If n = 0 or 0 < α ≤ 1, then (see also [1]) b

Iα bα T ( f ) (t) = f (t) − f (b) .

(16.16)

In conclusion we derive Theorem 16.7 ([8, p. 155]) 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 (16.17) 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 (16.18) ∀ t ∈ [a, b]. We make

16.2 Background—I

359

Remark 16.8 ([8, p. 155]) 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 (

(16.19)

   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) .

(16.20)

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,

(16.21)

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,

(16.22)

∀ t ∈ [a, b]. If f (k) (b) = 0, k = 1, . . . , n, then 1 f (t) − f (b) = − n!



b

(b − x)β−1 (x − t)n

t

b

αT (

 f ) (x) d x,

(16.23)

∀ t ∈ [a, b].

16.3 Background—II We need Definition 16.9 Here (, F, P) is a probability space, ω ∈ . We define the relative q-mean first modulus of continuity of stochastic process X (t, ω) by 1 (X, δ) L q ,[c,d] :=

360

16 Conformable Fractional Quantitative Approximation …

 sup





1 q

|X (x, ω) − X (y, ω)|q P (dω)

: x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ ,

(16.24) where δ > 0, 1 ≤ q < ∞, t ∈ [a, b] ⊂ R. Definition 16.10 Let 1 ≤ q < ∞. Let X (x, ω) be a stochastic process. We call X a q-mean uniformly continuous stochastic process over [a, b], iff ∀ ε > 0 ∃ δ > 0 : whenever |x − y| ≤ δ; x, y ∈ [a, b] implies that  

|X (x, s) − X (y, s)|q P (ds) ≤ ε.

(16.25)

U

We denote it as X ∈ CR q ([a, b]) . It holds U

Proposition 16.11 ([5]) Let X ∈ CR q ([a, b]), then 1 (X, δ) L q ,[a,b] < ∞, any δ > 0. Also it holds Proposition 16.12 ([5]) Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R. Then following are true ([c, d] ⊆ [a, b]): (i) 1 (X, δ) L q ,[c,d] is nonnegative and nondecreasing in δ > 0, U (ii) lim1 (X, δ) L q ,[c,d] = 1 (X, 0) L q ,[c,d] = 0, iff X ∈ CR q ([c, d]) , δ↓0

(iii) 1 (X, δ1 + δ2 ) L q ,[c,d] ≤ 1 (X, δ1 ) L q ,[c,d] + 1 (X, δ2 ) L q ,[c,d] , δ1 , δ2 > 0, (iv) 1 (X, mδ) L q ,[c,d] ≤ m1 (X, δ) L q ,[c,d] , δ > 0, m ∈ N, (v) 1 (X, λδ) L q ,[c,d] ≤ λ 1 (X, δ) L q ,[c,d] ≤ (λ + 1) 1 (X, δ) L q ,[c,d] , λ > 0, δ > 0, · is the ceiling of the number, (vi) 1 (X + Y, δ) L q ,[c,d] ≤ 1 (X, δ) L q ,[c,d] + 1 (Y, δ) L q ,[c,d] , δ > 0, U (vii) 1 (X, ·) L q ,[c,d] is continuous on R+ for X ∈ CR q ([c, d]) . We give Remark 16.13 (to Proposition 16.12) By Proposition 16.12(v) we get  1 (X, |x − y|) L q ,[c,d] ≤

|x − y| 1 (X, δ) L q ,[c,d] , δ

(16.26)

∀ x, y ∈ [c, d], any δ > 0. We give n+1 Remark 16.14 Let α ∈ (n, ([a, b]), ∀ ω ∈ . We

n +∗1], n ∈ Z+ , X (·, ω) ∈ C

(n+1)

assume that X (t, ω) ≤ M , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Let δ > 0, 1 ≤ q < ∞. Then

16.3 Background—II

361

  1 Tcα X, δ L q ,[c,d] = sup

 

 c   

T X (x, ω) − Tc X (y, ω) q P (dω) α α

q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤  sup 



   c  

T X (x, ω) + Tc X (y, ω) q P (dω) α α

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ}  sup 

q1

:

(16.19.)

≤

(β:=α−n)



q  (x − c)1−β X (n+1) (x, ω) + (y − c)1−β X (n+1) (y, ω) P (dω)

q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ ∗



M sup 

 q (x − c)1−β + (y − c)1−β P (dω)

q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ 2M ∗ (b − a)1−β . That is

  1 Tcα X, δ L q ,[c,d] ≤ 2M ∗ (b − a)1−β ,

(16.27)

for any [c, d] ⊆ [a, b] . Similarly, it holds 1

d



α TX, δ L q ,[c,d]

≤ 2M ∗ (b − a)1−β ,

(16.28)

for any [c, d] ⊆ [a, b] . In particular it holds   sup 1 Ttα X, δ L q ,[t,b] ≤ 2M ∗ (b − a)1−β ,

(16.29)

t∈[a,b]

and sup 1 t∈[a,b]

t



α TX, δ L q ,[a,t]

≤ 2M ∗ (b − a)1−β .

(16.30)

Above, it is not strange to assume that Tcα X,dα TX, Ttα X,tα TX are stochastic processes, see Remark 16.8. We need

362

16 Conformable Fractional Quantitative Approximation …

Concepts 16.15 Let  L be a positive linear operator from C ([a, b]) into itself. Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R, where (, F, P) is a probability space. Let α ∈ (n, n + 1], n ∈ Z+ , and X (·, ω) ∈ C n+1 ([a, b]), ∀ ω ∈ . Also we assume for each t ∈ [a, b] that X (k) (t, ·) is measurable over (, F) for all k = 1, . . . , n + 1. Define M (X ) (t, ω) :=  L (X (·, ω)) (t) , ∀ ω ∈ , ∀ t ∈ [a, b] ,

(16.31)

and assume that it is a random variable in ω. Clearly M is a positive linear operator on stochastic processes. We make Assumption 16.16 (i) For any t ∈ [a, b] we assume that Ttα X (z, ω) is continuous in z ∈ [t, b], uniformly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ > 0 : whenever |z 1 − z 2 | ≤ δ; z 1 , z 2 ∈ [t, b], then Ttα X (z 1 , ω) − Ttα X (z 2 , ω) ≤ ε, ∀ ω ∈ . We denote this by Ttα X ∈ CRU ([t, b]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [t, b] . (ii) For any t ∈ [a, b] we assume that tα TX (z, ω) is continuous in z ∈ [a, t], uniformly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ > 0 : whenever |z 1 − z 2 | ≤ δ; z 1 , z 2 ∈ [a, t], then tα TX (z 1 , ω) −tα TX (z 2 , ω) ≤ ε, ∀ ω ∈ . We denote this by tα TX ∈ CRU ([a, t]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [a, t] . Remark 16.17 Assumption 16.16 implies: (i) Ttα X (·, ω) ∈ C ([t, b]), ∀ ω ∈ , and Ttα X is q-mean uniformly continuous U in z ∈ [t, b], that is Ttα X ∈ CR q ([t, b]), for any 1 ≤ q < ∞. t (ii) α TX (·, ω) ∈ C ([a, t]), ∀ ω ∈ , and tα TX is q-mean uniformly continuous U in z ∈ [a, t], that is tα TX ∈ CR q ([a, t]), for any 1 ≤ q < ∞. We need Definition 16.18 Denote by  (E X ) (t) :=



X (t, ω) P (dω) , ∀ t ∈ [a, b] ,

(16.32)

the expectation operator. We make Assumption 16.19 We assume that 

q  E X (k) (t) < ∞, ∀ t ∈ [a, b] , q > 1, for all k = 0, 1, . . . , n.

(16.33)

16.3 Background—II

363

We make Assumption 16.20 We assume that  (k)  E X (t) < ∞, ∀ t ∈ [a, b] ,

(16.34)

for all k = 0, 1, . . . , n. We give Remark 16.21 By the Riesz representation theorem [12] we have that there exists μt unique, completed Borel measure on [a, b] with m t := μt ([a, b]) =  L (1) (t) > 0,

(16.35)

f (x) dμt (x) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(16.36)

such that  L ( f ) (t) =

 [a,b]

Consequently we have that  X (x, ω) dμt (x) , ∀ (t, ω) ∈ [a, b] × ,

M (X ) (t, ω) =

(16.37)

[a,b]

and X as in Concepts 16.15. Here χ[γ,δ] (s) stands for the characteristic function on [γ, δ] ⊆ [a, b] . Notice that (r > 0)     χ[t,b] (s) |s − t|r μt (ds) =  L |· − t|r χ[t,b] (·) (t) , (s − t)r μt (ds) = [t,b]

[a,b]

(16.38) and 

 (t − s) μt (ds) = r

[a,t]

  χ[a,t] (s) |s − t|r μt (ds) =  L |· − t|r χ[a,t] (·) (t) .

[a,b]

(16.39) Let α ∈ (n, n + 1], n ∈ N and k = 1, . . . , n. Then by Hö lder’s inequality we obtain





[a,b]

 ≤ [a,b]

Therefore it holds



(x − t) dμt (x)

≤

|x − t|k dμt (x)

k

|x − t|α+1 dμt (x)

[a,b]

k ( α+1 )

(μt ([a, b]))(

α+1−k α+1

).

(16.40)

364

16 Conformable Fractional Quantitative Approximation …

         L |· − t|k (t)∞,[a,b] ≤ L (· − t)k (t)∞,[a,b] ≤  k   ( α+1−k )  ( α+1 )  α+1   , L |· − t|α+1 (t)∞,[a,b] L (1)∞,[a,b]

(16.41)

all k = 1, . . . , n. Also, we observe that

and

C ([a, b])  |· − t|α+1 χ[a,t] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] ,

(16.42)

C ([a, b])  |· − t|α+1 χ[t,b] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] .

(16.43)

By positivity of  L we obtain          L |· − t|α+1 (t)∞,[a,b] < ∞, L |· − t|α+1 χ[a,t] (·) (t)∞,[a,b] ≤ 

(16.44)

  by  L |· − t|α+1 (t) being continuous in t ∈ [a, b], see p. 388 of [4], and          L |· − t|α+1 (t)∞,[a,b] . L |· − t|α+1 χ[t,b] (·) (t)∞,[a,b] ≤ 

(16.45)

Above (16.38)–(16.41) and (16.44), (16.45) can be used to derive convergences from our main results next.   L (1) (t), and by In this work we denote by  L χ[a,t] (·) (t) := μt ([a, t]) ≤     L χ[t,b] (·) (t) := μt ([t, b]) ≤  L (1) (t) . We make

Assumption 16.22 Assume that X (n+1) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. And suppose that X (n+1) (·, ω) is continuous over [a, b], uniformly with respect to ω ∈ . Clearly, Assumption 16.22 implies Assumption 16.16, see also Remark 16.8.

16.4 Main Results Next we present our first main result on the quantitative stochastic fractional approximation regarding stochastic processes: Theorem 16.23 Let n ∈ Z+ , α ∈ (n, n + 1) ; p, q > 1 : 1p + q1 = 1, with β := α − (q−1)  p(β−1)+1) n > q1 ; set λ1 := (np+1)( . Suppose Concepts 16.15, Assumptions ( p(α−1)+2) 16.16 and 16.19.

16.4 Main Results

365

Then, ∀ t ∈ [a, b], we have:

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + 1  (k) q   q1 1  E X (t)  2 p λ1q k

 L (· − t) (t) + α k! (q + 1) q(α+1) n!

n  k=1



 1



Ttα X,

 1 q(α+1)  1  q(α+1) L |· − t| χ[t,b] (·) (t) (q + 1)

L q ,[t,b]

   1     α  L χ[t,b] (·) (t) p  L |· − t|q(α+1) χ[t,b] (·) (t) q(α+1)    q1  1   α+1 α  (α+1) + L χ[t,b] (·) (t) +1 (q + 1) 



1



t α TX,

 1 q(α+1)  1  q(α+1) L |· − t| χ[a,t] (·) (t) (q + 1)

(16.46)

L q ,[a,t]

   1     α  L χ[a,t] (·) (t) p  L |· − t|q(α+1) χ[a,t] (·) (t) q(α+1)    q1    1 α  L χ[a,t] (·) (t) α+1 (q + 1) (α+1) + 1 Proof By Theorem 16.7 we have X (s, ω) = 1 n!



s

t

n  X (k) (t, ω) (s − t)k + k! k=0

  (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz,

(16.47)

for all t ≤ s ≤ b, ∀ ω ∈ ; α ∈ (n, n + 1), see also (16.21). Furthermore it holds from Theorem 16.7 that X (s, ω) = 1 n!



t s

n  X (k) (t, ω) (s − t)k − k! k=0

(t − z)β−1 (z − s)n

t

α TX

 (z, ω) −tα TX (t, ω) dz,

(16.48)

366

16 Conformable Fractional Quantitative Approximation …

for all a ≤ s ≤ t, ∀ ω ∈ ; α ∈ (n, n + 1), see also (16.21). Let p, q > 1 : 1p + q1 = 1, and α > n + q1 , i.e. β := α − n > q1 , implying p (β − 1) + 1 > 0. Next, we work on the remainder of (16.47). Using Hölder’s inequality we have 

s t

(s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz ≤ 

t

T X (z, ω) − Tt X (t, ω) q dz α α

s t



s

q1

(s − z)(np+1)−1 (z − t)( p(β−1)+1)−1 dz

1p

=

(16.49)

t



t

T X (z, ω) − Tt X (t, ω) q dz α α

s t



 (np + 1)  ( p (β − 1) + 1)  ( p (α − 1) + 2)

1p

q1

(s − t)

qα−1 q

,

t ≤ s ≤ b. Hence we have q  s

n β−1 t t

Tα X (z, ω) − Tα X (t, ω) dz ≤ (s − z) (z − t) t



s t



t

T X (z, ω) − Tt X (t, ω) q dz α α

 (np + 1)  ( p (β − 1) + 1)  ( p (α − 1) + 2)

t ≤ s ≤ b. We call

 λ1 :=

(q−1)



(s − t)(qα−1) ,

 (np + 1)  ( p (β − 1) + 1)  ( p (α − 1) + 2)

(q−1)

.

Applying again Hölder’s inequality we obtain: 1 :=

1 n!

  

b t



s

(s − z)n (z − t)β−1

t



t 1 

T X (z, ω) − Tt X (t, ω) dz μt (ds) q P (dω) q α α

(16.50)

(16.51)

16.4 Main Results

367 1

≤

(μt ([t, b])) p n!

  



b

t

s

(s − z)n (z − t)β−1

t



t  1

T X (z, ω) − Tt X (t, ω) dz q μt (ds) P (dω) q α α 1

1

λ1q (μt ([t, b])) p n!

((16.50.),(16.51.))

≤

  



b t

t

s

(16.52)

t

T X (z, ω) − Tt X (t, ω) q dz α α



  1 (s − t)(qα−1) μt (ds) P (dω) q =: (∗) ,

q where ϕ (z, ω) := Ttα X (z, ω) − Ttα X (t, ω) ≥ 0, is a real valued random variable for each z ∈ [t, b] (by assumptions Ttα X (z, ω), t ≤ z ≤ b, is continuous in z and measurable in ω). Hence ϕ (z, is continuous in z and jointly measurable, see  ω) s [6, p. 353]. Thus δ (s, ω) := t ϕ (z, ω) dz (s − t)qα−1 is a real valued random variable in ω and continuous in s ∈ [t, b]. Again by [6, p. 353], δ (s, ω) is jointly measurable in (s, ω) . Therefore by applying twice Tonelli’s theorem, [12, p. 270], we get 1

1

λ q (μt ([t, b])) p (∗) = 1 n! 

b

t

 s  

t

 q1

t

T X (z, ω) − Tt X (t, ω) q P (dω) dz (s − t)(qα−1) μt (ds) . α α (16.53)

We have proved that 1 1 := n!

  

b t



s

(s − z)n (z − t)β−1

t



t 1 

T X (z, ω) − Tt X (t, ω) dz μt (ds) q P (dω) q α α 1

1

λ q (μt ([t, b])) p ≤ 1 n! 

b t

 s  t



 q1

q

t t (qα−1)

T X (z, ω) − T X (t, ω) P (dω) dz (s − t) μt (ds) α α (16.54)

(continuing estimation) 1

1

λ q (μt ([t, b])) p ≤ 1 n!

368

16 Conformable Fractional Quantitative Approximation …



b



t

s t

 q1  q 1 Ttα X, z − t L q ,[t,b] dz (s − t)(qα−1) μt (ds)

(let h 1 > 0)

1

(16.26.)

≤

1

  λ q (μt ([t, b])) p 1 Ttα X, h 1 L q ,[t,b] 1 n! 

b

 s 

t

t

z−t h1

q

dz (s − t)

(qα−1)

1

μt (ds)

 q1

≤

(16.55)

1

  λ q (μt ([t, b])) p 1 Ttα X, h 1 L q ,[t,b] 1 n! 

b

t



b

τ t

 q1  s  z−t q (qα−1) dz (s − t) μt (ds) ≤ 1+ h1 t  q1  s  (z − t)q (qα−1) 1+ dz (s − t) μt (ds) = q h1 t

(where 1

1

  λ q (μt ([t, b])) p τ := 2 1 Ttα X, h 1 L q ,[t,b] 1 n! 1 p

)

(16.56)

 τ

 q1   s  1 qα q (qα−1) μt (ds) = (z − t) dz (s − t) (s − t) + q h1 t [t,b]  q1   1 (s − t)q+1 qα (qα−1) μt (ds) = (s − t) (s − t) + q h 1 (q + 1) [t,b]

 τ





τ [t,b]

q1  q(α+1) 1 − t) (s = μt (ds) (s − t)qα + q h 1 (q + 1)

 τ

(s − t)qα μt (ds) + [t,b]

1 q h 1 (q + 1)

 τ

(s − t)

q(α+1)

[t,b]

μt (ds)

 (s − t)q(α+1) μt (ds) [t,b] α α+1

1

(μt ([t, b])) α+1 +

(16.57)  q1

≤

16.4 Main Results

369

1 q h 1 (q + 1)

 (s − t)q(α+1) μt (ds)

 q1

=: (∗∗) .

[t,b]

We set and assume  h 1 :=

1 (q + 1)

I.e. q(α+1)

h1

=

 (s − t)q(α+1) μt (ds)

1 q(α+1)

> 0.

(16.58)

[t,b]

1 (q + 1)

 (s − t)q(α+1) μt (ds) > 0.

(16.59)

[t,b]

Therefore we get  1 α qα qα (∗∗) = τ (μt ([t, b])) α+1 h 1 (q + 1) α+1 + h 1  1 α τ h α1 (μt ([t, b])) α+1 (q + 1) α+1 + 1

1 q

1 q

=

.

(16.60)

We have proved that  1 α 1 ≤ τ h α1 (μt ([t, b])) α+1 (q + 1) α+1 + 1

1 q

.

(16.61)

The last means that 1 n!

1 :=

  

b



t

s

(s − z)n (z − t)β−1

t



t 1 

T X (z, ω) − Tt X (t, ω) dz μt (ds) q P (dω) q α α  1 p

≤ 2 1



1 Ttα X, (q + 1)

1

1

λ1q (μt ([t, b])) p n! 



 (s − t)q(α+1) μt (ds)

 1 q(α+1)

[t,b]

1 (q + 1) 1

L q ,[t,b]

 (s − t)q(α+1) μt (ds)

α q(α+1)

[t,b] α

(μt ([t, b])) α+1 (q + 1) α+1 + 1

1 q

.

Next we work on the remainder related to (16.48) (a ≤ s ≤ t), we have

(16.62)

370

16 Conformable Fractional Quantitative Approximation …



t

s

(t − z)β−1 (z − s)n tα TX (z, ω) −tα TX (t, ω) dz ≤ 

t

s



t

t

TX (z, ω) −t TX (t, ω) q dz α α

(t − z)

( p(β−1)+1)−1

( pn+1)−1

(z − s)

q1

1p dz

=

(16.63)

s



t

s



t

TX (z, ω) −t TX (t, ω) q dz α α

 ( p (β − 1) + 1)  ( pn + 1)  ( p (α − 1) + 2)

1p

q1

(t − s)

qα−1 q

,

a ≤ s ≤ t. Hence we have q  t

β−1 n t t

(t − z) (z − s) α TX (z, ω) −α TX (t, ω) dz ≤ s



t s



t

TX (z, ω) −t TX (t, ω) q dz α α

 ( p (β − 1) + 1)  ( pn + 1)  ( p (α − 1) + 2)

(q−1)



(t − s)(qα−1) ,

a ≤ s ≤ t. Applying again Hölder’s inequality we obtain 2 :=

1 n!

  t  

a

t

(t − z)β−1 (z − s)n

s



t 1 

TX (z, ω) −t TX (t, ω) dz μt (ds) q P (dω) q α α 1

(μt ([a, t])) p ≤ n!

  t  

a

t

(t − z)β−1 (z − s)n

s



t  1

TX (z, ω) −t TX (t, ω) dz q μt (ds) P (dω) q α α 1

((16.51.),(16.64.))

≤

1

λ1q (μt ([a, t])) p n!

(16.64)

16.4 Main Results

371

 1   t  t  q

t

TX (z, ω) −t TX (t, ω) q dz (t − s)(qα−1) μt (ds) P (dω) α α 

a

s

(16.65)

(as before) 1

1

λ q (μt ([a, t])) p = 1 n!  t  t  a



s

 q1

t

TX (z, ω) −t TX (t, ω) q P (dω) dz (t − s)(qα−1) μt (ds) α α 1

1

λ q (μt ([a, t])) p ≤ 1 n!  t  a

t

s

q 1 tα TX, t

−z



 L q ,[a,t]

(let h 2 > 0) 1

t

a

s

1

1



α TX, h 2 L q ,[a,t]

 t  t 

t −z h2

t

q

dz (t − s)

(qα−1)

μt (ds)

 q1

≤

1

λ1q (μt ([a, t])) p n!

 q1 dz (t − s)(qα−1) μt (ds) ≤ 1



α TX, h 2 L q ,[a,t]

1

λ1q (μt ([a, t])) p n!

 q1  t  t  t−z q (qα−1) 1+ dz (t − s) μt (ds) ≤ h2 a s  q1  t  t  (t − z)q (qα−1) 1+ dz (t − s) ρ μt (ds) = q h2 a s (where 1 p

ρ := 2 1 )

t



α TX, h 2 L q ,[a,t]

1

1

λ1q (μt ([a, t])) p n!

(16.66)

372

16 Conformable Fractional Quantitative Approximation …



 ρ

(t − s)qα + [a,t]

1 q h2



t

 q1  (t − z)q dz (t − s)(qα−1) μt (ds) =

s

 q1   1 (t − s)q+1 qα (qα−1) μt (ds) = (t − s) (t − s) + q h 2 (q + 1) [a,t]

 ρ





ρ [a,t]

 ρ

(t − s)

qα

[a,t]

q1  q(α+1) 1 − s) (t = μt (ds) (t − s)qα + q h 2 (q + 1)

1 μt (ds) + q h 2 (q + 1)

 ρ

(t − s)

q(α+1)

μt (ds)

 μt (ds)

(t − s)

q(α+1)

(16.67)  q1

≤

[a,t] α α+1

1

(μt ([a, t])) α+1 +

[a,t]

1 q h 2 (q + 1)

 (t − s)

q(α+1)

μt (ds)

 q1

=: (∗ ∗ ∗) .

[a,t]

We set and assume  h 2 :=

1 (q + 1)

I.e. q(α+1) h2

 (t − s)

q(α+1)

μt (ds)

1 q(α+1)

> 0.

(16.68)

[a,t]

1 = + (q 1)

 (t − s)q(α+1) μt (ds) > 0.

(16.69)

[a,t]

Therefore we get  1 α qα qα (∗ ∗ ∗) = ρ (μt ([a, t])) α+1 h 2 (q + 1) α+1 + h 2  1 α ρh α2 (μt ([a, t])) α+1 (q + 1) α+1 + 1 That is

1 q

 1 α 2 ≤ ρh α2 (μt ([a, t])) α+1 (q + 1) α+1 + 1

The last means 2 :=

1 n!

  t  

a

t s

1 q

=

(16.70)

.

(16.71)

. 1 q

(t − z)β−1 (z − s)n

16.4 Main Results

373



t 1 

TX (z, ω) −t TX (t, ω) dz μt (ds) q P (dω) q α α  1 p

≤ 2 1



1 (q + 1)

t α TX,

1

1

λ1q (μt ([a, t])) p n!



 (t − s)

q(α+1)

μt (ds)

 1 q(α+1)

[a,t]

1 (q + 1)

L q ,[a,t]

 (t − s)q(α+1) μt (ds)

α q(α+1)

[a,t]

 1 α (μt ([a, t])) α+1 (q + 1) α+1 + 1

1 q

.

(16.72)

Next we connect facts together. We have that M (X ) (t, ω) − X (t, ω)  L (1) (t) =  X (s, ω) μt (ds) − X (t, ω)  L (1) (t) = [a,b]



 X (s, ω) μt (ds) +

(t,b]

[a,t]



 [a,t]

1 n!



t s

 (t,b]



((16.47.),(16.48.))

=

(16.73)

n  X (k) (t, ω) (s − t)k − k! k=0

(t − z)β−1 (z − s)n 

1 n!

X (s, ω) μt (ds) − X (t, ω)  L (1) (t)

  t TX ω) − TX ω) dz μt (ds) + (z, (t, α α

t

n  X (k) (t, ω) (s − t)k + k! k=0

   (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz μt (ds)

s

t

−X (t, ω)  L (1) (t) =  n  X (k) (t, ω) (s − t)k μt (ds) − k! [a,t] k=0 1 n!



 [a,t]

s

t

β−1

(t − z)

(z −



s)n tα TX

(z, ω)

−tα





TX (t, ω) dz μt (ds) +

374

16 Conformable Fractional Quantitative Approximation …

 n  X (k) (t, ω) (s − t)k μt (ds) + k! (t,b] k=0 1 n!



 (t,b]

s t

  (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz μt (ds) − X (t, ω)  L (1) (t) =

(16.74)

n    X (k) (t, ω)   L (· − t)k (t) + k! k=1

1 n!



 (t,b]





t

s

t

  (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz μt (ds) −

(t − z)β−1 (z − s)n

s

[a,t]

  t . (16.75) TX ω) − TX ω) dz μ (z, (t, (ds) t α α

t

Furthermore we have

M (X ) (t, ω) − X (t, ω)  L (1) (t) ≤

n

(k)   X (t, ω) 

 L (· − t)k (t) + k! k=1 1 n!



 [t,b]

t



 [a,t]

s

t



(s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz μt (ds) +

(t − z)

β−1

s

(z −

s)n tα TX

(z, ω)

−tα

(16.76)

TX (t, ω) dz μt (ds) .

We have eventually that |M (X ) (t, ω) − X (t, ω)| ≤



M (X ) (t, ω) − X (t, ω)  L (1) (t) − 1 ≤ L (1) (t) + |X (t, ω)| 

n (k)

  X (t, ω) 



 |X (t, ω)| L (1) (t) − 1 + L (· − t)k (t) + k! k=1 1 n!





s

[t,b]

t

β−1

(s − z) (z − t) n



t

T X (z, ω) − Tt X (t, ω) dz μt (ds) + α α (16.77)

16.4 Main Results







(t − z)β−1 (z − s)n tα TX (z, ω) −tα TX (t, ω) dz μt (ds) .

t s

[a,t]

375

Hence it holds

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 +  (k) q   q1  E X (t) 

 L (· − t)k (t) + k!

n  k=1

1 n!

  

 [t,b]

s

(s − z)n (z − t)β−1

t



t 1 

T X (z, ω) − Tt X (t, ω) dz μt (ds) q P (dω) q α α   +





t

(16.78)

(t − z)β−1 (z − s)n

s

[a,t]



t 1 

TX (z, ω) −t TX (t, ω) dz μt (ds) q P (dω) q . α α By (16.62) and (16.72) we obtain

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 +  (k) q   q1  E X (t) 

 L (· − t)k (t) + k!

n  k=1 1





1

2 p λ1q

α

(q + 1) q(α+1) n!

1



Ttα X,

1

1 (q + 1)

 (s − t)

q(α+1)

[t,b]

L q ,[t,b]

 (s − t)q(α+1) μt (ds)

(μt ([t, b])) p

μt (ds)

 1 q(α+1)

α q(α+1)

[t,b]

  1

t α TX,



α

1

(μt ([t, b])) α+1 (q + 1) α+1 + 1

1 (q + 1)

1 q

 (t − s)

q(α+1)

[a,t]

μt (ds)

+  1 q(α+1) L q ,[a,t]

376

16 Conformable Fractional Quantitative Approximation …

(μt ([a, t]))



1 p

(t − s)

q(α+1)

μt (ds)

α q(α+1)

[a,t]

 1 α (μt ([a, t])) α+1 (q + 1) α+1 + 1

1 q

 .

(16.79)

Clearly (16.79) is trivially valid when t = a or t = b, from what follows, see (16.81) and (16.85), just replace [t, b] by [a, b] or [a, t] by [a, b], respectively. Next assume that  (16.80) (s − t)q(α+1) μt (ds) = 0, [t,b]

then (s − t)q(α+1) = 0, a.e. on [t, b], thus s = t, a.e. on [t, b], i.e. μt {s ∈ [t, b] : s = t} = 0, hence μt (t, b] = 0. Therefore μt concentrates on [a, t]. In that case inequality (16.79) becomes

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n  k=1 1



1

2 p λ1q

α

 (k) q   q1  E X (t) 

 L (· − t)k (t) + k!

(q + 1) q(α+1) n!

1



1 t α TX, (q + 1)

(μt ([a, t]))

1 p

 (t − s)q(α+1) μt (ds)

 1 q(α+1)

[a,t]

 (t − s)q(α+1) μt (ds)

L q ,[a,t] α q(α+1)

(16.81)

[a,t]

  q1 1 α (μt ([a, t])) α+1 (q + 1) α+1 + 1 . So far inequality (16.81) is valid when  (t − s)q(α+1) μt (ds) > 0. [a,t]

 In the case of [a,t] (t − s)q(α+1) μt (ds) = 0, we have (t − s)q(α+1) = 0, a.e. on [a, t], thus s = t, a.e. on [a, t], i.e. μt {s ∈ [a, t] : s = t} = 0, hence μt [a, t) = 0. Therefore μt = δt μt ([a, b]) = δt m t , where δt is the unit Dirac measure at t. Clearly then  L (· − t)k (t) = 0, all k = 1, . . . , n. Then (16.81) collapses to

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 ,

(16.82)

16.4 Main Results

377

equivalently,

q     L (1) (t) − 1 . E |M (X ) − X |q (t) ≤ E |X |q (t) 

(16.83)

 By (16.37) we have that  M (X ) (t, ω) = X (t, ω) L (1) (t). Therefore M (X ) (t, ω) − X (t, ω) = X (t, ω)  L (1) (t) − 1 . Hence, obviously, (16.83) holds as equality. Thus inequality (16.81) is valid trivially. Finally, let us go the other way around. Let us assume that  (t − s)q(α+1) μt (ds) = 0,

(16.84)

[a,t]

then (t − s)q(α+1) = 0, a.e. on [a, t], thus s = t, a.e. on [a, t], i.e. μt {s ∈ [a, t] : s = t} = 0, hence μt [a, t) = 0. Therefore μt concentrates on [t, b] . In that case inequality (16.79) becomes

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + n  k=1 1



1

2 p λ1q

α

 (k) q   q1  E X (t) 

 L (· − t)k (t) + k!

(q + 1) q(α+1) n!

1



Ttα X,

(μt ([t, b]))

1 p

1 (q + 1)

 (s − t)

q(α+1)

μt (ds)

 1 q(α+1)

[t,b]

L q ,[t,b]

 (s − t)

q(α+1)

μt (ds)

α q(α+1)

(16.85)

[t,b]

  q1 1 α (μt ([t, b])) α+1 (q + 1) α+1 + 1 . So inequality (16.85) is valid when  (s − t)q(α+1) μt (ds) > 0. [t,b]

 In the case of [t,b] (s − t)q(α+1) μt (ds) = 0, we have (s − t)q(α+1) = 0, a.e. on [t, b], hence s = t, a.e. on [t, b]. Therefore μt {s ∈ [t, b] : s = t} = 0, thus μt (t, b] = 0. L (· − t)k (t) = 0, all Consequently μt = δt μt ([a, b]) = δt m t . Clearly then  k = 1, . . . , n. Therefore (16.85) collapses to

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 .

(16.86)

378

16 Conformable Fractional Quantitative Approximation …

The last is an equality as earlier, see (16.83), etc. Thus inequality (16.85) is valid trivially. Inequality (16.79) has been proved in all possible cases. Inequality (16.46) is equivalent to (16.79) via (16.36). Theorem 16.23 now is proved completely.  It follows our second main result, the L 1 -quantitative stochastic fractional approximation of stochastic processes. Theorem 16.24 Let n ∈ Z+ and α ∈ (n, n + 1), β := α − n. Concepts 16.15, Assumptions 16.16, 16.20. Then, ∀ t ∈ [a, b], we have:

Suppose

E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) − 1 + n 

(k)

   E X (t)  1

 L (· − t)k (t) + n ! k! k=1 (β + j) j=1

⎧⎡   ⎤   1 ⎨    α L χ[a,t] (·) (t) α+1 1 ⎦   ⎣ + L |· − t|α+1 χ[a,t] (·) (t) α+1 ⎩ α−n α+1 1



t α TX,

    1   L |· − t|α+1 χ[a,t] (·) (t) α+1

L 1 ,[a,t]

+

⎤ ⎡    1      α L χ[t,b] (·) (t) α+1 1 ⎦  ⎣ L |· − t|α+1 χ[t,b] (·) (t) α+1 + α−n (α + 1)      1  1 Ttα X,  L |· − t|α+1 χ[t,b] (·) (t) α+1

 L 1 ,[t,b]

.

(16.87)

Proof We have that

E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 +

n  (k)    E X (t) 

 L (· − t)k (t) + k! k=1 1 n!

  t  

a

t

(t − z)β−1 (z − s)n

s

t

  

TX (z, ω) −t TX (t, ω) dz μt (ds) P (dω) + α α

(16.88)

16.4 Main Results

 

379



b

s

β−1

(s − z) (z − t) n



t

t

t

T X (z, ω) − Tt X (t, ω) dz α α



μt (ds)) P (dω)]} = (by applying Tonelli’s theorem twice) n 

(k)



  E X (t) 



 L (· − t)k (t) + (E |X |) (t) L (1) (t) − 1 + k! k=1

1 n!

 t  t  a



s

t

TX (z, ω) −t TX (t, ω) P (dω) α α



  (t − z)β−1 (z − s)n dz μt (ds) + 

b

 s 

t



t

t

T X (z, ω) − Tt X (t, ω) P (dω) α α



  (s − z)n (z − t)β−1 dz μt (ds) ≤ n 

(k)



  E X (t) 





L (· − t)k (t) + (E |X |) (t) L (1) (t) − 1 + k! k=1

1 n!

 t  a



b 

t

t

s s

t

1

t

α TX, t − z



(16.89)

 β−1 n + dz μ − z) − s) (t (z (ds) t L 1 ,[a,t]

  t  n β−1 1 Tα X, z − t L 1 ,[t,b] (s − z) (z − t) dz μt (ds)

≤

(h 1 ,h 2 >0)

n 

(k)



  E X (t) 





L (· − t)k (t) + (E |X |) (t) L (1) (t) − 1 + k! k=1

  t  t   t  1 t −z β−1 n 1+ 1 α TX, h 1 L 1 ,[a,t] (z − s) dz μt (ds) + (t − z) n! h1 a s (16.90)   b  s   t  z−t n β−1 1 Tα X, h 2 L 1 ,[t,b] 1+ dz μt (ds) = (s − z) (z − t) h2 t t n

)   ( E | X (k) |)(t)

 (set λ2 := (E |X |) (t)  L (· − t)k (t) ) L (1) (t) − 1 + k! k=1

380

16 Conformable Fractional Quantitative Approximation …

 1  t 1 α TX, h 1 L 1 ,[a,t] n!

λ2 +  t  a

1 h1



t

t

(t − z)β−1 (z − s)(n+1)−1 dz+

s

  (t − z)(β+1)−1 (z − s)(n+1)−1 dz μt (ds)

s

  +1 Ttα X, h 2 L 1 ,[t,b] 1 h2



s

(s − z)



a



t

(n+1)−1

s

(s − z)(n+1)−1 (z − t)β−1 dz+

t (β+1)−1

(z − t)





dz μt (ds)

t

= λ2 +  t 

b

 1  t 1 α TX, h 1 L 1 ,[a,t] n!

   (β) n! 1  (β + 1) n! α α+1 μt (ds) + (t − s) + (t − s)  (β + 1 + 1) h 1  (β + n + 2)   1 Ttα X, h 2 L 1 ,[t,b]



b



t

 n! (β) 1 n! (β + 1) α α+1 μt (ds) (s − t) + (s − t)  (β + n + 1) h 2  (β + n + 2)    = λ2 + 1 tα TX, h 1 L 1 ,[a,t]

⎤⎫ ⎪ ⎪ ⎪ ⎬ ⎥ ⎢ t ⎢ ⎥ 1 1 1 ⎥ ⎢ ⎢ α α+1 ⎥ + μ − s) (t − s) + (t (ds) ⎥ ⎢ ⎢ n ⎥ t n ⎦⎪ ⎣ a ⎣! ⎦ h1 ! ⎪ ⎪ (β + j) (β + 1 + j) ⎭ ⎡

⎡

⎤

j=0

j=0



(16.91)

  1 Ttα X, h 2 L 1 ,[t,b]

⎤⎫ ⎪ ⎪ ⎬ ⎥⎪ ⎢ b ⎢ ⎥ 1 1 1 ⎥ ⎢ ⎢ α α+1 ⎥ μ − t) (s − t) + (s (ds) ⎥ ⎢ ⎢ n ⎥ t n ⎦⎪ ⎣ t ⎣! ⎦ h2 ! ⎪ ⎪ (β + j) (β + 1 + j) ⎭ ⎡

⎡

⎤

j=0

j=0

   = λ2 + 1 tα TX, h 1 L 1 ,[a,t]

16.4 Main Results

381

⎤⎫ ⎪ ⎪  t  t ⎬ ⎥⎪ ⎢ 1 1 1 ⎥ ⎢ α α+1 μt (ds)⎥ (t − s) μt (ds) + n (t − s) ⎢ n ! ⎦⎪ ⎣! h ⎪ ⎪ (β + j) a (β + 1 + j) 1 a ⎭ ⎡

j=0

j=0

   + 1 Ttα X, h 2 L 1 ,[t,b] ⎡ ⎢ 1 ⎢ ⎢ n ⎣! (β + j)



b

1

(s − t)α μt (ds) +

n !

t

j=0

(β + 1 + j)



1 h2

j=0

⎤⎫ ⎪ ⎪ ⎥⎪ ⎬ b ⎥ α+1 μt (ds)⎥ =: (ξ) . (s − t) ⎦⎪ t ⎪ ⎪ ⎭

(16.92) (Choose



t

h 1 :=

(t − s)

α+1

μt (ds)

1 α+1

> 0,

a

and



b

h 2 :=

α+1

(s − t)

μt (ds)

1 α+1

> 0.

t



Then h α+1 = 1 and h α+1 2

t

(t − s)α+1 μt (ds) > 0,

a



b

=

(s − t)α+1 μt (ds) > 0.

t

) Hence it holds 



(ξ) ≤ λ2 + 1



t α TX,

t

α+1

(t − s)

μt (ds)

1  α+1

a

L 1 ,[a,t]

⎡

⎤

α α+1  t 1 ⎢ (μ ([a, t])) α+1 ⎥ h α1 ⎢ t ⎥ α+1 μt (ds) + n (t − s) ⎢ n ⎥+ ! ⎣ ! ⎦ a (β + j) (β + 1 + j)

j=0

j=0

 1 Ttα X,



b t

(s − t)α+1 μt (ds)

1  α+1

(16.93) L 1 ,[t,b]

382

16 Conformable Fractional Quantitative Approximation …

⎡  b 1 ⎢ (μ ([t, b])) α+1 ⎢ t (s − t)α+1 μt (ds) ⎢ n ⎣ ! t (β + j) j=0

α α+1

⎤⎫ ⎪ ⎪ ⎬ ⎥⎪ h α2 ⎥ + n ⎥ = ! ⎦⎪ ⎪ (β + 1 + j) ⎪ ⎭ j=0





λ2 + 1



t α TX,

t

(t − s)

α+1

μt (ds)

1  α+1

a

L 1 ,[a,t]

⎡

⎤

α α+1 1 ⎢ (μ ([a, t])) α+1 ⎥ t 1 ⎢ t ⎥ α+1 + n μt (ds) (t − s) ⎢ n ⎥+ ! ⎣ ! ⎦ a (β + j) (β + 1 + j)



j=0

 1



Ttα X,

b

j=0

α+1

(s − t)

μt (ds)

1  α+1

(16.94)

t

L 1 ,[t,b]

⎤⎫ ⎪ ⎪ ⎪ 1 ⎢ (μ ([t, b])) α+1 ⎥⎬ 1 ⎢ t ⎥ + n ⎢ n ⎥ . ! ⎣ ! ⎦⎪ ⎪ (β + j) (β + 1 + j) ⎪ ⎭ ⎡



b

(s − t)α+1 μt (ds)



α α+1

t

j=0

j=0

We have proved that

E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 + n 

(k)

   E X (t)  1

 L (· − t)k (t) + n ! k! k=1 (β + j) j=1





t

(t − s)

α+1

μt (ds)

1

1 (μt ([a, t])) α+1 + α−n (α + 1)

α α+1

a

 +

 1



t α TX,

(μt ([t, b])) α−n

1 α+1

t



(t − s)

α+1

a

+

μt (ds)

1  α+1

L 1 ,[a,t]

1 (α + 1)



(16.95)

16.4 Main Results



b

383

α+1

(s − t)

μt (ds)

α α+1

 1

t



Ttα X,

b

α+1

(s − t)

μt (ds)

1  α+1

t

. L 1 ,[t,b]

The trivial cases of h 1 = 0 or h 2 = 0 or h 1 = h 2 = 0 are treated similarly as in the proof of Theorem 16.23. Theorem 16.24 is now proved.  Next we give several corollaries: Corollary 16.25 (to Theorem 16.23, case of n = 0) Let α ∈ (0, 1) ; p, q > 1 : 1p + 1 = 1, with α > q1 . Suppose Concepts 16.15, Assumptions 16.16 and 16.19 . Then, q ∀ t ∈ [a, b], we have:

    1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 + 1

2p α

1

(q + 1) q(α+1) ( p (α − 1) + 1) p 



1

 Tαt X,

 1 q(α+1)  1  q(α+1) L |· − t| χ[t,b] (·) (t) (q + 1)

L q ,[t,b]

   1     α  L χ[t,b] (·) (t) p  L |· − t|q(α+1) χ[t,b] (·) (t) q(α+1)    q1  1   α+1 α  (α+1) + L χ[t,b] (·) (t) +1 (q + 1)  1





t αT

 1  L |· − t|q(α+1) χ[a,t] (·) (t) X, (q + 1)

(16.96)

 1 q(α+1) L q ,[a,t]

   1     α  L χ[a,t] (·) (t) p  L |· − t|q(α+1) χ[a,t] (·) (t) q(α+1)    q1  1   α+1 α  (α+1) . L χ[a,t] (·) (t) +1 (q + 1) Corollary 16.26 (to Theorem 16.24, case of n = 0) Let α ∈ (0, 1). Suppose Concepts 16.15, Assumptions 16.16, 16.20. Then, ∀ t ∈ [a, b], we have:

E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) − 1 + ⎧⎡   ⎤   1 ⎨    α   L χ[a,t] (·) (t) α+1 1 ⎦  ⎣ + L |· − t|α+1 χ[a,t] (·) (t) α+1 ⎩ α α+1

(16.97)

384

16 Conformable Fractional Quantitative Approximation …



1

t αT

    1  X,  L |· − t|α+1 χ[a,t] (·) (t) α+1

L 1 ,[a,t]

+

⎤ ⎡    1      α L χ[t,b] (·) (t) α+1 1 ⎦  ⎣ + L |· − t|α+1 χ[t,b] (·) (t) α+1 α (α + 1) 1



Tαt X,

    1   L |· − t|α+1 χ[t,b] (·) (t) α+1

 L 1 ,[t,b]

.

Corollary 16.27 All as in Theorem 16.23. Case of  L (1) = 1 and X (k) (t0 , ω) = 0, ∀ ω ∈ , all k = 1, . . . , n, for a fixed t0 ∈ [a, b]. Then    1 E |M (X ) − X |q (t0 ) q ≤ 



1

1

2 p λ1q

α

(q + 1) q(α+1) n!



 1  L |· − t0 |q(α+1) χ[t0 ,b] (·) (t0 ) Ttα0 X, + 1) (q

1

 1 q(α+1) L q ,[t0 ,b]

   1     α  L χ[t0 ,b] (·) (t0 ) p  L |· − t0 |q(α+1) χ[t0 ,b] (·) (t0 ) q(α+1)    q1    1 α  + L χ[t0 ,b] (·) (t0 ) α+1 (q + 1) (α+1) + 1  1





t0 α TX,

 1  L |· − t0 |q(α+1) χ[a,t0 ] (·) (t0 ) (q + 1)

(16.98)

 1 q(α+1) L q ,[a,t0 ]

   1     α  L χ[a,t0 ] (·) (t0 ) p  L |· − t0 |q(α+1) χ[a,t0 ] (·) (t0 ) q(α+1)    q1    1 α  . L χ[a,t0 ] (·) (t0 ) α+1 (q + 1) (α+1) + 1 Corollary 16.28 All as in Theorem 16.24. Case of  L (1) = 1 and X (k) (t0 , ω) = 0, ∀ ω ∈ , all k = 1, . . . , n, for a fixed t0 ∈ [a, b]. Then E (|M (X ) − X |) (t0 ) ≤

1 n ! j=1

(β + j)

16.4 Main Results

385

⎧⎡   ⎤   1 ⎨    α   L χ[a,t0 ] (·) (t0 ) α+1 1 ⎣ ⎦  + L |· − t0 |α+1 χ[a,t0 ] (·) (t0 ) α+1 ⎩ α−n α+1 1



t0 α TX,

    1   L |· − t0 |α+1 χ[a,t0 ] (·) (t0 ) α+1

L 1 ,[a,t0 ]

+

⎤ ⎡    1    α L χ[t0 ,b] (·) (t0 ) α+1 1 ⎦   ⎣ + L |· − t0 |α+1 χ[t0 ,b] (·) (t0 ) α+1 α−n (α + 1) 1



    1   L |· − t0 |α+1 χ[t0 ,b] (·) (t0 ) α+1

Ttα0 X,

 L 1 ,[t0 ,b]

.

(16.99)

Corollary 16.29 All as in Theorem 16.24, α = 21 and  L (1) = 1. Then, ∀ t ∈ [a, b] , we have: E (|M (X ) − X |) (t) ≤     13    2 2   3 L |· − t| 2 χ[a,t] (·) (t) 2  L χ[a,t] (·) (t) 3 + 3  1

    23 3  2 χ |· T X, L − t| (·) (t) 1 [a,t]

t

2

L 1 ,[a,t]

+

    13    2 2   3 2  L χ[t,b] (·) (t) 3 + L |· − t| 2 χ[t,b] (·) (t) 3  1

    23 3 T 1 X,  L |· − t| 2 χ[t,b] (·) (t)



t

2

L 1 ,[t,b]

.

(16.100)

We continue with uniform estimates. First is an application of Theorem 16.23. Theorem 16.30 Let n ∈ Z+ , α ∈ (n, n + 1) ; p, q > 1 : 1p + q1 = 1, with β := α − (q−1)  p(β−1)+1) n > q1 ; set λ1 := (np+1)( . Suppose Concepts 16.15, Assumptions ( p(α−1)+2) 16.19 and 16.22. Then       1  1   E |M (X ) − X |q  q ≤  E |X |q  q  L (1) − 1∞ + ∞ ∞ 1   1  1 

q n  

(k) q  q     p    E X λ 1 2 L (1) ∞ k ∞   L (· − t) (t) ∞ + α k! (q + 1) q(α+1) n! k=1

  q1  1 α α+1  L (1)∞ (q + 1) (α+1) + 1

386

16 Conformable Fractional Quantitative Approximation …



 sup 1 Ttα X,

t∈[a,b]

1     q(α+1) 1  L |· − t|q(α+1) χ[t,b] (·) (t)∞ (q + 1)

L q ,[t,b]

 α     q(α+1)  + L |· − t|q(α+1) χ[t,b] (·) (t)∞ 

 sup 1 t∈[a,b]

t α TX,

1     q(α+1) 1  L |· − t|q(α+1) χ[a,t] (·) (t)∞ (q + 1)

L q ,[a,t]

 α     q(α+1)  L |· − t|q(α+1) χ[a,t] (·) (t)∞ .

(16.101)

The next is an application of Theorem 16.24. Theorem 16.31 Let n ∈ Z+ , α ∈ (n, n + 1) , β := α − n. Suppose Concepts 16.15, Assumptions 16.20, 16.22. Then   E (|M (X ) − X |)∞ ≤ E (|X |)∞  L (1) − 1∞ + 

n  

(k)    E X k!

k=1

1 n !

(β + j)

∞

     L (· − t)k (t)∞ +

⎡ ⎤  1 α+1  L (1)∞ 1 ⎦ ⎣ + α−n α+1

j=1

  α   α+1  L |· − t|α+1 χ[a,t] (·) (t)∞ sup 1 t∈[a,b]



1      α+1 |· − t|α+1 χ[a,t] (·) (t)∞

t  α TX, L

L 1 ,[a,t]

+

α     α+1  L |· − t|α+1 χ[t,b] (·) (t)∞

 1      α+1 sup 1 Ttα X,  L |· − t|α+1 χ[t,b] (·) (t)∞

t∈[a,b]

L 1 ,[t,b]

.

(16.102)

Some corollaries follow: Corollary 16.32 (to Theorem 16.30, case of n = 0) Let α ∈ (0, 1) ; p, q > 1 : 1p + 1 = 1, with α > q1 . Suppose Concepts 16.15, Assumptions 16.19, 16.22 and  L (1) = q 1. Then

16.4 Main Results

387

   1  E |M (X ) − X |q  q ≤ ∞ 

  sup 1 Tαt X,

t∈[a,b]

1 1  α 2 p (q + 1) α+1 + 1 q α

1

(q + 1) q(α+1) ( p (α − 1) + 1) p

1     q(α+1) 1  L |· − t|q(α+1) χ[t,b] (·) (t)∞ (q + 1)

L q ,[t,b]

 α     q(α+1)  + L |· − t|q(α+1) χ[t,b] (·) (t)∞ 

 sup 1

t∈[a,b]



t αT

1     q(α+1) 1  L |· − t|q(α+1) χ[a,t] (·) (t)∞ X, (q + 1)

L q ,[a,t]

 α     q(α+1)  . L |· − t|q(α+1) χ[a,t] (·) (t)∞

(16.103)

The next is an application of Theorem 16.31. Corollary 16.33 (to Theorem 16.31, case of n = 0) Here α ∈ (0, 1). Suppose Concepts 16.15, Assumptions 16.20, 16.22 and  L (1) = 1. Then  E (|M (X ) − X |)∞ ≤

2α + 1 α (α + 1)



  α   α+1  L |· − t|α+1 χ[a,t] (·) (t)∞ sup 1



t∈[a,b]

t αT

1      α+1 X,  L |· − t|α+1 χ[a,t] (·) (t)∞

L 1 ,[a,t]

+

α     α+1  L |· − t|α+1 χ[t,b] (·) (t)∞

sup 1 t∈[a,b]



1      α+1 Tαt X,  L |· − t|α+1 χ[t,b] (·) (t)∞

L 1 ,[t,b]

.

(16.104)

Corollary 16.34 (to Theorem 16.31, case of n = 0) Here α = 21 . Suppose Concepts 16.15, Assumptions 16.20, 16.22 and  L (1) = 1. Then E (|M (X ) − X |)∞  sup 1 t∈[a,b]

t αT

    13 8  3   ≤  L |· − t| 2 χ[a,t] (·) (t) ∞ 3

    23 3   L |· − t| 2 χ[a,t] (·) (t) X,  ∞

L 1 ,[a,t]

+

388

16 Conformable Fractional Quantitative Approximation …

    13 3    L |· − t| 2 χ[t,b] (·) (t)

∞

 sup 1 t∈[a,b]

    23 3   t 2 Tα X,  L |· − t| χ[t,b] (·) (t)

.

(16.105)

t k (1 − t) N −k ,

(16.106)

∞

L 1 ,[t,b]

16.5 Application Let f ∈ C ([0, 1]) and the Bernstein polynomials B N ( f ) (t) :=

N  k=0

 f

k N



N k



∀ t ∈ [0, 1], ∀ N ∈ N. We have that B N 1 = 1 and B N is a positive linear operator. We have that   t (1 − t) , ∀ t ∈ [0, 1] , B N (· − t)2 (t) = N and

   1 1  B N (· − t)2 (t) 2 ≤ √ , ∀ N ∈ N. ∞ 2 N

(16.107)

(16.108)

L in Concepts 16.15. So B N is like  Define the corresponding application of M by  B N (X ) (t, ω) := B N (X (·, ω)) (t) =

N  k=0

 X

k ,ω N



N k



t k (1 − t) N −k ,

(16.109) ∀ t ∈ [0, 1], ∀ ω ∈ , N ∈ N, where X is a stochastic process. B N is a positive linear operator. Clearly  B N X is a stochastic process and  We give Proposition 16.35 Let α ∈ (0, 1). Let X (t, ω) be a stochastic process from [0, 1] × (, F, P) into R, where (, F, P) is a probability space. Here X (·, ω) ∈ C 1 ([0, 1]), ∀ ω ∈ . Also we assume for each t ∈ [0, 1] that X (1) (t, ·) is measurable over (, F). For any t ∈ [0, 1] we assume that Tαt X (z, ω) is continuous in z ∈ [t, 1], uniformly with respect to ω ∈ , and for any t ∈ [0, 1] we assume that tα T X (z, ω) is continuous in z ∈ [0, t], uniformly with respect to ω ∈ . Suppose also that (E |X |) (t) < ∞, ∀ t ∈ [0, 1]. Then, ∀ t ∈ [0, 1], we have:

  E  B N (X ) − X (t) ≤

16.5 Application

389

⎧⎡   ⎤ 1   α+1 ⎨ B χ   α   1 N [0,t] (·) (t) ⎣ ⎦ B N |· − t|α+1 χ[0,t] (·) (t) α+1 + ⎩ α α+1 1



t αT

    1  X, B N |· − t|α+1 χ[0,t] (·) (t) α+1

L 1 ,[0,t]

+

⎡

⎤    1   α B N χ[t,1] (·) (t) α+1 1 ⎦  ⎣ + B N |· − t|α+1 χ[t,1] (·) (t) α+1 α α+1 1



Tαt X,



   1  B N |· − t|α+1 χ[t,1] (·) (t) α+1

(16.110)

 .

L 1 ,[t,1]



Proof By Corollary 16.26.

Proposition 16.36 All as in Proposition 16.35 with α = 21 . Then, ∀ t ∈ [0, 1], we have:

  E  B N (X ) − X (t) ≤     13     23 2   3 2 B N χ[0,t] (·) (t) + B N |· − t| 2 χ[0,t] (·) (t) 3  1 

    23 3 2 χ |· − t| T X, B (·) (t) 1 N [0,t]

t

2

L 1 ,[0,t]

(16.111)

+

   13    2 2   3 2 B N χ[t,1] (·) (t) 3 + B N |· − t| 2 χ[t,1] (·) (t) 3 





3 2



1 T 1 X, B N |· − t| χ[t,1] (·) (t) t



 23

2

L 1 ,[t,1]

. 

Proof By Proposition 16.35. Proposition 16.37 All as in Proposition 16.35 with α = 21 . Then   13

  8  3 B N |· − t| 2 (t) E  B N (X ) − X (t) ≤ 3     23   3 + 1 t1 T X, B N |· − t| 2 (t) 2

 1

L 1 ,[0,t]

   23 3 2 T 1 X, B N |· − t| (t) t

2





L 1 ,[t,1]

,

(16.112)

390

16 Conformable Fractional Quantitative Approximation …

∀ t ∈ [0, 1], ∀ N ∈ N. Proof From (16.111) and the positivity of B N .



We further have Proposition 16.38 All as in Proposition 16.35 with α = 21 . Then

  8 E  B N (X ) − X (t) ≤ √ 3 4 4N 

 1

1 √ 1 T X, 2 2 N





t

L 1 ,[0,t]

+ 1

1 T 1 X, √ 2 2 N





,

t

L 1 ,[t,1]

(16.113)

∀ t ∈ [0, 1], ∀ N ∈ N. Proof By discrete Hölder’s inequality we get that      3 3 B N |· − t| 2 (t) ≤ B N (· − t)2 (t) 4 

1 t (1 − t) N

34

≤

1 3

(4N ) 4

(16.107.)

=

,

(16.114)

∀ t ∈ [0, 1], ∀ N ∈ N. Hence, the right hand side of (16.112.) ≤ 8 1 √ 3 4 4N



 1

t 1 2

1



T X, √ 4N

 L 1 ,[0,t]

+ 1 T 1t X, √ 2

1 4N



L 1 ,[t,1]

.

(16.115) 

Consequently we obtain Proposition = 21 . Additionally assume that

16.39∗ All as in Proposition 16.35 with α

(1) ∗

X (t, ω) ≤ M , ∀ (t, ω) ∈ [0, 1] × , where M > 0. Then

    E  B N (X ) − X 

∞



 sup 1 t∈[0,1]

1 t √ 1 T X, 2 2 N



8 ≤ √ 4 3 4N 

L 1 ,[0,t]

+ sup 1 t∈[0,1]

1 T 1t X, √ 2 2 N



L 1 ,[t,1]

< ∞, (16.116)

∀ N ∈ N.

16.5 Application

391

As N → +∞, we obtain

    E  B N (X ) − X 

∞

That is  BN

“1-mean”

→

→ 0.

(16.117)

I (unit operator) with rates.

Proof By (16.113); see also Remark 16.14, in particular see (16.29), (16.30).



Remark 16.40 (to Proposition 16.39) Assume that  1 K1 1 T 1t X, √ ≤ √ , ∀ t ∈ [0, 1] , ∀ N ∈ N, 2 2 N L 1 ,[0,t] 2 N where K 1 > 0, and  1 K2 t ≤ √ , ∀ t ∈ [0, 1] , ∀ N ∈ N, 1 1 T X, √ 2 2 N L 1 ,[t,1] 2 N

(16.118)

(16.119)

where K 2 > 0. Conditions (16.118), (16.119) are of Lipschitz type of order 1. Then, by (16.116), (16.118), (16.119), we get:

    E  B N (X ) − X 

∞

= That is

8 ≤ √ 4 3 4N



K1 K2 √ + √ 2 N 2 N

4 (K 1 + K 2 ) √ 3 , ∀ N ∈ N. 3 4 4N 4

    E  B N (X ) − X 

∞

≤

4 (K 1 + K 2 ) √ 3 , ∀ N ∈ N. 3 4 4N 4



(16.120)

(16.121)

Thus, at fractional smoothness of order 21 we achieve speed of convergence 13 , N4 N ∈ N. Without any smoothness, in [5], we proved that the speed of convergence was √1N , N ∈ N. In the presence of the ordinary first derivative, see [5], in deterministic approximation the rate of convergence was N1 , N ∈ N. Naturally, we have as expected: 1 1 1 < 3 < √ , ∀ N ∈ N − {1}. N N N4

392

16 Conformable Fractional Quantitative Approximation …

16.6 Stochastic Korovkin Results We give at first pointwise results. Theorem 16.41 Assume all as in Theorem 16.23. Here L, M are meant  as sequences q(α+1)   |· of operators. Assume further that L → 1 and L − t| (1) (t) (t) → 0, point  wise in t ∈ [a, b]. Then E |M (X ) − X |q (t) → 0, pointwise in t ∈ [a, b], that is M → I (stochastic unit operator) in q-mean-pointwise with rates, quantitatively. Proof We use (16.46). First let n ∈ N. We take into account  L (1) → 1 and that

      k   q(α+1)−k

 L |· − t|q(α+1) (t) q(α+1)  L (1) (t) q(α+1) , L (· − t)k (t) ≤ 

(16.122)

k = 1, . . . , n.     L χ[a,t] (·) (t) ≤  L (1) (t), which  L (1) (t) is bounded, Here it is  L χ[t,b] (·) (t),   we get that and to (16.42), (16.43) and by the positivity   similarly    of L q(α+1) q(α+1) q(α+1)    χ[t,b] (·) (t) , L |· − t| χ[a,t] (·) (t) ≤ L |· − t| L |· − t| (t) → 0. Finally, we use Proposition 16.12(ii) for the 1 (·, ·)’s to go to zero. The case of n = 0 is now obvious.  We continue with Theorem 16.42 Assume all as in Theorem 16.24. Here  L, M are meant  as sequences of operators. Assume further that  L (1) (t) → 1 and  L |· − t|(α+1) (t) → 0, pointwise in t ∈ [a, b]. Then E (|M (X ) − X |) (t) → 0, pointwise in t ∈ [a, b], that is M → I in 1-mean-pointwise with rates, quantitatively.  Proof We use (16.87).First let n ∈ N. We  take into  account L (1) → 1, also (16.36) and (16.40). Notice  L χ[t,b] (·) (t),  L χ[a,t] (·) (t) ≤  L (1) (t), which  L (1) (t) is   α+1   bounded, and by (16.42), (16.43) and positivity of L we get that L |· − t| χ[t,b] (·)     L |· − t|α+1 (t) → 0. L |· − t|α+1 χ[a,t] (·) (t) ≤  (t) ,  Finally, we use Proposition 16.12(ii) for the 1 (·, ·)’s to go to zero. The case of n = 0 is now obvious.  We continue with uniform convergence results. Theorem 16.43 All as in Theorem 16.30. Here  L, M are as sequences  meant q(α+1)   of L |· − t| operators. Assume further that  L (1) → 1 uniformly, and  (t)∞ →    “q-mean” 0. Then  E |M (X ) − X |q ∞ → 0 over [a, b] , that is M → I, the stochastic unit operator with rates, quantitatively.   Proof Use of (16.101). Here  L (1) − 1∞ → 0 and k       q(α+1)−k     q(α+1)   L (· − t)k (t)∞ ≤  L |· − t|q(α+1) (t)∞ L (1)∞q(α+1) ,

(16.123)

16.6 Stochastic Korovkin Results

393

k = 1, . .. , n.      L |· − t|q(α+1) χ[t,b] (·) (t)∞ , L (1)∞ is bounded, see further that  Also           L |· − t|q(α+1) χ[a,t] (·) (t)∞ ≤  L |· − t|q(α+1) (t)∞ < ∞, similarly to (16.44), (16.45). The rest are clear.  We continue with Theorem 16.44 All as in Theorem 16.31. Here  L, M are meant   of oper as sequences ators. Assume further that  L (1) → 1 uniformly, and  L |· − t|(α+1) (t)∞ → 0. “1 -mean”

Then E (|M (X ) − X |)∞ → 0 over [a, b] , that is M → I with rates quantitatively.   Proof Use of (16.102) and  L (1) − 1∞ → 0, (16.41), (16.44), ( 16.45), also     L (1) is bounded. The rest are clear. ∞

We finish with Remark 16.45 An amazing fact/observation follows: In all convergence results here, see Theorems 16.41–16.44, the forcing conditions for convergences are based only on  L and basic real valued continuous functions on [a, b] and are not related to stochastic processes, but they are giving convergence results on stochastic processes! As related work also see [7, pp. 355–376].

References 1. Abdeljawad, T.: On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 3. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 4. Anastassiou, G.: Quantitative Approximations. Chapman & Hall/CRC, Boca Raton (2001) 5. Anastassiou, G.: Stochastic Korovkin theory given quantitatively, Facta Universitatis (Nis). Ser. Math. Inform. 22(1), 43–60 (2007) 6. Anastassiou, G.A.: Fuzzy Mathematics: Approximation Theory. Springer, Heildelberg (2010) 7. Anastassiou, G.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 8. Anastassiou, G.: Nonlinearity: Ordinary and Fractional Approximations by Sublinear and MaxProduct Operators. Springer, Heidelberg (2018) 9. Anastassiou, G.: Conformable fractional approximation of stochastic processes. Submitted (2020) 10. Khalil, R., Al Horani, M., Yousef, A., Sababheh, M.: A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 11. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 12. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 13. Shisha, O., Mond, B.: The degree of convergence of sequences of linear positive operators. Natl. Acad. Sci. U.S. 60, 1196–1200 (1968) 14. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192, 73–80 (1986) 15. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61, 74–87 (1990)

Chapter 17

Trigonometric Conformable Fractional Approximation of Stochastic Processes

Here we consider very general stochastic positive linear operators induced by general positive linear operators that are acting on continuous functions in the trigonometric sense. These are acting on the space of real conformable fractionally differentiable stochastic processes. Under some very mild, general and natural assumptions on the stochastic processes we produce related trigonometric conformable fractional stochastic Shisha-Mond type inequalities of L q -type 1 ≤ q < ∞ and corresponding trigonometric conformable fractional stochastic Korovkin type theorems. These are regarding the trigonometric stochastic q-mean conformable fractional convergence of a sequence of stochastic positive linear operators to the stochastic unit operator for various cases. All convergences are derived with rates and are given via the trigonometric conformable fractional stochastic inequalities involving the stochastic modulus of continuity of the αth conformable fractional derivatives of the engaged stochastic process, α ∈ (n, n + 1), n ∈ Z+ . The impressive fact is that only two basic real Korovkin test functions assumptions, one of them trigonometric, are enough for the conclusions of our trigonometric conformable fractional stochastic Korovkin theorems. We give applications to stochastic Bernstein operators in the trigonometric sense. See also [8].

17.1 Introduction Motivation for this work comes from [2, 3, 12–14]. This work continues our earlier work [5], now at the stochastic conformable fractional level. In Sect. 17.2 we talk briefly about conformable fractional calculus. In the Sect. 17.3, we talk about the q-mean (1 ≤ q < ∞) first modulus of continuity of a stochastic process and its upper bounds. There we describe completely our setting by introducing our stochastic positive linear operator M, see (17.19), which is based on the positive linear operator  L from C ([−π, π]) into itself. The operator M is acting on a wide space of conformable fractional differentiable real valued stochastic © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_17

395

396

17 Trigonometric Conformable Fractional Approximation …

processes X . See there Assumptions 17.11, 17.14, 17.15, 17.17. We first give the main trigonometric pointwise conformable fractional stochastic Shisha-Mond type inequalities [12], see Theorems 17.23, 17.24, and their several corollaries covering important trigonometric special cases. We continue with trigonometric conformable fractional q-mean uniform ShishaMond type inequalities, see Theorems 17.25, 17.26, and their interesting corollaries. All this theory is regarding the trigonometric conformable fractional stochastic convergence of operators M to I (stochastic unit operator) given quantitatively with rates. An extensive trigonometric application about the stochastic Bernstein operators follows in full details. Based on our Shisha-Mond type inequalities of our main Theorems 17.23–17.26 we derive trigonometric pointwise and uniform Stochastic Korovkin theorems [10] on stochastic processes, see Theorems 17.41–17.44. The amazing fact here is, that basic conditions on operator  L regarding two simple real valued functions, one of them trigonometric, that are not stochastic, are able to imply conformable fractional stochastic convergence on all stochastic processes we are dealing with; see Concepts 17.10 and Assumptions 17.11, 17.14, 17.15, 17.17 on [−π, π]. This work is based on [7].

17.2 Background—I Here we follow [1], see also [9]. We need Definition 17.1 ([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 

(17.1)

  If Tαa f (t) exists on (a, b), then 

   Tαa f (a) = lim Tαa f (t) . t→a+

(17.2)

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



(17.3)

 f (t) exists on (a, b), then b

αT

   f (b) = lim bα T f (t) . t→b−

(17.4)

17.2 Background—I

397

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) .

(17.5)

(17.6)

In the higher order case we can generalize things as follows: Definition 17.2 ([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) .

(17.7)

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) .

(17.8)

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 (17.7), (17.8) collapse to {(17.1), (17.2)}, {(17.3), (17.4)} respectively. We make Remark 17.3 ([6, p. 155]) 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 (

(17.9)

   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) .

(17.10)

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, when 0 < β < 1, i.e. when α ∈ (n, n + 1).

(17.11)

398

17 Trigonometric Conformable Fractional Approximation …

17.3 Background—II We need Definition 17.4 Here (, F, P) is a probability space, ω ∈ . We define the relative q-mean first modulus of continuity of stochastic process X (t, ω) by 1 (X, δ) L q ,[c,d] := 

1 q |X (x, ω) − X (y, ω)| P (dω) : x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ , q

sup



(17.12) where δ > 0, 1 ≤ q < ∞, t ∈ [a, b] ⊂ R. Definition 17.5 Let 1 ≤ q < ∞. Let X (x, ω) be a stochastic process. We call X a q-mean uniformly continuous stochastic process over [a, b], iff ∀ ε > 0 ∃ δ > 0 : whenever |x − y| ≤ δ; x, y ∈ [a, b] implies that  

|X (x, s) − X (y, s)|q P (ds) ≤ ε.

(17.13)

U

We denote it as X ∈ CR q ([a, b]) . It holds U

Proposition 17.6 ([5]) Let X ∈ CR q ([a, b]), then 1 (X, δ) L q ,[a,b] < ∞, any δ > 0. Also it holds Proposition 17.7 ([5]) Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R. Then following are true ([c, d] ⊆ [a, b]): (i) 1 (X, δ) L q ,[c,d] is nonnegative and nondecreasing in δ > 0, U (ii) lim1 (X, δ) L q ,[c,d] = 1 (X, 0) L q ,[c,d] = 0, iff X ∈ CR q ([c, d]) , δ↓0

(iii) 1 (X, δ1 + δ2 ) L q ,[c,d] ≤ 1 (X, δ1 ) L q ,[c,d] + 1 (X, δ2 ) L q ,[c,d] , δ1 , δ2 > 0, (iv) 1 (X, mδ) L q ,[c,d] ≤ m1 (X, δ) L q ,[c,d] , δ > 0, m ∈ N, (v) 1 (X, λδ) L q ,[c,d] ≤ λ 1 (X, δ) L q ,[c,d] ≤ (λ + 1) 1 (X, δ) L q ,[c,d] , λ > 0, δ > 0, · is the ceiling of the number, (vi) 1 (X + Y, δ) L q ,[c,d] ≤ 1 (X, δ) L q ,[c,d] + 1 (Y, δ) L q ,[c,d] , δ > 0, U (vii) 1 (X, ·) L q ,[c,d] is continuous on R+ for X ∈ CR q ([c, d]) . We give Remark 17.8 (to Proposition 17.7) By Proposition 17.7 (v) we get

1 (X, |x − y|) L q ,[c,d] ≤ ∀ x, y ∈ [c, d], any δ > 0.

 |x − y| 1 (X, δ) L q ,[c,d] , δ

(17.14)

17.3 Background—II

399

We give Remark 17.9 Let α ∈ (n, n ∈ Z+ , X (·, ω) ∈ C n+1 ([a, b]), ∀ ω ∈ . We n + 1], (n+1) ∗ assume that X (t, ω) ≤ M , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. Let δ > 0, 1 ≤ q < ∞. Then   1 Tcα X, δ L q ,[c,d] = sup

 

 c    T X (x, ω) − Tc X (y, ω) q P (dω) α α

 q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤  sup 

    c   T X (x, ω) + Tc X (y, ω) q P (dω) α α

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ}  sup 

 q1

:

(17.9.)

≤

(β:=α−n)

q

(x − c)1−β X (n+1) (x, ω) + (y − c)1−β X (n+1) (y, ω) P (dω)

 q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ ∗



M sup 

 q (x − c)1−β + (y − c)1−β P (dω)

 q1

:

x, y ∈ [c, d] ⊆ [a, b] , |x − y| ≤ δ} ≤ 2M ∗ (b − a)1−β . That is

  1 Tcα X, δ L q ,[c,d] ≤ 2M ∗ (b − a)1−β ,

(17.15)

for any [c, d] ⊆ [a, b] . Similarly, it holds 1

d



α TX, δ L q ,[c,d]

≤ 2M ∗ (b − a)1−β ,

(17.16)

for any [c, d] ⊆ [a, b] . In particular it holds   sup 1 Ttα X, δ L q ,[t,b] ≤ 2M ∗ (b − a)1−β ,

(17.17)

t∈[a,b]

and sup 1 t∈[a,b]

t



α TX, δ L q ,[a,t]

≤ 2M ∗ (b − a)1−β .

(17.18)

400

17 Trigonometric Conformable Fractional Approximation …

Above, it is not strange to assume that Tcα X,dα TX, Ttα X,tα TX are stochastic processes, see Remark 17.3. We need Concepts 17.10 Let  L be a positive linear operator from C ([a, b]) into itself. Let X (t, ω) be a stochastic process from [a, b] × (, F, P) into R, where (, F, P) is a probability space. Let α ∈ (n, n + 1], n ∈ Z+ , and X (·, ω) ∈ C n+1 ([a, b]), ∀ ω ∈ . Also we assume for each t ∈ [a, b] that X (k) (t, ·) is measurable over (, F) for all k = 1, . . . , n + 1. Define M (X ) (t, ω) :=  L (X (·, ω)) (t) , ∀ ω ∈ , ∀ t ∈ [a, b] ,

(17.19)

and assume that it is a random variable in ω. Clearly M is a positive linear operator on stochastic processes. We make Assumption 17.11 (i) For any t ∈ [a, b] we assume that Ttα X (z, ω) is continuous in z ∈ [t, b], uniformly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ > 0 : whenever |z 1 − z 2 | ≤ δ; z 1 , z 2 ∈ [t, b], then Ttα X (z 1 , ω) − Ttα X (z 2 , ω) ≤ ε, ∀ ω ∈ . We denote this by Ttα X ∈ CRU ([t, b]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [t, b] . (ii) For any t ∈ [a, b] we assume that tα TX (z, ω) is continuous in z ∈ [a, t], uniformly with respect to ω ∈ . I.e. ∀ ε > 0 ∃ δ > 0 : whenever |z 1 − z 2 | ≤ δ; z 1 , z 2 ∈ [a, t], then tα TX (z 1 , ω) −tα TX (z 2 , ω) ≤ ε, ∀ ω ∈ . We denote this by tα TX ∈ CRU ([a, t]), the space of continuous in x, uniformly with respect to ω, stochastic processes over [a, t] . Remark 17.12 Assumption 17.11 implies: (i) Ttα X (·, ω) ∈ C ([t, b]), ∀ ω ∈ , and Ttα X is q-mean uniformly continuous U in z ∈ [t, b], that is Ttα X ∈ CR q ([t, b]), for any 1 ≤ q < ∞. (ii) tα TX (·, ω) ∈ C ([a, t]), ∀ ω ∈ , and tα TX is q-mean uniformly continuous U in z ∈ [a, t], that is tα TX ∈ CR q ([a, t]), for any 1 ≤ q < ∞. We need Definition 17.13 Denote by  (E X ) (t) := the expectation operator. We make



X (t, ω) P (dω) , ∀ t ∈ [a, b] ,

(17.20)

17.3 Background—II

401

Assumption 17.14 We assume that  q  E X (k) (t) < ∞, ∀ t ∈ [a, b] ,

(17.21)

q > 1, for all k = 0, 1, . . . , n. We make Assumption 17.15 We assume that  (k)  E X (t) < ∞, ∀ t ∈ [a, b] ,

(17.22)

for all k = 0, 1, . . . , n. We give Remark 17.16 By the Riesz representation theorem [11] we have that there exists μt unique, completed Borel measure on [a, b] with L (1) (t) > 0, m t := μt ([a, b]) = 

(17.23)

f (x) dμt (x) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(17.24)

such that  L ( f ) (t) =

 [a,b]

Consequently we have that  X (x, ω) dμt (x) , ∀ (t, ω) ∈ [a, b] × ,

M (X ) (t, ω) =

(17.25)

[a,b]

and X as in Concepts 17.10. Here χ[γ,δ] (s) stands for the characteristic function on [γ, δ] ⊆ [a, b] . Notice that (r > 0)     r χ[t,b] (s) |s − t|r μt (ds) =  L |· − t|r χ[t,b] (·) (t) , (s − t) μt (ds) = [t,b]

[a,b]

(17.26) and 

 (t − s) μt (ds) = r

[a,t]

  χ[a,t] (s) |s − t|r μt (ds) =  L |· − t|r χ[a,t] (·) (t) .

[a,b]

(17.27) Let α ∈ (n, n + 1], n ∈ N and k = 1, . . . , n. Then by Hölder’s inequality we obtain

402

17 Trigonometric Conformable Fractional Approximation …



[a,b]

 (x − t)k dμt (x) ≤

|x − t|k dμt (x) ≤

[a,b]



α+1

|x − t|

dμt (x)

k ( α+1 )

(μt ([a, b]))(

α+1−k α+1

).

(17.28)

[a,b]

Therefore it holds          L (· − t)k (t)∞,[a,b] ≤  L |· − t|k (t)∞,[a,b] ≤ k   ( α+1−k )  ( α+1 )  α+1   , L |· − t|α+1 (t)∞,[a,b] L (1)∞,[a,b]

(17.29)

all k = 1, . . . , n. Also, we observe that

and

C ([a, b])  |· − t|α+1 χ[a,t] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] ,

(17.30)

C ([a, b])  |· − t|α+1 χ[t,b] (·) ≤ |· − t|α+1 , ∀ t ∈ [a, b] .

(17.31)

By positivity of  L we obtain          L |· − t|α+1 (t)∞,[a,b] < ∞, L |· − t|α+1 χ[a,t] (·) (t)∞,[a,b] ≤ 

(17.32)

  by  L |· − t|α+1 (t) being continuous in t ∈ [a, b], see p. 388 of [4], and          L |· − t|α+1 (t)∞,[a,b] . L |· − t|α+1 χ[t,b] (·) (t)∞,[a,b] ≤ 

(17.33)

Above (17.26)–(17.29) and (17.32), (17.33) were used to derive convergences in our earlier results mentioned next.   L (1) (t), and by In this work we denote by  L χ[a,t] (·) (t) := μt ([a, t]) ≤     L χ[t,b] (·) (t) := μt ([t, b]) ≤  L (1) (t) . We make Assumption 17.17 Assume that X (n+1) (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [a, b] × , where M ∗ > 0. And suppose that X (n+1) (·, ω) is continuous over [a, b], uniformly with respect to ω ∈ . Clearly, Assumption 17.17 implies Assumption 17.11, see also Remark 17.3. We mention the following pointwise result on the quantitative stochastic conformable fractional approximation regarding stochastic processes:

17.3 Background—II

403

Theorem 17.18 ([7]) Let n ∈ Z+ , α ∈ (n, n + 1) ; p, q > 1 : 1p + q1 = 1, with (q−1)  p(β−1)+1) β := α − n > q1 ; set λ1 := (np+1)( . Suppose Concepts 17.10, ( p(α−1)+2) Assumptions 17.11 and 17.14. Then, ∀ t ∈ [a, b], we have:   1   1   L (1) (t) − 1 + E |M (X ) − X |q (t) q ≤ E |X |q (t) q  1  (k) q   q1 1 q pλ  E X (t)  2 1 k  L (· − t) (t) + α k! (q + 1) q(α+1) n!

n  k=1



 1



Ttα X,

 1  q(α+1)  1  q(α+1) L |· − t| χ[t,b] (·) (t) (q + 1)

L q ,[t,b]

   1     α  L χ[t,b] (·) (t) p  L |· − t|q(α+1) χ[t,b] (·) (t) q(α+1)    q1    1 α  + L χ[t,b] (·) (t) α+1 (q + 1) (α+1) + 1  1





t α TX,

 1  q(α+1)  1  q(α+1) χ[a,t] (·) (t) L |· − t| (q + 1)

(17.34)

L q ,[a,t]

   1     α  L χ[a,t] (·) (t) p  L |· − t|q(α+1) χ[a,t] (·) (t) q(α+1)    q1  1   α+1 α  (α+1) L χ[a,t] (·) (t) +1 (q + 1) We mention also the following pointwise results, the L 1 -quantitative stochastic conformable fractional approximation of stochastic processes. Theorem 17.19 ([7]) Let n ∈ Z+ and α ∈ (n, n + 1), β := α − n. Suppose Concepts 17.10, Assumptions 17.11, 17.15. Then, ∀ t ∈ [a, b], we have: E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 + n  (k)    E X (t)  1  L (· − t)k (t) + n  k! k=1 (β + j) j=1

404

17 Trigonometric Conformable Fractional Approximation …

⎧⎡   ⎤   1 ⎨    α   L χ[a,t] (·) (t) α+1 1 ⎣ ⎦  + L |· − t|α+1 χ[a,t] (·) (t) α+1 ⎩ α−n α+1 1



t α TX,

    1   L |· − t|α+1 χ[a,t] (·) (t) α+1

L 1 ,[a,t]

+

⎤ ⎡    1    α L χ[t,b] (·) (t) α+1 1 ⎦   ⎣ + L |· − t|α+1 χ[t,b] (·) (t) α+1 α−n (α + 1)      1  1 Ttα X,  L |· − t|α+1 χ[t,b] (·) (t) α+1

 .

L 1 ,[t,b]

(17.35)

Uniform estimates follow: Theorem 17.18 implies Theorem 17.20 ([7]) Let n ∈ Z+ , α ∈ (n, n + 1) ; p, q > 1 : 1p + q1 = 1, with β := (q−1)  p(β−1)+1) α − n > q1 ; set λ1 := (np+1)( . Suppose Concepts 17.10, Assump( p(α−1)+2) tions 17.14 and 17.17. Then       1  1   E |M (X ) − X |q (t) q ≤  E |X |q  q  L (1) − 1∞ + ∞ ∞ 1   1  1  n   (k) q  q      E X λ1q 2  L (1)∞ p k ∞  L (· − t) (t)∞ + α k! (q + 1) q(α+1) n! k=1

  q1  1 α α+1  L (1)∞ (q + 1) (α+1) + 1 

 sup 1 Ttα X,

t∈[a,b]

1     q(α+1) 1  L |· − t|q(α+1) χ[t,b] (·) (t)∞ (q + 1)

 L q ,[t,b]

 α     q(α+1)  + L |· − t|q(α+1) χ[t,b] (·) (t)∞ 

 sup 1

t∈[a,b]

t α TX,

1     q(α+1) 1  L |· − t|q(α+1) χ[a,t] (·) (t)∞ (q + 1)

" α     q(α+1)  L |· − t|q(α+1) χ[a,t] (·) (t)∞ . Theorem 17.19 implies:

 (17.36) L q ,[a,t]

17.3 Background—II

405

Theorem 17.21 ([7]) Let n ∈ Z+ , α ∈ (n, n + 1) , Concepts 17.10, Assumptions 17.15, 17.17. Then

β := α − n.

Suppose

  E (|M (X ) − X |)∞ ≤ E (|X |)∞  L (1) − 1∞ +  n   (k)    E X k!

k=1

1 n 

(β + j)

∞

     L (· − t)k (t)∞ +

⎡ ⎤  1 α+1  L (1)∞ 1 ⎣ ⎦ + α−n α+1

j=1

#  α   α+1  L |· − t|α+1 χ[a,t] (·) (t)∞ sup 1 t∈[a,b]



1      α+1 |· − t|α+1 χ[a,t] (·) (t)∞

t  α TX, L

L 1 ,[a,t]

+

α     α+1  L |· − t|α+1 χ[t,b] (·) (t)∞

 1      α+1 sup 1 Ttα X,  L |· − t|α+1 χ[t,b] (·) (t)∞

t∈[a,b]

L 1 ,[t,b]

.

(17.37)

Remark 17.22 Next we specify [a, b] as [−π, π]. Clearly then  L : C ([−π, π]) → C ([−π, π]) is the positive linear operator on hand. Here α ∈ (n, n + 1), n ∈ Z+ , k = 1, . . . , n. Next we use of Hölder’s inequality we notice that         |· − t| k |x − t| k  sin sin dμt (x) ≤ L (t) = 4 4 [−π,π] 

k  α+1    |x − t| α+1 α+1−k sin dμt (x) (μt ([−π, π])) α+1 = 4 [−π,π]

(17.38)

k       α+1   α+1−k |· − t| α+1   sin L (1) (t) α+1 . L (t) 4

That is

    |· − t| k  sin L (t) ≤ 4

(17.39)

406

17 Trigonometric Conformable Fractional Approximation … k       α+1 |· − t| α+1 α+1−k  sin L (t) (L (1) (t)) α+1 , 4

for k = 1, . . . , n; true also for q (α + 1) instead of (α + 1), for any 1 < q < ∞. Next ·∞ denotes ·∞,[−π,π] . Consequently, it holds         |· − t| k   sin (t) L   4

≤

(17.40)

∞

k        α+1    α+1−k |· − t| α+1    sin L (1) (t)∞α+1 , (t) L   4

∞

for k = 1, . . . , n; true also for q (α + 1) instead of (α + 1), for any 1 < q < ∞. In this work we use a lot the following well known inequality:  |z| ≤ π sin

 |z| , ∀ z ∈ [−π, π] . 2

(17.41)

Notice that, for any t ∈ [−π, π], we have C ([−π, π])  |· − t| χ[−π,t] (·) ≤ |· − t| ∈ C ([−π, π]), therefore       |· − t| χ[−π,t] (·) α+1 |· − t| α+1 C ([−π, π])  sin ≤ sin ∈ C ([−π, π]) . 4 4

(17.42)

Consequently, by positivity of  L we obtain         |· − t| χ[−π,t] (·) α+1   sin (t) L   4

∞

        |· − t| α+1   sin ≤ L (t) .   4 ∞

(17.43) Similarly, for any t ∈ [−π, π], we have C ([−π, π])  |· − t| χ[t,π] (·) ≤ |· − t| ∈ C ([−π, π]), thus 



|· − t| χ[t,π] (·) C ([−π, π])  sin 4

α+1

Hence         |· − t| χ[t,π] (·) α+1   sin (t) L   4

∞

   |· − t| α+1 ≤ sin ∈ C ([−π, π]) . 4 (17.44)         |· − t| α+1   sin ≤ L (t) .   4 ∞

(17.45)

17.3 Background—II

407

So, if the right hand side of (17.43), (17.45) goes to zero, so do their left hand sides. Above in (17.42)–(17.45), one can use q (α + 1) instead of (α + 1), 1 < q < ∞. A further detailed analysis reveals: We have that (1 ≤ q < ∞)   (17.24.)  L |· − t|q(α+1) χ[t,π] (·) (t) =

 (x − t)q(α+1) dμt (x) = [t,π]



 2q(α+1) [t,π]

x −t 2

q(α+1) dμt (x)

(17.41.)

≤

  q(α+1) x −t sin dμt (x) = 4 [t,π]

 (2π)q(α+1)



 (2π)

q(α+1) [t,π]

 (2π)

q(α+1)



|x − t| sin 4

(17.46)

q(α+1) dμt (x) =

   |x − t| q(α+1) sin χ[t,π] (x) dμt (x) = 4 [−π,π]

q(α+1)   |x − t| (17.24.) sin χ[t,π] (x) dμt (x) = 4 [−π,π]

 (2π)

q(α+1)

(2π)

q(α+1)

 L





|· − t| sin χ[t,π] (·) 4

q(α+1)  (t) .

That is, we have obtained    L |· − t|q(α+1) χ[t,π] (·) (t) ≤ (2π)q(α+1)  L

  q(α+1)  |· − t| sin χ[t,π] (·) (t) . 4 (17.47)

Similarly, it holds 



 L |· − t|q(α+1) χ[−π,t] (·) (t) ≤ (2π)q(α+1)  L



q(α+1)  |· − t| sin χ[−π,t] (·) (t) . 4 

(17.48) Above inequalities (17.47), (17.48) are valid for any 1 ≤ q < ∞. Furthermore, we observe that     k k  |x − t|k dμt (x) = L (· − t) (t) = (x − t) dμt (x) ≤ [−π,π]

[−π,π]

408

17 Trigonometric Conformable Fractional Approximation …



 2

k [−π,π]

That is

|x − t| 2

   |x − t| k sin dμt (x) ≤ (2π) dμt (x) 4 [−π,π] (17.49)   k  |· − t| sin L = (2π)k  (t) . 4

k

(17.41.)

   L (· − t)k (t) ≤ (2π)k  L



k

    |· − t| k sin (t) , 4

(17.50)

∀ t ∈ [−π, π], all k = 1, . . . , n.

17.4 Main Results Next we give our first main result on the trigonometric quantitative stochastic conformable fractional approximation of stochastic processes, a pointwise result. Theorem 17.23 All as in Theorem 17.18 for [a, b] = [−π, π]. Then, ∀ t ∈ [−π, π], we have     1   1 E |M (X ) − X |q (t) q ≤ E |X |q (t) q  L (1) (t) − 1 +   1   α+ 1p q α |· − t| k λ π 2 1 sin · L (t) + (2π)  α k! 4 (q + 1) q(α+1) n! k=1 ⎧⎡ ⎛ ⎞(17.51) 1    q(α+1)   q(α+1) ⎨ |· − t| 1  ⎣1 ⎝Ttα X, 2π ⎠ sin χ[t,π] (·) L (t) ⎩ 4 (q + 1) n 

 (k) q   q1 E X (t)

k

L q ,[t,π]

   1  L χ[t,π] (·) (t) p

α    q(α+1)   q(α+1) |· − t|  sin χ[t,π] (·) L (t) 4

   q1  1   α+1 α  (α+1) + L χ[t,π] (·) (t) +1 (q + 1) ⎡

⎛

⎣1 ⎝tα TX, 2π



1  L (q + 1)

⎞ 1   q(α+1)   q(α+1) |· − t| ⎠ sin χ[−π,t] (·) (t) 4

L q ,[−π,t]

17.4 Main Results

409

   1  L χ[−π,t] (·) (t) p

α    q(α+1)   q(α+1) |· − t|  χ[−π,t] (·) L sin (t) 4

   q1  1   α+1 α  (α+1) . L χ[−π,t] (·) (t) +1 (q + 1) Proof Apply Theorem 17.18, upper bound right hand side of (17.34) by using (17.50), (17.47) and (17.48).  We continue with the trigonometric L 1 -quantitative stochastic conformable fractional pointwise approximation of stochastic processes. Theorem 17.24 All as in Theorem 17.19 for [a, b] = [−π, π]. Then, ∀ t ∈ [−π, π], we have: E (|M (X ) − X |) (t) ≤ (E |X |) (t)  L (1) (t) − 1 +     n  (k)   E X (t) |· − t| k (2π)α k sin (t) + n (2π) L  k! 4 k=1 (β + j) j=1

⎧⎡   ⎤  α   1   α+1   α+1 ⎨  L χ[−π,t] (·) (t) α+1 |· − t| 1 ⎦  ⎣ sin L + χ[−π,t] (·) (t) ⎩ α−n 4 (α + 1)

⎛

1 ⎞    α+1   α+1 |· − t| ⎠ sin χ[−π,t] (·) 1 ⎝tα TX, 2π  L (t) 4

+ L 1 ,[−π,t]

⎤  ⎡  α   1   α+1   α+1  L χ[t,π] (·) (t) α+1 |· − t| 1 ⎦  ⎣ sin + χ[t,π] (·) L (t) α−n 4 (α + 1) ⎛

1 ⎞    α+1   α+1 |· − t| ⎠ sin χ[t,π] (·) L 1 ⎝Ttα X, 2π  (t) 4

L 1 ,[t,π]

⎫ (17.52) ⎬ . ⎭

Proof Apply Theorem 17.19, upper bound right hand side of (17.35) by using (17.50), (17.47) and (17.48) for q = 1.  We continue with trigonometric conformable fractional uniform estimates (·∞,[−π,π] := ·∞ ) in L q -mean (1 ≤ q < ∞). Theorem 17.25 All as in Theorem 17.20 for [a, b] = [−π, π]. Then       1  1   E |M (X ) − X |q  q ≤  E |X |q  q  L (1) − 1∞ + ∞ ∞

410

17 Trigonometric Conformable Fractional Approximation …

   k     |· − t|   sin L (t) + (2π)k    4

1  n   (k) q  q  E X

∞

k!

k=1

∞

1  1 1 p   q1   1 L (1)∞ 2α+ p λ1q π α  α α+1  L (1)∞ (q + 1) (α+1) + 1 α (q + 1) q(α+1) n!

⎡

⎛

 

⎣ sup 1 ⎝Ttα X, 2π    (q t∈[−π,π]

1  L + 1)

(17.53)

⎞ 1   q(α+1)    q(α+1) |· − t|  ⎠ sin χ[t,π] (·) (t)  4 ∞

L q ,[t,π]

⎤ α    q(α+1)   q(α+1)   |· − t|   ⎦+ sin χ[t,π] (·) (t) L   4 ∞

⎡

⎛

 

⎣ sup 1 ⎝tα TX, 2π    (q t∈[−π,π]

1  L + 1)

⎞ 1   q(α+1)    q(α+1) |· − t|  ⎠ sin χ[−π,t] (·) (t)  4 ∞

L q ,[−π,t]

⎤⎫ α    q(α+1)    q(α+1) ⎬  |· − t|   ⎦ . sin χ[−π,t] (·) (t) L   ⎭ 4 ∞



Proof By (17.51). Theorem 17.26 All as in Theorem 17.21 for [a, b] = [−π, π]. Then   E (|M (X ) − X |)∞ ≤ E (|X |)∞  L (1) − 1∞ +  n   (k)    E X k=1

k!

∞

   k     |· − t|   sin L (t) + (2π)k    4 ∞

(2π)α

n 

(β + j)

⎡ ⎤  1 α+1  L (1)∞ 1 ⎦ ⎣ + α−n α+1

(17.54)

j=1

⎧  α   α+1    α+1 ⎨ |· − t|   sin χ[−π,t] (·) (t) L  ⎩ 4 ∞

⎛

1 ⎞    α+1    α+1  |· − t|   sin L χ[−π,t] (·) sup 1 ⎝tα TX, 2π  (t) ⎠   4 t∈[−π,π]

∞

+ L 1 ,[−π,t]

17.4 Main Results

411 α    α+1    α+1  |· − t|   sin χ[t,π] (·) (t) L   4

∞

1 ⎞    α+1   α+1   |· − t|   sin L χ[t,π] (·) sup 1 ⎝Ttα X, 2π  (t) ⎠   4 t∈[−π,π]

⎫ ⎬

⎛

∞

L 1 ,[t,π]

⎭

. 

Proof By (17.52). We continue with interesting pointwise corollaries.

Corollary 17.27 All as in Theorem 17.18 for [a, b] = [−π, π]. Further assume that  L (1) (t0 ) = 1, and X (k) (t0 , ω) = 0, ∀ ω ∈ , all k = 1, . . . , n, for a fixed t0 ∈ [−π, π]. Then    1 E |M (X ) − X |q (t0 ) q ≤

1

2α+ p λ1q π α 1

(17.55)

α

(q + 1) q(α+1) n!

⎧⎡ ⎛     1 ⎞ q(α+1)  ⎨ q(α+1) |· | − t 1 0  ⎠ ⎣1 ⎝Ttα0 X, 2π sin χ[t0 ,π] (·) L (t0 ) ⎩ 4 (q + 1)

   1  L χ[t0 ,π] (·) (t0 ) p

L q ,[t0 ,π]

α  q(α+1)    q(α+1)  |· − t0 |  sin χ[t0 ,π] (·) L (t0 ) 4

   q1   1 α  L χ[t0 ,π] (·) (t0 ) α+1 (q + 1) (α+1) + 1 + ⎛ 1 ⎝tα0 TX, 2π



1  L (q + 1)



   1  L χ[−π,t0 ] (·) (t0 ) p



|· − t0 | sin χ[−π,t0 ] (·) 4



q(α+1)  (t0 )

⎞ 1 q(α+1) ⎠ L q ,[−π,t0 ]

α  q(α+1)    q(α+1)  |· − t0 |  sin χ[−π,t0 ] (·) L (t0 ) 4

   q1    1 α  . L χ[−π,t0 ] (·) (t0 ) α+1 (q + 1) (α+1) + 1 Proof By (17.51).



Corollary 17.28 All as in Theorem 17.18 for [a, b] = [−π, π]. Further assume that  L (1) = 1 and q1 < α < 1 (i.e. n = 0). Then, ∀ t ∈ [−π, π], we have

412

17 Trigonometric Conformable Fractional Approximation …

   1 E |M (X ) − X |q (t) q ≤

2α+ p π α 1

1

(17.56)

α

( p (α − 1) + 1) p (q + 1) q(α+1)

⎧⎡ ⎛ ⎞ 1    q(α+1)   q(α+1) ⎨ |· − t| 1  ⎠ ⎣1 ⎝Tαt X, 2π sin χ[t,π] (·) L (t) ⎩ 4 (q + 1)    1  L χ[t,π] (·) (t) p

L q ,[t,π]

α    q(α+1)   q(α+1) |· − t|  sin χ[t,π] (·) L (t) 4

   q1   1 α  L χ[t,π] (·) (t) α+1 (q + 1) (α+1) + 1 + ⎡

⎛



+ ⎣1 ⎝tα T X, 2π

1  L (q + 1)

   1  L χ[−π,t] (·) (t) p

⎞ 1 q(α+1)   q(α+1) |· − t| ⎠ χ[−π,t] (·) sin (t) 4





L q ,[−π,t]

α    q(α+1)   q(α+1) |· − t|  sin L χ[−π,t] (·) (t) 4

   q1    1 α  . L χ[−π,t] (·) (t) α+1 (q + 1) (α+1) + 1 

Proof By (17.51).

Corollary 17.29 All as in Theorem 17.19 for [a, b] = [−π, π]. Further assume that  L (1) (t0 ) = 1, and X (k) (t0 , ω) = 0, ∀ ω ∈ , all k = 1, . . . , n, for a fixed t0 ∈ [−π, π]. Then (2π)α E (|M (X ) − X |) (t0 ) ≤ n (17.57)  (β + j) j=1

⎧⎡   ⎤   1 ⎨  L χ[−π,t0 ] (·) (t0 ) α+1 1 ⎦ ⎣ + ⎩ α−n (α + 1) α     α+1 α+1  |· − t0 |  sin L χ[−π,t0 ] (·) (t0 ) 4

⎛

1 ⎞     α+1 α+1  |· | − t 0 ⎠ sin χ[−π,t0 ] (·) 1 ⎝tα0 TX, 2π  L (t0 ) 4

+ L 1 ,[−π,t0 ]

17.4 Main Results

413

⎤ ⎡    1  L χ[t0 ,π] (·) (t0 ) α+1 1 ⎦ ⎣ + α−n (α + 1) α     α+1 α+1  |· − t0 |  sin L χ[t0 ,π] (·) (t0 ) 4 1 ⎞     α+1 α+1  |· − t0 | ⎠ sin χ[t0 ,π] (·) 1 ⎝Ttα0 X, 2π  L (t0 ) 4

⎫ ⎬

⎛

L 1 ,[t0 ,π]

⎭

. 

Proof By (17.52).

Corollary 17.30 All as in Theorem 17.19 for [a, b] = [−π, π]. Further assume that  L (1) = 1 and 0 < α < 1. Then, ∀ t ∈ [−π, π], we have E (|M (X ) − X |) (t) ≤ (2π)α

(17.58)

⎧⎡   ⎤  α   1   α+1   α+1 ⎨  L χ[−π,t] (·) (t) α+1 |· − t| 1  ⎦ ⎣ sin L + χ[−π,t] (·) (t) ⎩ α 4 (α + 1)

⎛

1 ⎞    α+1   α+1 |· − t| ⎠ sin χ[−π,t] (·) 1 ⎝tα T X, 2π  L (t) 4

+ L 1 ,[−π,t]

⎤  ⎡  α 1   α+1   α+1   α+1  L χ[t,π] (·) (t) |· − t| 1 ⎦  ⎣ sin + χ[t,π] (·) L (t) α 4 (α + 1) 1 ⎞    α+1   α+1 |· − t| ⎠ sin χ[t,π] (·) 1 ⎝Tαt X, 2π  L (t) 4

⎫ ⎬

⎛

L 1 ,[t,π]

⎭

. 

Proof By (17.52).

Corollary 17.31 All as in Theorem 17.19 for [a, b] = [−π, π]. Further assume that  L (1) = 1 and α = 21 . Then, ∀ t ∈ [−π, π] , we have E (|M (X ) − X |) (t) ≤

√ 2π

(17.59)

⎧  23   13     ⎨+     23 |· − t| 2  sin 2  L χ[−π,t] (·) (t) + L χ[−π,t] (·) (t) ⎩ 3 4

414

17 Trigonometric Conformable Fractional Approximation …

⎛

⎞     23   23 |· − t| sin χ[−π,t] (·) 1 ⎝t1 T X, 2π  L (t) ⎠ 2 4

+ L 1 ,[−π,t]

+  23   13         23 |· − t| 2  χ[t,π] (·) 2  L χ[t,π] (·) (t) + sin L (t) 3 4 ⎞     23   23 |· − t| sin χ[t,π] (·) 1 ⎝T 1t X, 2π  L (t) ⎠ 2 4

⎫ ⎬

⎛

L 1 ,[t,π]

⎭

. 

Proof By (17.58). We continue with interesting uniform corollaries:

Corollary 17.32 All as in Theorem 17.20 for [a, b] = [−π, π]. Further assume that  L (1) = 1 and q1 < α < 1 (i.e. n = 0). Then    1  E |M (X ) − X |q  q ≤ ∞ +

⎡

⎛

 

⎣ sup 1 ⎝Tαt X, 2π    (q t∈[−π,π]

2

α+ 1p

 1 p L (1)∞ π α  1

α

( p (α − 1) + 1) p (q + 1) q(α+1)

1 q   1 α α+1  L (1)∞ (q + 1) α+1 + 1

1  L + 1)

(17.60)

⎞ 1   q(α+1)    q(α+1) |· − t|  ⎠ sin χ[t,π] (·) (t)  4 ∞

L q ,[t,π]

⎤ α    q(α+1)    q(α+1)  |· − t|   ⎦+ sin χ[t,π] (·) (t) L   4 ∞

⎡

⎛

⎞ 1    q(α+1)    q(α+1)  1 |· − t|    ⎣ sup 1 ⎝tα T X, 2π  ⎠ L sin χ[−π,t] (·) (t)   (q + 1) 4 t∈[−π,π] ∞

L q ,[−π,t]

⎤⎫ α    q(α+1)     q(α+1) ⎬ |· − t|   ⎦ . sin χ[−π,t] (·) (t) L ⎭   4 ∞

Proof By (17.53).



Corollary 17.33 All as in Theorem 17.21 for [a, b] = [−π, π]. Further assume that  L (1) = 1, and 0 < α < 1 (i.e. n = 0). Then

17.4 Main Results

415

E (|M (X ) − X |)∞

⎡ ⎤  1   α+1 L (1) 1 ∞ ⎦ ≤ (2π)α ⎣ + α α+1

⎧  α   α+1    α+1 ⎨ |· − t|   sin χ[−π,t] (·) (t) L  ⎩ 4 ∞

⎛

1 ⎞    α+1   α+1   |· − t|   sin L sup 1 ⎝tα T X, 2π  χ[−π,t] (·) (t) ⎠   4 t∈[−π,π]

∞

L 1 ,[−π,t]

(17.61)

α    α+1    α+1  |· − t|   sin χ[t,π] (·) + L (t)   4

∞

1 ⎞    α+1    α+1  |· − t|   sin L χ[t,π] (·) sup 1 ⎝Tαt X, 2π  (t) ⎠   4 t∈[−π,π]

⎫ ⎬

⎛

∞

L 1 ,[t,π]

⎭

. 

Proof By (17.54).

Corollary 17.34 All as in Theorem 17.21 for [a, b] = [−π, π]. Further assume that  L (1) = 1, and α = 21 . Then E (|M (X ) − X |)∞ ≤

√

 +  2 2 3   2π 2 L (1) ∞ + 3

⎧  1    23   3 ⎨ |· − t|   sin χ[−π,t] (·) (t) L  ⎩ 4

∞

⎛

2 ⎞     23   3  |· − t|   sin L χ[−π,t] (·) sup 1 ⎝t1 T X, 2π  (t) ⎠ 2   4 t∈[−π,π]

∞

(17.62) L 1 ,[−π,t]

1     23   3  |· − t|   sin χ[t,π] (·) + L (t)   4

∞

2 ⎞     23   3  |· − t|   sin L χ[t,π] (·) sup 1 ⎝T 1t X, 2π  (t) ⎠ 2   4 t∈[−π,π]

⎫ ⎬

⎛

∞

L 1 ,[t,π]

⎭

.

416

17 Trigonometric Conformable Fractional Approximation …



Proof By (17.61).

17.5 Application Consider the Bernstein polynomials on [−π, π] for f ∈ C ([−π, π]) : B N ( f ) (x) =

 N   N k=0

k

 f

−π +

2πk N



x +π 2π

k 

π−x 2π

 N −k

,

(17.63)

N ∈ N, any x ∈ [−π, π]. There are positive linear operators from C ([−π, π]) into itself. Setting g (t) = f (2πt − π), t ∈ [0, 1], we have g (0) = f (−π), g (1) = f (π), and    N   k N g t k (1 − t) N −k = (B N f ) (x) , x ∈ [−π, π] . (B N g) (t) = k N k=0 (17.64) Here x = ϕ (t) = 2πt − π is an 1 − 1 and onto map from [0, 1] onto [−π, π]. Clearly here g ∈ C ([0, 1]). Notice also that  

    (2π)2 t (1 − t) B N (· − x)2 (x) = B N (· − t)2 (t) (2π)2 = N    π−x 1 π2 (2π)2 x + π = = , ∀ x ∈ [−π, π] . (x + π) (π − x) ≤ N 2π 2π N N 

I.e.



  π2 , ∀ x ∈ [−π, π] . B N (· − x)2 (x) ≤ N

(17.65)

(B N 1) (x) = 1, ∀ x ∈ [−π, π] .

(17.66)

In particular

Define the corresponding application of M by  B N (X ) (t, ω) := B N (X (·, ω)) (t) =  N   N k=0

k

     t + π k π − t N −k 2πk ,ω , X −π + N 2π 2π

(17.67)

17.5 Application

417

∀ N ∈ N, ∀ t ∈ [−π, π], ∀ ω ∈ , where X is a stochastic process. Clearly  B N is a stochastic process. We give Proposition 17.35 Let X (t, ω) be a stochastic process from [−π, π] × (, F, P) into R, where (, F, P) is a probability space. Here 0 < α < 1 (i.e. n = 0) and X (·, ω) ∈ C 1 ([−π, π]), ∀ ω ∈  and X (1) (t, ·) is measurable over (, F), ∀ t ∈ [−π, π] . For any t ∈ [−π, π] we assume that Tαt X (z, ω) is continuous in z ∈ [t, π], uniformly with respect to ω ∈ . And for any t ∈ [−π, π] we assume that tα T X (z, ω) is continuous in z ∈ [−π, t], uniformly with respect to ω ∈ . Finally, we assume that (E |X |) (t) < ∞, ∀ t ∈ [−π, π]. Then, for any t ∈ [−π, π], we have:   B N (X ) − X (t) ≤ (2π)α E  ⎧⎡  ⎤ α     1 α+1   α+1 ⎨ B χ (·) (t) α+1 |· − t| 1 ⎦ ⎣ N [−π,t] sin + χ[−π,t] (·) BN (t) ⎩ α 4 (α + 1) ⎛



1 ⎝tα T X, 2π

 BN



α+1 

|· − t| sin χ[−π,t] (·) 4



1 α+1

(t)

⎞ ⎠

+ L 1 ,[−π,t]

⎡

⎤ α      1 α+1   α+1 B N χ[t,π] (·) (t) α+1 |· − t| 1 ⎣ ⎦ BN sin + χ[t,π] (·) (t) α 4 (α + 1) ⎛ 1 ⎝Tαt X, 2π



 BN

1 ⎞ α+1   α+1 |· − t| ⎠ χ[t,π] (·) sin (t) 4

⎫ ⎬



L 1 ,[t,π]

⎭

,

(17.68)

∀ N ∈ N. 

Proof By Corollary 17.30. We give Proposition 17.36 All as in Proposition 17.35 with α = [−π, π], we have: √   E  B N (X ) − X (t) ≤ 2π

1 2

. Then, for any t ∈

⎧  23   13     ⎨+     23 |· − t| 2 sin 2 B N χ[−π,t] (·) (t) + χ[−π,t] (·) BN (t) ⎩ 3 4 ⎛

⎞    23   23 |· − t| sin 1 ⎝t1 T X, 2π B N χ[−π,t] (·) (t) ⎠ 2 4 

+ L 1 ,[−π,t]

418

17 Trigonometric Conformable Fractional Approximation …

 23   13 +         23 |· − t| 2 sin 2 B N χ[t,π] (·) (t) + χ[t,π] (·) BN (t) 3 4 ⎛

⎞    23   23 |· − t| 1 ⎝T 1t X, 2π B N sin χ[t,π] (·) (t) ⎠ 2 4

⎫ ⎬



L 1 ,[t,π]

⎭

,

(17.69)

∀ N ∈ N. 

Proof By Proposition 17.35. We continue with Proposition 17.37 All as in Proposition 17.35 with α = [−π, π] , we have:

1 2

. Then, for any t ∈

√     23   13   |· 5 − t| 2π sin E  B N (X ) − X (t) ≤ BN (t) 3 4 ⎡

⎛

⎞    23   23 |· − t| ⎣1 ⎝t1 T X, 2π B N sin (t) ⎠ 2 4 ⎛





1 ⎝T 1t X, 2π B N 2





|· − t| sin 4

 23 

+ L 1 ,[−π,t]

 23 ⎞ (t) ⎠

⎤ ⎦,

(17.70)

L 1 ,[t,π]

∀ N ∈ N. Proof By (17.69) and the positivity of B N , see also (17.42) and (17.44).



We make Remark 17.38 By |sin x| < |x| , ∀ x ∈ R − {0}, in particular sin x ≤ x, for x ≥ 0, we get    23  3 |· − t| |· − t| 2 1 3 sin ≤ = |· − t| 2 . 4 4 8 Hence     23     |· − t|   sin (t)  BN   4

∞

≤

   1 3   B N |· − t| 2 (t) . ∞ 8

(17.71)

17.5 Application

419

We observe that 3       N    2πk 2 N t + π k π − t N −k 3 t + π − B N |· − t| 2 (t) = k N 2π 2π k=0 (by discrete Hölder’s inequality)  N  , 43        2πk 2 N t + π k π − t N −k ≤ t +π− k N 2π 2π k=0  3   = B N (· − t)2 (t) 4 Consequently it holds

3

π2

(17.65.)

≤

3

N4

    3   B N |· − t| 2 (t)

    23     |· − t|   sin (t)  BN   4

(17.72)

3

∞

and

, ∀ t ∈ [−π, π] .

π2

≤

3

N4

,

(17.73)

3

≤

∞

π2

3

8N 4

, ∀ N ∈ N.

(17.74)

We further have Proposition 17.39 All as in Proposition 17.35 with α = 21 . Then, for any t ∈ [−π, π] we have: √   5 2π  E B N (X ) − X (t) ≤ √ 64N ,      π2 π2 t t + 1 T 1 X, √ , (17.75) 1 1 T X, √ 2 2 2 N L 1 ,[−π,t] 2 N L 1 ,[t,π] ∀ N ∈ N.   As N → +∞, we get E  B N (X ) − X (t) → 0. Proof By positivity of B N , (17.70) and (17.74). See also Proposition 17.6.



Consequently we obtain 1 Proposition 17.40 All as in Proposition 17.35 with α = 2 . Assume further that (1) X (t, ω) ≤ M ∗ , ∀ (t, ω) ∈ [−π, π] × , where M ∗ > 0. Then

    E  B N (X ) − X 

∞

√ 5 2π ≤ √ 64N

(17.76)

420

17 Trigonometric Conformable Fractional Approximation …



 sup 1

t∈[−π,π]

π2 √ 1 T X, 2 2 N





t

L 1 ,[−π,t]

+ sup 1 t∈[−π,π]

π2 T 1 X, √ 2 2 N

,



t

L 1 ,[t,π]

,

∀ N ∈ N.    B N (X ) − X ∞ → 0, i.e.  B N → I (stochastic unit As N → +∞, then  E  operator) in 1-mean. Proof By (17.75) and Remark 17.9, see (17.17), (17.18).



17.6 Trigonometric Conformable Fractional Stochastic Korovkin Results In this section  L, M are meant as sequences of operators. We give first pointwise results: Theorem Here all as in Theorem 17.23. Assume further that  L (1) (t) → 1  17.41  q(α+1)  |·−t| and  L sin 4 (t) → 0, pointwise in t ∈ [−π, π].   Then E |M (X ) − X |q (t) → 0, pointwise in t ∈ [−π, π], that is M → I (stochastic unit operator) in q-mean-pointwise with rates, quantitatively. Proof We use (17.51), we take into account  L (1) (t) → 1, (17.39); and    L χ[t,π] (·) (t),  L χ[−π,t] (·) (t) ≤  L (1) (t), which  L (1) (t) is bounded, and by  (17.42), (17.44) and the positivity of L we get that     q(α+1)  q(α+1)  |·−t|   χ χ L sin |·−t| L sin (·) (·) (t), (t) [t,π] [−π,t] 4 4    q(α+1) ≤ L sin |·−t| (t) → 0. 4 Finally, we use Proposition 17.7(ii) for the 1 (·, ·)’s to go to zero.



We continue with Theorem Here all as in Theorem 17.24. Assume further that  L (1) (t) → 1  17.42  (α+1) |·−t| and  L sin 4 (t) → 0, pointwise in t ∈ [−π, π]. Then E (|M (X ) − X |) (t) → 0, pointwise in t ∈ [−π, π], that is M → I in 1mean-pointwise with rates, quantitatively. Proof Based on (17.52), similar to the proof of Theorem 17.41, just take q = 1 there. 

17.6 Trigonometric Conformable Fractional Stochastic Korovkin Results

421

Next we give uniform results: Theorem 17.43 in Theorem Assume further that  L (1) → 1, uni Hereall as   17.25. q(α+1)   |·−t|  formly, and  (t)  L sin 4  → 0. ∞    Then  E |M (X ) − X |q ∞ → 0 over [−π, π] , that is M → I in the q-mean, quantitatively with rates.   Proof We use (17.53), we take into account  L (1) → 1 uniformly, (17.40);  L (1)∞ is bounded, use of (17.43), (17.45) and Remark 17.9, see there (17.17), (17.18).  Next we give the L 1 -mean uniform result  Theorem 17.44 in Theorem  Hereall as  17.26. Assume further that L (1) → 1, uni(α+1)   |·−t|  formly, and  (t)  L sin 4  → 0. ∞ Then E (|M (X ) − X |)∞ → 0 over [−π, π] , that is M → I in the 1-mean, quantitatively with rates. Proof Use of (17.54), similar to the proof of Theorem 17.43, just take q = 1 there.  We finish with Remark 17.45 An amazing fact/observation follows: In all trigonometric convergence results here, see Theorems 17.41–17.44, the forcing conditions for convergences are based only on  L and basic real valued continuous functions on [−π, π] and are not related to stochastic processes, but they are giving trigonometric convergence results on stochastic processes!

References 1. Abdeljawad, T.: On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 3. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 4. Anastassiou, G.: Quantitative Approximations. Chapman & Hall/CRC, Boca Raton (2001) 5. Anastassiou, G.: Stochastic Korovkin theory given quantitatively, Facta Universitatis (Nis). Ser. Math. Inform. 22(1), 43–60 (2007) 6. Anastassiou, G.: Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Maxproduct operators. Springer, Heidelberg (2018) 7. Anastassiou, G.: Conformable fractional approximation of stochastic processes (2020). Submitted for publication 8. Anastassiou, G.: Trigonometric conformable fractional quantitative approximation of stochastic processes. In: Progress in Fractional Differentiation and Applications (2020). Accepted

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17 Trigonometric Conformable Fractional Approximation …

9. Khalil, R., Al Horani, M., Yousef, A., Sababheh, M.: A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 10. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi, India (1960) 11. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 12. Shisha, O., Mond, B.: The degree of approximation to periodic functions by linear positive Operators. J. Approx. Theory 1, 335–339 (1968) 13. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192(1), 73–80 (1986) 14. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61(1), 74–87 (1990)

Chapter 18

Commutative Caputo Fractional Korovkin Approximation for Stochastic Processes

Here we consider and study expectation commutative stochastic positive linear operators acting on L 1 -continuous stochastic processes which are Caputo fractional differentiable. Under some mild, general and natural assumptions on the stochastic processes we produce related Caputo fractional stochastic Shisha-Mond type inequalities pointwise and uniform. All convergences are produced with rates and are given by the fractional stochastic inequalities involving the first modulus of continuity of the expectation of the αth right and left fractional derivatives of the engaged stochastic process, α > 0, α ∈ / N. The amazing fact here is that the basic real Korovkin test functions assumptions impose the conclusions of our Caputo fractional stochastic Korovkin theory. We include also a detailed application to stochastic Bernstein operators. See also [11].

18.1 Introduction Our work is motivated by the following:   Korovkin’s Theorem [13, 1960] Let T j j∈N be a sequence of positive linear   operators from C ([a, b]) into itself, [a, b] ⊂ R. In order to have lim T j f (t) = j→∞

f (t) (in the sup-norm) for all f ∈ C ([a, b]), it is enough to prove it for f 0 (t) = 1, f 1 (t) = t and f 2 (t) = t 2 . The rate of the above convergence for arbitrary f ∈ C ([a, b]) can be determined exactly from the rates of convergence for f 0 , f 1 , f 2 . The above theorem was put in an inequality form: Shisha-Mond inequality [15] We have          T j ( f ) − f  ≤  f  · T j (1) − 1 + ω1 f, ρ j · 1 + T j (1) , where

  1   ρ j = T j (x − y)2 (y) 2 .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_18

423

424

18 Commutative Caputo Fractional Korovkin Approximation …

In the last inequality · stands for the supremum norm and ω1 for the first modulus of continuity. This inequality gives the rate of convergence of T j to the unit operator I. Anastassiou in [2–4] established a series of sharp inequalities for various cases of the parameters of the problem. However, Weba in [16–19] was the first, among many workers in quantitative results of Shisha-Mond type, to produce inequalities for stochastic processes. He assumed that T j are E-commutative (E means expectation) and stochastically simple. According to his work, if a stochastic process X (t, ω), t ∈ Q—a compact convex subset of a real normed vector space, ω ∈ Q—probability space, is to be approximated by positive linear operators T j , then the maximal error in the qth mean is (q ≥ 1)   1     T j X − X  = sup E  T j X (t, ω) − X (t, ω)q q . t∈Q

  So, Weba established upper bounds for T j X − X  involving his own natural general first modulus of continuity of X with several interesting applications. Anastassiou met [5] the pointwise case of q = 1. Without stochastic simplicity     of T j he found nearly best and best upper bounds for  E T j X (x0 ) − (E X ) (x0 ), x0 ∈ Q. The author here continues his above work on the approximation of stochastic processes, now at the Caputo stochastic fractional level. He derives pointwise and uniform Caputo fractional stochastic Shisha-Mond type inequalities, see the main Theorems 18.4, 18.7 and the several related corollaries. He gives an extensive application to stochastic Bernstein operators. He finishes with a pointwise and a uniform fractional stochastic Korovkin theorem, derived by Theorems 18.4, 18.7. The stochastic convergences, about stochastic processes, of our fractional Korovkin Theorems 18.15, 18.16 are enforced only by the convergences of real basic non-stochastic functions.

18.2 Background We need Definition 18.1 ([10]) Let non-integer α > 0, n = α ( · is the ceiling of the number), t ∈ [a, b] ⊂ R, ω ∈ , where (, F, P) is a general probability space. Here X (t, ω) stands for a stochastic process. Assume that X (·, ω) ∈ AC n ([a, b]) (spaces of functions X (·, ω) with X (n−1) (·, ω) ∈ AC ([a, b]) absolutely continuous functions), ∀ ω ∈ . We call stochastic left Caputo fractional derivative α X (x, ω) = D∗a

∀ x ∈ [a, b], ∀ ω ∈ .

1  (n − α)

 a

x

(x − t)n−α−1 X (n) (t, ω) dt,

(18.1)

18.2 Background

425

And, we call stochastic right Caputo fractional derivative α Db− X

(−1)n (x, ω) =  (n − α)



b

(z − x)n−α−1 X (n) (z, ω) dz,

(18.2)

x

∀ x ∈ [a, b], ∀ ω ∈ . Above  stands for the gamma function. We make Remark 18.2 (to Definition 18.1) We further assume here that  (n)   X (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0. Then, by (18.1), we have   α  D X (x, ω) ≤ ∗a M  (n − α)

1  (n − α) 

x



x

  (x − t)n−α−1  X (n) (t, ω) dt ≤

a

(x − t)n−α−1 dt =

a

M (x − a)n−α .  (n − α + 1)

That is n−α  α   D X (x, ω) ≤ M (x − a) , ∀ x ∈ [a, b] , any ω ∈ . ∗a  (n − α + 1)

(18.3)

Also, from (18.2) we get  α   D X (x, ω) ≤ b− M  (n − α)

1  (n − α) 

b x



b

  (z − x)n−α−1  X (n) (z, ω) dz ≤

x

(z − x)n−α−1 dz =

M (b − x)n−α .  (n − α + 1)

That is n−α  α   D X (x, ω) ≤ M (b − x) , ∀ x ∈ [a, b] , any ω ∈ . b−  (n − α + 1)

(18.4)

α α It is not strange to assume that D∗a X , Db− X are stochastic processes. α By [7, p. 388], we get that D∗a X (·, ω) ∈ C ([a, b]), ∀ ω ∈ . And by [8], we get α that Db− X (·, ω) ∈ C ([a, b]), ∀ ω ∈ .

426

18 Commutative Caputo Fractional Korovkin Approximation …

Similarly, we obtain n−α   α M (b − t)n−α  D X (x, ω) ≤ M (x − t) ≤ , ∗t  (n − α + 1)  (n − α + 1)

∀ x ∈ [t, b] , any t ∈ [a, b] , ∀ ω ∈ , and n−α   α M (t − a)n−α  D X (x, ω) ≤ M (t − x) ≤ , t−  (n − α + 1)  (n − α + 1)

(18.5)

(18.6)

∀ x ∈ [a, t] , any t ∈ [a, b] , ∀ ω ∈ . α α Above D∗t X , Dt− X are assumed to be stochastic processes for any t ∈ [a, b], α α X (·, ω) ∈ C ([a, t]), ∀ ω ∈ . and it holds D∗t X (·, ω) ∈ C ([t, b]), Dt− Clearly, then n−α   α    E D X (x) ≤ M (b − t) , (18.7) ∗t  (n − α + 1)  ∀ x ∈ [t, b], any t ∈ [a, b], where E is the expectation operator (E X ) (t) =  X (t, ω) P (dω) and similarly, n−α   α    E D X (x) ≤ M (t − a) , t−  (n − α + 1)

(18.8)

∀ x ∈ [a, t], any t ∈ [a, b] . We observe that the first modulus of continuity (δ > 0)   α   ω1 E D∗t X , δ [t,b] :=

  α   α   sup  E D∗t X (x) − E D∗t X (y)

x,y∈[t,b]: |x−y|≤δ

(18.7)

≤

2M (b − t)n−α ,  (n − α + 1)

(18.9)

any t ∈ [a, b] . Hence, it holds (δ > 0)   α   2M (b − a)n−α . sup ω1 E D∗t X , δ [t,b] ≤  (n − α + 1) t∈[a,b]

(18.10)

Similarly, it holds (δ > 0)   α   2M (b − a)n−α sup ω1 E Dt− . X , δ [a,t] ≤  (n − α + 1) t∈[a,b]

(18.11)

18.2 Background

427

By [6, p. 209], we have that (δ1 > 0) 

  α          E D X (z) − E D α X (t) ≤ ω1 E D α X , δ1 ∗t

∗t

∗t

[t,b]

|z − t| δ1

   α   (z − t) ω1 E D∗t X , δ1 [t,b] 1 + , δ1

≤

(18.12)

∀ z ∈ [t, b] , and similarly (δ2 > 0),

   α         t −z  E D X (z) − E D α X (t) ≤ ω1 E D α X , δ2 1 + , t− t− t− [a,t] δ2 (18.13) ∀ z ∈ [a, t] . We also set           α   α X , δ [t,b] , ω1 E Dt− X , δ [a,t] , ω1 E Dtα X , δ := max ω1 E D∗t (18.14) where δ > 0. We make Remark 18.3 Let the positive linear operator L mapping C ([a, b]) into B ([a, b]) (the bounded functions). By the Riesz representation theorem [14] we have that there exists μt unique, completed Borel measure on [a, b] with μt ([a, b]) = L (1) (t) > 0,

(18.15)

such that  L ( f ) (t) =

f (s) dμt (s) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(18.16)

[a,b]

Let now n = α , α ∈ / N, α > 0, k = 1, . . . , n − 1. Then by Hölder’s inequality we obtain      k ≤  |s − t|k dμt (s) ≤ dμ − t) (s (s) t   [a,b]



[a,b]

|s − t|α+1 dμt (s)

k ( α+1 )

(μt ([a, b]))(

α+1−k α+1

).

(18.17)

[a,b]

The last means  α+1−k            k  L (s − t)k (t) ≤ L |s − t|k (t) ≤ L |s − t|α+1 (t) ( α+1 ) (L (1) (t)) α+1 , (18.18)

428

18 Commutative Caputo Fractional Korovkin Approximation …

all k = 1, . . . , n − 1. It is clear that      L |s − t|α+1 (t)

∞,[a,b]

< ∞.

Furthermore we derive      L (s − t)k (t)

∞,[a,b]

    ≤  L |s − t|k (t)∞,[a,b] ≤

α+1−k     k  L |s − t|α+1 (t)( α+1 ) L (1)( α+1 ) , ∞,[a,b] ∞,[a,b]

(18.19)

all k = 1, . . . , n − 1. From now on we will denote ·∞,[a,b] = ·∞ the supremum norm.

18.3 Preliminaries Let (, F, P) be a probabilistic space and L 1 (, F, P) be the space of all realvalued random variables Y = Y (ω) with  |Y (ω)| P (dω) < ∞. 

Let X = X (t, ω) denote a stochastic process with index set [a, b] ⊂ R and real state space (R, B), where B is the σ-field of Borel subsets of R. Here C ([a, b]) is the space of continuous real-valued functions on [a, b] and B ([a, b]) is the space of bounded real-valued functions on [a, b]. Also C ([a, b]) = C [a, b] , L 1 (, F, P) processes in t and B ([a, b]) = is the space of L 1 -continuous stochastic    X : sup  |X (t, ω)| P (dω) < ∞ , obviously C ([a, b]) ⊂ B ([a, b]). t∈[a,b]

Let α > 0, α ∈ / N, α = n, and consider the subspace of stochastic processes Cα,n ([a, b]) := {X : X (·, ω) ∈ AC n ([a, b]) , ∀ ω ∈  and  X (n) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0; X (k) (t, ω) ∈ C ([a, b]), k = 0, 1, . . . , n − 1; α α X , Dt− X are stochastic processes for any t ∈ [a, b]}. That is, for every also D∗t ω ∈  we have X (t, ω) ∈ C n−1 ([a, b]). Consider the linear operator L : C ([a, b]) → B ([a, b]) . If X ∈ C ([a, b]) is nonnegative and L X , too, then L is called positive. If E L = L E, then L is called E-commutative.

18.4 Main Results

429

18.4 Main Results Following 3. Preliminaries we state Theorem 18.4 Consider the positive E-commutative linear operator L : C / N, α = n, and let X ∈ Cα,n ([a, b]), with ([a, b]) → B ([a, b]), and α > 0, α ∈ δ > 0. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +

(18.20)

      n−1     ω E Dtα X , δ E X (k) (t)    L (s − t)k (t) + 1 k!  (α + 1) k=1 ⎤ ⎡ 1   α+1   α+1 α     |s L − t| (t) 1 ⎦, L |s − t|α+1 (t) α+1 ⎣(L (1) (t)) α+1 + δ (α + 1) ∀ t ∈ [a, b] .     Above ω1 E Dtα X , δ is as in (18.14). Proof We have that

α α D∗t X (t, ω) = Dt− X (t, ω) = 0,

(18.21)

∀ ω ∈ , see [9, pp. 358–359]. We also assume α X (s, ω) = 0, for s < t, D∗t and

α X (s, ω) = 0, for s > t, Dt−

∀ ω ∈ . We get by left Caputo fractional Taylor’s formula that [12, p. 54] X (s, ω) = 1  (α)

 t

s

n−1  X (k) (t, ω) (s − t)k + k! k=0

  α α X (z, ω) − D∗t X (t, ω) dz, (s − z)α−1 D∗t

for all t ≤ s ≤ b, ∀ ω ∈ .

(18.22)

430

18 Commutative Caputo Fractional Korovkin Approximation …

Also, from [9, p. 341], using the right Caputo fractional Taylor formula we get n−1  X (k) (t, ω) X (s, ω) = (s − t)k + k! k=0

1  (α)



t s

(18.23)

 α  α X (z, ω) − Dt− X (t, ω) dz, (z − s)α−1 Dt−

for all a ≤ s ≤ t, ∀ ω ∈ . Therefore we get  n−1   E X (k) (t) (E X ) (s) = (s − t)k + k! k=0 1 E  (α)



s t

  α  α X (z, ω) − D∗t X (t, ω) dz , (s − z)α−1 D∗t

all t ≤ s ≤ b, and (E X ) (s) = 1 E  (α)



t s

α−1

(z − s)



(18.24)

 n−1   E X (k) (t) (s − t)k + k! k=0 α Dt− X

(z, ω) −

α Dt− X





(t, ω) dz ,

(18.25)

all a ≤ s ≤ t. α α X and Dt− X are jointly measurBy [1, p. 156, Theorem 20.15], the functions D∗t able.   s α−1  α α By (18.5) and  (18.6) we obtain that E t (s − z)  D  ∗t X (z, ω) − D∗t t X (t, ω)| dz), E (z − s)α−1  D α X (z, ω) − D α X (t, ω) dz are finite. s

t−

t−

Therefore, by Fubini-Tonelli’s theorem [14], we get  n−1   E X (k) (t) (E X ) (s) = (s − t)k + k! k=0 1  (α) all t ≤ s ≤ b,

 t

s

  α   α   X (z) − E D∗t X (t) dz, (s − z)α−1 E D∗t

(18.26)

18.4 Main Results

431

and

 n−1   E X (k) (t) (s − t)k + (E X ) (s) = k! k=0 1  (α)



t s

  α   α   X (z) − E Dt− X (t) dz, (z − s)α−1 E Dt−

all a ≤ s ≤ t. Call the remainders  s   α   α   1 R1 (t, s) := X (z) − E D∗t X (t) dz, (s − z)α−1 E D∗t  (α) t

(18.27)

(18.28)

all t ≤ s ≤ b, and R2 (s, t) =

1  (α)



t

s

  α   α   X (z) − E Dt− X (t) dz, (z − s)α−1 E Dt−

(18.29)

all a ≤ s ≤ t. We observe that (t ≤ s ≤ b, δ1 > 0) |R1 (t, s)| ≤

1  (α)



s

t

  α   α   (18.12) X (z) − E D∗t X (t) dz ≤ (s − z)α−1  E D∗t

  α   ω1 E D∗t X , δ1 [t,b]   (α)   α   ω1 E D∗t X , δ1 [t,b] 

s t

s

 (z − t) dz = (s − z)α−1 1 + δ1

(s − z)α−1 dz +

1 δ1



s

 (s − z)α−1 (z − t)2−1 dz =

 (α) t t (18.30)   α     ω1 E D∗t X , δ1 [t,b] (s − t)α 1  (α)  (2) + (s − t)α+1 =  (α) α δ1  (α + 2)   α   ω1 E D∗t X , δ1 [t,b]   (α + 1)

 (s − t)α+1 . (s − t) + δ1 (α + 1) α

That is   α    ω1 E D∗t X , δ1 [t,b]  (s − t)α+1 α |R1 (t, s)| ≤ , (s − t) +  (α + 1) δ1 (α + 1) where t ≤ s ≤ b, any t ∈ [a, b], δ1 > 0.

(18.31)

432

18 Commutative Caputo Fractional Korovkin Approximation …

Similarly, we have that (a ≤ s ≤ t, δ2 > 0) 1 |R2 (s, t)| ≤  (α)

 s

t

  α   α   (18.13) X (z) − E Dt− X (t) dz ≤ (z − s)α−1  E Dt−

  α   ω1 E Dt− X , δ2 [a,t]   (α)   α   ω1 E Dt− X , δ2 [a,t] 

t

α−1

(z − s)

s t

α−1

1 dz + δ2

(z − s)  (α) s   α   ω1 E Dt− X , δ2 [a,t]  (t − s)α  (α)

α

  α   ω1 E Dt− X , δ2 [a,t]   (α + 1)

 t −z 1+ dz = δ2



t

(t − z)

2−1

α−1

(z − s)

s

 1  (2)  (α) α+1 = + (t − s) δ2  (α + 2)

 dz = (18.32)

 (t − s)α+1 . (t − s) + δ2 (α + 1) α

That is   α    ω1 E Dt− X , δ2 [a,t]  (t − s)α+1 α |R2 (s, t)| ≤ , (t − s) +  (α + 1) δ2 (α + 1)

(18.33)

where a ≤ s ≤ t, any t ∈ [a, b], δ2 > 0. So, we have  n−1   E X (k) (t) (E X ) (s) − (E X ) (t) = (s − t)k + R1 (t, s) , k! k=1

(18.34)

all t ≤ s ≤ b, and  n−1   E X (k) (t) (E X ) (s) − (E X ) (t) = (s − t)k + R2 (s, t) , k! k=1 all a ≤ s ≤ t. From now on we take δ1 = δ2 =: δ > 0. Therefore, it holds      ω1 E Dtα X , δ |t − s|α+1 α |t − s| + |R1 (t, s)| , |R2 (s, t)| ≤ ,  (α + 1) δ (α + 1) for any s, t ∈ [a, b] .

(18.35)

(18.36)

18.4 Main Results

433

We have that

(18.16)

L (E X ) (t) − (E X ) (t) L (1) (t) = 

(18.15)

(E X ) (s) dμt (s) − (E X ) (t) L (1) (t) = [a,b]



 [a,t)

(E X ) (s) dμt (s) +

(E X ) (s) dμt (s) − [t,b]



 [a,t)

(E X ) (t) dμt (s) −

(E X ) (t) dμt (s) = [t,b]



 [a,t)

((E X ) (s) − (E X ) (t)) dμt (s) +

((E X ) (s) − (E X ) (t)) dμt (s)

(by (18.34), (18.35))

=

[t,b]

(18.37)    n−1  (k)   EX (t)  L (s − t)k (t) + R2 (s, t) dμt (s) + R1 (t, s) dμt (s) . k! [a,t) [t,b] k=1 That is  n−1    E X (k) (t)  L (s − t)k (t) L (E X ) (t) − (E X ) (t) L (1) (t) = k! k=1  +

[a,t)

 R2 (s, t) dμt (s) +

R1 (t, s) dμt (s) .

(18.38)

[t,b]

From [18, pp. 3–5] we have the following results (i) C ([a, b]) ⊂ C ([a, b]) , (ii) if X ∈ C ([a, b]), then E X ∈ C ([a, b]), and (iii) if L is E-commutative, then L maps the subspace C ([a, b]) into B ([a, b]) . One can rewrite L (E X ) (t) − (E X ) (t) = (E X ) (t) [(L (1)) (t) − 1] +   n−1    E X (k) (t)  L (s − t)k (t) + R2 (s, t) dμt (s) k! [a,t) k=1  R1 (t, s) dμt (s) .

+ [t,b]

(18.39)

434

18 Commutative Caputo Fractional Korovkin Approximation …

Consequently, by E-commutativity of L we find |E [(L X ) (t, ω) − X (t, ω)]| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +     n−1     E X (k) (t)    L (s − t)k (t) + |R2 (s, t)| dμt (s) k! [a,t) k=1 

(18.36)

|R1 (t, s)| dμt (s) ≤

+ [t,b]

  n−1     E X (k) (t)    L (s − t)k (t) + |(E X ) (t)| |(L (1)) (t) − 1| + k! k=1        ω1 E Dtα X , δ 1 |t − s|α dμt (s) + |t − s|α+1 dμt (s) ≤  (α + 1) δ (α + 1) [a,b] [a,b] (by Hölder’s inequality)   n−1     E X (k) (t)    L (s − t)k (t) + |(E X ) (t)| |(L (1)) (t) − 1| + k! k=1       α  α+1 ω1 E Dtα X , δ 1 α+1 |t − s| dμt (s) (L (1) (t)) α+1  (α + 1) [a,b] 1 + δ (α + 1)



α+1

|t − s|

(18.40)

 dμt (s) .

[a,b]

We have proved that |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +       n−1     ω E Dtα X , δ E X (k) (t)    L (s − t)k (t) + 1 k!  (α + 1) k=1      α 1 L |s − t|α+1 (t) α+1 (L (1) (t)) α+1 +

     1 α+1 L |s − t| (t) = δ (α + 1) (18.41)    n−1     E X (k) (t)    L (s − t)k (t) + |(E X ) (t)| |(L (1)) (t) − 1| + k! k=1

18.4 Main Results

435

      α ω1 E Dtα X , δ   L |s − t|α+1 (t) α+1  (α + 1) ⎡ 1 ⎣(L (1) (t)) α+1

  1 ⎤   L |s − t|α+1 (t) α+1 ⎦. + δ (α + 1) 

The theorem now is valid. We need

Definition 18.5 If 0 < α < then n = 1, and Cα,1 ([a, b]) := {X : X (·, ω) ∈  1, AC ([a, b]) , ∀ ω ∈  and  X (1) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0; α α X (t, ω) ∈ C ([a, b]) ; also D∗t X , Dt− X are stochastic processes for any t ∈ [a, b]}. We give Corollary 18.6 Consider the positive E-commutative linear operator L : C ([a, b]) → B ([a, b]), and 0 < α < 1 and let X ∈ Cα,1 ([a, b]), with δ > 0. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +       α ω1 E Dtα X , δ   L |s − t|α+1 (t) α+1  (α + 1) ⎡ ⎣(L (1) (t))

1 α+1

  1 ⎤   L |s − t|α+1 (t) α+1 ⎦, + δ (α + 1)

(18.42)

∀ t ∈ [a, b] . 

Proof By Theorem 18.4. We further present Theorem 18.7 All as in Theorem 18.4. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +   n−1   E X (k) 

∞

k=1

k!

        ω E Dtα X , δ  L (s − t)k (t) + sup 1 ∞  (α + 1) t∈[a,b]

⎡     1 ⎤  L |s − t|α+1 (t) α+1 1 α     ∞ ⎦ α+1  L |s − t|α+1 (t) α+1 ⎣L (1)∞ < ∞. (18.43) + ∞ δ (α + 1)

436

18 Commutative Caputo Fractional Korovkin Approximation …



Proof By Theorem 18.4, see (18.10), (18.11) and (18.14). Corollary 18.8 All as in Corollary 18.6. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +       α ω1 E Dtα X , δ    L |s − t|α+1 (t) α+1 sup ∞  (α + 1) t∈[a,b] ⎡ 1 α+1

⎣L (1)∞

    1 ⎤  L |s − t|α+1 (t) α+1 ∞ ⎦ . + δ (α + 1)

(18.44) 

Proof By Corollary 18.6. Corollary 18.9 All as in Corollary 18.6, and L (1) = 1. Then E (L X ) − E X ∞

    ω1 E Dtα X , δ ≤ sup  (α + 1) t∈[a,b]

⎡     1 ⎤  L |s − t|α+1 (t) α+1 α     ∞ ⎦  L |s − t|α+1 (t) α+1 ⎣1 + . ∞ δ (α + 1)

(18.45) 

Proof By (18.44). We give Corollary 18.10 All as in Corollary 18.6, and L (1) = 1. Then

E (L X ) − E X ∞ ≤

  1      α+1  1  L |s − t|α+1 (t)∞ 2 sup ω1 E Dtα X , (α+1) t∈[a,b]

 (α + 1)     α  L |s − t|α+1 (t) α+1 . ∞

(18.46)

Proof By (18.45): we take there δ=

    1 1  L |s − t|α+1 (t) α+1 > 0. ∞ (α + 1)

      In case of  L |s − t|α+1 (t)∞ = 0, we have that L |s − t|α+1 (t) = 0, ∀ t ∈  [a, b]. That is, by (18.16) [a,b] |s − t|α+1 dμt (s) = 0, ∀ t ∈ [a, b], where μt is a probability measure, any t ∈ [a, b] .

18.4 Main Results

437

The last implies |s − t|α+1 = 0, a.e., hence |s − t| = 0, a.e., thus s = t, a.e., which means μt {s ∈ [a, b] : s = t} = 0, i.e. μt = δt , ∀ t ∈ [a, b], where δt is the unit Dirac measure.  Consequently we have E (L X ) (t) = L (E X ) (t) = [a,b] (E X ) (s) dδt (s) = (E X ) (t), ∀ t ∈ [a, b] . That is E (L X ) = E X over [a, b]. Therefore both sides of inequality (18.46) equal to zero. Hence (18.46) is always true. 

18.5 Application Let f ∈ C ([0, 1]) and the Bernstein polynomials B N ( f ) (t) :=

N 

f

k=0

k N



N k



t k (1 − t) N −k ,

∀ t ∈ [0, 1], ∀ N ∈ N. We have that B N 1 = 1 and B N is a positive linear operator. We have that   t (1 − t) , ∀ t ∈ [0, 1] , B N (· − t)2 (t) = N and

   1 1  B N (· − t)2 (t) 2 ≤ √ , ∀ N ∈ N. ∞ 2 N

(18.47)

(18.48)

(18.49)

Define the corresponding stochastic application of B N by B N (X ) (t, ω) := B N (X (·, ω)) (t) =

N  k=0

X

k ,ω N



N k



t k (1 − t) N −k ,

(18.50) ∀ t ∈ [0, 1], ∀ ω ∈ , N ∈ N, where X is a stochastic process. Clearly B N (X ) is a stochastic process and B N : C ([0, 1]) → C ([0, 1]) . Notice that (E B N (X )) (t) =

N  k=0

(E X )

k N



N k



t k (1 − t) N −k = (B N (E X )) (t) ,

(18.51) ∀ t ∈ [0, 1] . That is E B N = B N E, i.e. B N is an E-commutative positive linear operator. We give

438

18 Commutative Caputo Fractional Korovkin Approximation …

Proposition 18.11 Let 0 < α < 1 and X ∈ Cα,1 ([0, 1]). Then

E (B N X ) − E X ∞ ≤

  1      α+1  1  B N |s − t|α+1 (t)∞ 2 sup ω1 E Dtα X , (α+1) t∈[0,1]

 (α + 1)     α  B N |s − t|α+1 (t) α+1 . ∞

(18.52) 

Proof By Corollary 18.10. In particular we get: 1

,1

Corollary 18.12 Let X ∈ C2 ([0, 1]). Then E (B N X ) − E X ∞

   23   2  1 4 3   2 ≤ √ sup ω1 E Dt X ,  B N |s − t| 2 (t) ∞ 3 π t∈[0,1]     13 3   B N |s − t| 2 (t) , ∀ N ∈ N. ∞

Proof Apply (18.52) for α = 21 .

(18.53) 

We make Remark 18.13 We notice that 3  N      k  2 N k 3  2 B N |s − t| (t) = t (1 − t) N −k t − N  k k=0 (by discrete Hölder’s inequality)  N   43 2    N k k N −k t −  ≤ t (1 − t)  k N k=0 (18.48)

=

That is



 43 1 1 t (1 − t) ≤ 3 , ∀ t ∈ [0, 1] . N (4N ) 4

    3    B N |s − t| 2 (t)

∞

and

(18.54)

≤

    13 3    B N |s − t| 2 (t) ≤ ∞

1 3

,

(18.55)

1

,

(18.56)

(4N ) 4 1 (4N ) 4

18.6 Caputo Fractional Stochastic Korovkin Theory

and

    23 1 3    B N |s − t| 2 (t) ≤ √ , ∞ 2 N

439

(18.57)

∀ N ∈ N. We derive 1

,1

Proposition 18.14 Let X ∈ C2 ([0, 1]). Then E (B N X ) − E X ∞

√ 3 

  2 1 1 2 , E D ≤√ √ ω X , sup √ 1 t π 4 N t∈[0,1] 3 N

(18.58)

∀ N ∈ N. Hence lim E (B N X ) = E X , uniformly. N →∞

Proof By (18.53) and Remark 18.13.



18.6 Caputo Fractional Stochastic Korovkin Theory Here L is meant as a sequence of positive E-commutative linear operators and all assumptions are as in Theorem 18.4. We give   Theorem 18.15 We further assume that L (1) (t) → 1 and L |s − t|α+1 (t) → 0, then (E (L X )) (t) → (E X ) (t), for any X ∈ Cα,n ([a, b]), ∀ t ∈ [a, b], a pointwise convergence; where α > 0, α ∈ / N, α = n. Proof Based on (18.20),   (18.18),   and that L (1) (t) is bounded as a sequence of functions. Also ω1 E Dtα X , δ is bounded, see (18.10), (18.11) and (18.14).  We continue with Theorem 18.16  further assume that L (1) (t) → 1, uniformly and    We  L |s − t|α+1 (t) → 0, then E (L X ) → E X , uniformly over [a, b], for any X ∈ ∞ Cα,n ([a, b]); where α > 0, α ∈ / N, α = n. Proof Based (18.19), and that L (1)∞ is bounded. Also it is   on (18.43),   sup ω1 E Dtα X , δ < ∞, by (18.10), (18.11) and (18.14). t∈[a,b]

We finish with Remark 18.17 The stochastic convergences of Theorems 18.15, 18.16 are derived by of the basic and simple real non-stochastic functions   the convergences 1, |s − t|α+1 , an amazing fact!

440

18 Commutative Caputo Fractional Korovkin Approximation …

References 1. Aliprantis, C., Burkinshaw, O.: Principles of Real Analysis, 3rd edn. Academic, San Diego (1998) 2. Anastassiou, G.: A study of positive linear operators by the method of moments, onedimensional case. J. Approx. Theory 45, 247–270 (1985) 3. Anastassiou, G.: Korovkin type inequalities in real normed vector spaces. Approx. Theory Appl. 2, 39–53 (1986) 4. Anastassiou, G.: Multi-dimensional quantitative results for probability measures approximating the unit measure. Approx. Theory Appl. 2, 93–103 (1986) 5. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 6. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 7. Anastassiou, G.: Fractional Differentiation Inequalities. Springer, Heildelberg (2009) 8. Anastassiou, G.: Fractional representation formulae and right fractional inequalities. Math. Comput. Model. 54(11–12), 3098–3115 (2011) 9. Anastassiou, G.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 10. Anastassiou, G.A.: Foundation of stochastic fractional calculus with fractional approximation of stochastic processes. Rev. R. Acad. Cienc. Exactas Fis. Nat. Ser. A Mat. RACSAM 114(2), Paper No. 89 (2020) 11. Anastassiou, G.: Commutative caputo fractional Korovkin inequalities for stochastic processes (2020). Submitted 12. Diethelm, K.: The Analysis of Fractional Differential Equations. Springer, New York (2010) 13. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 14. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 15. Shisha, O., Mond, B.: The degree of convergence of sequences of linear positive operators. Natl. Acad. of Sci. U.S., 60, 1196–1200 (1968) 16. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192(1), 73–80 (1986) 17. Weba, M.: Quantitative results on monotone approximation of stochastic processes. Probab. Math. Stat. 11(1), 109–120 (1990) 18. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61(1), 74–87 (1990) 19. Weba, M.: Monotone approximation of random functions with multivariate domains in respect of lattice semi-norms. Results Math. 20(1–2), 554–576 (1991)

Chapter 19

Trigonometric Commutative Caputo Fractional Korovkin Approximation for Stochastic Processes

Here we consider and study from the trigonometric point of view expectation commutative stochastic positive linear operators acting on L 1 -continuous stochastic processes which are Caputo fractional differentiable. Under some mild, general and natural assumptions on the stochastic processes we produce related trigonometric Caputo fractional stochastic Shisha-Mond type inequalities pointwise and uniform. All convergences are produced with rates and are given by the trigonometric fractional stochastic inequalities involving the first modulus of continuity of the expectation of the αth right and left fractional derivatives of the engaged stochastic process, α > 0, α∈ / N. The amazing fact here is that the basic non-stochastic real Korovkin test functions assumptions impose the conclusions of our trigonometric Caputo fractional stochastic Korovkin theory. We include also a detailed trigonometric application to stochastic Bernstein operators. See also [11].

19.1 Introduction In this work among others we are motivated by the following results. Theorem A (Korovkin [12], (1960)) Let L n : C ([−π, π]) → C ([−π, π]), n ∈ u N, be a sequence of positive linear operators. Assume L n (1) → 1 (uniformly), u u u L n (cos t) → cos t, L n (sin t) → sin t, as n → ∞. Then L n f → f , for every f ∈ C ([−π, π]) that is 2π-periodic. Let f ∈ C ([a, b]) and 0 ≤ δ ≤ b − a. The first modulus of continuity of f at δ is given by ω1 ( f, δ) = sup {| f (x) − f (y)| ; x, y ∈ [a, b] , |x − y| ≤ δ} . If δ > b − a, then we define ω1 ( f, δ) = ω1 ( f, b − a) . © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_19

441

442

19 Trigonometric Commutative Caputo Fractional Korovkin …

Another motivation is the following. Theorem B (Shisha and Mond [14], (1968)) Let L 1 , L 2 , . . . , be linear positive operators, whose common domain D consists of real functions with domain (−∞, ∞). Suppose 1, cos x, sin x, f belong to D, where f is an everywhere continuous, 2π-periodic function, with modulus of continuity ω1 . Let −∞ < a < b < ∞, and suppose that for n = 1, 2, . . . , L n (1) is bounded in [a, b]. Then for n = 1, 2, . . . , L n ( f ) − f ∞ ≤  f ∞ L n (1) − 1∞ + L n (1) + 1∞ ω1 ( f, μn ) , where

(∗ )

     21   2 t −x  μn = π  L n sin (x)  , 2 ∞

and ·∞ stands for the sup norm over [a, b] . In particular, if L n (1) = 1, then (∗ ) reduces to L n ( f ) − f ∞ ≤ 2ω1 ( f, μn ) . One can easily see that, for m = 1, 2, . . . ,  μ2n

≤

π2 2



 L n (1) − 1∞ +

 (L n (cos t)) (x) − cos x∞ + (L n (sin t)) (x) − sin x∞ , so the last along with (∗ ) prove Korovkin’s Theorem A in a quantitative way and with rates of convergence. One more motivation follows. Theorem C (see [5, p. 217]) Let f ∈ C n ([−π, π]) , n ≥ 1, and μ a measure on [−π, π] of mass m > 0. Put 1   n+1  |t| n+1 sin β := · μ (dt) 2

 and denote by ω := ω1 f (n) , β the modulus of continuity of f (n) at β. Then  n (k)

f (0) f dμ − f (0) ≤ | f (0)| · |m − 1| + k! k=1

 1 w m n+1 +

 n n π β π · . n+1 n!

 k · t μ (dt) +

19.1 Introduction

443

Anastassiou in [1–3] established a series of sharp inequalities for various cases of the parameters of the problem. However, Weba in [15–18] was the first, among many workers in quantitative results of Shisha-Mond type, to produce inequalities for stochastic processes. He assumed that the positive linear operators L j are Ecommutative (E means expectation) and stochastically simple. According to his work, if a stochastic process X (t, ω), t ∈ Q—a compact convex subset of a real normed vector space, ω ∈ Q—probability space, is to be approximated by positive linear operators L j , then the maximal error in the qth mean is (q ≥ 1)  1 



  L j X − X  = sup E L j X (t, ω) − X (t, ω) q q . t∈Q

  So, Weba established upper bounds for  L j X − X  involving his own natural general first modulus of continuity of X with several interesting applications. Anastassiou met [4] the pointwise case of q = 1. Without stochastic simplicity

 of L j he found nearly best and best upper bounds for E L j X (x0 ) − (E X ) (x0 ) , x0 ∈ Q. The author here continues his above work on the trigonometric approximation of stochastic processes, now at the Caputo stochastic fractional level. He derives pointwise and uniform trigonometric Caputo fractional stochastic Shisha-Mond type inequalities, see the main Theorems 19.8, 19.9 and the several related corollaries. He gives an extensive trigonometric application to stochastic Bernstein operators. He finishes with a pointwise and a uniform fractional trigonometric stochastic Korovkin theorem, derived by Theorems 19.8, 19.9. The stochastic convergences, about stochastic processes, of our trigonometric fractional Korovkin Theorems 19.18, 19.19 are implied only by the convergences of real basic non-stochastic functions. Our results here are built on [10].

19.2 Background—I We need Definition 19.1 ([9]) Let non-integer α > 0, n = α ( · is the ceiling of the number), t ∈ [a, b] ⊂ R, ω ∈ , where (, F, P) is a general probability space. Here X (t, ω) stands for a stochastic process. Assume that X (·, ω) ∈ AC n ([a, b]) (spaces of functions X (·, ω) with X (n−1) (·, ω) ∈ AC ([a, b]) absolutely continuous functions), ∀ ω ∈ . We call stochastic left Caputo fractional derivative α X (x, ω) = D∗a

∀ x ∈ [a, b], ∀ ω ∈ .

1  (n − α)

 a

x

(x − t)n−α−1 X (n) (t, ω) dt,

(19.1)

444

19 Trigonometric Commutative Caputo Fractional Korovkin …

And, we call stochastic right Caputo fractional derivative α Db− X

(−1)n (x, ω) =  (n − α)



b

(z − x)n−α−1 X (n) (z, ω) dz,

(19.2)

x

∀ x ∈ [a, b], ∀ ω ∈ . Above  stands for the gamma function. We make Remark 19.2 (to Definition 19.1) We further assume here that (n) X (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0. Then, by (19.1), we have α D X (x, ω) ≤ ∗a M  (n − α)

1  (n − α) 

x



x

(x − t)n−α−1 X (n) (t, ω) dt ≤

a

(x − t)n−α−1 dt =

a

M (x − a)n−α .  (n − α + 1)

That is n−α α D X (x, ω) ≤ M (x − a) , ∀ x ∈ [a, b] , any ω ∈ . ∗a  (n − α + 1)

(19.3)

Also, from (19.2) we get α D X (x, ω) ≤ b− M  (n − α)

1  (n − α) 

b x



b

(z − x)n−α−1 X (n) (z, ω) dz ≤

x

(z − x)n−α−1 dz =

M (b − x)n−α .  (n − α + 1)

That is n−α α D X (x, ω) ≤ M (b − x) , ∀ x ∈ [a, b] , any ω ∈ . b−  (n − α + 1)

(19.4)

α α It is not strange to assume that D∗a X , Db− X are stochastic processes. α By [6], p. 388, we get that D∗a X (·, ω) ∈ C ([a, b]), ∀ ω ∈ . And by [7], we get α that Db− X (·, ω) ∈ C ([a, b]), ∀ ω ∈ . Similarly, we obtain

19.2 Background—I

445

n−α α M (b − t)n−α D X (x, ω) ≤ M (x − t) ≤ , ∗t  (n − α + 1)  (n − α + 1)

∀ x ∈ [t, b] , any t ∈ [a, b] , ∀ ω ∈ , and n−α α M (t − a)n−α D X (x, ω) ≤ M (t − x) ≤ , t−  (n − α + 1)  (n − α + 1)

(19.5)

(19.6)

∀ x ∈ [a, t] , any t ∈ [a, b] , ∀ ω ∈ . α α Above D∗t X , Dt− X are assumed to be stochastic processes for any t ∈ [a, b], α α X (·, ω) ∈ C ([a, t]), ∀ ω ∈ . and it holds D∗t X (·, ω) ∈ C ([t, b]), Dt− Clearly, then n−α α  E D X (x) ≤ M (b − t) , (19.7) ∗t  (n − α + 1) ∀ x ∈ [t, b], any t ∈ [a, b], where E is the expectation operator (E X ) (t) =  X (t, ω) P (dω) and similarly, n−α α  E D X (x) ≤ M (t − a) , t−  (n − α + 1)

(19.8)

∀ x ∈ [a, t], any t ∈ [a, b] . We observe that the first modulus of continuity (δ > 0)

α   ω1 E D∗t X , δ [t,b] :=

α 

α  sup E D∗t X (x) − E D∗t X (y)

x,y∈[t,b]: |x−y|≤δ

(19.7)

≤

2M (b − t)n−α ,  (n − α + 1)

(19.9)

any t ∈ [a, b] . Hence, it holds (δ > 0)

α   2M (b − a)n−α . sup ω1 E D∗t X , δ [t,b] ≤  (n − α + 1) t∈[a,b]

(19.10)

Similarly, it holds (δ > 0)

α   2M (b − a)n−α sup ω1 E Dt− . X , δ [a,t] ≤  (n − α + 1) t∈[a,b] We also set

(19.11)

446

19 Trigonometric Commutative Caputo Fractional Korovkin …





α        α ω1 E Dtα X , δ := max ω1 E D∗t X , δ [t,b] , ω1 E Dt− X , δ [a,t] , (19.12) where δ > 0. We make Remark 19.3 Let the positive linear operator L mapping C ([a, b]) into B ([a, b]) (the bounded functions). By the Riesz representation theorem [13] we have that there exists μt unique, completed Borel measure on [a, b] with μt ([a, b]) = L (1) (t) > 0,

(19.13)

such that  L ( f ) (t) =

f (s) dμt (s) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(19.14)

[a,b]

We denote ·∞ = ·∞,[−π,π] the supremum norm. Next we specify [a, b] as [−π, π]. Clearly then L : C ([−π, π]) → B ([−π, π]) is the positive linear operator on hand. Here n = α , α ∈ / N, α > 0, k = 1, . . . , n − 1. By the use of Hölder’s inequality we notice that        |s − t| k |s − t| k sin sin dμt (s) ≤ L (t) = 4 4 [−π,π] 

k α+1    |s − t| α+1 α+1−k sin dμt (s) (μt ([−π, π])) α+1 = 4 [−π,π]

(19.15)

k     α+1 |s − t| α+1 α+1−k sin L (L (1) (t)) α+1 . (t) 4

   |s − t| k sin L (t) ≤ 4

That is

k     α+1 |s − t| α+1 α+1−k sin L (L (1) (t)) α+1 , (t) 4

for k = 1, . . . , n − 1. Consequently, it holds

(19.16)

19.2 Background—I

447

       |s − t| k   sin (t) L   4

≤

(19.17)

∞

k       α+1  α+1−k |s − t| α+1   sin (t) L (1)∞α+1 , L   4

∞

for k = 1, . . . , n − 1. In this work we will use a lot the following well known inequality:  |z| ≤ π sin

 |z| , ∀ z ∈ [−π, π] . 2

Furthermore, we observe that  

 k L (s − t)k (t) = (s − t) dμt (s) ≤ [−π,π]



 2

k [−π,π]

|s − t| 2

(19.18)

|s − t|k dμt (s) = [−π,π]

   |s − t| k sin dμt (s) ≤ (2π) dμt (s) 4 [−π,π] (19.19)   k |s − t| sin = (2π)k L (t) . 4

k

(19.18.)



k

That is

 L (s − t)k (t) ≤ (2π)k L

   |s − t| k sin (t) , 4

(19.20)

∀ t ∈ [−π, π], and 

   L (s − t)k (t)

∞

   k    |s − t|   sin ≤ (2π)k  L (t) ,   4

(19.21)

∞

all k = 1, . . . , n − 1. Then, by (19.16) and (19.20) we get k     α+1 |s − t| α+1 α+1−k sin L (t) (L (1) (t)) α+1 , 4 (19.22) k = 1, . . . , n − 1, ∀ t ∈ [−π, π], and by (19.17) and (19.21), we find



 L (s − t)k (t) ≤ (2π)k

448

19 Trigonometric Commutative Caputo Fractional Korovkin …



   L (s − t)k (t)

∞

k       α+1  α+1−k |s − t| α+1   sin ≤ (2π)  L (t) L (1)∞α+1 ,   4 (19.23)

k

k = 1, . . . , n − 1. We also have

 L |s − t|α+1 (t) =



|s − t|α+1 dμt (s) =

[−π,π]

2α+1



 [−π,π]

     (19.18) |s − t| α+1 |s − t| α+1 sin dμt (s) ≤ (2π)α+1 dμt (s) 2 4 [−π,π]

(19.24) α+1

= (2π)

   |s − t| α+1 sin L (t) . 4

That is

α+1

L |s − t|



α+1

   |s − t| α+1 sin L (t) , 4

α+1

       |s − t| α+1   sin (t) . L   4

(t) ≤ (2π)

(19.25)

∀ t ∈ [−π, π] , and 

   L |s − t|α+1 (t)

∞

Also we have

≤ (2π)

α α D∗t X (t, ω) = Dt− X (t, ω) = 0,

∀ ω ∈ , see [8], pp. 358–359. We assume that α X (s, ω) = 0, for s < t, D∗t and ∀ ω ∈ .

(19.26)

∞

α X (s, ω) = 0, for s > t, Dt−

(19.27)

19.3 Preliminaries

449

19.3 Preliminaries Let (, F, P) be a probabilistic space and L 1 (, F, P) be the space of all realvalued random variables Y = Y (ω) with  |Y (ω)| P (dω) < ∞. 

Let X = X (t, ω) denote a stochastic process with index set [a, b] ⊂ R and real state space (R, B), where B is the σ-field of Borel subsets of R. Here C ([a, b]) is the space of continuous real-valued functions on [a, b] and B ([a,

b]) is the space of bounded  real-valued functions on [a, b]. Also C ([a, b]) = C [a, b] , L 1 (, F, P) is the space of L 1 -continuous stochastic  processes in t and B ([a, b]) =   X : sup  |X (t, ω)| P (dω) < ∞ , obviously C ([a, b]) ⊂ B ([a, b]). t∈[a,b]

Let α > 0, α ∈ / N, α = n, and consider the subspace of stochastic processes Cα,n ([a, b]) := {X : X (·, ω) ∈ AC n ([a, b]) , ∀ ω ∈  and X (n) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0; X (k) (t, ω) ∈ C ([a, b]), k = 0, 1, . . . , n − 1; α α X , Dt− X are stochastic processes for any t ∈ [a, b]}. That is, for every also D∗t ω ∈  we have X (t, ω) ∈ C n−1 ([a, b]). Consider the linear operator L : C ([a, b]) → B ([a, b]) . If X ∈ C ([a, b]) is nonnegative and L X , too, then L is called positive. If E L = L E, then L is called E-commutative.

19.4 Background—II Following 3. Preliminaries we proved Theorem 19.4 ([10]) Consider the positive E-commutative linear operator L : C ([a, b]) → B ([a, b]), and α > 0, α ∈ / N, α = n, and let X ∈ Cα,n ([a, b]), with δ > 0. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| + 



  n−1

 ω1 E Dtα X , δ E X (k) (t)

k L (s − t) (t) + k!  (α + 1) k=1

(19.28)

450

19 Trigonometric Commutative Caputo Fractional Korovkin …

⎤ ⎡ 1   α+1



α+1 α



  |s L − t| (t) 1 ⎦, L |s − t|α+1 (t) α+1 ⎣(L (1) (t)) α+1 + δ (α + 1) ∀ t ∈ [a, b] .

  Above ω1 E Dtα X , δ is as in (19.12). We also mention Theorem 19.5 ([10]) All as in Theorem 19.4. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +   n−1 

E X (k) 

∞

k=1

k!



  

  ω1 E Dtα X , δ k  L (s − t) (t) + sup ∞  (α + 1) t∈[a,b]

(19.29)

⎡ 

  1 ⎤  L |s − t|α+1 (t) α+1 1 α  

 ∞ ⎦ α+1  L |s − t|α+1 (t) α+1 ⎣L (1)∞ < ∞. + ∞ δ (α + 1) We specify: Definition 19.6 If 0 < α < 1, then n = 1, and Cα,1 ([a, b]) := {X : X (·, ω) ∈ AC ([a, b]) , ∀ ω ∈  and X (1) (t, ω) ≤ M , ∀ (t, ω) ∈ [a, b] × , where M > 0; α α X (t, ω) ∈ C ([a, b]) ; also D∗t X , Dt− X are stochastic processes for any t ∈ [a, b]}. α,1 We will specialize on C ([−π, π]) . Note 19.7 From [17, pp. 3–5] we have the following results (i) C ([a, b]) ⊂ C ([a, b]) , (ii) if X ∈ C ([a, b]), then E X ∈ C ([a, b]), and (iii) if L is E-commutative, then L maps the subspace C ([a, b]) into B ([a, b]) .

19.5 Main Results We present the following pointwise result over [−π, π] . Theorem 19.8 Consider the 3. Preliminaries for [a, b] = [−π, π] and the positive E-commutative linear operator L : C ([−π, π]) → B ([−π, π]), and α > 0, α ∈ / N, α = n, and let X ∈ Cα,n ([−π, π]), with δ > 0. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +

19.5 Main Results

451

 n−1

E X (k) (t) k=1

k!





|s − t| sin 4

(2π) L k

k (t) +

α   



  α+1 α+1 ω1 E Dtα X , δ |s − t| sin (2π)α L (t)  (α + 1) 4

⎡ 1 ⎣(L (1) (t)) α+1

1 ⎤    α+1 α+1 |s − t| 2π ⎦, sin + L (t) δ (α + 1) 4

(19.30)

∀ t ∈ [−π, π] .

  Above ω1 E Dtα X , δ is over [−π, π]. Proof From (19.28) we have (∀ t ∈ [−π, π]) |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +   



n−1

 ω E Dtα X , δ E X (k) (t)

L (s − t)k (t) + 1 k!  (α + 1) k=1 ⎤ ⎡ 1   α+1



α+1 α  



|s L − t| (t) 1 ⎦ L |s − t|α+1 (t) α+1 ⎣(L (1) (t)) α+1 + δ (α + 1) (by (19.20), (19.25))

≤

 n−1

E X (k) (t) k=1

k!

|(E X ) (t)| |(L (1)) (t) − 1| + 

(2π) L k



|s − t| sin 4

k (t) +

α   



   α+1 ω1 E Dtα X , δ |s − t| α+1 α sin (2π) L (t)  (α + 1) 4

⎡ ⎣(L (1) (t))

1 α+1

2π + δ (α + 1)

1 ⎤     α+1 |s − t| α+1 ⎦, sin L (t) 4

proving the claim. By (19.30) we obtain the following uniform estimate Theorem 19.9 All as in Theorem 19.8. Then

(19.31) 

452

19 Trigonometric Commutative Caputo Fractional Korovkin …

E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +   n−1 

E X (k) 

∞

k!

k=1

   k    |s − t|   sin (t) + (2π)k  L   4



  sup ω1 E Dtα X , δ

t∈[−π,π]

 (α + 1) ⎡

∞

α    α+1   α+1  |s − t|   sin (2π)α  L (t)   4

∞

1 ⎤       α+1  |s − t| α+1 2π  ⎦  sin + < ∞. (t) L  δ (α + 1)  4

1 α+1

⎣L (1)∞

(19.32)

∞

We give Corollary 19.10 Consider the positive E-commutative linear operator L : C ([−π, π]) → B ([−π, π]), and 0 < α < 1 and let X ∈ Cα,1 ([−π, π]), with δ > 0. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| + α   



  α+1 α+1 ω1 E Dtα X , δ |s − t| sin (2π)α L (t)  (α + 1) 4

⎡ ⎣(L (1) (t))

1 α+1

2π + δ (α + 1)

1 ⎤     α+1 |s − t| α+1 ⎦, sin L (t) 4

(19.33)

∀ t ∈ [−π, π] . 

Proof By (19.30). We continue with Corollary 19.11 All as in Corollary 19.10. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +



  sup ω1 E Dtα X , δ

t∈[−π,π]

 (α + 1) ⎡

α    α+1   α+1  |s − t|   sin (2π)α  L (t)   4

∞

⎤

1 α+1

⎣L (1)∞

1       α+1  |s − t| α+1 2π  ⎦  sin . + (t) L  δ (α + 1)  4

∞

(19.34)

19.5 Main Results

453



Proof From (19.33). Corollary 19.12 All as in Corollary 19.10 and L (1) = 1. Then E (L X ) − E X ∞ ≤



  sup ω1 E Dtα X , δ

t∈[−π,π]

 (α + 1)

α       α+1  |s − t| α+1   sin (2π)  L (t)   4

α

∞

⎡

1 ⎤       α+1  |s − t| α+1 2π   ⎣1 + sin (t) ⎦ . L  δ (α + 1)  4

(19.35)

∞



Proof From (19.34). In particular we give Corollary 19.13 All as in Corollary 19.10 and L (1) = 1. Then E (L X ) − E X ∞ ≤ 1 ⎞    α+1  α+1   |s − t| 2π  2 π  sin sup ω1 ⎝ E Dtα X , (t) ⎠ L   (α + 1) t∈[−π,π] α+1 4

α+1 α

⎛





∞

α       α+1  |s − t| α+1   sin (t) . L   4

(19.36)

∞

Proof By (19.35): we take there 1       α+1 |s − t| α+1 2π    sin > 0. δ= (t) L  α+1 4

∞

   α+1     |s−t|  In case of  L sin 4 (t) 

  α+1  |s−t| = 0, we have that L sin 4 (t) ∞    α+1  = 0, ∀ t ∈ [−π, π] . That is, by ( 19.14), [−π,π] sin |s−t| dμt (s) = 0, ∀ t ∈ 4 is a probability measure on π] . μ [−π, [−π, π], where    t |s−t| ≥ 0, ∀ s ∈ π], we get sin = 0, a.e., that is |s − t| = Since sin |s−t| [−π, 4 4 0, a.e., and s = t, a.e., which means μt {s ∈ [−π, π] : s = t} = 0, i.e. μt = δt , ∀ t ∈ [−π, π] , where δt is the unit Dirac measure. Consequently we have

454

19 Trigonometric Commutative Caputo Fractional Korovkin …

 E (L X ) (t) = L (E X ) (t) =

(E X ) (s) dδt (s) = (E X ) (t) , [−π,π]

∀ t ∈ [−π, π] . That is E (L X ) = E X over [−π, π]. Therefore both sides of inequality (19.36) equal to zero. Hence (19.36) is always true. 

19.6 Application Consider the Bernstein polynomials on [−π, π] for f ∈ C ([−π, π]) : B N ( f ) (x) =

 N 

N k=0

k

 f

2πk −π + N



x +π 2π

k 

π−x 2π

 N −k

,

(19.37)

N ∈ N, any x ∈ [−π, π]. There are positive linear operators from C ([−π, π]) into itself. Setting g (t) = f (2πt − π), t ∈ [0, 1], we have g (0) = f (−π), g (1) = f (π), and    N 

k N t k (1 − t) N −k = (B N f ) (x) , x ∈ [−π, π] . g (B N g) (t) = k N k=0 (19.38) Here x = ϕ (t) = 2πt − π is an 1 − 1 and onto map from [0, 1] onto [−π, π]. Clearly here g ∈ C ([0, 1]). Notice also that

 

  (2π)2 t (1 − t) B N (· − x)2 (x) = B N (· − t)2 (t) (2π)2 = N    π−x 1 π2 (2π)2 x + π = = , ∀ x ∈ [−π, π] . (x + π) (π − x) ≤ N 2π 2π N N

I.e.



 π2 , ∀ x ∈ [−π, π] . B N (· − x)2 (x) ≤ N

(19.39)

(B N 1) (x) = 1, ∀ x ∈ [−π, π] .

(19.40)

In particular

Define the corresponding stochastic application of B N as follows:

19.6 Application

455

     t + π k π − t N −k 2πk , X −π + ,ω k N 2π 2π k=0 (19.41) ∀ N ∈ N, ∀ t ∈ [−π, π], ∀ ω ∈ , where X is a stochastic process. Clearly B N X is a stochastic process. Notice that B N (X (·, ω)) (t) =

E (B N X ) (t) =

 N 

N

 N 

N k=0

k

     t + π k π − t N −k 2πk (E X ) −π + N 2π 2π = B N (E X ) (t) ,

(19.42)

i.e. E B N = B N E, that is B N is an E-commutative positive linear operator from C ([−π, π]) into itself. We give Corollary 19.14 Let 0 < α < 1 and X ∈ Cα,1 ([−π, π]). Then E (B N X ) − E X ∞ ≤ ⎛

2α+1 π α  (α + 1)

1 ⎞    α+1   α+1  |s − t| 2π   sup ω1 ⎝ E Dtα X , sin (t) ⎠  BN  α+1 4 t∈[−π,π]





∞

α       α+1  |s − t| α+1   sin (t) ,  BN   4

(19.43)

∞

∀ N ∈ N. 

Proof By (19.36). In particular we get: 1

,1

Corollary 19.15 Let X ∈ C2 ([−π, π]). Then √ E (B N X ) − E X ∞ ≤ 4 2 ⎛

 4π sup ω1 ⎝ E Dt X , 3 t∈[−π,π] 

1 2

2 ⎞     23  3   |s − t|   sin (t) ⎠  BN   4

∞

1     23  3  |s − t|   sin (t) ,  BN   4

∞

(19.44)

456

19 Trigonometric Commutative Caputo Fractional Korovkin …

∀ N ∈ N. Proof By (19.43) for α = 21 .



We make Remark 19.16 By |sin x| < |x| , ∀ x ∈ R − {0}, in particular sin x ≤ x, for x ≥ 0, we get    23  3 |· − t| |· − t| 2 1 3 sin ≤ = |· − t| 2 . 4 4 8 Hence     23    |· − t|   sin (t)  BN   4

≤

∞

   1 3    B N |· − t| 2 (t) . ∞ 8

(19.45)

We observe that 3       N  

2 N t + π k π − t N −k 3 t + π − 2πk B N |· − t| 2 (t) = k N 2π 2π k=0 (by discrete Hölder’s inequality)  N       

2πk 2 N t + π k π − t N −k t +π− ≤ k N 2π 2π k=0

 3 = B N (· − t)2 (t) 4

Consequently it holds

(19.39)

≤

3

N4

    3   B N |· − t| 2 (t)

    23    |· − t|   sin (t)  BN   4

∞

(19.46)

3

π2

, ∀ t ∈ [−π, π] .

3

∞

and

3 4

π2

≤

3

N4

,

(19.47)

3

≤

π2

3

8N 4

, ∀ N ∈ N.

(19.48)

We derive 1

,1

Proposition 19.17 Let X ∈ C2 ([−π, π]). Then E (B N X ) − E X ∞

√    π2  1 2 2π 2 , ≤ √ sup ω1 E Dt X , √ 4 N t∈[−π,π] 3 N

(19.49)

19.6 Application

457

∀ N ∈ N. Hence lim E (B N X ) = E X , uniformly. N →∞

Proof By (19.44) and (19.48).



19.7 Commutative Trigonometric Caputo Fractional Stochastic Korovkin Results Here L is meant as a sequence of positive E-commutative linear operators and all assumptions are as in Theorem 19.8. We give   α+1  |s−t| Theorem 19.18 We further assume that L (1) (t) → 1 and L sin 4 (t) → 0, then (E (L X )) (t) → (E X ) (t), for any X ∈ Cα,n ([−π, π]), ∀ t ∈ [−π, π], a pointwise convergence; where α > 0, α ∈ / N, α = n. Proof Based on (19.30),

(19.16),   and that L (1) (t) is bounded as a sequence of functions. Also ω1 E Dtα X , δ over [−π, π] is bounded, see (19.10)–(19.12).  We continue with Theorem   19.19We further   assume that L (1) (t) → 1, uniformly and α+1   |s−t|  L sin (t)   → 0, then E (L X ) → E X , uniformly over [−π, π], 4 ∞

for any X ∈ Cα,n ([−π, π]); where α > 0, α ∈ / N, α = n. Proof Based

on (19.32),   (19.17), and that L (1)∞ is bounded. Also it is  sup ω1 E Dtα X , δ < ∞, by (19.10)–(19.12).

t∈[−π,π]

We finish with Remark 19.20 The stochastic convergences of Theorems 19.18, 19.19 are derived by the convergences of the basic and simple real non-stochastic functions !  α+1 "  |s−t| , an amazing fact! 1, sin 4

References 1. Anastassiou, G.: A study of positive linear operators by the method of moments, onedimensional case. J. Approx. Theory 45, 247–270 (1985) 2. Anastassiou, G.: Korovkin type inequalities in real normed vector spaces. Approx. Theory Appl. 2, 39–53 (1986)

458

19 Trigonometric Commutative Caputo Fractional Korovkin …

3. Anastassiou, G.: Multi-dimensional quantitative results for probability measures approximating the unit measure. Approx. Theory Appl. 2, 93–103 (1986) 4. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 5. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 6. Anastassiou, G.: Fractional Differentiation Inequalities. Springer, Heildelberg (2009) 7. Anastassiou, G.: Fractional representation formulae and right fractional inequalities. Math. Comput. Model. 54(11–12), 3098–3115 (2011) 8. Anastassiou, G.: Intelligent Mathematics: Computational Analysis. Springer, Heidelberg (2011) 9. Anastassiou, G.A.: Foundation of stochastic fractional calculus with fractional approximation of stochastic processes. Rev. R. Acad. Cienc. Exactas Fis. Nat. Ser. A Mat. RACSAM 114(2), Paper No. 89 (2020) 10. Anastassiou, G.A.: Commutative caputo fractional Korovkin inequalities for Stochastic processes (2020). Submitted for publication 11. Anastassiou, G.A.: Trigonometric commutative caputo fractional Korovkin theory for stochastic processes. Progr. Fract. Differ. Appl. (2020). Accepted 12. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 13. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 14. Shisha, O., Mond, B.: The degree of approximation to periodic functions by linear positive Operators. J. Approx. Theory 1, 335–339 (1968) 15. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192(1), 73–80 (1986) 16. Weba, M.: Quantitative results on monotone approximation of stochastic processes. Probab. Math. Stat. 11(1), 109–120 (1990) 17. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61(1), 74–87 (1990) 18. Weba, M.: Monotone approximation of random functions with multivariate domains in respect of lattice semi-norms. Results Math. 20(1–2), 554–576 (1991)

Chapter 20

Commutative Conformable Fractional Korovkin Approximation for Stochastic Processes

Here we research the expectation commutative stochastic positive linear operators acting on L 1 -continuous stochastic processes which are conformable fractional differentiable. Under some mild, general and natural assumptions on the stochastic processes we produce related conformable fractional stochastic Shisha-Mond type inequalities pointwise and uniform. All convergences are produced quantitatively and are given by the conformable fractional stochastic inequalities involving the first modulus of continuity of the expectation of the αth right and left conformable fractional derivatives of the engaged stochastic process, α ∈ (n, n + 1), n ∈ Z+ . The amazing fact here is that the simple real Korovkin test functions assumptions imply the conclusions of our conformable fractional stochastic Korovkin theory. We include also a full details application to stochastic Bernstein operators. See also [9].

20.1 Introduction Our work is inspired by the following:   Korovkin’s Theorem ([11], 1960) Let T j j∈N be a sequence of positive linear   operators from C ([a, b]) into itself, [a, b] ⊂ R. In order to have lim T j f (t) = j→∞

f (t) (in the sup-norm) for all f ∈ C ([a, b]), it is enough to prove it for f 0 (t) = 1, f 1 (t) = t and f 2 (t) = t 2 . The rate of the above convergence for arbitrary f ∈ C ([a, b]) can be determined exactly from the rates of convergence for f 0 , f 1 , f 2 . The above theorem was put in an inequality form: Shisha-Mond inequality ([13]) We have          T j ( f ) − f  ≤  f  · T j (1) − 1 + ω1 f, ρ j · 1 + T j (1) , where

  1   ρ j = T j (x − y)2 (y) 2 .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_20

459

460

20 Commutative Conformable Fractional Korovkin …

In the last inequality · stands for the supremum norm and ω1 for the first modulus of continuity. This inequality gives the rate of convergence of T j to the unit operator I. Anastassiou in [3–5] established a series of sharp inequalities for various cases of the parameters of the problem. However, Weba in [14–17] was the first, among many workers in quantitative results of Shisha-Mond type, to produce inequalities for stochastic processes. He assumed that T j are E-commutative (E means expectation) and stochastically simple. According to his work, if a stochastic process X (t, ω), t ∈ Q - a compact convex subset of a real normed vector space, ω ∈ Q - probability space, is to be approximated by positive linear operators T j , then the maximal error in the qth mean is (q ≥ 1)   1     T j X − X  = sup E  T j X (t, ω) − X (t, ω)q q . t∈Q

  So, Weba established upper bounds for T j X − X  involving his own natural general first modulus of continuity of X with several interesting applications. Anastassiou continued [6] with the pointwise case of q = 1. Without stochastic simplicity nearly best and best upper bounds for     of T j he presented  E T j X (x0 ) − (E X ) (x0 ), x0 ∈ Q. The author here continues his above work on the approximation of stochastic processes, now at the conformable stochastic fractional level. He derives pointwise and uniform conformable fractional stochastic Shisha-Mond type inequalities, see the main Theorems 20.10, 20.11 and the several related corollaries. He gives an extensive application to stochastic Bernstein operators. He finishes with a pointwise and a uniform conformable fractional stochastic Korovkin theorem, derived by Theorems 20.10, 20.11. The stochastic convergences, about stochastic processes, of our conformable fractional Korovkin Theorems 20.21, 20.22 are implied only by the convergences of real basic non-stochastic functions.

20.2 Background—I Here we follow [1], see also [10]. We need Definition 20.1 ([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 

  If Tαa f (t) exists on (a, b), then

(20.1)

20.2 Background—I

461



   Tαa f (a) = lim Tαa f (t) . t→a+

(20.2)

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



(20.3)

 f (t) exists on (a, b), then b

αT

   f (b) = lim bα T f (t) . t→b−

(20.4)

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,

(20.5)

(20.6)

(20.7)

a

 Iα f (t) =



b

(b − x)α−1 f (x) d x,

(20.8)

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 20.2 ([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) .

(20.9)

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) .

(20.10)

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) .

462

20 Commutative Conformable Fractional Korovkin …

When n = 0 (or α ∈ (0, 1]), then β = α, and (20.9), (20.10) collapse to (20.1), (20.2), (20.3), (20.4) respectively. Lemma 20.3 ([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) .

(20.11)

We need Definition 20.4 (see also [1]) If α ∈ (n, n + 1], then the left fractional integral of order α starting at a is defined by a  1 Iα f (t) = n!



t

(t − x)n (x − a)α−1 f (x) d x.

(20.12)

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)α−1 f (x) d x.

(20.13)

t

We need Proposition 20.5 ([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

(20.14)

We also have Proposition 20.6 ([8], p. 154) 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

(20.15)

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

(20.16)

20.2 Background—I

463

Theorem 20.7 ([8], p. 155) 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 (20.17) 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 (20.18) ∀ t ∈ [a, b]. We make Remark 20.8 ([8], p. 155) 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 (

(20.19)

   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) .

(20.20)

Consequently we get that 

   Taα ( f ) (x) , bα T ( f ) (x) ∈ C ([a, b]) .

Furthermore it is obvious that     a Tα ( f ) (a) = bα T ( f ) (b) = 0,

(20.21)

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,

(20.22)

∀ t ∈ [a, b]. If f (k) (b) = 0, k = 1, . . . , n, then 1 f (t) − f (b) = − n!



b t

(b − x)β−1 (x − t)n

b

αT (

 f ) (x) d x,

(20.23)

464

20 Commutative Conformable Fractional Korovkin …

∀ t ∈ [a, b]. We make Remark 20.9 Let the positive linear operator L mapping C ([a, b]) into B ([a, b]) (the bounded functions). By the Riesz representation theorem [12] we have that there exists μt unique, completed Borel measure on [a, b] with μt ([a, b]) = L (1) (t) > 0,

(20.24)

such that  L ( f ) (t) =

f (s) dμt (s) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(20.25)

[a,b]

Let α ∈ (n, n + 1), n ∈ Z+ , k = 1, . . . , n. Then by Hölder’s inequality we obtain    

   (s − t) dμt (s) ≤

|s − t|k dμt (s) ≤

k

[a,b]



[a,b]

|s − t|α+1 dμt (s)

k ( α+1 )

(μt ([a, b]))(

α+1−k α+1

).

(20.26)

[a,b]

The last means

α+1−k            k  L (s − t)k (t) ≤ L |s − t|k (t) ≤ L |s − t|α+1 (t) ( α+1 ) (L (1) (t)) α+1 , (20.27) all k = 1, . . . , n. It is clear that

     L |s − t|α+1 (t)

∞,[a,b]

< ∞.

Furthermore we derive      L (s − t)k (t)

∞,[a,b]

    ≤  L |s − t|k (t)∞,[a,b] ≤

α+1−k     k  L |s − t|α+1 (t)( α+1 ) L (1)( α+1 ) , ∞,[a,b] ∞,[a,b]

all k = 1, . . . , n. From now on we will denote ·∞,[a,b] = ·∞ the supremum norm.

(20.28)

20.3 Preliminaries

465

20.3 Preliminaries Let (, F, P) be a probabilistic space and L 1 (, F, P) be the space of all realvalued random variables Y = Y (ω) with  |Y (ω)| P (dω) < ∞. 

Let X = X (t, ω) denote a stochastic process with index set [a, b] ⊂ R and real state space (R, B), where B is the σ -field of Borel subsets of R. Here C ([a, b]) is the space of continuous real-valued functions on [a, b] and B ([a, b])  is the space of bounded  real-valued functions on [a, b]. Also C ([a, b]) = C [a, b] , L 1 (, F, P) is processes in t and B ([a, b]) = the space of L 1 -continuous stochastic 

X : sup  |X (t, ω)| P (dω) < ∞ , obviously C ([a, b]) ⊂ B ([a, b]). t∈[a,b]

Let α ∈ (n, n + 1), n ∈ Z+ , and consider the subspace of stochastic processes  Cα,n+1 ([a, b]) := {X : X (·, ω) ∈ C n+1 ([a, b]) , ∀ ω ∈  and  X (n+1) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] ×  , where M > 0; X (k) (t, ω) ∈ C ([a, b]), k = 0, 1, . . . , n; also X (n+1) (t, ω), (t, ω) ∈ [a, b] × , is a stochastic process}. Clearly by (20.19) and (20.20), respectively, we get that Ttα X (z, ω), z ∈ [t, b], and tα TX (z, ω), z ∈ [a, t], are stochastic processes for any t ∈ [a, b]. Consider the linear operator L : C ([a, b]) → B ([a, b]) . If X ∈ C ([a, b]) is nonnegative and L X , too, then L is called positive. If E L = L E, then L is called E -commutative, where E is the expectation operator. From [16, pp. 3–5] we have the following results (i) C ([a, b]) ⊂ C ([a, b]) , (ii) if X ∈ C ([a, b]), then E X ∈ C ([a, b]), and (iii) if L is E-commutative, then L maps the subspace C ([a, b]) into B ([a, b]) .

20.4 Background—II   Let X ∈ Cα,n+1 ([a, b]). We have that  X (n+1) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × . We notice that (20.19)

Ttα X (z, ω) = (z − t)1−β X (n+1) (z, ω) , z ∈ [t, b] , and t α TX

(20.20 )

(z, ω) = (−1) (t − z) n

1−β

X

(n+1)

(z, ω) , z ∈ [a, t] ,

(20.29)

466

20 Commutative Conformable Fractional Korovkin …

∀ ω ∈ , where β = α − n, α ∈ (n, n + 1), n ∈ Z+ ; ∀ t ∈ [a, b] . We have that Ttα X (t, ω) =tα TX (t, ω) = 0. It is

    t T X (z, ω) = (z − t)1−β X (n+1) (z, ω) ≤ (b − a)1−β M, α

(20.30)

(20.31)

∀ (z, ω) ∈ [t, b] × , and    t  TX (z, ω) = (−1)n (t − z)1−β X (n+1) (z, ω) ≤ (b − a)1−β M, α

(20.32)

∀ (z, ω) ∈ [a, t] × . Here E is the expectation operator  (E X ) (t) = Clearly, we get that ∀ z ∈ [t, b] , and



X (t, ω) P (dω) .

  t    E T X (z) ≤ (b − a)1−β M, α

(20.33)

 t    E TX (z) ≤ (b − a)1−β M, α

(20.34)

∀ z ∈ [a, t] . We observe that the first modulus of continuity (δ > 0) ω1

    t        E Tα X , δ [t,b] := sup  E Ttα X (x) − E Ttα X (y) x,y∈[t,b] |x−y|≤δ

(20.33)

≤ 2 (b − a)1−β M, ∀ t ∈ [a, b] .

(20.35)

Hence it holds (δ > 0)   t   E Tα X , δ [t,b] ≤ 2 (b − a)1−β M,

(20.36)

 t   (20.34) E α TX , δ [a,t] ≤ 2 (b − a)1−β M.

(20.37)

sup ω1 t∈[a,b]

and similarly, it holds sup ω1 t∈[a,b]

By [7], p. 209, we have that (δ1 > 0)

20.4 Background—II

467

    t         |z − t|  E T X (z) − E Tt X (t) ≤ ω1 E Tt X , δ1 ≤ α α α [t,b] δ1

(20.38)

   t   (z − t) ω1 E Tα X , δ1 [t,b] 1 + , δ1 ∀ z ∈ [t, b] , where · is the ceiling of the number, and similarly (δ2 > 0 )  (t − z) 1 + , (20.39) [a,t] δ2

 t          E TX (z) − E t TX (t) ≤ ω1 E t TX , δ2 α

α

α

∀ z ∈ [a, t] . We also set               ω1 E α Tt X , δ := max ω1 E Ttα X , δ1 [t,b] , ω1 E tα TX , δ2 [a,t] , (20.40) where δ > 0.

20.5 Main Results Following 3. Preliminaries we state Theorem 20.10 Let α ∈ (n, n + 1), n ∈ Z+ . Consider the positive E-commutative α,n+1 linear operator ([a, b]), δ > 0.  α t L : C ([a, b]) → B ([a, b]), and let X ∈ C Here ω1 E T X , δ as in (20.40). Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +        n     ω E α Tt X , δ E X (k) (t)    L (s − t)k (t) + 1 n  k! k=1 (α + j − n) j=1

⎤ ⎡ 1   α+1   1 α+1 α α+1     |s L − t| (t) (L (1) (t)) ⎦, + L |s − t|α+1 (t) α+1 ⎣ α−n δ (α + 1) ∀ t ∈ [a, b] .

(20.41)

468

20 Commutative Conformable Fractional Korovkin …

Proof By Theorem 20.7 we have X (s, ω) = 1 n!



n  X (k) (t, ω) (s − t)k + k! k=0

  (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz,

s

t

(20.42)

for all t ≤ s ≤ b, ∀ ω ∈ ; α ∈ (n, n + 1), see also (20.30). Furthermore it holds from Theorem 20.7 that X (s, ω) = 1 n!



t

n  X (k) (t, ω) (s − t)k − k! k=0

(t − z)β−1 (z − s)n

t

s

α TX

 (z, ω) −tα TX (t, ω) dz,

(20.43)

for all a ≤ s ≤ t, ∀ ω ∈ ; α ∈ (n, n + 1), see also (20.30). Therefore we get  n   E X (k) (t) (E X ) (s) = (s − t)k + k! k=0 1 E n!



s t

all t ≤ s ≤ b, and

1 E n!

  (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz ,

(20.44)

 n   E X (k) (t) (E X ) (s) = (s − t)k − k! k=0

 s

t

(t − z)β−1 (z − s)n

 t TX ω) − TX ω) dz , (z, (t, α α

t

(20.45)

all a ≤ s ≤ t. By [2], p. 156, Theorem 20.15, the functions Ttα X and tα TX are jointly measurable. By (20.31) and (20.32)we obtain that    s E t (s − z)n (z − t)β−1 Ttα X (z, ω) − Ttα X (t, ω) dz ,

t   E s (t − z)β−1 (z − s)n tα TX (z, ω) −tα TX (t, ω) dz are finite. Therefore, by Fubini-Tonelli’s theorem [12], we get

20.5 Main Results

469

 n   E X (k) (t) (E X ) (s) − (E X ) (t) = (s − t)k + R1 (t, s) , k! k=0

(20.46)

where 1 R1 (t, s) := n!



      (s − z)n (z − t)β−1 E Ttα X (z) − E Ttα X (t) dz, (20.47)

s t

all t ≤ s ≤ b, and (E X ) (s) − (E X ) (t) =

 n   E X (k) (t) (s − t)k − R2 (s, t) , k! k=0

(20.48)

where 1 R2 (s, t) := n!



      (t − z)β−1 (z − s)n E tα TX (z) − E tα TX (t) dz, (20.49)

t

s

all a ≤ s ≤ t. We observe that (t ≤ s ≤ b, δ1 > 0) 1 |R1 (t, s)| ≤ n!

 t

s

      (s − z)n (z − t)β−1  E Ttα X (z) − E Ttα X (t) dz

    ω1 E Ttα X , δ1 [t,b]  n!

s t

 (z − t) dz = (s − z)n (z − t)β−1 1 + δ1

(20.38)

≤

(20.50)

    ω1 E Ttα X , δ1 [t,b] 

s

(s − z)(n+1)−1 (z − t)β−1 dz +

t

n! 

1 δ1

s

 (s − z)(n+1)−1 (z − t)(β+1)−1 dz =

t

    ω1 E Ttα X , δ1 [t,b] n! 

  (n + 1)  (β) 1  (n + 1)  (β + 1) (s − t)n+β + (s − t)n+β+1 =  (n + 1 + β) δ1  (n + β + 2) ⎡

⎤

⎢ ⎥     1 1 1 ⎢ α+1 ⎥ − t) ω1 E Ttα X , δ1 [t,b] ⎢ n (s − t)α + (s ⎥. n ⎣ ⎦ δ1  (β + j) (β + 1 + j) j=0

j=0

(20.51)

470

20 Commutative Conformable Fractional Korovkin …

That is ⎡

⎤

⎢ (s − t)α ⎥     (s − t)α+1 ⎢ ⎥ |R1 (t, s)| ≤ ω1 E Ttα X , δ1 [t,b] ⎢ n + ⎥, n   ⎣ ⎦ (β + j) δ1 (β + 1 + j) j=0

j=0

(20.52) all t ≤ s ≤ b, δ1 > 0. Next we also observe (a ≤ s ≤ t, δ2 > 0) |R2 (s, t)| ≤

1 n!

 s

t

      (20.39) (t − z)β−1 (z − s)n  E tα TX (z) − E tα TX (t) dz ≤

    ω1 E tα TX , δ2 [a,t]  n!

t s

 (t − z) dz = (t − z)β−1 (z − s)n 1 + δ2

(20.53)

    ω1 E tα TX , δ2 [a,t] 

t

(t − z)

β−1

(n+1)−1

(z − s)

s

n! 

1 dz + δ2

t

(β+1)−1

(t − z)

(n+1)−1

(z − s)

 dz =

s

    ω1 E tα TX , δ2 [a,t] n! 

  (β)  (n + 1) 1  (β + 1)  (n + 1) (t − s)β+n + (t − s)β+n+1 =  (β + n + 1) δ2  (β + n + 2) ⎡

⎤

⎢ ⎥     1 1 1 ⎢ α+1 ⎥ ω1 E tα TX , δ2 [a,t] ⎢ n − s) (t − s)α + (t ⎥. n ⎣ ⎦ δ2  (β + j) (β + 1 + j) j=0

j=0

(20.54) That is ⎡

⎤

⎢ (t − s)α ⎥     (t − s)α+1 ⎢ ⎥ |R2 (s, t)| ≤ ω1 E tα TX , δ2 [a,t] ⎢ n + ⎥, n  ⎣ ⎦ (β + j) δ2 (β + 1 + j) j=0

j=0

(20.55) all a ≤ s ≤ t, δ2 > 0. From now on we choose δ1 = δ2 = δ > 0. In conclusion we have (see (20.52), (20.55), (20.40))

20.5 Main Results

471

|R1 (t, s)| , |R2 (s, t)| ≤ ⎡

⎤

⎥    ⎢ |s − t|α |s − t|α+1 ⎢ ⎥ ω1 E α Tt X , δ ⎢ n + n ⎥,  ⎣ ⎦ (α + j − n) δ (α + j + 1 − n) j=0

(20.56)

j=0

for any s, t ∈ [a, b] . We further have that (20.25)

L (E X ) (t) − (E X ) (t) L (1) (t) = 

(20.24)

(E X ) (s) dμt (s) − (E X ) (t) L (1) (t) = [a,b]





[a,t)

(E X ) (s) dμt (s) + 



[a,t)

(E X ) (s) dμt (s) − [t,b]

(E X ) (t) dμt (s) −

(E X ) (t) dμt (s) = [t,b]



 [a,t)

n 



((E X ) (s) − (E X ) (t)) dμt (s) +

EX

k=1

((E X ) (s) − (E X ) (t)) dμt (s)

(by (20.46), (20.48))

=

[t,b]

(k)

k!



 (t)  L (s − t)k (t) −



(20.57)



[a,t)

R2 (s, t) dμt (s) +

R1 (t, s) dμt (s) . [t,b]

That is  n    E X (k) (t)  L (s − t)k (t) L (E X ) (t) − (E X ) (t) L (1) (t) = k! k=1   − R2 (s, t) dμt (s) + R1 (t, s) dμt (s) . (20.58) [a,t)

[t,b]

One can rewrite L (E X ) (t) − (E X ) (t) = (E X ) (t) [(L (1)) (t) − 1] +   n    E X (k) (t)  R2 (s, t) dμt (s) L (s − t)k (t) − k! [a,t) k=1  R1 (t, s) dμt (s) .

+ [t,b]

(20.59)

472

20 Commutative Conformable Fractional Korovkin …

Consequently, by E-commutativity of L we find |E [(L X ) (t, ω) − X (t, ω)]| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +      E X (k) (t)      L (s − t)k (t) + |R2 (s, t)| dμt (s) k! [a,t) k=1

n 



(20.60)

(20.56)

|R1 (t, s)| dμt (s) ≤

+ [t,b]

|(E X ) (t)| |(L (1)) (t) − 1| +

  n         E X (k) (t) 

 L (s − t)k (t) + ω1 E α Tt X , δ k! k=1

⎡

⎤

⎢ ⎢ ⎢ n ⎣



1 (α + j − n)

[a,b]



1

|s − t|α dμt (s) + δ

j=0

n 

(α + j + 1 − n)

⎥ ⎥ |s − t|α+1 dμt (s)⎥ ⎦ [a,b]

j=0

(by Hölder’s inequality)   n     E X (k) (t)    L (s − t)k (t) + ≤ |(E X ) (t)| |(L (1)) (t) − 1| + k! k=1 ⎡    ⎢ ⎢ ω1 E α Tt X , δ ⎢ n ⎣

1



(α + j − n)

α+1

|t − s|

dμt (s)

α α+1

1

(L (1) (t)) α+1

[a,b]

j=0

⎤ 1

+ δ

n 

(α + j + 1 − n)



⎥ ⎥ |t − s|α+1 dμt (s)⎥ . ⎦ [a,b]

j=0

We have proved that |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +   n         E X (k) (t)    L (s − t)k (t) + ω1 E α Tt X , δ k! k=1

(20.61)

20.5 Main Results

473

⎡ ⎢ ⎢ ⎢ n ⎣

1 (α + j − n)

    α 1 L |s − t|α+1 (t) α+1 (L (1) (t)) α+1 +

j=0

⎤ 1 δ

n 

(α + j + 1 − n)

 ⎥   ⎥ L |s − t|α+1 (t) ⎥ ⎦

(20.62)

j=0

  n     E X (k) (t)    L (s − t)k (t) + = |(E X ) (t)| |(L (1)) (t) − 1| + k! k=1       α ω1 E α Tt X , δ   L |s − t|α+1 (t) α+1 n  (α + j − n) j=1

⎡ ⎣ (L (1) (t)) α−n

1 α+1

  1 ⎤   L |s − t|α+1 (t) α+1 ⎦, + δ (α + 1) 

establishing the claim. We continue with a uniform estimate: Theorem 20.11 All as in Theorem 20.10. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +   n   E X (k) 

∞

k=1

k!

     L (s − t)k (t) + ∞

    sup ω1 E α Tt X , δ t∈[a,b] n 

(α + j − n)

j=1

⎤ ⎡ 1   1   α+1 α+1 α+1   α    |s  L − t| (t) L (1) ∞ ∞ ⎦  L |s − t|α+1 (t) α+1 ⎣ + < ∞. (20.63) ∞ α−n δ (α + 1) Proof By (20.36), (20.37), (20.40) and (20.41). Next, we cover the case n = 0.



474

20 Commutative Conformable Fractional Korovkin …

Definition 20.12 We call the subspace process: Cα,1 ([a, b]) := {X :  (1)of stochastic  1 X (·, ω) ∈ C ([a, b]) , ∀ ω ∈  and  X (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0; and X (t, ω) ∈ C ([a, b]) ; also X (1) (t, ω), (t, ω) ∈ [a, b] ×  is a stochastic process}. Corollary 20.13 Let 0 < α < 1. Consider the positive E-commutative linear operator L : C([a,b]) → B ([a, b]), and let X ∈ Cα,1 ([a, b]), δ > 0. Here ω1 E α T t X , δ as in (20.40). Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +

(20.64)

    α    ω1 E α T t X , δ L |s − t|α+1 (t) α+1     1 ⎤ 1 α+1 L |s − t|α+1 (t) α+1 (L (1) (t)) ⎦, ⎣ + α δ (α + 1) ⎡

∀ t ∈ [a, b] . 

Proof By (20.41). Corollary 20.14 All as in Corollary 20.13. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ + α        α+1 sup ω1 E α T t X , δ  L |s − t|α+1 (t)∞

(20.65)

t∈[a,b]

  1   1 ⎤ α+1  L |s − t|α+1 (t) α+1 L (1)∞ ∞ ⎦ ⎣ + < ∞. α δ (α + 1) ⎡



Proof By (20.63). Corollary 20.15 All as in Corollary 20.13, and L (1) = 1. Then     E (L X ) − E X ∞ ≤ sup ω1 E α T t X , δ

(20.66)

t∈[a,b]

⎡     1 ⎤  L |s − t|α+1 (t) α+1 α     1 ∞ ⎦  L |s − t|α+1 (t) α+1 ⎣ + . ∞ α δ (α + 1) Proof From (20.65).



20.5 Main Results

475

We continue with Corollary 20.16 All as in Corollary 20.13, and L (1) = 1. Then

E (L X ) − E X ∞ ≤

     1   α+1 1  L |s − t|α+1 (t) α+1 sup ω1 E α T t X , ∞ α t∈[a,b] (α + 1)     α  L |s − t|α+1 (t) α+1 . ∞

(20.67)

Proof By (20.66): we take there δ=

    1 1  L |s − t|α+1 (t) α+1 > 0. ∞ (α + 1)

      In case of  L |s − t|α+1 (t)∞ = 0, we have that L |s − t|α+1 (t) = 0, ∀ t ∈

[a, b]. That is, by (20.25) [a,b] |s − t|α+1 dμt (s) = 0, ∀ t ∈ [a, b], where μt is a probability measure, any t ∈ [a, b] . The last implies |s − t|α+1 = 0, a.e., hence |s − t| = 0, a.e., thus s = t, a.e., which means μt {s ∈ [a, b] : s = t} = 0, i.e. μt = δt , ∀ t ∈ [a, b], where δt is the unit Dirac measure.

Consequently we have E (L X ) (t) = L (E X ) (t) = [a,b] (E X ) (s) dδt (s) = (E X ) (t), ∀ t ∈ [a, b] . That is E (L X ) = E X over [a, b]. Therefore both sides of inequality (20.67) equal to zero. Hence (20.67) is always true. 

20.6 Application Let f ∈ C ([0, 1]) and the Bernstein polynomials B N ( f ) (t) :=

N  k=0

 f

k N



N k



t k (1 − t) N −k ,

∀ t ∈ [0, 1], ∀ N ∈ N. We have that B N 1 = 1 and B N is a positive linear operator. We have that   t (1 − t) , ∀ t ∈ [0, 1] , B N (· − t)2 (t) = N and

   1 1  B N (· − t)2 (t) 2 ≤ √ , ∀ N ∈ N. ∞ 2 N

(20.68)

(20.69)

(20.70)

476

20 Commutative Conformable Fractional Korovkin …

Define the corresponding stochastic application of B N by B N (X ) (t, ω) := B N (X (·, ω)) (t) =

N 

 X

k=0

k ,ω N



N k



t k (1 − t) N −k ,

(20.71) ∀ t ∈ [0, 1], ∀ ω ∈ , N ∈ N, where X is a stochastic process. Clearly B N (X ) is a stochastic process and B N : C ([0, 1]) → C ([0, 1]) . Notice that (E B N (X )) (t) =

N 

 (E X )

k=0

k N



N k



t k (1 − t) N −k = (B N (E X )) (t) ,

(20.72) ∀ t ∈ [0, 1] . That is E B N = B N E, i.e. B N is an E-commutative positive linear operator. We give Proposition 20.17 Let 0 < α < 1 and X ∈ Cα,1 ([0, 1]). Then E (B N X ) − E X ∞ ≤       1  1 α+1  B N |s − t|α+1 (t) α+1 sup ω1 E α T t X , ∞ α t∈[0,1] (α + 1)     α  B N |s − t|α+1 (t) α+1 , ∀ N ∈ N. ∞

(20.73) 

Proof By (20.67). In particular we get: 1

,1

Proposition 20.18 Let X ∈ C2 ([0, 1]). Then 

  23  2

1 3   E (B N X ) − E X ∞ ≤ 3 sup ω1 E 2 T t X , B N |s − t| 2 (t) ∞ 3 t∈[0,1] 

  13 3   B N |s − t| 2 (t) , ∀ N ∈ N. ∞

Proof By (20.73) and α = 21 . We make Remark 20.19 We notice that 3  N 

   k  2 N k 3  B N |s − t| 2 (t) = t (1 − t) N −k t − N  k k=0

(20.74) 

20.6 Application

477

(by discrete Hölder’s inequality)  N   43 2    N k k N −k t −  ≤ t (1 − t)  k N k=0 (20.69)

=

That is



43 1 1 t (1 − t) ≤ 3 , ∀ t ∈ [0, 1] . N (4N ) 4



  3   B N |s − t| 2 (t)

≤

,

(20.76)

1

,

(20.77)



  23 1 3   B N |s − t| 2 (t) ≤ √ , ∞ 2 N

(20.78)



  13 3   B N |s − t| 2 (t) ≤ ∞

and

1 3

∞

and

(20.75)

(4N ) 4 1 (4N ) 4

∀ N ∈ N. We derive 1

,1

Proposition 20.20 Let X ∈ C2 ([0, 1]). Then 

 3 1 1 2TtX E (B N X ) − E X ∞ ≤ √ , E ω , sup √ 1 4 4N t∈[0,1] 3 N

(20.79)

∀ N ∈ N. Hence lim E (B N X ) = E X , uniformly. N →∞

Proof By (20.74) and Remark 20.19.



20.7 Conformable Fractional Stochastic Korovkin Theory Here L is meant as a sequence of positive E-commutative linear operators and all assumptions are as in Theorem 20.10. We give   Theorem 20.21 We further assume that L (1) (t) → 1 and L |s − t|α+1 (t) → 0, then (E (L X )) (t) → (E X ) (t), for any X ∈ Cα,n+1 ([a, b]), ∀ t ∈ [a, b], a pointwise convergence; where α ∈ (n, n + 1) , n ∈ Z+ .

478

20 Commutative Conformable Fractional Korovkin …

Proof Based on (20.41),   (20.27),   and that L (1) (t) is bounded as a sequence of functions. Also ω1 E α Tt X , δ is bounded, see (20.36), (20.37) and (20.40).  We continue with Theorem 20.22  further assume that L (1) (t) → 1, uniformly and    We  L |s − t|α+1 (t) → 0, then E (L X ) → E X , uniformly over [a, b], for any X ∈ ∞ Cα,n+1 ([a, b]); where α ∈ (n, n + 1) , n ∈ Z+ . Proof Based (20.28), and that L (1)∞ is bounded. Also it is   on (20.63),   sup ω1 E α Tt X , δ < ∞, by (20.36), (20.37) and (20.40). t∈[a,b]

We finish with Remark 20.23 The stochastic convergences of Theorems 20.21, 20.22 are derived by of the basic and simple real non-stochastic functions  the convergences  1, |s − t|α+1 , an amazing fact!

References 1. Abdeljawad, T.: On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. Aliprantis, C., Burkinshaw, O.: Principles of Real Analysis, 3rd edn. Academic, San Diego (1998) 3. Anastassiou, G.: A study of positive linear operators by the method of moments, onedimensional case. J. Approx. Theory 45, 247–270 (1985) 4. Anastassiou, G.: Korovkin type inequalities in real normed vector spaces. Approx. Theory Appl. 2, 39–53 (1986) 5. Anastassiou, G.: Multi-dimensional quantitative results for probability measures approximating the unit measure. Approx. Theory Appl. 2, 93–103 (1986) 6. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 7. Anastassiou, G.A.: Moments in Probability and Approximation Theory. Pitman/Longman, # 287, UK (1993) 8. Anastassiou, G.: Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Maxproduct operators. Springer, Heidelberg (2018) 9. Anastassiou, G.: Commutative conformable fractional Korovkin properties for Stochastic processes (2020). Submitted for publication 10. Khalil, R., Al Horani, M., Yousef, A., Sababheh, M.: A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 11. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 12. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 13. Shisha, O., Mond, B.: The degree of convergence of sequences of linear positive operators. Natl. Acad. Sci. U.S. 60, 1196–1200 (1968) 14. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192(1), 73–80 (1986) 15. Weba, M.: Quantitative results on monotone approximation of stochastic processes. Probab. Math. Stat. 11(1), 109–120 (1990) 16. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61(1), 74–87 (1990) 17. Weba, M.: Monotone approximation of random functions with multivariate domains in respect of lattice semi-norms. Results Math. 20(1–2), 554–576 (1991)

Chapter 21

Trigonometric Commutative Conformable Fractional Korovkin Approximation for Stochastic Processes

Here we research from the trigonometric point of view expectation commutative stochastic positive linear operators acting on L 1 -continuous stochastic processes which are Conformable fractional differentiable. Under some mild, general and natural assumptions on the stochastic processes we produce related trigonometric Conformable fractional stochastic Shisha-Mond type inequalities pointwise and uniform. All convergences are produced with rates and are given by the trigonometric Conformable fractional stochastic inequalities involving the first modulus of continuity of the expectation of the αth right and left Conformable fractional derivatives of the engaged stochastic process, α ∈ (n, n + 1), n ∈ Z+ . The amazing fact here is that the basic non-stochastic real Korovkin test functions assumptions imply the conclusions of our trigonometric Conformable fractional stochastic Korovkin theory. We include also a detailed trigonometric application to stochastic Bernstein operators. See also [8].

21.1 Introduction In this work among others we are motivated by the following results. Theorem A (Korovkin [10], (1960)) Let L n : C ([−π, π]) → C ([−π, π]), n ∈ u N, be a sequence of positive linear operators. Assume L n (1) → 1 (uniformly), u u u L n (cos t) → cos t, L n (sin t) → sin t, as n → ∞. Then L n f → f , for every f ∈ C ([−π, π]) that is 2π-periodic. Let f ∈ C ([a, b]) and 0 ≤ δ ≤ b − a. The first modulus of continuity of f at δ is given by ω1 ( f, δ) = sup {| f (x) − f (y)| ; x, y ∈ [a, b] , |x − y| ≤ δ} . © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_21

479

480

21 Trigonometric Commutative Conformable Fractional Korovkin …

If δ > b − a, then we define ω1 ( f, δ) = ω1 ( f, b − a) . Another motivation is the following. Theorem B (Shisha and Mond [12], (1968)) Let L 1 , L 2 , . . . , be linear positive operators, whose common domain D consists of real functions with domain (−∞, ∞). Suppose 1, cos x, sin x, f belong to D, where f is an everywhere continuous, 2π-periodic function, with modulus of continuity ω1 . Let −∞ < a < b < ∞, and suppose that for n = 1, 2, . . . , L n (1) is bounded in [a, b]. Then for n = 1, 2, . . . , L n ( f ) − f ∞ ≤  f ∞ L n (1) − 1∞ + L n (1) + 1∞ ω1 ( f, μn ) , where

(∗ )

     21   2 t −x  μn = π  L n sin (x)  , 2 ∞

and ·∞ stands for the sup norm over [a, b]. In particular, if L n (1) = 1, then (∗ ) reduces to L n ( f ) − f ∞ ≤ 2ω1 ( f, μn ) . One can easily see that, for m = 1, 2, . . . ,  μ2n

≤

π2 2



 L n (1) − 1∞ +

 (L n (cos t)) (x) − cos x∞ + (L n (sin t)) (x) − sin x∞ , so the last along with (∗ ) prove Korovkin’s Theorem A in a quantitative way and with rates of convergence. Anastassiou in [2–4] established a series of sharp inequalities for various cases of the parameters of the problem. However, Weba in [13–16] was the first, among many workers in quantitative results of Shisha-Mond type, to produce inequalities for stochastic processes. He assumed that the positive linear operators L j acting on stochastic processes X , are E-commutative (E means expectation) and stochastically simple. Anastassiou continued in [5] the pointwise case andwithout stochastic simplicity   of L j he found nearly best and best upper bounds for  E L j X (x0 ) − (E X ) (x0 ), x0 ∈ Q—a compact convex subset of a real normed vector space. The author here continues his above work on the trigonometric approximation of stochastic processes, now at the Conformable stochastic fractional level. He derives pointwise and uniform trigonometric Conformable fractional stochastic Shisha-

21.1 Introduction

481

Mond type inequalities, see the main Theorems 21.8, 21.9 and the several related corollaries. He gives an extensive trigonometric application to stochastic Bernstein operators. He finishes with a pointwise and a uniform Conformable fractional trigonometric stochastic Korovkin theorem, derived by Theorems 21.8, 21.9. The stochastic convergences, about stochastic processes, of our trigonometric Conformable fractional Korovkin Theorems 21.18, 21.19 are derived only by the convergences of real basic non-stochastic functions. Our results here are relied on [7].

21.2 Background—I Here we follow [1], see also [9]. We need Definition 21.1 ([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

(21.1)

 If Tαa f (t) exists on (a, b), then 

 Tαa f (a) = lim Tαa f (t) . t→a+

(21.2)

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) . T f = −lim (t) α ε→0 ε

b If

b

αT

(21.3)

f (t) exists on (a, b), then b

αT

 f (b) = lim bα T f (t) . t→b−

(21.4)

Note that if f is differentiable then  and

b

Tαa f (t) = (t − a)1−α f  (t) ,

αT

We need:

f (t) = − (b − t)1−α f  (t) .

(21.5)

(21.6)

482

21 Trigonometric Commutative Conformable Fractional Korovkin …

Definition 21.2 ([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) .

(21.7)

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) .

(21.8)

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 (21.7), (21.8) collapse to (21.1), (21.2), (21.3), (21.4) respectively. We make Remark 21.3 ([6], p. 155) 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 (

(21.9)

 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) .

(21.10)

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,

(21.11)

when 0 < β < 1, i.e. when α ∈ (n, n + 1).

21.3 Preliminaries Let (, F, P) be a probabilistic space and L 1 (, F, P) be the space of all realvalued random variables Y = Y (ω) with

21.3 Preliminaries

483



|Y (ω)| P (dω) < ∞.

Let X = X (t, ω) denote a stochastic process with index set [a, b] ⊂ R and real state space (R, B), where B is the σ-field of Borel subsets of R. Here C ([a, b]) is the space of continuous real-valued functions on [a, b] and B ([a,  b]) is the space of bounded real-valued functions on [a, b]. Also C ([a, b]) = C [a, b] , L 1 (, F, P) is the stochastic processes in t and space of  L 1 -continuous

B ([a, b]) = X : sup  |X (t, ω)| P (dω) < ∞ , obviously C ([a, b]) t∈[a,b]

⊂ B ([a, b]). Let α ∈ (n, n + 1), n ∈ Z+ , and consider the subspace of stochastic processes  Cα,n+1 ([a, b]) := {X : X (·, ω) ∈ C n+1 ([a, b]) , ∀ ω ∈  and  X (n+1) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] ×  , where M > 0; X (k) (t, ω) ∈ C ([a, b]), k = 0, 1, . . . , n; also X (n+1) (t, ω), (t, ω) ∈ [a, b] × , is a stochastic process}. Clearly by (21.9) and (21.10), respectively, we get that Ttα X (z, ω), z ∈ [t, b], and tα TX (z, ω), z ∈ [a, t], are stochastic processes for any t ∈ [a, b]. Consider the linear operator L : C ([a, b]) → B ([a, b]) . If X ∈ C ([a, b]) is nonnegative and L X , too, then L is called positive. If E L = L E, then L is called E -commutative, where E is the expectation operator. From [15, pp. 3–5] we have the following results (i) C ([a, b]) ⊂ C ([a, b]) , (ii) if X ∈ C ([a, b]), then E X ∈ C ([a, b]), and (iii) if L is E-commutative, then L maps the subspace C ([a, b]) into B ([a, b]) .

21.4 Background—II   Let X ∈ Cα,n+1 ([a, b]). We have that  X (n+1) (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × . We notice that (21.9)

Ttα X (z, ω) = (z − t)1−β X (n+1) (z, ω) , z ∈ [t, b] , and t α TX

(21.12)

(21.10)

(z, ω) = (−1)n (t − z)1−β X (n+1) (z, ω) , z ∈ [a, t] ,

∀ ω ∈ , where β = α − n, α ∈ (n, n + 1), n ∈ Z+ ; ∀ t ∈ [a, b] . We have that Ttα X (t, ω) =tα TX (t, ω) = 0.

(21.13)

484

It is

21 Trigonometric Commutative Conformable Fractional Korovkin …

   t  T X (z, ω) = (z − t)1−β X (n+1) (z, ω) ≤ (b − a)1−β M, α

(21.14)

∀ (z, ω) ∈ [t, b] × , and    t  TX (z, ω) = (−1)n (t − z)1−β X (n+1) (z, ω) ≤ (b − a)1−β M, α

(21.15)

∀ (z, ω) ∈ [a, t] × . Here E is the expectation operator

(E X ) (t) = Clearly, we get that ∀ z ∈ [t, b] , and



X (t, ω) P (dω) .

  t   E T X (z) ≤ (b − a)1−β M, α

(21.16)

 t   E TX (z) ≤ (b − a)1−β M, α

(21.17)

∀ z ∈ [a, t] . We observe that the first modulus of continuity (δ > 0) ω1

    t    E Tα X , δ [t,b] := sup  E Ttα X (x) − E Ttα X (y) x,y∈[t,b] |x−y|≤δ

(21.16)

≤ 2 (b − a)1−β M, ∀ t ∈ [a, b] .

(21.18)

  t E Tα X , δ [t,b] ≤ 2 (b − a)1−β M,

(21.19)

 t (21.17) E α TX , δ [a,t] ≤ 2 (b − a)1−β M.

(21.20)

Hence it holds (δ > 0) sup ω1 t∈[a,b]

and similarly, it holds sup ω1 t∈[a,b]

We also set         ω1 E α Tt X , δ := max ω1 E Ttα X , δ [t,b] , ω1 E tα TX , δ [a,t] , (21.21) where δ > 0.

21.4 Background—II

485

We make Remark 21.4 Let the positive linear operator L mapping C ([a, b]) into B ([a, b]) (the bounded functions). By the Riesz representation theorem [11] we have that there exists μt unique, completed Borel measure on [a, b] with μt ([a, b]) = L (1) (t) > 0,

(21.22)

such that

L ( f ) (t) =

f (s) dμt (s) , ∀ t ∈ [a, b] , ∀ f ∈ C ([a, b]) .

(21.23)

[a,b]

We denote ·∞ = ·∞,[−π,π] the supremum norm. Next we specify [a, b] as [−π, π]. Clearly then L : C ([−π, π]) → B ([−π, π]) is the positive linear operator on hand. Here α ∈ (n, n + 1), n ∈ Z+ , k = 1, . . . , n. By the use of Hölder’s inequality we notice that       

|s − t| k |s − t| k sin sin dμt (s) ≤ L (t) = 4 4 [−π,π] 

 [−π,π]



|s − t| sin 4

α+1

k  α+1

dμt (s)

(μt ([−π, π]))

α+1−k α+1

=

(21.24)

k       α+1 |s − t| α+1 α+1−k sin L (L (1) (t)) α+1 . (t) 4

    |s − t| k sin L (t) ≤ 4

That is

(21.25)

k       α+1 |s − t| α+1 α+1−k sin L (L (1) (t)) α+1 , (t) 4

for k = 1, . . . , n. Consequently, it holds         |s − t| k   sin (t) L   4

∞

≤

(21.26)

486

21 Trigonometric Commutative Conformable Fractional Korovkin … k        α+1  α+1−k |s − t| α+1   sin (t) L (1)∞α+1 , L   4

∞

for k = 1, . . . , n. In this work we will use a lot the following well known inequality:  |z| ≤ π sin

 |z| , ∀ z ∈ [−π, π] . 2

Furthermore, we observe that 



     k  L (s − t)k (t) =  (s − t) dμt (s) ≤  [−π,π]



2

k [−π,π]

|s − t| 2

k



(21.27)

|s − t|k dμt (s) = [−π,π]





|s − t| sin dμt (s) ≤ (2π) 4 [−π,π]     |s − t| k k sin = (2π) L (t) . 4 (21.27)

k

k

dμt (s) (21.28)

That is     L (s − t)k (t) ≤ (2π)k L





|s − t| sin 4

k  (t) ,

(21.29)

∀ t ∈ [−π, π], and     L (s − t)k (t)

∞

   k     |s − t|   sin ≤ (2π)k  L (t) ,   4

(21.30)

∞

all k = 1, . . . , n. Then, by (21.25) and (21.29) we get     L (s − t)k (t) ≤ (2π)k

  L



|s − t| sin 4

α+1 

k  α+1

(t)

(L (1) (t))

α+1−k α+1

,

(21.31) k = 1, . . . , n, ∀ t ∈ [−π, π], and by (21.26 ) and (21.30), we find     L (s − t)k (t)

∞

k = 1, . . . , n.

k    α+1    α+1  α+1−k |s − t|   sin ≤ (2π)k  L (t) L (1)∞α+1 ,   4 (21.32)

21.4 Background—II

487

We also have

  L |s − t|α+1 (t) =

[−π,π]

2α+1



[−π,π]

|s − t|α+1 dμt (s) =

   

(21.27) |s − t| α+1 |s − t| α+1 α+1 sin dμt (s) ≤ (2π) dμt (s) 2 4 [−π,π] = (2π)α+1 L





|s − t| sin 4

α+1 

(21.33) (t) .

That is 

L |s − t|α+1 (t) ≤ (2π)α+1 L





|s − t| sin 4

α+1 

(t) ,

(21.34)

∀ t ∈ [−π, π] , and     L |s − t|α+1 (t)

∞

≤ (2π)

α+1

        |s − t| α+1   sin (t) . L   4

(21.35)

∞

Following 3. Preliminaries we mention Theorem 21.5 ([7]) Let α ∈ (n, n + 1), n ∈ Z+ . Consider the positive E-commutative linear operator L  : C ([a, b]) → B ([a, b]), and let X ∈ Cα,n+1 ([a, b]), δ > 0. Here ω1 E α Tt X , δ as in (21.21). Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +     n    ω E α Tt X , δ E X (k) (t)    L (s − t)k (t) + 1 n  k! k=1 (α + j − n) j=1

⎤ ⎡ 1 α+1   1 α+1 α α+1   |s L − t| (t) (L (1) (t)) ⎦, + L |s − t|α+1 (t) α+1 ⎣ α−n δ (α + 1) ∀ t ∈ [a, b] .

(21.36)

488

21 Trigonometric Commutative Conformable Fractional Korovkin …

We also need Theorem 21.6 ([7]) All as in Theorem 21.5. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ + n 

   E X (k) 

∞

k!

k=1

    L (s − t)k (t) + ∞

  sup ω1 E α Tt X , δ t∈[a,b] n 

(α + j − n)

j=1

⎤ ⎡ 1   1  α+1 α+1 α+1   α    |s L − t| (t) ∞ ⎦  L |s − t|α+1 (t) α+1 ⎣ L (1)∞ + < ∞. (21.37) ∞ α−n δ (α + 1) We will use (case of n = 0): Definition 21.7 ([7]) We call the subspace stochastic process: Cα,1 ([a, b]) :=  of  1 (1) {X : X (·, ω) ∈ C ([a, b]) , ∀ ω ∈  and  X (t, ω) ≤ M, ∀ (t, ω) ∈ [a, b] × , where M > 0; and X (t, ω) ∈ C ([a, b]) ; also X (1) (t, ω), (t, ω) ∈ [a, b] ×  is a stochastic process}. Here we will consider Cα,1 ([−π, π]) .

21.5 Main Results We present our main pointwise approximation result Theorem 21.8 Let α ∈ (n, n + 1), n ∈ Z+ . Consider the positive E-commutative α,n+1 linear operator L : C ([−π, ([−π, π]), π]) → B ([−π, π]), and let X ∈ C α t δ > 0. Here ω1 E T X , δ , as in (21.21) it is over [−π, π]. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +  n   E X (k) (t) k!

k=1

 (2π) L k



|s − t| sin 4

k  (t) +

α      α+1   α+1 ω1 E α Tt X , δ |s − t| sin (2π)α L (t) n  4 (α + j − n)

j=1

⎡ ⎢ (L (1) (t)) ⎢ ⎢ ⎣ α−n

1 α+1

1 ⎤    α+1   α+1 (2π) L sin |s−t| (t) ⎥ 4 ⎥ + ⎥, ⎦ δ (α + 1)

(21.38)

21.5 Main Results

489

∀ t ∈ [−π, π] . Proof By (21.36), applied over [−π, π], we get |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| +     n    ω E α Tt X , δ E X (k) (t)    L (s − t)k (t) + 1 n  k! k=1 (α + j − n) j=1

⎤ ⎡ 1 α+1   1 α+1 α α+1   |s L − t| (t) (L (1) (t)) ⎦ + L |s − t|α+1 (t) α+1 ⎣ α−n δ (α + 1) (by (21.29), (21.34))

≤

 n   E X (k) (t) k!

k=1

|(E X ) (t)| |(L (1)) (t) − 1| + 

(2π) L k



|s − t| sin 4

k  (t) +

α         α+1 ω1 E α Tt X , δ |s − t| α+1 α sin (2π) L (t) n  4 (α + j − n)

j=1

⎡ 1 ⎢ (L (1) (t)) α+1 ⎢ ⎢ ⎣ α−n

1 ⎤    α+1   α+1 |s−t| (2π) L sin 4 (t) ⎥ ⎥ + ⎥, ⎦ δ (α + 1)

(21.39)

∀ t ∈ [−π, π] .



It follows the main uniform related estimate Theorem 21.9 All as in Theorem 21.8. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ +   n   E X (k) 

∞

k=1

k!

  sup ω1 E α Tt X , δ t∈[−π,π] n  j=1

(α + j − n)

        |s − t| k   sin (t) + (2π)  L   4 k

∞

α    α+1    α+1  |s − t|   sin (2π)α  L (t)   4

∞

(21.40)

490

21 Trigonometric Commutative Conformable Fractional Korovkin …

⎡ 1 ⎢ L (1) α+1 ∞ ⎢ ⎢ ⎣ α−n

1 ⎤    α+1     α+1 |s−t| (2π)  (t)  L sin 4  ⎥ ∞ ⎥ + ⎥ < +∞. ⎦ δ (α + 1)



Proof Directly from (21.38). We give the following special results:

Corollary 21.10 Let 0 < α < 1. Consider the positive E-commutative linear operα,1 ator L : C ([−π, π]) → B ([−π, π]), and let X ∈ C ([−π, π]), δ > 0. Here α t ω1 E T X , δ , as in (21.21) it is over [−π, π]. Then |(E (L X )) (t) − (E X ) (t)| ≤ |(E X ) (t)| |(L (1)) (t) − 1| + α       α+1  α t |s − t| α+1 α sin ω1 E T X , δ (2π) L (t) 4

⎡ 1 ⎢ (L (1) (t)) α+1 ⎢ ⎢ ⎣ α

1 ⎤    α+1   α+1 |s−t| (2π) L sin 4 (t) ⎥ ⎥ + ⎥, ⎦ δ (α + 1)

(21.41)

∀ t ∈ [−π, π] . 

Proof From (21.38). Corollary 21.11 All as in Corollary 21.10. Then E (L X ) − E X ∞ ≤ E X ∞ L (1) − 1∞ + α    α+1    α+1   α t |s − t|   sin sup ω1 E T X , δ (2π)α  L (t)   4 t∈[−π,π]

∞

⎡ ⎢ L (1) ∞ ⎢ ⎢ ⎣ α

1 α+1

Proof From (21.40).

1 ⎤    α+1     α+1 |s−t|   (2π)  L sin 4 (t) ⎥ ∞ ⎥ + ⎥. ⎦ δ (α + 1)

(21.42)



21.5 Main Results

491

Corollary 21.12 All as in Corollary 21.10, plus L (1) = 1. Then E (L X ) − E X ∞ ≤ α    α+1    α+1   α t |s − t|   sin sup ω1 E T X , δ (2π)α  L (t)   4 t∈[−π,π]

∞

⎡

1 ⎤    α+1     α+1 |s−t|   (2π)  L sin 4 (t) ⎥ ⎢1 ⎢ ∞ ⎥ ⎢ + ⎥. ⎣α ⎦ δ (α + 1)

(21.43)



Proof From (21.42). In particular we have Corollary 21.13 All as in Corollary 21.10 and L (1) = 1. Then  E (L X ) − E X ∞ ≤ ⎛ sup ω1 ⎝ E t∈[−π,π]

α

 α+1 (2π)α α

1 ⎞    α+1   α+1   |s − t| 2π   sin TtX , (t) ⎠ L  α+1 4



∞

α        α+1  |s − t| α+1   sin (t) . L   4

(21.44)

∞

Proof By (21.43): we take there 1        α+1 |s − t| α+1 2π    sin > 0. δ= (t) L  α+1 4

∞

   α+1     |s−t|  In case of  L sin 4 (t) 

  α+1  |s−t| = 0, we have that L sin 4 (t) ∞    α+1 = 0, ∀ t ∈ [−π, π] . That is, by (21.23), [−π,π] sin |s−t| dμt (s) = 0, ∀ t ∈ 4 is a probability measure on π] . μ [−π, [−π, π], where    t |s−t| ≥ 0, ∀ s ∈ π], we get sin = 0, a.e., that is |s − t| = Since sin |s−t| [−π, 4 4 0, a.e., and s = t, a.e., which means μt {s ∈ [−π, π] : s = t} = 0, i.e. μt = δt , ∀ t ∈ [−π, π] , where δt is the unit Dirac measure.

492

21 Trigonometric Commutative Conformable Fractional Korovkin …

Consequently we have

(E X ) (s) dδt (s) = (E X ) (t) ,

E (L X ) (t) = L (E X ) (t) = [−π,π]

∀ t ∈ [−π, π] . That is E (L X ) = E X over [−π, π]. Therefore both sides of inequality (21.44) equal to zero. Hence (21.44) is always true. 

21.6 Application Consider the Bernstein polynomials on [−π, π] for f ∈ C ([−π, π]) : B N ( f ) (x) =

 N   N k=0

k

 f

−π +

2πk N



x +π 2π

k 

π−x 2π

 N −k

,

(21.45)

N ∈ N, any x ∈ [−π, π]. There are positive linear operators from C ([−π, π]) into itself. Setting g (t) = f (2πt − π), t ∈ [0, 1], we have g (0) = f (−π), g (1) = f (π), and    N   k N t k (1 − t) N −k = (B N f ) (x) , x ∈ [−π, π] . g k N k=0 (21.46) Here x = ϕ (t) = 2πt − π is an 1 − 1 and onto map from [0, 1] onto [−π, π]. Clearly here g ∈ C ([0, 1]). Notice also that (B N g) (t) =

    (2π)2 t (1 − t) B N (· − x)2 (x) = B N (· − t)2 (t) (2π)2 = N    π−x 1 π2 (2π)2 x + π = = , ∀ x ∈ [−π, π] . (x + π) (π − x) ≤ N 2π 2π N N 

I.e.



 π2 , ∀ x ∈ [−π, π] . B N (· − x)2 (x) ≤ N

(21.47)

21.6 Application

493

In particular (B N 1) (x) = 1, ∀ x ∈ [−π, π] .

(21.48)

Define the corresponding stochastic application of B N as follows:  N   N

     2πk t + π k π − t N −k X −π + B N (X (·, ω)) (t) = ,ω , k N 2π 2π k=0 (21.49) ∀ N ∈ N, ∀ t ∈ [−π, π], ∀ ω ∈ , where X is a stochastic process. Clearly B N X is a stochastic process. Notice that E (B N X ) (t) =

 N   N k=0

k

     t + π k π − t N −k 2πk −π + X (E ) N 2π 2π = B N (E X ) (t) ,

(21.50)

i.e. E B N = B N E, that is B N is an E-commutative positive linear operator from C ([−π, π]) into itself. We give Corollary 21.14 Let 0 < α < 1 and X ∈ Cα,1 ([−π, π]). Then  E (B N X ) − E X ∞ ≤ ⎛ sup ω1 ⎝ E

α

t∈[−π,π]

 α+1 (2π)α α

1 ⎞    α+1    α+1  |s − t| 2π   sin TtX , (t) ⎠  BN  α+1 4



∞

α        α+1  |s − t| α+1   sin (t) ,  BN   4

(21.51)

∞

∀ N ∈ N. 

Proof By (21.44). In particular we get: 1

,1

Corollary 21.15 Let X ∈ C2 ([−π, π]). Then

494

21 Trigonometric Commutative Conformable Fractional Korovkin …

√ E (B N X ) − E X ∞ ≤ 3 2π ⎛

 4π sup ω1 ⎝ E 2 T t X , 3 t∈[−π,π] 1

2 ⎞     23   3   |s − t|   sin (t) ⎠  BN   4

∞

1     23   3  |s − t|   sin (t) ,  BN   4

(21.52)

∞

∀ N ∈ N. Proof By (21.51) for α = 21 .



We make Remark 21.16 By |sin x| < |x| , ∀ x ∈ R − {0}, in particular sin x ≤ x, for x ≥ 0, we get    23  3 |· − t| |· − t| 2 1 3 sin ≤ = |· − t| 2 . 4 4 8 Hence     23     |· − t|   sin (t)  BN   4

≤

∞

   1 3    B N |· − t| 2 (t) . ∞ 8

(21.53)

We observe that 3       N     2 N  t + π k π − t N −k 3 t + π − 2πk  B N |· − t| 2 (t) =  k N  2π 2π k=0 (by discrete Hölder’s inequality)       N   2πk 2 N t + π k π − t N −k ≤ t +π− k N 2π 2π k=0 3   = B N (· − t)2 (t) 4 Consequently it holds

(21.47)

≤

! 34 (21.54)

3

π2

3

N4

    3   B N |· − t| 2 (t)

∞

, ∀ t ∈ [−π, π] .

3

≤

π2

3

N4

,

(21.55)

21.6 Application

495

    23     |· − t|   sin (t)  BN   4

and

3

≤

∞

π2

3

8N 4

, ∀ N ∈ N.

(21.56)

We derive 1

,1

Proposition 21.17 Let X ∈ C2 ([−π, π]). Then E (B N X ) − E X ∞

√    π2  3 2π 1 2TtX , E ≤ √ ω , √ sup 1 2 4 N t∈[−π,π] 3 N

(21.57)

∀ N ∈ N. Hence lim E (B N X ) = E X , uniformly. N →∞

Proof By (21.52) and (21.56).



21.7 Commutative Trigonometric Conformable Fractional Stochastic Korovkin Properties Here L is meant as a sequence of positive E-commutative linear operators and all assumptions are as in Theorem 21.8. We give   α+1  Theorem 21.18 We further assume that L (1) (t) → 1 and L sin |s−t| (t) 4 → 0, then (E (L X )) (t) → (E X ) (t), for any X ∈ Cα,n+1 ([−π, π]), ∀ t ∈ [−π, π], a pointwise convergence; where α ∈ (n, n + 1) , n ∈ Z+ . Proof Based on (21.38),   (21.31), and that L (1) (t) is bounded as a sequence of functions. Also ω1 E α Tt X , δ over [−π, π] is bounded, see (21.19)–(21.21).  We continue with Theorem   21.19We further   assume that L (1) (t) → 1, uniformly and α+1    L sin |s−t| (t)   → 0, then E (L X ) → E X , uniformly over [−π, π], 4 ∞

for any X ∈ Cα,n+1 ([−π, π]); where α ∈ (n, n + 1) , n ∈ Z+ . Proof Based   on (21.40), (21.32), and that L (1)∞ is bounded. Also it is  sup ω1 E α Tt X , δ < ∞, by (21.19)–(21.21). t∈[−π,π]

496

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We finish with Remark 21.20 The stochastic convergences of Theorems 21.18, 21.19 are implied by the convergences of the basic and simple real non-stochastic functions "  α+1 #  |s−t| , an amazing fact! 1, sin 4

References 1. Abdeljawad, T.: On conformable fractional calculus. J. Comput. Appl. Math. 279, 57–66 (2015) 2. Anastassiou, G.: A study of positive linear operators by the method of moments, onedimensional case. J. Approx. Theory 45, 247–270 (1985) 3. Anastassiou, G.: Korovkin type inequalities in real normed vector spaces. Approx. Theory Appl. 2, 39–53 (1986) 4. Anastassiou, G.: Multi-dimensional quantitative results for probability measures approximating the unit measure. Approx. Theory Appl. 2, 93–103 (1986) 5. Anastassiou, G.A.: Korovkin inequalities for stochastic processes. J. Math. Anal. Appl. 157(2), 366–384 (1991) 6. Anastassiou, G.: Nonlinearity: Ordinary and Fractional Approximations by Sublinear and Maxproduct operators. Springer, Heidelberg (2018) 7. Anastassiou, G.A.: Commutative conformable fractional Korovkin properties for Stochastic processes (2020). Submitted for publication 8. Anastassiou, G.: Trigonometric Commutative Conformable fractional Korovkin properties for Stochastic processes, Analele Universitatii Oradea Fasc. Matematica (2020). Accepted for publication 9. Khalil, R., Al Horani, M., Yousef, A., Sababheh, M.: A new definition of fractional derivative. J. Comput. Appl. Math. 264, 65–70 (2014) 10. Korovkin, P.P.: Linear Operators and Approximation Theory. Hindustan Publ. Corp., Delhi (1960) 11. Royden, H.L.: Real Analysis, 2nd edn. MacMillan Publishing Co., Inc., New York (1968) 12. Shisha, O., Mond, B.: The degree of approximation to periodic functions by linear positive operators. J. Approx. Theory 1, 335–339 (1968) 13. Weba, M.: Korovkin systems of stochastic processes. Math. Z. 192(1), 73–80 (1986) 14. Weba, M.: Quantitative results on monotone approximation of stochastic processes. Probab. Math. Stat. 11(1), 109–120 (1990) 15. Weba, M.: A quantitative Korovkin theorem for random functions with multivariate domains. J. Approx. Theory 61(1), 74–87 (1990) 16. Weba, M.: Monotone approximation of random functions with multivariate domains in respect of lattice semi-norms. Results Math. 20(1–2), 554–576 (1991)

Chapter 22

Concluding Remarks

In a 1695 letter to L’Hospital, Leibniz asked, “Can we generalize ordinary\derivatives to ones of arbitrary order?” L’Hospital replied to Leibniz with another question, “What if the order is 1/2?” Then Leibniz, in a letter dated September 30, 1695 replied, “It will lead to a paradox, from which one day many useful consequences will be drawn.” That was the beginning of fractional calculus. The subject has been ongoing for more than 300 years now. Many famous mathematicians contributed to the topic over the years such as Liouville, Euler, Laplace, Lagrange, Riemann, Weyl, Fourier, Abel, Lacroix, Grunwald, and Letnikov. The first attempt to give a rigorous definition of fractional derivatives and study the subject in depth was by Liouville during 1832–1855. The concept grew out of his earlier work in electromagnetism. Fractional derivatives describe solutions of fractional integral equations, many times arising from physics, as done by Abel in 1823 to solve the brachistochrone problem. Fractional calculus has a long history but, from the applicative point of view, it fell into oblivion for hundreds of years because of lack of applications to other sciences such as physics and engineering. This oblivion was also due to its complexity and lack of physical and geometric connection. Fortunately, that was then as things have changed dramatically. Fractional calculus has developed a lot in the last fifty years and especially since 1974 when the first international conference in the field took place, organized by B. Ross, in New Haven, Connecticut, USA. Now we frequently see such conferences around the world. There exist a lot of related research activities resulting in many interesting articles and books. AMS/Mathscinet lists about 50,000 fractional publications on 4/19/20. We now see many important applications: in acoustic wave propagation in inhomogeneous porous material, diffusive transport, fluid flow, dynamical processes in self-similar structures, dynamics of earthquakes, optics, geology, viscoelastic materials, biosciences, bioengineering, medicine, economics, probability and statistics, astrophysics, chemical engineering, physics, splines, tomography, fluid mechanics, electromagnetic waves, non-linear control, signal processing, control of electronic power, converters, chaotic dynamics, © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 G. A. Anastassiou, Generalized Fractional Calculus, Studies in Systems, Decision and Control 305, https://doi.org/10.1007/978-3-030-56962-4_22

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polymer science, proteins, polymer physics, electrochemistry, statistical physics, rheology, thermodynamics, neural networks, and many others. By now almost all fields of research in science and engineering use fractional calculus to better describe them. The current mathematical theory of fractional calculus seems to lag behind the needs for mathematical modeling of all the above applications. Therefore there exists a lot of room for mathematical theoretical expansion on the subject. The recent development of fractal theory has shown great connections to fractional calculus. In general, fractional calculus currently provides the best description for fractals and chaotic situations. All in all fractional calculus provides the best expression for the complexity of our modern science. In recent years researchers came up with more efficient and practical fractional derivatives, e.g. Caputo (1967) inspired by studies in Geophysics, Canavati (1987) and Khalil et al. (2014) with the conformable fractional derivative. The author was the first to study fractional inequalities and fractional approximation theory by publishing numerous of articles and monographs. In this monograph the author continuous his studies by employing his generalized fractional calculus (2016) methods. He uses also the, in some cases equivalent, generalized fractional calculus by Almeida (2017). The idea is to cover as many as possible cases in the applications since now the fractional calculus is weighted, that is the produced theory is as rich as possible. In the second part of the book the author lays down for the first time the foundations of stochastic fractional calculus and studies the fractional approximation of stochastic processes in detail. Definitely what is developed in this monograph can have great applications and can lead to more interesting avenues of research.