Higher Order Derivatives

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CHAPMAN & HALL/CRC Monographs and Surveys in Pure and Applied Mathematics

HIGHER ORDER DERIVATIVES

144

CHAPMAN & HALL/CRC Monographs and Surveys in Pure and Applied Mathematics Main Editors H. Brezis, Université de Paris R.G. Douglas, Texas A&M University A. Jeffrey, University of Newcastle upon Tyne (Founding Editor)

Editorial Board R. Aris, University of Minnesota G.I. Barenblatt, University of California at Berkeley H. Begehr, Freie Universität Berlin P. Bullen, University of British Columbia R.J. Elliott, University of Alberta R.P. Gilbert, University of Delaware R. Glowinski, University of Houston D. Jerison, Massachusetts Institute of Technology K. Kirchgässner, Universität Stuttgart B. Lawson, State University of New York B. Moodie, University of Alberta L.E. Payne, Cornell University D.B. Pearson, University of Hull G.F. Roach, University of Strathclyde I. Stakgold, University of Delaware W.A. Strauss, Brown University J. van der Hoek, University of Adelaide

CHAPMAN & HALL/CRC Monographs and Surveys in Pure and Applied Mathematics

HIGHER ORDER DERIVATIVES

Satya N. Mukhopadhyay in collaboration with P. S. Bullen

144

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2012 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Version Date: 20111215 International Standard Book Number-13: 978-1-4398-8048-7 (eBook - PDF) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright. com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

This book is dedicated to

My Family

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Contents

Preface

xi

Introduction

xiii

1 Higher Order Derivatives 1.1 1.2 1.3 1.4

Divided Differences of Order n . . . . . . . . . . . . . General Derivatives of Order n . . . . . . . . . . . . . Generalized Riemann Derivatives of Order n . . . . . Peano Derivatives . . . . . . . . . . . . . . . . . . . . 1.4.1 Bilateral Peano Derivatives . . . . . . . . . . . 1.4.2 Unilateral Peano Derivatives . . . . . . . . . . 1.4.3 Peano Boundedness . . . . . . . . . . . . . . . 1.4.4 Generalized Peano Derivatives . . . . . . . . . 1.4.5 Absolute Peano Derivatives . . . . . . . . . . . 1.5 Riemann∗ Derivatives . . . . . . . . . . . . . . . . . . 1.5.1 Bilateral Riemann∗ Derivatives . . . . . . . . . 1.5.2 Unilateral Riemann∗ Derivatives . . . . . . . . 1.6 Symmetric de la Vallée Poussin Derivatives . . . . . . 1.6.1 Symmetric de la Vallée Poussin Derivative and Symmetric Continuity . . . . . . . . . . . . . . 1.6.2 Smoothness . . . . . . . . . . . . . . . . . . . . 1.6.3 de la Vallée Poussin Boundedness . . . . . . . . 1.7 Symmetric Riemann∗ Derivatives . . . . . . . . . . . 1.8 Ces` aro Derivatives . . . . . . . . . . . . . . . . . . . . 1.8.1 Ces` aro Continuity and Cesàro Derivative . . . 1.8.2 Ces` aro Boundedness . . . . . . . . . . . . . . . 1.9 Symmetric Ces` aro Derivatives . . . . . . . . . . . . . 1.9.1 Symmetric Cesàro Continuity and Symmetric Derivative . . . . . . . . . . . . . . . . . . . . . 1.9.2 Symmetric Cesàro Boundedness . . . . . . . . . 1.10 Borel Derivatives . . . . . . . . . . . . . . . . . . . . . 1.10.1 Bilateral Borel Derivatives . . . . . . . . . . . . 1.10.2 Unilateral Borel Derivatives . . . . . . . . . . . 1.10.3 Borel Boundedness . . . . . . . . . . . . . . . . 1.11 Symmetric Borel Derivatives . . . . . . . . . . . . . .

1 . . . . . . . . . . . . .

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

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

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

1 5 9 16 16 19 20 21 23 24 24 27 28

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cesàro . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28 30 31 33 33 34 38 40 40 43 44 44 47 47 49 vii

viii

Contents

1.12

1.13

1.14

1.15

1.16

1.11.1 Symmetric Borel Derivatives and Symmetric Borel Continuity . . . . . . . . . . . . . . . . . . . . . . 1.11.2 Borel Smoothness . . . . . . . . . . . . . . . . . . 1.11.3 Symmetric Borel Boundedness . . . . . . . . . . . Lp -Derivatives . . . . . . . . . . . . . . . . . . . . . . . . 1.12.1 Lp -Derivatives and Lp -Continuity . . . . . . . . . . 1.12.2 Lp -Boundedness . . . . . . . . . . . . . . . . . . . Symmetric Lp -Derivatives . . . . . . . . . . . . . . . . . . 1.13.1 Symmetric Lp -Derivatives and Symmetric Lp -Continuity . . . . . . . . . . . . . . . . . . . . . 1.13.2 Lp -Smoothness . . . . . . . . . . . . . . . . . . . . 1.13.3 Symmetric Lp -Boundedness . . . . . . . . . . . . . Abel Derivatives . . . . . . . . . . . . . . . . . . . . . . . 1.14.1 Abel Summability . . . . . . . . . . . . . . . . . . 1.14.2 Abel Derivatives . . . . . . . . . . . . . . . . . . . 1.14.3 Abel Continuity . . . . . . . . . . . . . . . . . . . 1.14.4 Abel Smoothness . . . . . . . . . . . . . . . . . . . Laplace Derivatives . . . . . . . . . . . . . . . . . . . . . 1.15.1 Laplace Derivatives and Laplace Continuity . . . . 1.15.2 Laplace Boundedness . . . . . . . . . . . . . . . . 1.15.3 Bilateral Laplace Derivatives . . . . . . . . . . . . Symmetric Laplace Derivatives and Laplace Smoothness 1.16.1 Symmetric Laplace Derivatives and Symmetric Laplace Continuity . . . . . . . . . . . . . . . . . . 1.16.2 Symmetric Laplace Boundedness . . . . . . . . . . 1.16.3 Laplace Smoothness . . . . . . . . . . . . . . . . .

. . . . . . .

. . . . . . .

49 52 53 55 55 59 60

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

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

60 62 63 63 63 64 68 70 70 70 75 76 77

. . . . . .

77 82 84

2 Relations between Derivatives 2.1 2.2 2.3 2.4

2.5 2.6 2.7 2.8

Ordinary and Peano Derivatives, f (k) and f(k) . . . . . . . . ∗ Riemann∗ and Peano Derivatives, f(k) and f(k) . . . . . . . . ∗ Symmetric Riemann and Symmetric de la Vallée Poussin (s) ∗(s) Derivatives, f(k) and f(k) . . . . . . . . . . . . . . . . . . . Ces` aro and Peano Derivatives . . . . . . . . . . . . . . . . . 2.4.1 Ces` aro and Peano derivatives, Ck Df and f(k) . . . . . 2.4.2 Ces` aro and absolute Peano derivatives, Ck Df and f ∗ Peano and Symmetric de la Vallée Poussin Derivatives, f(k) and (s) f(k) , and Smoothness of Order k . . . . . . . . . . . . . . . . Symmetric Ces` aro and Symmetric de la Vallée Poussin (s) Derivatives, SCk Df and f(k) . . . . . . . . . . . . . . . . . . Borel and Peano Derivatives, BDk f and f(k) . . . . . . . . . Symmetric Borel and Symmetric de la Vallée Poussin (s) Derivatives, SBDk f and f(k) . . . . . . . . . . . . . . . . . .

85 85 86 88 98 98 101 102 106 108 116

Contents

ix

2.9

Borel and Symmetric Borel Derivatives, BDk f and SBDk f , and Borel Smoothness of Order k . . . . . . . . . . . . . . . 2.10 Peano and Lp -Derivatives, f(k) and f(k),p . . . . . . . . . . . (s)

2.11 Lp - and Symmetric Lp -Derivatives, f(k),p and f(k),p . . . . . 2.12 Symmetric de la Vallée Poussin and Symmetric Lp -Derivatives, (s) (s) f(k) and f(k),p . . . . . . . . . . . . . . . . . . . . . . . . . . 2.13 Borel and Lp -Derivatives, BD(k) f and f(k),p . . . . . . . . . 2.14 Symmetric Borel and Symmetric Lp -Derivatives, SBD(k) f and (s)

2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23

f(k),p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ces` aro and Borel Derivatives, Ck Df and BDk f . . . . . . . Symmetric Ces` aro and Symmetric Borel Derivatives, SCk Df and SBDk f . . . . . . . . . . . . . . . . . . . . . . . . . . . Abel and Symmetric de la Vallée Poussin Derivatives, ADk f (s) and f(k) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Laplace, Peano and Generalized Peano Derivatives, LDk f , f(k) and f[k] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Laplace and Borel Derivatives, LDk f and BDk f . . . . . . . Symmetric Laplace and Symmetric de la Vallée Poussin (s) Derivatives, SLDk f and f(k) . . . . . . . . . . . . . . . . . . Laplace and Symmetric Laplace Derivatives, LDk f and SLDk f . . . . . . . . . . . . . . . . . . . . . . . . . . . Peano and Unsymmetric Riemann Derivatives, f(k) and RDk f . . . . . . . . . . . . . . . . . . . . . . . . . . . . Symmetric de la Vallée Poussin and Symmetric Riemann (s) Derivatives, f(k) (s)

and RDk f . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.24 Generalized Riemann and Peano Derivatives, GRDk f and f(k) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e k f and f(k) . . . . . . . . . . . 2.25 MZ- and Peano Derivatives, D

127 130 133 136 140 144 146 149 151 165 169 172 175 176

178 182 184

Bibliography

187

Index

199


Preface

The concept of the first-order ordinary derivative is as old as the invention of the calculus. Since then the concept of the first-order ordinary derivative has been extended to higher order derivatives in various directions. The nthorder ordinary derivative f (n) of a function f is the first order derivative of its (n−1)st-order derivative f (n−1) and has no special importance. But, the higher order derivatives other than the ordinary one are particularly interesting in general because they are derivatives for which the nth order derivative can exist without the (n − 1)st-order derivative existing. For instance, the classical Riemann derivative is an example of this type and plays an important role in the theory of trigonometric series. Higher order derivatives of different types have been considered by several authors with the results appearing in various journals over a long period of time. We consider these higher order derivatives and study the relations between them. It is hoped that the resulting monograph will be particularly helpful to those young mathematicians who wish to pursue their studies in this branch of real analysis. We cannot claim to have considered all known types of higher order derivatives and suggestions for further inclusion and improvements in general are welcome.

xi


Introduction

The concept of higher order derivatives is useful in many branches of mathematics and its applications. Among the higher order derivatives, the Peano derivative is the most well known. This notion started from the viewpoint of approximating a function by polynomials and its origin is in the paper of G. Peano [139]. This derivative was subsequently considered, under various different names, by A. Genocchi, [68] and C.J. de la Vallée Poussin [174], and later by J.C. Burkill [28], A. Denjoy [45], J. Marcinkiewicz & A. Zygmund [106], J. Marcinkiewicz [105], A. Zygmund [194], and E. Corominas [37]. Finally in 1954, H.W. Oliver [134] gave the concept the name “Peano derivative” and made a systematic study of the Peano derivative. After this C.E. Weil [184] began to work on this derivative and since then extensive work has been done by Weil and many other authors [3,10,13–15,17,19,23,24,26,27,33,41,52,57,59,60,63,65,70,71,73,74,82,93–95, 97, 100–102, 124, 127, 128,135,137, 138, 142, 151, 159,161,181, 183, 185, 195, 196]. The Riemann* derivative is considered by A. Denjoy [45] and E. Corominas [37] under the name generalized derivative. They proved that this derivative exists finitely if and only if the Peano derivative exists finitely, and then the values are equal. The infinite case is considered in [26]; in this paper, there are some lacunæ pointed out in [94], which are filled up in [197]. A.M. Russel [145] gave this derivative its present name, Riemann* derivative, and used it to study functions of bounded variations of higher order. This derivative has been found to be useful as well for the study of functions of higher-order absolute continuity [43, 114]. The de la Vallée Poussin derivative, also called the symmetric Peano derivative, was introduced under the name generalized derivative by C.J. de la Vallée Poussin himself [174]; he used it to study various properties of Fourier series. Since then, this derivative has been studied and used by many authors [7, 9, 16, 20, 22, 25, 30, 34, 36, 38, 47, 72, 77, 78, 84, 87–92, 105, 108–110, 115, 116,119,121–123,128,144,146,147,155,159,162,165–168,172,173,193,194,196]. The Cesàro derivative originated from the Cesàro summability of series. Introduced by J.C. Burkill to define the Cesàro–Perron integral [28, 29], these derivatives were studied in detail by W.L.C. Sargent [150–152]. In addition,

xiii

xiv

Introduction

there is the work of J.A. Bergin [10], who studied the Cesàro derivatives proving, in certain cases, their equivalence to the Peano derivatives (see also [99]). J.C. Burkill later introduced symmetric Cesàro derivatives [30] to define the symmetric Ces` aro–Perron integral, which is useful in solving the so-called coefficient problem in the theory of trigonometric series; see also [25]. Higher order symmetric Ces` aro derivatives were introduced in [22] to define the SCn P -integral (see also [38]). The concept of the Borel derivative was introduced by E. Borel, calling it the average derivative (derivée moyenne.) First-order Borel derivatives, both unsymmetric and symmetric, were studied by A. Khintchine [86] and by J. Marcinkiewicz & A. Zygmund [106]. Then W.L.C. Sargent [149] made an intensive study of the first-order unsymmetric Borel derivative (see also C.J. Neugebauer [133]). The symmetric Borel derivative was used by J. Marcinkiewicz & A. Zygmund [106] and by P.S. Bullen & S.N. Mukhopadhyay [25] to consider a trigonometric integral. The Lp -derivative originated from the work of A.P. Calder´ on & A. Zygmund [32] and was considered in [159]. This derivative appears in [2, 5, 50, 58, 133,136,189,196]. The symmetric Lp -derivative is considered in [2,52,131,190]. The Abel derivative has a very special nature. Its origin can be found in a dormant state in the theory of trigonometric series and occurred in the work of A. Rajchman and A. Zygmund when they considered the Riemann and Abel summability of trigonometric series (see [196; p. 353, Lemma 7.6]). The definition of the Abel derivative is not given there, but is defined by S.J. Taylor [163]. Taylor only defined the second-order derivative and used it to introduce his Abel–Perron integral, which is helpful in solving the coefficient problem for Abel summable trigonometric series. As in the case of the de la Vallée Poussin derivative, the existence of the second-order Abel derivative of a function at a point does not imply the existence of the first-order Abel derivative at that point. To help with this, the concept of Abel smoothness is defined. The seeds of this and of higher-order Abel derivatives are in [122] and [121], respectively, and they are fully defined in the text. The Laplace derivative is introduced in [160, 161]; its theory is still being developed [198]. Symmetric Laplace derivatives are introduced in the text in a very natural way. The symmetric Riemann derivative was found to be convenient for studying the Riemann summability of trigonometric series. Both unsymmetric and symmetric Riemann derivatives of higher order are considered and studied by J. Marcinkiewicz [105] and J. Marcinkiewicz & A. Zygmund [106]. Symmetric Riemann derivatives of higher order are the special study of P.L. Butzer & W. Kozakiewicz [31] and T.K. Dutta & S.N. Mukhopadhyay [48]. Further work on both unsymmetric and symmetric Riemann derivatives of higher order can be found in [71,73,74,81–83,125,148,175–180,182,196]. J. Marcinkiewicz & A. Zygmund [106] defined another derivative of higher order that is of interest, which we call the MZ-derivative.

Introduction

xv

J.M. Ash [2], following a suggestion of A. Zygmund, introduces a system, a finite set of real numbers satisfying certain conditions, and uses this system to define a generalized derivative of higher order. Suitable specializations of the system lead to the symmetric and unsymmetric Riemann derivatives and to the MZ-derivative. This work is continued in [4,6,64,75,76,126]. The derivative of Ash is such that, if any of the derivatives of Peano or de la Vallée Poussin exists finitely, then the appropriate derivative of Ash also exists with the same value. This equivalence does not extend to the case of derivates, however, although a suitable modification of the Ash definition can be made to allow for this. That is, the modified Ash definition not only includes the Peano and de la Vallée Poussin derivatives, but their corresponding derivates as well. As has been mentioned, higher order derivatives occur in the study of convexity and bounded variation of higher order (see [1,16,21,35,39,40,69,70, 85, 111–113, 119, 129, 140, 141, 143, 154, 188, 192] and [11, 43, 114, 130, 145, 166], respectively). In the case of symmetric derivatives, smoothness is sometimes helpful in various discussions (see above paragraph on the Abel derivative). Thus, if a function f has a de la Vallée Poussin derivative of order n at x, then while f has a derivative of order n − 2 at x it may not have a derivative of order n − 1 at x. Then smoothness of order n may compensate for the nonexistence of this derivative of order n − 1, or if f is not smooth, then it may be quasi-smooth; see [1, 18, 46, 49, 51, 55, 56, 80, 104, 132, 153, 156, 158, 169–171]. We have not considered the approximate analogues of the derivatives and smoothness discussed above, but papers on these topics are included in the bibliography. This bibliography is surely not complete and any suggestion for further inclusion is welcome. Although much work has been done on the Peano and de la Vallée Poussin derivatives, there is a large amount of work to be done on the other higher order derivatives as their properties remain often virtually unexplored. The purpose of this book is to introduce to any newcomer interested in the field of higher order derivatives to the present state of knowledge. The background required is that of only basic advanced real analysis and, although the special Denjoy integral has been used, a knowledge of the Lebesgue integral should suffice. The book contains two chapters. Chapter 1 contains 16 subsections where the various higher order derivatives are introduced. Chapter 2 contains 24 subsections where the relations between these derivatives are given.


Chapter 1 Higher Order Derivatives

1.1

Divided Differences of Order n

Let n be a fixed positive integer and let f be a real valued function defined on the set {x0 , . . . , xn } of n + 1 distinct points. The nth order difference is defined by Qn (f ; x0 , . . . , xn ) =

Qn−1 (f ; x0 , . . . , xn−1 ) − Qn−1 (f ; x1 , . . . , xn ) , n ≥ 2, x0 − xn (1.1.1)

with

f (xi ) − f (xj ) , i 6= j, i, j = 0, . . . , n. xi − xj A simple inductive argument, given below, shows that Q1 (f ; xi , xj ) =

n X

Qn (f ; x0 , . . . , xn ) =

i=0

where

(1.1.2)

n

X f (xi ) f (xi ) = , j=0 (xi − xj ) ω ′ (xi ) i=0

Qn

(1.1.3)

i6=j

ω(x) =

n Y

(x − xj ).

j=0

In fact, if n = 1, then (1.1.3) follows from (1.1.2). Suppose that (1.1.3) holds for n = m and let n = m + 1. Then, from (1.1.1), we have, since (1.1.3) holds for n = m, (x0 − xm+1 )Qm+1 (f ; x0 , . . . , xm+1 ) = Qm (f ; x0 , . . . , xm ) − Qm (f ; x1 , . . . , xm+1 ) =

m X i=0

m+1 X f (xi ) f (xi ) − Qm+1 j=0 (xi − xj ) j=1 (xi − xj ) i=1

Qm

i6=j

i6=j

m

X f (xi ) f (x0 ) Qm + = Qm j=0 (xi − xj ) j=1 (x0 − xj ) i=1 i6=j

−

m X i=1

f (xi ) f (xm+1 ) − Qm Qm+1 j=1 (xi − xj ) j=1 (xm+1 − xj ) i6=j

1

2

Higher Order Derivatives

m X f (xi )(xi − xm+1 − xi + x0 ) = (x0 − xm+1 ) Qm+1 + Qm+1 j=0 (xi − xj ) j=1 (x0 − xj ) i=1

f (x0 )

i6=j

f (xm+1 )(xm+1 − x0 ) − Qm j=0 (xm+1 − xj ) " # m X f (x0 ) f (xi ) f (xm+1 ) = (x0 − xm+1) Qm+1 + + Qm Qm+1 j=0 (xi − xj ) j=0 (xm+1 − xj ) j=1 (x0 − xj ) i=1 i6=j

= (x0 − xm+1 )

m+1 X i=0

f (xi ) Qm+1 j=0 (xi − xj ) i6=j

and so (1.1.3) is established for n = m + 1.

It follows, from (1.1.3), that for any two functions f and g and for any two real numbers α and β Qn (αf + βg; x0 , . . . , xn ) = αQn (f ; x0 , . . . , xn ) + βQn (g; x0 , . . . , xn ). (1.1.4) It follows, also from (1.1.3), that Qn (f ; x0 , . . . , xn ) is independent of the order of the points x0 , . . . , xn . We show that 1 1 ... 1 x0 x1 ... xn .......................... .......................... n−1 x x1n−1 . . . xnn−1 0 f (x0 ) f (x1 ) . . . f (xn ) D f (x) = Qn (f ; x0 , . . . , xn ) = , say. (1.1.5) 1 D(xn ) 1 ... 1 x0 x1 ... xn ........................ ........................ n−1 x x1n−1 . . . xnn−1 0n x xn ... xn 0

1

n

If Dr is the co-factor of f (xr ) in D f (x) , then 1 1 ... 1 1 ... 1 x0 x . . . x x . . . x 1 r−1 r+1 n n+r+2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Dr = (−1) ........................................... ........................................... n−1 n−1 n−1 x x1n−1 . . . xr−1 xr+1 . . . xnn−1 0

.

(1.1.6)


3

Since the determinants Dr and D(xn ) are van der Monde determinants, we have Y (xj − xi ), (1.1.7) D(xn ) = i<j

Dr = (−1)n+r+2

Y′

(xj − xi ),

(1.1.8)

i<j

where in (1.1.8) xr never occurs. Hence, writing all the factors in these two expressions and cancelling equal factors we get D(xn ) = (−1)n+r+2 (xn − xr )(xn−1 − xr ) . . . (1.1.9) Dr × (xr+1 − xr )(xr − x0 )(xr − x1 ) . . . (xr − xr−1 ). Since Dr is the co-factor of f (xr ) in D f (x) , we have from the above that n n X X D f (x) f (xr )Dr f (xr ) Qn = = = Qn (f ; x0 , . . . , xn ), n) D(xn ) D(x j=0 (xr − xj ) r=0 r=0 r6=j

which completes the proof of (1.1.5). From (1.1.5) it follows that ( 0, if p = 0, 1, . . . , n − 1, p Qn (x ; x0 , . . . , xn ) = 1, if p = n.

(1.1.10)

(1.1.11)

Theorem 1.1.1 If f : R 7→ R is a polynomial of degree at most n, then for every choice of distinct points x0 , . . . , xn , Qn (f ; x0 , . . . , xn ) = 0 or an according, as the degree of f is less than n or equal to n; an being the coefficient of xn in f .

The proof follows from (1.1.4) and (1.1.11).

Theorem 1.1.2 If f : R 7→ R, then for every choice of distinct points x0 , . . . , xn , Qn (f ; x0 , . . . , xn ) = xn − y y − x0 Qn (f ; x0 , . . . , xn−1 , y) + Qn (f ; y, . . . , xn ), (1.1.12) xn − x0 xn − x0 where y is a point distinct from x0 , . . . , xn .

From (1.1.1), (x0 − y)Qn (f ; x0 , . . . , xn−1 , y) = Qn−1 (f ; x0 , . . . , xn−1 ) − Qn−1 (f ; x1 , . . . , xn−1 , y)

(1.1.13)

(y − xn )Qn (f ; y, x1 , . . . , xn ) = Qn−1 (f ; y, x1 , . . . , xn−1 ) − Qn−1 (f ; x1 , . . . , xn ).

(1.1.14)

4


Adding (1.1.13) and (1.1.14) and using (1.1.1), we get (1.1.12).

Theorem 1.1.3 Let f : R 7→ R and x0 , . . . , xn be distinct points. If y0 , . . . , yr−1 is another set of distinct points, each of which is different from x0 , . . . , xn , rename x0 , . . . , xn , y0 , . . . , yr−1 as z0 , . . . , zn+r . Then there are real numbers α1 , . . . , αr independent of f such that Qn (f ; x0 , . . . , xn ) =

r X

αi Qn (f ; zi , . . . , zi+n )

(1.1.15)

i=0

and

r X

αi = 1.

(1.1.16)

i=0

If r = 1, the prooffollows from Theorem 1.1.2 with α0 = (y0 −x0 ) (xn − x0 ) and α1 = (xn − y0 ) (xn − x0 ). We suppose that the theorem is true for r = r0 and consider r0 + 1 distinct points y0 , . . . , yr0 different from x0 , . . . , xn . Let the points x0 , . . . , xn , y0 , . . . , yr0 −1 be renamed as z0 , . . . , zn+r0 . Since the theorem is true for r = r0 , there are numbers α0 , . . . , αr0 such that Qn (f ; x0 , . . . , xn ) =

r0 X

αi Qn (f ; zi , . . . , zi+n )

(1.1.17)

i=0

and

r0 X

αi = 1.

(1.1.18)

i=0

Now applying (1.1.12) replacing x0 , . . . , xn , y by zr0 , . . . , zr0 +n , yr0 , respectively, we have Qn (f ; zr0 , . . . , zr0 +n ) =

y r0 − z r0 Qn (f ; zr0 , . . . , zr0 +n−1 , yr0 ) (1.1.19) zr0 +n − zr0 zr +n − yr0 + 0 Qn (f ; yr0 , zr0 , . . . , zr0 +n ), zr0 +n − zr0

and so from (1.1.17) and (1.1.19) Qn (f ; x0 , . . . , xn ) =

rX 0 −1

αi Qn (f ; zi , . . . , zi+n ) + αr0 Qn (f ; zr0 , . . . , zr0 +n )

i=0

=

rX 0 −1

αi Qn (f ; zi , . . . , zi+n ) + αr0

i=0

+ αr0

y r0 − z r0 Qn (f ; zr0 , . . . , zr0 +n−1 , yr0 ) zr0 +n − zr0

zr0 +n − yr0 Qn (f ; yr0 , zr0 +1 , . . . , zr0 +n ). zr0 +n − zr0

(1.1.20)

Higher Order Derivatives y

5

−z

z

−y

r0 r0 +n r0 0 Now putting βi = αi , 0 ≤ i ≤ r0 −1, βr0 = αr0 zr r+n −zr0 , βr0 +1 = αr0 zr0 +n −zr0 0 we have from (1.1.18)

rX 0 +1 i=0

βi =

rX 0 −1

αi + αr0

i=0

y r0 − z r0 zr +n − yr0 + 0 zr0 +n − zr0 zr0 +n − zr0

=

r0 X

αi = 1. (1.1.21)

i=0

So, renaming the points z0 , . . . , zr0 +n−1 , y0 , zr0 +n as ω0 , . . . , ωn+r0 +1 we have from (1.1.20) Qn (f ; x0 . . . . , xn ) =

rX 0 +1

βi Qn (f ; ωi , . . . , ωi+n ).

(1.1.22)

i=0

The relations (1.1.22) and (1.1.21) prove the theorem for r = r0 + 1 and so the theorem is proved by induction. Remark. If the points y0 , . . . , yr−1 in Theorem 1.1.3 satisfy min xi ≤ yj ≤ max xi , j = 0, . . . , r − 1,

0≤i≤n

0≤i≤n

then all the αj s are positive. This is true for Theorem 1.1.2 and so the remark follows by the induction above; [145].

1.2

General Derivatives of Order n

Let n be a fixed positive integer and let B = {a0 , . . . , an } where a0 , . . . , an are distinct real numbers. Let f : R 7→ R, the general derivative of f at x ∈ R with respect to B is defined by GDn f (x, B) = lim n!Qn (f ; x + ha0 , . . . , x + han ), h→0

(1.2.1)

provided the limit on the right-hand side of (1.2.1) exists. So by (1.1.3) n n! X f (x + ai h) Qn h→0 hn j=0 (ai − aj ) i=0

GDn f (x, B) = lim

j6=i

n 1 X Ai f (x + ai h), = lim n h→0 h

(1.2.2)

i=0

where Ai = Qn

j=0 j6=i

n! . (ai − aj )

(1.2.3)

6


By (1.1.3) and (1.1.11) n X

Ai api = n!

i=0

(

= Special Cases

n X i=0

0,

api = n!Qn (xp ; a0 , . . . , an ) j=0 (ai − aj )

Qn

(1.2.4)

j6=i

if p = 0, . . . , n − 1,

n!, if p = n.

Case I Let ai = i, 1 = 0, . . . , n. Then

n Y

n Y

(ai − aj ) =

(i − j)

j=0 j6=i

j=0 j6=i

= i(i − 1) · · · 1(−1)(−2) · · · (i − n) = (−1)(n−i) i!(n − i)! and so Ai =

(1.2.5)

n! (n−i) n . = (−1) i (−1)(n−i) i!(n − i)!

In this case, we get the unsymmetric Riemann derivative of f at x of order n: n n 1 X 1 X (n−i) n RDn f (x) = lim n f (x + ih). Ai f (x + ai h) = lim n (−1) h→0 h h→0 h i i=0 i=0 (1.2.6) For n = 1, this is f (x + h) − f (x) RD1 f (x) = lim , (1.2.7) h→0 h the ordinary derivative of f at x. For n = 2, we get f (x + 2h) − 2f (x + h) + f (x) . h→0 h2

RD2 f (x) = lim

(1.2.8)

Example. If f (x) = |x|, then RD2 f (x) = 0 for all x, although RD1 f (0) does not exist. So, if the nth derivative RDn f (x) exists, the previous derivatives may not exist. Case II. Let ai = 2i − n, i = 0, 1, . . . , n. Then, as in (1.2.5), n Y

(ai − aj ) =

j=0 j6=i

and thus Ai =

n Y j=0 j6=i

(2i − 2j) = 2

n

n Y

(i − j)

j=0 j6=i

= 2n (−1)n−i i!(n − i)!

(1.2.9)

n! 1 n−i n . (−1) = 2n (−1)n−i i!(n − i)! 2n i

(1.2.10)


7

In this case, we get the symmetric Riemann derivative of f at x of order n: n 1 Ai f (x + ai h) (1.2.11) h→0 hn i=0 n 1 (n−i) n = lim f (x + 2ih − nh). (−1) h→0 (2h)n i i=0

RDns f (x) = lim

For n = 1, this is RD1s f (x) = lim

h→0

f (x + h) − f (x − h) , 2h

(1.2.12)

which is the first symmetric derivative of f at x. For n = 2, this is f (x + 2h) − 2f (x) + f (x − 2h) , h→0 (2h)2

RD2s f (x) = lim

(1.2.13)

which is the second symmetric derivative of f at x. Example. If f (x) = |x|, then RD1s f (x) = 1 for x > 0, RD1s f (x) = −1 for x < 0, RD1s f (0) = 0 while RD2s f (x) = 0 for x = 0, and RD2s f (0) = ∞. If the limits in (1.2.1) and consequently in (1.2.6) and (1.2.11) do not exist then the corresponding upper and lower limits are denoted s by GDn f (x; B), GDn f (x; B), RDn f (x), RDn f (x), RDn f (x), RDsn f (x), respectively. The unilateral notion of these are obtained by suitably restricting h while taking the limits. The derivatives RDn f (x) and RDns f (x) also can be defined with the help of difference operators. For RDn f (x), consider the operator, defined by induction Δ1 (f, x, h) = f (x + h) − f (x)

(1.2.14)

Δk (f, x, h) = Δk−1 (f, x + h, h) − Δk−1 (f, x, h), k = 2, 3, · · · . Then, Δn (f, x, h) =

n i=0

(−1)n−i

n f (x + ih), n = 1, 2, 3, · · · . i

(1.2.15)

8 Then,

Higher Order Derivatives For (1.2.15) is true for n = 1. Suppose (1.2.15) is true for n = r.

∆r+1 (f, x, h) = ∆r (f, x + h, h) − ∆r (f, x, h) r r X X r−i r r−i r f x + ih f x + (i + 1)h − (−1) (−1) = i i i=0 i=0 r r+1 X X r r f x + ih f (x + jh)− (−1)r−i (−1)r+1−j = i j−1 i=0 j=1 r X r r + f (x + jh) + (−1)r+1 f (x) (−1)r+1−j = f x + (r + 1)h) + j − 1 j j=1 r X r+1−j r + 1 f (x + jh) + (−1)r+1 f (x) (−1) = f x + (r + 1)h) + j j=1 =

r+1 X j=0

r+1−j

(−1)

r+1 f (x + jh), j

and so (1.2.15) is true for n = r + 1, proving (1.2.15). So, (1.2.6) can be written as RDn f (x) = lim

h→0

∆n (f, x, h) hn

(1.2.16)

where ∆n (f, x, h) is given by (1.2.14). For RDns , consider the operator, defined by induction 1 1 ∆s1 (f, x, h) = f x + h − f x − h ; (1.2.17) 2 2 1 1 ∆sk (f, x, h) = ∆sk−1 (f, x + h, h) − ∆sk−1 (f, x − h, h), k = 2, 3, · · · . 2 2 Then the operators ∆k (·, ·, ·) and ∆sk (·, ·, ·) have the following relation: 1 ∆sk (f, x, h) = ∆k (f, x − kh, h). 2

(1.2.18)

In fact, (1.2.18) is true for k = 1. Suppose that (1.2.18) is true for k = r. Then by (1.2.14), (1.2.17) and by the induction hypothesis ∆r+1 (f, x − 21 (r + 1)h, h) = ∆r (f, x − 21 (r − 1)h, h) − ∆r (f, x − 12 (r + 1)h, h) = ∆sr (f, x + 12 h, h) − ∆sr (f, x − 21 h, h) = ∆sr+1 (f, x, h), showing that (1.2.18) is true for k = r + 1 and therefore it is true for all k.


9

Hence, by (1.2.18) and (1.2.15) ∆sn (f, x, h) =

n X

(−1)n−i

i=0

n f (x − 21 nh + ih). i

(1.2.19)

∆sn (f, x, h) . h→0 hn

(1.2.20)

Hence, (1.2.11) can be written RDns f (x) = lim

If the limit in (1.2.16) does not exist we write RDn f (x) = lim sup h→0

∆n (f, x, h) ∆n (f, x, h) and RDn f (x) = lim inf (1.2.21) n h→0 h hn

and call them the upper and lower unsymmetric Riemann derivatives of f at x of order n, respectively. If the limit (1.2.20) does not exist, we write s

RDn f (x) = lim sup h→0

∆sn (f, x, h) ∆sn (f, x, h) s f (x) = lim inf and RD (1.2.22) n h→0 hn hn

and call them the upper and lower symmetric Riemann derivatives of f at x of order n, respectively.

1.3

Generalized Riemann Derivatives of Order n

Let n be a fixed positive integer and ℓ a nonnegative integer. Let S = {a0 , . . . , an+ℓ ; A0 , . . . , An+ℓ ; L} be a system of real numbers such that ai 6= aj for i 6= j, i, j = 0, . . . , n + ℓ and L 6= 0 satisfying n+ℓ X i=0

Ai api

=

(

0, for p = 0, . . . , n − 1, L, for p = n.

(1.3.1)

If f is defined in some neighbourhood of a point x ∈ R, then the generalized Riemann derivative of f at x of order n determined by S is defined to be n+ℓ n! X Ai f (x + ai h), h→0 Lhn i=0

GRDn f (x, S) = lim

(1.3.2)

provided the limit exists. If in particular ℓ = 0 and L = n!, then the derivative GRDn f (x, S) becomes the derivative GDn f (x, B) defined in (1.2.1). To see this (although in

10


this case the expressions (1.2.2) and (1.3.2) are the same), we verify that the constants Ai in (1.3.2) satisfy (1.2.3). If a0 , . . . , an are known, then the relations (1.3.1) determine the Ai uniquely. The coefficient matrix api , 0 ≤ i ≤ n, 0 ≤ p ≤ n, of the system of linear equations in Q (1.3.1) is a van der Monde matrix whose determinant is given by det api = i<j (aj −ai ). If cnr is the co-factor of anr in det api , then cnr Q′ is (−1)n+r times a van der Monde determinant and so cnr = (−1)n+r i<j (aj − Q Q Q det api = ni=0 (ar − ai ). ai ) where ′ means ar never occurs in ′ . Thus i6=r cnr Hence n n hY i−1 hY i−1 Ar = L (ar − ai ) = n! (ar − ai ) , i=0 i6=r

i=0 i6=r

which is the same as (1.2.3).

Thus, in the special case when ℓ = 0 and L = n! the relations (1.2.4) and (1.3.1) agree and so the derivatives GRDn f and GDn f , which were defined in (1.3.2) and (1.2.2), respectively, are the same. The nonnegative number ℓ considered above is called the excess if ℓ 6= 0. The excess terms were first considered by Ash [2]. Excess terms are sometimes necessary in numerical analysis and elsewhere. If the limit in (1.3.2) does not exist, the upper and lower generalized Riemann derivatives of f at x of order n with respect to S are defined by GRDn f (x, S) = lim sup h→0

and GRDn f (x, S) = lim inf h→0

n+ℓ n! X Ai f (x + ai h) Lhn i=0 n+ℓ n! X Ai f (x + ai h). Lhn i=0

Now, consider the following system of n + 1 linear equations in the n + 1 unknowns A0 , . . . , An : n X

i=0 n X

Ai = 0, 2(i−1)p Ai = 0 for p = 1, . . . , n − 1,

(1.3.3)

i=1

An = 1.

The coefficient matrix of the system of linear equations (1.3.3) is nonsingular and so A0 , . . . P , An can be determined uniquely. Pn n Note that i=1 2(i−1)n Ai 6= 0. For if i=1 2(i−1)n i = 0, then this equaPA n tion together with the n − 1 equations in (1.3.3), i=1 2(i−1)p Ai = 0, p =


11

1, . . . , n − 1 form the system of n linear homogeneous equations in the n unknowns A1 , . . . , An and since the determinant of the coefficient matrix of this last system does not vanish, Ai = 0, i = 1, . . . , n, which is a contradiction because An = 1. Let n X 2(i−1)n Ai . (1.3.4) L= i=1

Now consider the system of real numbers

S = {0, 1, 2, 22, . . . , 2n−1 ; A0 , . . . , An ; L}.

(1.3.5)

Then this system satisfies the conditions (1.3.1) with ℓ = 0. So, the system (1.3.5) will give a derivative of order n as defined in (1.3.2) with ℓ = 0. This derivative will be called the MZ-derivative after the names of its inventors [106] e n f . That is, by (1.3.2), and will be denoted by D e n f (x) = lim D

h→0

h

n X n! i−1 A f (x + 2 h) . A f (x) + i 0 (i−1)n A i i=1 2 i=1

Pn n

(1.3.6)

e n f also can be defined with the help of difference operators. The derivative D Consider the difference operator, defined by induction ˜ 1 (f, x, h) = f (x + h) − f (x) ∆ (1.3.7) k−1 ˜ ˜ ˜ ∆k (f, x, h) = ∆k−1 (f, x, 2h) − 2 ∆k−1 (f, x, h), k = 2, 3, · · · .

Then, for any fixed k, ˜ k (f, x, h) = ∆ ˜ k−1 (f, x, 2h) − 2k−1 ∆ ˜ k−1 (f, x, h) ∆ 2 k−2 ˜ k−2 (f, x, 2 h) − 2 ˜ k−2 (f, x, 2h) = ∆ ∆ h i ˜ k−2 (f, x, 2h) − 2k−2 ∆ ˜ k−2 (f, x, h) − 2k−1 ∆

(1.3.8)

··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· ··· = αk f (x + 2k−1 h) + αk−1 f (x + 2k−2 h) + · · · + α1 f (x + h) + α0 f (x),

where αk = 1 and αi depends on i and k only. We show that the αi s satisfy the conditions k X

k X

αi = 0,

(1.3.9)

i=1

i=0

2(i−1)s αi = 0, for s = 1, . . . , k − 1.

We first claim that if f has continuous kth derivative, then ˜ s (f, x, h) ∆ exists, for s = 1, . . . , k. h→0 hs lim

(1.3.10)

12


Clearly (1.3.10) is true for s = 1. So, suppose 1 ≤ r < k and that (1.3.10) is true for s = r. By L’Hˆ opital’s rule r X (r + 1)! h hj (j) i f (x + h) − f (x) − f (x) = f (r+1) (x), r+1 h→0 h j! j=1

lim

and, so f (x + h) = f (x) +

r+1 j X h j=1

j!

f (j) (x) + o(hr+1 ).

(1.3.11)

Since (1.3.8) is true for all k, putting k = r + 1 in (1.3.8) and substituting in (1.3.11) we have ˜ r+1 (f, x, h) = α0 f (x) + ∆ = α0 f (x) +

=

r+1 X

r+1 X

i=1 r+1 X

αi f (x + 2i−1 h)

(1.3.12)

r+1 h i X (2i−1 h)j (j) f (x) + o(hr+1 ) αi f (x) + j! j=1 i=1

αi f (x) +

r+1 j X h j=1

i=0

j!

f (j) (x)

r+1 X

αi 2(i−1)j + o(hr+1 ).

i=1

By the induction hypothesis, (1.3.10) is true for s = r and, therefore, since ˜ r (f, x, 2h) ∆ ˜ r (f, x, h) i ˜ r+1 (f, x, h) 2r h ∆ ∆ , = − hr+1 h (2h)r hr

˜ r+1 (f, x, h) = o(hr ). Hence, from (1.3.12) we have using (1.3.10) that ∆ r+1 X

αi f (x) +

i=0

r+1 j X h j=1

j!

f (j) (x)

r+1 X

αi 2(i−1)j = o(hr ).

(1.3.13)

i=1

Since (1.3.13) is true for all functions f that have a continuous kth derivative, (1.3.13) gives r+1 X

αi = 0;

i=0

r+1 X

αi 2(i−1)j = 0, for j = 1, . . . , r.

(1.3.14)

i=1

So, (1.3.12) and (1.3.14) imply ˜ r+1 (f, x, h) = ∆

r+1 X hr+1 (r+1) f (x) αi 2(i−1)(r+1) + o(hr+1 ), (r + 1)! i=1

showing that r+1 X ˜ r+1 (f, x, h) ∆ 1 (r+1) αi 2(i−1)(r+1) . = f (x) h→0 hr+1 (r + 1)! i=1

lim

(1.3.15)


13

Clearly (1.3.15) shows that (1.3.10) is also true for s = r + 1. So, by induction, (1.3.10) is true for all s = 1, . . . , k, whenever f has a continuous kth derivative. Consequently, it follows from (1.3.14) that (1.3.9) is true. ˜ n (f, x, h) where f is any function. Then from (1.3.8) and Now consider ∆ (1.3.9) ˜ n (f, x, h) = αn f (x + 2n−1 h) + αn−1 f (x + 2n−2 h) + · · · + α1 f (x + h) + αo f (x) ∆ (1.3.16) where n X

αi = 0;

n X

αi 2(i−1)s = 0, for s = 1, . . . , n − 1;

αn = 1. (1.3.17)

i=1

i=0

The system of linear equations (1.3.3) and (1.3.17) are the same. Since the coefficient matrix of this system is nonsingular it will give a unique solution and, therefore, αi = Ai for i = 0, 1, . . . , n. So, from (1.3.6) and (1.3.16) ˜ n (f, x, h) n! ∆ . (i−1)n hn αi i=1 2

e n f (x) = lim Pn D h→0

(1.3.18)

e n f (x) can be defined with The relation (1.3.18) shows that this derivative D the help of the difference operators (1.3.7). The relation between ∆k and ∆sk being given in (1.2.18), we now show ˜ k can be expressed in terms of ∆s . that ∆ k Theorem 1.3.1 For each positive integer k ≥ 2 there are constants β0 , β1 , . . . , βλ with βλ = 1, where λ = 2k−1 − k, such that ˜ k (f, x, h) = ∆

λ X

βi ∆sk (f, x + 12 kh + ih, h).

(1.3.19)

i=0

Since ∆sk (f, x + 12 kh + ih, h) =

k X

(−1)k−j

j=0

=

k X j=0

(−1)k−j

k 1 f x + 21 kh + ih + jh − kh j 2

k f x + (i + j)h , j

∆sk (f, x + 12 kh + ih, h) is a linear form in the variables f (x + ih), f x + (i + 1)h , . . . , f x + (i + k)h .

(1.3.20)

Let Li denote this linear form. Since k ≥ 2 there are members of (1.3.20) having the form f (x + 2s−1 h); s is a positive integer. Let L′i denote the linear form obtained from Li by omitting those variables not of the form f (x+2s−1 h).

14


Considering all the Li , i = 0, 1, . . . , λ, the variables, 2k−1 +1 in number, are by (1.3.20) f (x+ph), p = 0, 1, . . . , 2k−1 . Let L′′i = Li −L′i . Since the total number of linear forms Li is λ + 1, that is also the total number of the linear forms L′i and L′′i . Considering all the linear forms L′i , i = 0, 1, . . . , λ, the variables are f (x + ph), p = 0, 1, 2, 22, . . . , 2k−1

(1.3.21)

of number k + 1. Therefore, considering all the linear forms L′′i , i = 0, 1, . . . , λ, the total number of variables is (2k−1 + 1) − (k + 1) = λ. Since the total number of linear forms L′′i is λ + 1, we can find constants c0 , c1 , . . . , cλ , not all of which are zero, such that c0 L′′0 + · · · + cλ L′′λ = 0 and, hence, c0 L0 + c1 L1 + · · · + cλ Lλ = c0 L′0 + c1 L′0 + · · · + cλ L′λ .

(1.3.22)

It follows from (1.3.20) that the variable f (x + 2k−1 h) occurs only in Lλ and so only in L′i . Therefore, from (1.3.22) cλ 6= 0, and so again from (1.3.22) c0 c1 cλ−1 c0 c1 cλ−1 ′ L0 + L1 + · · · + Lλ−1 + Lλ = L′0 + L′1 + · · · + L + L′λ . cλ cλ cλ cλ cλ cλ λ−1 (1.3.23) Now we show that, if f has a kth derivative at x, then for all γ ∆sk (f, x + γh, h) = hk f (k) (x) + o(hk ).

(1.3.24)

In fact, we have as in (1.3.11) f (x + h) = f (x) +

k X hj j=1

j!

f (j) (x) + o(hk ),

and so applying the well known relation ( k X 0, if p = 0, 1, . . . , k − 1, p k−i k i = (−1) i k!, if p = k, i=0 we get ∆sk (f, x

k f (x + γh + ih − 12 kh) (−1) + γh, h) = i i=0 k k X X (γ + i − 12 k)j hj (j) k−i k f (x) + (−1) = f (x) + o(hk ) i j! j=1 i=0 k X

=

k X i=0

k−i

k−i

(−1)

X k (γ + i − 12 k)j hj (j) k f (x) + o(hk ) j! i j=1

Higher Order Derivatives =

k X hj j=1

=

k X j=1

=

j!

f

(j)

(x)

k−i

(−1)

i=0

j

h (j) f (x) j!

hk (k) f (x) k!

k X

ℓ=0

k X k ℓ k i + o(hk ) (−1)k−i (γ − 21 k)k−ℓ i ℓ i=0

k X ℓ=0

X j k j ℓ i (γ − 12 k)j−ℓ + o(hk ) i ℓ

k X j k−i k j−ℓ 1 iℓ + o(hk ) (−1) (γ − 2 k) i ℓ i=0

j X ℓ=0

15

hk (k) = f (x)k! + o(hk ) = hk f (k) (x) + o(hk ), k!

proving (1.3.24). So Li = hk f (k) (x) + o(hk ) for all i, i = 0, 1, . . . , λ. Hence, c0 c1 cλ−1 L0 + L1 + · · · + Lλ−1 + Lλ cλ cλ cλ i hc c1 cλ−1 0 + + ··· + + 1 + o(hk ). = hk f (k) (x) cλ cλ cλ

(1.3.25)

Since the variables of L′0 , L′1 , . . . , L′λ are as in (1.3.21),

c0 ′ c1 cλ−1 ′ L + L′ + · · · + L + L′λ cλ 0 cλ 1 cλ λ−1 = bk f (x + 2k−1 h) + bk−1 f (x + 2k−2 h) + · · · + b1 f (x + h) + b0 f (x) (1.3.26) where the bi are constants with bk = 1. If f has a kth derivative, then we have as in (1.3.11) f (x + 2i−1 h) = f (x) +

k X (2i−1 h)j j=1

j!

f (j) (x) + o(hk ),

and so from (1.3.26) c0 ′ c1 cλ−1 ′ L + L′ + · · · + L + L′λ cλ 0 cλ 1 cλ λ−1 k X bi f (x + 2i−1 h) = b0 f (x) + i=1

k k h i X X (2i−1 h)j (j) bi f (x) + = b0 f (x) + f (x) + o(hk ) j! j=1 i=1

= f (x)

k X i=0

bi +

k X hj j=1

j!

f (j) (x)

k X

bi 2(i−1)j + o(hk ).

(1.3.27)

i=1

Pk Pk From (1.3.23), (1.3.25) and (1.3.27): i=0 bi = 0 and i=1 bi 2(i−1)j = 0 for j = 1, 2, , . . . , k − 1. Since bk = 1, the bi satisfy the equations (1.3.17) if αi in

16


(1.3.17) is replaced by bi , 1 = 0, 1, . . . , k. So from (1.3.16), the right-hand side ˜ k (f, x, h). Therefore, from (1.3.26) and (1.3.23) of (1.3.26) is ∆ ˜ k (f, x, h) = c0 L0 + c1 L1 + · · · + cλ−1 Lλ−1 + Lλ . ∆ cλ cλ cλ

(1.3.28)

Putting βi = ci /cλ , i = 0, 1, . . . , λ − 1 and βλ = 1 and noting that Li = ∆sk (f, x + 12 kh + ih, h) (1.3.28) gives (1.3.19) and completes the proof. The above theorem is due to Marcinkiewicz and Zygmund; see Lemma 1.2 of [106, p. 11], the proof of the theorem being simplified here.

1.4

Peano Derivatives

1.4.1

Bilateral Peano Derivatives

Let f be defined in some neighbourhood of the point x on the real line and let k be a positive integer. If there is a polynomial in t, Q(t) = Qx (t), of degree at most k such that f (x + t) = Q(t) + o(tk ), as t → 0,

(1.4.1)

then f is said to have a Peano derivative at x of order k, and if ak /k! is the coefficient of tk in Q(t), then ak is called the kth Peano derivative of f at x and is denoted by f(k) (x). The Peano derivative, if it exists, is unique. For if Q1 (t) is another polynomial of degree at most k satisfying (1.4.1), then from (1.4.1) Q(t) = Q1 (t)+o(tk ), as t → 0, which implies that for Q(t) − Q1 (t) = 0, and so the polynomials p = 0, 1, . . . , k we have that limt→0 tp Q(t) and Q1 (t) are identical. It is also clear that, if k ≥ 2 and if f(k) (x) exists, then f(i) (x) exists, 0 ≤ i < k. For if ai /i! is the coefficient of ti , then (1.4.1) can be written k X aj j f (x + t) = S(t) + o(t ), as t → 0, where S(t) = Q(t) − t , j! j=i+1 i

and S is a polynomial of degree at most i; and so f(i) (x) exists and ai = f(i) (x). If a0 is the constant term in Q(t), then writing f(0) (x) = a0 we have from (1.4.1) k X ti f (x + t) = f(i) (x) + o(tk ), as t → 0. (1.4.2) i! i=0


17

So f(0) (x) = f (x) if f is continuous at x, otherwise it is limt→0 f (x + h). Thus if f is continuous at x, then the first Peano derivative f(1) (x) is the first ordinary derivative f ′ (x). Supposing the existence of f(k) (x) we write k X ti tk+1 γk+1 (f ; x, t) = f (x + t) − f(i) (x). (k + 1)! i! i=0

(1.4.3)

The upper and lower Peano derivates of f at x, of order k + 1 are obtained by taking the upper and lower limits of γk+1 (f, x, t) as t → 0; they are written f (k+1) (x), f (k+1) (x), respectively. +

−

The unilateral Peano derivates, f (k+1) (x), f (k+1) (x), f + (x), f − (x) (k+1) (k+1) of order k + 1 are obtained by suitably restricting t; i.e., t → 0+ or t → 0−. Clearly h + i − f (k+1) (x) = max f (k+1) (x), f (k+1) (x) , h i − f (k+1) (x) = min f + (x), f (x) . (k+1) (k+1) If f (k+1) (x) = f (k+1) (x), then this common value, possibly infinite, is the Peano derivative f(k+1) (x) of f at x of order k + 1. Note that if f(k+1) (x) is finite, then this definition of f(k+1) (x) agrees with that given above, as we now show. If f(k+1) (x) is finite, then since limt→0 γk+1 (f ; x, t) = f(k+1) (x) γk+1 (f ; x, t) = f(k+1) (x) + o(1) as

t → 0,

and so from (1.4.3) f (x + t) =

k X ti i=0

i!

f(i) (x) + o(tk+1 ) as t → 0,

which agrees with (1.4.2). The infinite Peano derivative f(k+1) (x), if it exists, is also unique. For if f(k+1) (x) exists, then f(0) (x), f(1) (x), . . . , f(k) (x) exist and are finite, and from previous observations they are unique and therefore, by definition, f(k+1) (x) is unique. Theorem 1.4.1 If the ordinary kth derivative, f (k) (x0 ), of f at x0 exists, possibly infinite, then the kth Peano derivative, f(k) (x0 ), also exists and they are equal. For k = 1 the result is trivially true since the first derivative and the first Peano derivative are the same. So, we suppose that k ≥ 2. Since f (k) (x0 )

18


exists then f (k−1) (x0 ) exists, finitely in some neighbourhood of x0 , and then f (k−2) (x0 ) exists and is continuous in some neighbourhood of x0 . Let F (t) = f (x0 + t) −

k−1 i X t i=0

tk G(t) = . k!

i!

f (i) (x0 ),

Then since F (0) = F ′ (0) = · · · = F (k−2) (0) = 0 and G(0) = G′ (0) = · · · = G(k−2) (0) = 0 for any t > 0, there is, by the mean value theorem, an ξ such that 0 < ξ < t and F (k−1) (ξ) F (t) = (k−1) . (1.4.4) G(t) G (ξ) To see this, for x in some neighbourhood of 0 write Φ(x) = F (x) +

k−2 X i=1

i X (t − x)i (t − x)i (i) F (t) h G(x) + F (x) − G(i) (x) . i! G(t) i! i=1 k−2

Observe that Φ(0) = Φ(t) = 0 and Φ′ (x) =

F (t) (t − x)k−2 (k−1) (t − x)k−2 (k−1) F (x) − G (x); (k − 2)! G(t) (k − 2)!

and so applying the mean value theorem to Φ, Φ′ (ξ) = 0, 0 < ξ < t, which gives (1.4.4). Since F (k−1) (ξ) = f (k−1) (x0 + ξ) − f (k−1) (x0 ) from (1.4.3) and (1.4.4), γk (f ; x0 , t) =

F (t) f (k−1) (x0 + ξ) − f (k−1) (x0 ) = . G(t) ξ

(1.4.5)

Letting t → 0+, f (k−1) (x0 + ξ) − f (k−1) (x0 ) . ξ→0+ ξ

lim γk (f ; x0 , t) = lim

t→0+

Similarly taking t < 0 and proceeding as above, we ultimately get lim γk (f ; x0 , t) = lim

t→0

completing the proof.

ξ→0

f (k−1) (x0 + ξ) − f (k−1) (x0 ) , ξ

Example. The converse of the above theorem is not true except for k = 1 when the Peano and ordinary derivative are the same. If n is any positive integer, let ( 1 xn+1 sin n , if x 6= 0, (1.4.6) f (x) = x 0, if x = 0.


19

Then f ′ exists everywhere and ( 1 1 (n + 1)xn sin n − n cos n , if x 6= 0, ′ f (x) = x x 0, if x = 0. f ′ is not differentiable at the origin and so f (k) (0) does not exist for k ≥ 2. But n X xi 1 f (x) = 0 + xn x sin n for x 6= 0. i! x i=0

1 → 0 as x → 0, it follows from (1.4.1) that f(n) (0) exists and xn that f(k) (0) = 0, for k = 2, . . . , n. Since (x sin

1.4.2

Unilateral Peano Derivatives

Unilateral Peano derivatives can be defined as in section 4.1. We need f to be defined on one side of x, say in some right neighbourhood of x, and (1.4.1) to be replaced by f (x + t) = Q(t) + o(tk ) as t → 0+.

(1.4.7)

In this case, f is said to have a right-hand Peano derivative at x of order k, denoted by Rf(k) (x). The relations (1.4.2) and (1.4.3) now take the forms f (x + t) =

k X ti i=0

i!

Rf(i) (x) + o(tk ), as t → 0+,

k X tk ti + γk+1 (f ; x, t) = f (x + t) + Rf(i) (x), t > 0. (k + 1!) i! i=0

(1.4.8)

(1.4.9)

+ The upper and lower limits of γk+1 (f, x, t) are called the right upper and lower Peano derivates of f at x of order k + 1, written Rf (k+1) (x), Rf (k+1) (x),

respectively. If Rf (k+1) (x) = Rf (k+1) (x), then this common value, possibly infinite, is the right Peano derivative of f at x of order k + 1. + Note that the right upper derivate f (k+1) (x) defined in section 4.1 is, in general, different from Rf (k+1) (x) except for k = 0, since for k 6= 0 we are assuming the existence of Rf(k) (x) only and not f(k) (x) and since the existence of f(k) (x) implies that of Rf(k) (x) and not the converse. In a similar way one can define left derivatives and derivates that are denoted by Lf(k) (x) and Lf (k) (x), Lf (k) (x), respectively. Note that analogues of Theorem 4.1 hold for Rf(k) (x), Lf(k) (x), and that if these two derivatives exist and are equal the bilateral Peano derivative, f(k) (x) may not exist.

20


Example. Consider for k ≥ 2, k+1 x , f (x) = xk−1 ,

for x ≥ 0, for x < 0.

Then f is continuous and Rf(k) (0) = 0, but f(k) (0) does not exist. In fact, if +

(0) = 0 and f(k) (0) exists, then f(k−1) (0) also exists. But f (k−1) (0) = f + (k−1) −

f (k−1) (0) = f − (0) = (k − 1)!. (k−1) However, we have the following result: Theorem 1.4.2 If Rf(i) (x) = Lf(i) (x) for i = 1, 2, . . . , k, then f(k) (x) exists. Clearly the result is true for k = 1. Suppose that is true for all k, 1 ≤ k ≤ r. Let Rf(i) (x) = Lf(i) (x) for i = 1, 2, . . . , r + 1. Since the result holds if 1 ≤ k ≤ r, then f(r) (x) exists and Rf(i) (x) = Lf(i) (x) = f(i) (x) for i = 1, 2, . . . , r. Then by (1.4.3) and (1.4.3+) +

+ (x). Rf(r+1) (x) = lim γr+1 (f, x, t) = lim γr+1 (f, x, t) = f (r+1) (x) = f + (r+1) t→0+

t→0+

−

Similarly Lf(r+1) (x) = f (r+1) (x) = f − (x). Since Rf(r+1) (x) = (r+1) Lf(r+1) (x), f(r+1) (x) exists and the result is true for k = r + 1, completing the proof by induction.

1.4.3

Peano Boundedness

If in (1.4.1) the condition o(tk ) as t → 0 is replaced by the less restrictive condition O(tk ) as t → 0 then f is said to be Peano bounded of order k at x. It is clear that if f is Peano bounded of order k at x, then f need not be Peano differentiable of order k at x. Consider the function f defined in the example in Section 4.1, (1.4.6). We have seen in that example that f(n) (0) exists and f(i) (0) = 0 for i = 1, 2, . . . , n. Clearly this function is Peano bounded of order n+1 at 0, but the Peano derivative of order n+1 at 0 does not exist. However, we have: Theorem 1.4.3 If f has finite upper and lower Peano derivates of order k at x, then f is Peano bounded of order k at x. Since f (k) (x) and f (k) (x) are finite, we have that γk (f ; x, t) = O(1) as t → 0. Hence k−1 X ti f(i) (x) + O(tk ) as t → 0. f (x + t) = i! i=1 This proves the theorem.


21

Theorem 1.4.4 If f is Peano bounded of order k at x, then f has a finite Peano derivative of order k − 1 at x and has finite upper and lower Peano derivates of order k at x.

Since f is Peano bounded of order k at x, f (x + t) = Q(t) + O(tk ) as t → 0,

where Q(t) is a polynomial of degree at most k. Let Q(t) = Q1 (t) + bk tk say. Then, from (1.4.6)

(1.4.6) Pk

i=0 bi t

f (x + t) = Q1 (t) + bk tk + O(tk ) as t → 0.

i

=

(1.4.7)

Since O(tk ) implies o(tk−1 ) as t → 0, f(k−1) (x) exists and f(i) (x) = i!bi , i = 0, 1, . . . , k − 1. Using (1.4.3) we have from (1.4.7) i X ti k! h f (x + t) − f(i) (x) = O(1) as t → 0, k t i! i=0 k−1

γk (f ; x, t) =

and, hence, f (k) x and f (k) x are finite.

Remark. Unilateral Peano boundedness may similarly be defined and an analogue of Theorem 1.4.3 obtained.

1.4.4

Generalized Peano Derivatives

Let f be continuous in some neighbourhood of x ∈ R. Then f is said to have a generalized Peano derivative at x of order n if there is a nonnegative integer k such that a kth primitive of f in that neighbourhood has a finite Peano derivative at x of order k+n. If f (−k) denotes the kth primitive of f and if (f (−k) )(n+k) is the (k + n)th Peano derivative of f (−k) , then (f (−k) )(n+k) (x) is called the generalized Peano derivative of f at x of order n and is denoted by f[n] (x). That is, f[n] (x) = (f (−k) )(n+k) (x). (1.4.8) This definition does not depend on k. For if k1 < k2 and if (1.4.8) is true for k = k1 then it is also true for k = k2 by L’Hˆ opital’s rule and so k may be taken to be the smallest of all k for which (1.4.8) holds. Clearly, if f[n] (x), exists then f[i] (x) also exists for i = 1, 2, . . . , n − 1 and f[i] (x) = (f (−k) )(k+i) (x). Now suppose for a fixed positive integer r that f[r−1] (x) exists. Then there is a k such that f[i] (x) = (f (−k) )(k+i) (x) for i = 0, 1, . . . , r − 1. Let

22


Γk,r (f ; x, h) = γk+r (f (−k) ; x, h) k+r−1 i X hi (k + r)! h (−k) (−k) = f (x + h) − (f ) (x) (i) hk+r i! i=0

i X hi X hk+i (k + r)! h (−k) (−k+i) f (x + h) − f (x) − f (x) . [i] hk+r i! (k + i)! i=0 i=0 k−1

=

r−1

Define Uk = Uk,r (f ; x) = lim sup Γk,r (f ; x, h), h→0

Lk = Lk,r (f ; x) = lim inf Γk,r (f ; x, h). h→0

If k1 < k2 then, by the mean value theorem, Γk2 ,r (f ; x, h) = Γk1 ,r (f ; x, θh), 0 < θ < 1, and so Uk2 ,r (f ; x) = lim sup Γk2 ,r (f ; x, h) ≤ lim sup Γk1 ,r (f ; x, h) = Uk1 ,r (f ; x), h→0

h→0

and similarly Lk2 ,r (f ; x) ≥ Lk1 ,r (f ; x). Also for arbitrary k1 , k2 , Lk1 ,r (f ; x) ≤ Uk2 ,r (f ; x). In fact if k = max{k1 , k2 }, then Lk1 ≤ Lk ≤ Uk ≤ Uk2 . The upper and lower generalized Peano derivates of f at x of order r are defined by f [r] (x) = inf Uk,r (f ; x), k

f [r] (x) = sup Lk,r (f ; x). k

If f [r] (x) = f [r] (x), then f is said to have a generalized Peano derivative, possibly infinite. It can be shown that if the generalized derivative exists finitely according to this definition, then it exists according to the previous definition; that is, there exists a nonnegative integer k such that (1.4.8) holds. Now if f is continuous in some neighbourhood of x and if the finite Peano derivative f(r−1) (x) exists, then for any k we have, by the mean value theorem, that Γk,r (f ; x, h) = γr (f ; x, θh), 0 < θ < 1, where γr is defined as in (1.4.3). Hence, Uk,r (f, x) = lim sup γr (f ; x, θh) h→0

≤ lim sup γr (f ; x, h) = f (r) (x), h→0

and so f [r] (x) ≤ f (r) (x). Similarly, f [r] (x) ≥ f (r) (x) and, hence, f (r) (x) ≤ f [r] (x) ≤ f [r] (x) ≤ f (r) (x).


23

Thus it follows that if f is continuous in some neighbourhood of x and if f has a Peano derivative at x of order n, then f has generalized Peano derivative at x of order n. The following example shows that the converse is not true. Example. Let f (x) =

(

x2 sin 0,

1 , x2

if x 6= 0, if x = 0.

Then f ′ exists everywhere, but f(2) (0) does not exist; in fact, f (2) (0) = Rx −2, f (2) (0) = 2. We show that f[2] (0) does exist. Let F (x) = f (−1) (x) = 0 f . Then Z Z x 1 ∞ sin u du. F (x) = t2 sin(t−2 ) dt = 2 x−2 u5/2 0 Let x−2 < X < ∞. Then, by the second mean value theorem Z X Z sin u 5 ξ sin u du ≤ 2|x5 | −2 u5/2 du = x −2 x x

and so |F (x)| ≤ |x5 |; and so, F(i) (0) = 0 for i = 1, 2, 3, 4, and so, f[i] (0) exists with value zero for i = 1, 2, 3.

It may be noted that the existence of f(1) (x) does not imply the continuity of f in a neighbourhood of x and so, in general, f(1) (x) may exist and f[1] (x)may not.

1.4.5

Absolute Peano Derivatives

Let f be defined in some neighbourhood N (x0 ) of x0 . The function f is said to have an absolute Peano derivative at x0 if there is a function g and nonnegative integer n such that g has a Peano derivative of order n for all x ∈ N (x0 ) and has a Peano derivative of order n + 1 at x0 and g(n) (x) = f (x), x ∈ N (x0 ). If g(n+1) (x0 ) = A, then A is called the absolute Peano derivative of f at x0 , written f ∗ (x0 ). Consider the case n = 0: If f has an ordinary derivative at x0 , then f ′ (x0 ) = f(1) (x0 ), and so take g = f and f ∗ (x0 ) = f ′ (x0 ). However, f ∗ (x0 ) may exist when f ′ (x0 ) does not. Example. Consider f (x) =

(

1 , if x 6= 0, x2 0, if x = 0,

sin

Rx and take g(x) = f (−1) (x0 ) = 0 f . Then by changing the variable and using the second mean value theorem, Z x Z Z 1 ∞ sin u x3 ξ g(x) = sin(t−2 ) dt = sin u du, du = 2 x−2 u3/2 2 x−2 0

24


and so |g(x)| ≤ |x3 | for all x. This implies that g(1) (0) = g(2) (0) = 0. Also 1 g(1) (x) = sin 2 , x 6= 0, and so for all x, g(1) (x) = f (x). Taking n = 1 in the x above definition, f ∗ (0) = g(2) (0) = 0. Note that if f has an nth Peano derivative in N (x0 ) and has an (n + 1)st Peano derivative at x0 , then f(n+1) (x0 ) is the absolute Peano derivative of f(n) at x0 . From the definition, it follows that if f has an absolute Peano derivative at x0 , then the absolute Peano derivative f ∗ (x0 ) is the Peano derivative of suitable order of some function g and, hence, is a generalized Peano derivative at x0 of some function. In fact if we use the Cr P-integral [see Section 8], then since f (x) = g(n) (x) for x ∈ N (x0 ), f is Cn−1 P-integrable in [a, b] ⊂ N (x0 ), x0 ∈ (a, b). After a finite number of integrations we can get a continuous function F defined on [a, b] such that F(n) (x) = f (x) for x ∈ [a, b] and F(n+1) (x0 ) = f ∗ (x0 ).

1.5 1.5.1

Riemann∗ Derivatives Bilateral Riemann∗ Derivatives

Let x0 be a fixed point on R and let x1 , . . . , xk be points in some neighbourhood of x0 such that 0 < |x0 − x1 | < · · · < |x0 − xk |. If Qk (f ; x0 , . . . , xk ) is as defined in (1.1.1) or (1.1.3) and if the iterated limit lim

lim

xk →x0 xk−1 →x0

· · · lim k!Qk (f ; x0 , . . . , xk ), x1 →x0

(1.5.1)

exists, the last limit being finite or infinite, while the inner limits are all finite, then the limit in (1.5.1) is called the Riemann∗ derivative of f at x0 ∗ of order k and is denoted by f(k) (x0 ). It is clear that if the last limit in ∗ (1.5.1) in finite, then all the inner limits are finite. Clearly f(1) (x0 ) is the ′ ordinary first derivative f (x0 ). Taking the upper limit, respectively lower limit, at each stage in (1.5.1), we get the definitions of the upper, respectively ∗ lower, Riemann∗ derivates of f at x0 of order k, which are denoted by f (k) (x0 ) and f ∗(k) (x0 ), respectively. In what follows we shall show that this kth order derivative has the properties of the ordinary derivative. Theorem 1.5.1 If k ≥ 2 and if lim

xk−1 →x0

· · · lim Qk (f ; x0 , . . . , xk ) x1 →x0

exists finitely for some fixed xk , then f is continuous at x0 and lim · · · lim Qi (f ; x0 , . . . , xi )

xi →x0

x1 →x0

exists finitely for i = 1, 2, . . . , k − 1.

(1.5.2)


25

We have from (1.1.3)

Qk (f ; x0 , . . . , xk ) =

(x0 − x1 ) +

k X i=2

Qk

f (x0 ) Qk

j=2 (x0

− xj )

+

(x1 − x0 )

f (xi )

→ j=0 (xi

f (x1 ) Qk

j=2 (x1

− xj )

j6=i

# " 1 f (x1 ) f (x0 ) = − Qk Qk (x0 − x1 ) j=2 (x0 − xj ) j=2 (x1 − xj ) +

k X i=2

− xj )

f (xi )

Qk

→ j=0 (xi

− xj )

.

(1.5.3)

j6=i

The limit in (1.5.2) exists finitely and so all the inner limits are finite and in particular limx1 →x0 Qk (f ; x0 , x1 , . . . , xk ) exists finitely. Since the last summation in (1.5.3) tends to a finite limit as x1 → x0 , we have from (1.5.3) that # " f (x1 ) f (x0 ) 1 − Qk lim Qk x1 →x0 (x0 − x1 ) j=2 (x0 − xj ) j=2 (x1 − xj ) exists finitely. Hence,

lim Qk

x1 →x0

f (x1 )

j=2 (x1

− xj )

= Qk

f (x0 )

j=2 (x0

− xj )

,

which shows that f (x1 ) → f (x0 ) as x1 → x0 ; that is, f is continuous at x0 . Now, since f is continuous at x0 , we have from (1.1.3) that lim Qk−1 (f ; x1 , x2 , . . . , xk ) = Qk−1 (f ; x0 , x2 , . . . , xk ).

x1 →x0

(1.5.4)

Since, see (1.1.1) ,

Qk−1 (f ; x0 , x1 , . . . , xk−1 ) − Qk−1 (f ; x1 , x2 , . . . , xk ) , x0 − xk (1.5.5) it follows from (1.5.2) and (1.5.4) that limx1 →x0 Qk−1 (f ; x0 , x1 , . . . , xk−1 ) exists finitely. If then k = 2, the proof is complete; so suppose that k > 2. Calling the last limit Qk−1 (f ; x0 , x0 , x2 , . . . , xk−1 ) we conclude that Qk (f ; x0 , x1 , . . . , xk ) =

lim Qk−1 (f ; x0 , x2 , . . . , xk ) = Qk−1 (f ; x0 , x0 , x3 , . . . , xk ).

x2 →x0

(1.5.6)

From (1.5.4) and (1.5.6), lim

lim Qk−1 (f ; x1 , x2 , . . . , xk ) = Qk−1 (f ; x0 , x0 , x3 , . . . .xk ).

x2 →x0 x1 →x0

(1.5.7)

26


From (1.5.2), (1.5.5) and (1.5.7), the limx2 →x0 limx1 →x0 Qk−1 (f ; x0 , x1 , . . . , xk−1 ) exists finitely. Writing this limit as Qk−1 (f ; x0 , x0 , x0 , x3 , . . . , xk−1 ), we conclude that lim

lim Qk−1 (f ; x0 , x2 , x3 , . . . , xk ) = Qk−1 (f ; x0 , x0 , x0 , x4 , . . . .xk ).

x3 →x0 x2 →x0

(1.5.8)

From the continuity of f at x0 , we have from (1.5.8) that lim

lim

lim Qk−1 (f ; x1 , x2 , x3 , . . . , xk )

x3 →x0 x2 →x0 x1 →x0

= Qk−1 (f ; x0 , x0 , x0 , x4 , . . . .xk ).

(1.5.9)

So, by (1.5.2), (1.5.5), and (1.5.9), limx3 →x0 limx2 →x0 limx1 →x0 Qk−1 (f : x0 , x1 , x2 , . . . , xk−1 ) exists finitely. Now supposing that limxp →x0 limxp−1 →x0 · · · limx1 →x0 Qk−1 (f :x0 , x1 , . . . , xk−1 ) exists finitely and writing this limit as Qk−1 (f :(x0 , x0 , . . . , x0 , xp+1 , . . . , xk−1 ) we conclude from this that lim

xp+1 →x0

· · · lim Qk−1 (f ; x0 , x2 , x3 . . . , xp+1 , xp+2 , . . . , xk ) x2 →x0

= Qk−1 (f ; x0 , x0 , x0 . . . , x0 , xp+2 , . . . , xk ).

(1.5.10)

From the continuity of f at x0 , we conclude from (1.5.10) that lim

xp+1 →x0

· · · lim Qk−1 (f ; x1 , x2 , x3 . . . , xp+1 , xp+2 , . . . , xk ) x1 →x0

= Qk−1 (f ; x0 , x0 , x0 . . . , x0 , xp+2 , . . . , xk ).

(1.5.11)

From (1.5.2), (1.5.5) and (1.5.11), it follows that lim

xp+1 →x0

· · · lim Qk−1 (f ; x0 , x1 , . . . , xk−1 ) exists finitely. x1 →x0

Continuing this process inductively, we prove that lim

xk−1 →x0

· · · lim Qk−1 (f ; x0 , x1 , . . . , xk−1 ) exists finitely, x1 →x0

thus proving the theorem for i = k−1. Then replacing k by k−1 and repeating the above argument, we get that lim

xk−2 →x0

· · · lim Qk−2 (f ; x0 , x1 , . . . , xk−2 ) exists finitely. x1 →x0

The proof now follows by a repeated application of this argument.

∗ Theorem 1.5.2 Let k ≥ 2 and suppose that f(k) (x0 ) exists, finitely or in∗ finitely; then all the previous derivatives f(i) (x0 ) for i = 1, 2, . . . , k − 1 exist finitely. ∗ Since f(k) (x0 ) exists, it follows that all the inner limits in (1.5.1) are finite and so the condition of Theorem 1.5.1 is satisfied and so, by that theorem, limxi →x0 . . . . . . limx1 →x0 i!Qi (f ; x0 , x1 , . . . , xi ) exists finitely for i = 1, 2, . . . , k − 1 and this completes the proof.


1.5.2

27

Unilateral Riemann∗ Derivatives

The unilateral Riemann∗ derivative and derivates of f at x0 of order k can be defined by taking points x1 , . . . , xk on the same side of x0 . That is, the ∗ right-hand Riemann∗ derivative of f at x0 of order k, denoted by Rf(k) (x0 ), is defined to be the iterated limit lim

xk →x0 +

···

lim

x1 →x0 +

k!Qk (f ; x0 , x1 , . . . , xk )

where the variables x1 , . . . , xk are such that 0 < x1 − x0 < x2 − x0 < · · · < xk − x0 . Taking lim sup and lim inf instead of lim in the above we get the definitions of the right-hand upper and lower Riemann∗ derivates of f at x0 of ∗ ∗ order k that are denoted by Rf(k) (x0 ) and Rf(k) (x0 ). ∗ The left-hand Riemann derivative and derivates of f at x0 of order k are ∗ ∗ ∗ (x0 ) and Lf(k) (x0 ), defined similarly and they are denoted by Lf(k) (x0 ), Lf(k) respectively. It can be verified that the analogues Theorem 1.5.1 and Theorem 1.5.2 hold for the one-sided Riemann∗ derivates; the continuity of f at x0 in Theorem 1.5.1 is, in this case, only one-sided. ∗ ∗ The existence and equality of Rf(k) (x0 ) and Lf(k) (x0 ) do not ensure the ∗ existence of f(k) (x0 ). The example given in the case of the Peano derivative can be cited here also. However, the analogue of Theorem 1.4.2 is true. ∗ ∗ ∗ Theorem 1.5.3 If Rf(i) (x0 ) = Lf(i) (x0 ) for i = 1, 2, . . . , k, then f(k) (x0 ) exists.

Clearly the theorem is true for k = 1. Suppose it is true for k ≤ r and let ∗ ∗ Rf(i) (x0 ) = Lf(i) (x0 ) for i = 1, 2, . . . , r+1. Since the theorem is true for k ≤ r, ∗ we conclude that f(i) (x0 ) exists for i = 1, 2, . . . , r. This shows in particular that f is continuous at x0 . Let 0 < |x1 − x0 | < |x2 − x0 | < · · · < |xr+1 − x0 |. So, lim · · · lim Qr (f ; x0 , x1 , . . . , xr ) xr →x0

exists and, therefore,

lim

xr+1 →x0

x1 →x0

· · · lim Qr (f ; x0 , x2 , . . . , xr+1 ) x2 →x0

exists and so, by the continuity of f at x0 , lim

xr+1 →x0

· · · lim

lim Qr (f ; x1 , x2 , . . . , xr+1 )

x2 →x0 x1 →x0

exists and, hence, lim · · · lim Qr (f ; x1 , x2 , . . . , xr+1 ).

xr →x0

x1 →x0

So, by (1.1.1), limxr →x0 · · · limx1 →x0 Qr+1 (f ; x0 , x1 , . . . , xr+1 ) exists. Hence, ∗ (x0 ) = Rf(r+1) ∗ (x0 ) = Lf(r+1)

lim

lim · · · lim (r + 1)!Qr+1 (f ; x0 , x1 , . . . , xr+1 ),

lim

lim · · · lim (r + 1)!Qr+1 (f ; x0 , x1 , . . . , xr+1 ).

xr+1 →x0 + xr →x0

xr+1 →x0 − xr →x0

x1 →x0

x1 →x0

28


∗ ∗ Since, by assumption, Rf(r+1) (x0 ) = Lf(r+1) (x0 ), we conclude from the last ∗ two relations that f(r+1) (x0 ) exists. The proof is now complete by induction.

1.6 1.6.1

Symmetric de la Vall´ ee Poussin Derivatives Symmetric de la Vall´ ee Poussin Derivative and Symmetric Continuity

Let f be defined in some neighbourhood of x and let k be a fixed positive integer. If there is a polynomial P (t) = Px (t) of degree at most k such that k k 1 (1.6.1) 2 f (x + t) + (−1) f (x − t) = P (t) + o(t ) as t → 0,

then f is said to have a kth symmetric de la Vallée Poussin derivative, briefly kth symmetric d.l.V.P. derivative, or just kth d.l.V.P. derivative, at x, and if ak /k! is the coefficient of tk in P (t), then ak is called the kth d.l.V.P derivative (s) of f at x of order k, and is denoted by f(k) (x). (s)

It can be shown as in the case of the Peano derivative that if f(k) (x) exists, then it is unique. It is clear that the polynomial P (t) in (1.6.1) has only even or odd powers of t according as k is even or odd, as we now show. For suppose that k = 2m. Then from (1.6.1) f (x + t) + f (x − t) = 2P (t) + o(t2m ) as t → 0. Since this is true for positive as well as negative values of t, P (t) = P (−t) + o(t2m ) as t → 0 and so P (t) has only even powers of t. If k = 2m + 1, then from (1.6.1) f (x + t) − f (x − t) = 2P (t) + o(t2m+1 ) as t → 0, and so as above P (t) = −P (−t) + o(t2m+1 ) as t → 0 and hence P (t) has only odd powers of t and there is no constant term in P (t) in this case. (s)

It is also clear, as we now show, that if k ≥ 2 and if f(k) (x) exists, then (s)

f(k−2) (x) also exists.


29

For if ak /k! is the coefficient of tk in P (t) and if S(t) = P (t) −

then S(t) is a polynomial of degree at most k − 2 and from (1.6.1) k−2 1 f (x − t) = S(t) + o(tk−2 ) as t → 0, 2 f (x + t) + (−1) (s)

and so f(k−2) (x) exists.

Therefore, if k is even, then (s)

(s) f(0) (x)

1 h→0 2

f(0) (x) = lim

ak k t , k!

exists and

f (x + t) + f (x − t) .

(1.6.2)

Definition. A function f is called symmetrically continuous at x of even order if f (x + t) + f (x − t) = 2f (x) + o(1) as t → 0, (1.6.3) and symmetrically continuous at x of odd order if f (x + t) − f (x − t) = o(1) as t → 0.

(1.6.4)

In the literature, the condition (1.6.3) is used to mean that f is symmetric at x and condition (1.6.4) is used to mean that f is symmetrically continuous at x; see [168, pp. 24–25]. Clearly, if f is continuous at x, then f is symmetrically continuous at x of both even and odd order. From (1.6.2) it follows that (s) f is symmetrically continuous at x of even order if and only if f(0) (x) = f (x). To define the upper and lower d.l.V.P. derivates of f at x of order k + 2, (s) we suppose that f(k) (x) exists and define f (x + t) − f (x − t) ,

k+2 (f ; x, t) = 12 f (x + t) + (−1)k f (x − t) − P (t) t 1 (f ; x, t) =

tk+2 (k + 2)!

1 2

for k ≥ 0,

(1.6.5)

where

P (t) =

⎧ k/2 t2i (s) ⎪ ⎪ ⎪ f (x), ⎪ ⎨ (2i)! (2i) ⎪ ⎪ ⎪ ⎪ ⎩

i=0 (k−1)/2

i=0

t2i+1 (s) f (x), (2i + 1)! (2i+1)

if k is even, (1.6.6) if k is odd.

The upper d.l.V.P. derivate of f at x of order k + 2 is defined by (s)

f (1) (x) = lim sup 1 (f ; x, t). t→0

(s) f (k+2) (x)

= lim sup k+2 (f ; x, t), k ≥ 0. t→0

(1.6.7)

30

Higher Order Derivatives Replacing lim sup by lim inf in (1.6.7), we get the definition of the lower (s)

d.l.V.P. derivate of f at x of order k + 2, f (s) (x). If f (k+2) (x) = f (s) (x), (k+2) (k+2) (s)

then this common value is the d.l.V.P. derivative f(k+2) (x), possibly infinite (s)

in this case. Note that if f(k+2) (x) is finite, then this definition agrees with the one given above; this we now show. (s) In fact, if f (k+2) (x) = f (s) (x) = λ, finite, then from (1.6.7) (k+2) ̟k+2 (f ; x, t) = λ + o(1) as t → 0, and so from (1.6.5) 1 2

tk+2 ̟k+2 (f ; x, t) + P (t) (k + 2)! tk+2 = P (t) + λ + o(tk+2 ) as t → 0. (k + 2)!

f (x + t) + (−1)k f (x − t) =

k+2

t λ Since from (1.6.6), P (t) is a polynomial of degree at most k, P (t) + (k+2)! is a polynomial of degree at most k + 2 with λ/(k + 2)! as the coefficient of tk+2 , and so the result follows.

As in the case of the Peano derivative, it can be shown that the d.l.V.P. derivative, if it exists, is unique. (s) (s) It may be noted that if f(k) (x) exists then f(k−1) (x) may not exist. Example. f (x) =

(

1 x cos , if x 6= 0, x 0, if x = 0.

1 t (s) (s) and so f(1) (0) does not exist and, therefore, f(k) (0) cannot exists if k is odd. (s)

Since f (t)+f (−t) = 0, f(k) (0) exists for all even k. But, f (t)−f (−t) = 2t cos

The d.l.V.P. derivatives of order 1 and 2 are respectively called the first and second symmetric derivatives and they coincide with the symmetricRiemann derivatives of order 1 and 2, respectively cf. (1.2.12) and (1.2.13) , when f is continuous at x.

1.6.2

Smoothness (s)

(s)

It has been noted that if f(k) (x) exists, then f(k−1) (x) might not exist, (s)

(s)

although f(k−2) (x) must exist finitely. A property fitted between f(k−2) (x) (s)

and f(k) (x) is the property of smoothness of f of order k at x.


31 (s)

Let f be defined in some neighbourhood of x and let f(k−2) (x) exist finitely. Define S k (f ; x) = lim sup h̟k (f ; x, h), h→0+

S k (f ; x) = lim inf h̟k (f ; x, h). h→0+

Then S k (f ; x) and S k (f ; x) are respectively called the upper and lower index of d.l.V.P. smoothness of f of order k at x. If S k (f ; x) = S k (f ; x), the common value, denoted by Sk (f ; x), is called the index of d.l.V.P. smoothness of f of order k at x. If Sk (f ; x) exists and Sk (f ; x) = 0, then f is said to be d.l.V.P. smooth or simply smooth, of order k at x. If f is smooth of order 2 at x, then f is said to be smooth at x, omitting the order [?, 169]. If, however, −∞ < S k (f ; x) ≤ S k (f ; x) < ∞, then f is said to be quasi-smooth of order k at x; for the case k = 2, called just quasi-smoothness, see [168, p. 124]. In other words, f is said to be smooth or quasi-smooth of order k at x accordingly as h̟k (f ; x, h) = o(1) or h̟k (f ; x, h) = O(1) as h → 0. (s)

Theorem 1.6.1 If f (k) (x) and f (s) (x) are finite, then f is smooth of order (k) k at x.

(s)

(x) are finite, ̟k (f ; x, t) remains bounded as t → Since f (k) (x) and f (s) (k)

0 and so limt→0 t̟k (f ; x, t) = 0, which shows that S k (f ; x) = S k (f ; x) = 0 and, hence, f is smooth of order k at x. Corollary 1.6.2 If f is smooth of order k at x, then f is smooth of order k − 2 at x. (s)

If f is smooth of order k at x, then by definition f(k−2) (x) exists finitely and so by the above, f is smooth of order k − 2 at x. Remark. If f is smooth of order k at x, then f may not be smooth of order (s) k − 1 at x because f(k−3) (x) may not exist.

1.6.3

de la Vall´ ee Poussin Boundedness

If in (1.6.1) the condition o(tk ) as t → 0 is replaced by the less restrictive condition O(tk ) as t → 0, then f is said to be d.l.V.P. bounded at x of order k. Example. Consider the function ( 1 x2k sin 2k , if x 6= 0, f (x) = x 0, if x = 0.

32

Higher Order Derivatives Then 12 f (t) + (−1)2k f (−t) = t2k sin(1/x2k ) = O(t2k ) as t → 0. Since O(t2k ) (s) (s) as t → 0 implies o(t2k−2 ) as t → 0, f(2i) (0) exists and f(2i) (0) = 0 for i = (s)

0, 1, . . . , k − 1. Also f(2k) (0) does not exist, but f is d.l.V.P. bounded of order 2k at 0. However, we have: Theorem 1.6.3 If f is d.l.V.P. bounded at x of order k, then f has a finite d.l.V.P. derivative at x of order k − 2 and has finite upper and lower d.l.V.P. derivates of order k at x. 2m,

Let k be even, say k = 2m. Since f is d.l.V.P. bounded at x of order 1 2

f (x + t) + f (x − t) = P (t) + O(t2m ) as t → 0,

(1.6.8)

where P (t) is a polynomial of degree at most 2m. It can Pmbe shown that, in this case, P (t) has only even powers of t. So, let P (t) = i=0 b2i t2i . Then writing Pm−1 P1 (t) = i=0 b2i t2i , we have from (1.6.8) that: 2m 1 + O(t2m ) = P1 (t) + O(t2m ) as t → 0. 2 f (x + t) + f (x − t) = P1 (t) + b2m t (1.6.9) Since O(t2m ) as t → 0 implies o(t2m−2 ) as t → 0, we have from (1.6.9) that (s) (s) f(2m−2) (x) exists and f(2i) (x) = (2i)!b2i for i = 0, 1, . . . , m − 1. Using (1.6.5), we have from (1.6.9) ̟2m (f ; x.t) =

m−1 i (2m)! h f (x + t) + f (x − t) X t2i (s) − f(2i) (x) = O(1) as t → 0, 2m t 2 (2i)! i=0

(s)

(x) are finite. This proves the theorem for even and, hence, f (2m) (x) and f (s) (2m) k. The proof for odd k is similar. Theorem 1.6.4 If f has finite upper and lower d.l.V.P. derivates at x of order k, then f is d.l.V.P. bounded at x of order k. (s)

Since f (k) (x) and f (s) (x) are finite, we have ̟k (f ; x.t) = O(1) as t → 0 (k) and hence from (1.6.5) k k 1 2 f (x + t) + (−1) f (x − t) = P (t) + O(t ) as t → 0, where P (t) is the polynomial defined in (1.6.6). This proves that f is d.l.V.P. bounded at x of order k.

Theorem 1.6.5 If f is d.l.V.P. bounded at x of order k, then f is smooth at x of order k.

The proof follows from Theorem 1.6.3 and Theorem 1.6.4.


1.7

33

Symmetric Riemann∗ Derivatives

Let x be a fixed point on R and let f be defined in some neighbourhood of x. Let n be a fixed positive even integer, n = 2k say, and consider 0 < h1 < h2 < · · · < hk . If the iterated limit lim · · · lim n!Qn (f ; x−hk , x−hk−1 , · · · , x−h1 , x, x+h1 , · · · , x+hk ) (1.7.1)

hk →0

h1 →0

exists, the last limit being finite or infinite, while the inner limits are finite, and where Qn (f ; x − hk , x − hk−1 , · · · , x − h1 , x, x + h1 , · · · , x + hk ) is the divided difference of f at the points x−hk , x−hk−1 , · · · , x−h1 , x, x+h1 , · · · , x+hk as defined in (1.1.1), then this limit is called the symmetric Riemann∗ derivative of f at x of order n, n = 2k. If n is a positive odd integer, n = 2k − 1 say, consider 0 < h1 < h2 < · · · < hk . If the iterated limit lim · · · lim n!Qn (f ; x − hk , x − hk−1 , · · · , x − h1 , x + h1 , · · · , x + hk ) (1.7.2)

hk →0

h1 →0

exists, the last limit being finite or infinite, while the inner limits are finite, then this limit is called the symmetric Riemann∗ derivative of f at x of order n, n = 2k − 1. The symmetric Riemann∗ derivative of f at x of order n is ∗(s) written f(n) (x). ∗(s)

∗(s)

Clearly f(1) (x) and f(2) (x) are respectively the first and second symmetric derivatives of f at x considered in (1.2.12) and (1.2.13). Taking “lim sup” and “lim inf” instead of “lim” at each stage of (1.7.1) or (1.7.2) we get the corresponding upper and lower symmetric Riemann∗ derivates of f at x of order ∗(s) n, denoted by f (n) (x) and f ∗(s) (x), respectively. (n)

1.8

Ces` aro Derivatives

The Ces` aro derivative of order r of a function is defined with the help of an integral known as the Cesàro–Perron integral of order r − 1. Interestingly, the Ces` aro–Perron integral of order r −1 is defined with the help of the Cesàro derivative of order r − 1. Thus, the definition of order r is obtained step-bystep starting from order 0, the Cesàro–Perron integral of order 0 being the special Denjoy integral or what is also the Perron integral.

34


1.8.1

Ces` aro Continuity and Ces` aro Derivative

The Ces` aro continuity and Cesàro derivative of order 0, briefly called C0 continuity and C0 -derivative, are respectively ordinary continuity and the ordinary first derivative. The Cesàro–Perron integral of order 0, briefly the C0 P integral, is the special Denjoy integral, or equivalently the Perron integral or more recently the Henstock integral; see [10, 28, 29]. Starting from this we get the definition of Cr -continuity and Cr -derivative as follows. Suppose that the definitions of Cr−1 -continuity, the Cr−1 - derivative and the Cr−1 P -integral are known. Also suppose that if f is Cr−1 P -integrable, then Qf is also Cr−1 P -integrable for every polynomial Q and that the integration by parts formula holds; this is known when r = 1, the C0 case. Let f be Cr−1 P -integrable on [a, b], then for x, x + h in [a, b] write C0 (f ; x, x + h) = f (x + h), Cr (f ; x, x + h) =

Z x+h r (C P (x + h − t)r−1 f (t) dt, r ≥ 1, r−1 ) hr x

(1.8.1)

where (Cr−1 P ) indicates that the integral is a Cr−1 P -integral. The quantity Cr (f ; x, x + h) is called the r-th Cesàro mean of f on the interval with endpoints x and x + h. The right upper and lower Cr -limits of f at x are defined by +

C r f (x) = lim sup Cr (f ; x, x + h), h→0+

C+ r f (x) = lim inf Cr (f ; x, x + h), h→0+

respectively. − The left Cr -limits, C r f (x) and C − r f (x), are defined by considering h → 0−. Define the upper and lower Cr -limits of f at x by +

−

C r f (x) = max{C r f (x), C r f (x)},

− C r f (x) = min{C + r f (x), C r f (x)}.

If C r f (x) = C r f (x), the common value, denoted by Cr f (x), is called the Cr -limit of f at x. If Cr f (x) exists and is equal to f (x), then f is said to be Cr -continuous at x. The right upper and lower Cr -derivates of f at x are defined by r + 1 + Cr D f (x) = lim sup Cr (f ; x, x + h) − f (x) , h h→0+ r + 1 + Cr (f ; x, x + h) − f (x) , Cr D f (x) = lim inf h→0+ h −

respectively. The left Cr -derivates of f at x, Cr D f (x) and Cr D− f (x), are defined by considering h → 0−. The upper and lower Cr -derivates of f at x are defined, respectively, by +

−

Cr Df (x) = max{Cr D f (x), Cr D f (x)}, Cr Df (x) = min{Cr D+ f (x), Cr D− f (x)}.


35

A function M is called a Cr -major function of f on [a, b] if (i) M is Cr -continuous, (ii) M (a) = 0, (iii) Cr DM (x) ≥ f (x) for all x in [a, b], (iv) Cr DM (x) > −∞ for all x in [a, b], where, at the end points a and b, one-sided continuity and one-sided derivates will be considered. A function m is a Cr -minor function of f on [a, b] if −m is a Cr -major function of −f on [a, b]. If {M } and {m} denote the family of Cr -major and Cr -minor functions respectively then it can be shown that M − m is increasing on [a, b] for any M ∈ {M } and m ∈ {m}. Since M (a) − m(a) = 0, it follows that M (b) − m(b) ≥ 0 and so supm∈{m} m(b) ≤ inf M∈{M} M (b). If supm∈{m} m(b) = inf M∈{M} M (b) 6= ±∞, then f is said to be Cr P -integrable and this common value is the Cr P - integral of f on [a, b] and is denoted by Rb (Cr P ) a f (t) dt. It is known that if f is Cr−1 P -integrable, then f is Cr P -integrable and the integrals are equal. Further, the Cr P -integral admits the following integration by parts formula: Theorem R x 1.8.1 Let f be Cr P -integrable on [a, b] and let F (x) = (Cr P ) a f (t) dt, x ∈ [a, b]. If G is an r times repeated integral of a function of bounded variation in [a, b], then f G is Cr P -integrable on [a, b] and Z b Z b b (Cr P ) f G = F G a − (Cr−1 P ) F G′ . a

a

Lemma 1.8.2 If r ≥ 1 and f is Cr−1 P -integrable in some neighbourhood of 0, then for small |h| 6= 0 Z h Z hZ ξ r (h − t) f (t) dt = r (ξ − t)r−1 f (t) dt dξ. (1.8.2) 0

0

0

We have, integrating by parts successively, Z ξr Z h Z ξ1 Z h f (ξ) dξ dξr · · · dξ1 ··· (h − t)r f (t) dt = r!

0

0

and similarly Z Z ξ1 (ξ1 − t)r−1 f (t) dt = (r − 1)!

ξ1

Integrating (1.8.4), we have Z Z h Z ξ1 (ξ1 −t)r−1 f (t) dt dξ1 = r! r 0

0

Z

ξ2

···

0

0

0

(1.8.3)

0

0

h 0

Z

Z

ξ1

···

0

and so (1.8.2) follows from (1.8.3) and (1.8.5).

ξr

f (ξ) dξ dξr · · · dξ2 . (1.8.4) 0

Z

ξr

f (ξ) dξ dξr · · · dξ1 , (1.8.5)

0

36


Theorem 1.8.3 If r ≥ 1 and f is Cr−1 P -integrable in [a, b], then for a ≤ x 0 such that Cr (f ; 0.h) < λ for 0 < h < δ. So,

Z

r

ξ

(ξ − t)r−1 f (t) dt < λξ r for 0 < ξ < δ.

0

Hence,

Z

r

h

0

Z

0

ξ

(ξ − t)r−1 f (t) dt dξ≤ λ

hr+1 for 0 < h < δ. r+1

So, by, Lemma 1.8.2 Z

h

(h − t)r f (t) dt ≤ λ

0

hr+1 for 0 < h < δ, r+1

which gives Cr+1 (f ; 0.h) =

r+1 hr+1

Z

h

(h − t)r f (t) dt ≤ λ for 0 < h < δ.

0

+

Hence, C r+1 f (x) ≤ λ. Since λ is arbitrary, the last inequality follows. The first inequality is proved in a similar manner. + (ii) As in (i), we suppose that Cr D f (0) < λ < ∞. We also suppose that f (0) = 0. Then there is a δ > 0 such that r+1 t Cr (f ; 0, t) < λ for 0 < t < δ. Hence, Z t tr+1 (t − ξ)r−1 f (ξ) dξ < λ for 0 < t < δ. r(r + 1) 0 Hence,

r

Z

0

h

Z

t

(t − ξ)r−1 f (ξ) dξ dt ≤ λ

0

hr+2 for 0 < h < δ. (r + 1)(r + 2)

So, by Lemma 1.8.2 Z

0

h

(h − ξ)r f (ξ) dξ ≤ λ

hr+2 for 0 < h < δ, (r + 1)(r + 2)

Higher Order Derivatives which gives

37

r+2 Cr+1 (f ; 0, h) ≤ λ for 0 < h < δ, h +

and so Cr+1 D f (0) ≤ λ, which completes the proof of (ii) as in (i).

If r = 0, we have the following theorem. Remember that for r = 0 the notation C0 refers to the ordinary limits and derivates, and the special Denjoy integral. Theorem 1.8.4 If f is C0 P -integrable in [a, b], then for a ≤ x ≤ b +

+

+ (i) C + 0 f (x) ≤ C 1 f (x) ≤ C 1 f (x) ≤ C 0 f (x); +

+

(ii) C0 D+ f (x) ≤ C1 D+ f (x) ≤ C1 D f (x) ≤ C0 D f (x). +

+

(i) Suppose x = 0 and C 0 f (0) < ∞. Choose C 0 f (0) < λ < ∞. Then there exists δ > 0 such that C0 (f ; 0, t) < λ for 0 < t < δ. Since C0 (f ; 0, t) = f (t), f (t) < λ for 0 < t < δ and so 1 C1 (f ; 0, h) = h

Z

h

f ≤ λ for 0 < h < δ.

0

+

Letting h → 0, C 1 f (0) ≤ λ and the result follows as above. + + (ii) Suppose that x = 0 = f (x) and C0 D f (0) < ∞. Choose C0 D f (0) < 1 λ < ∞. Then there is a δ > 0 such that C0 (f ; 0, t) < λ for 0 < t < δ. Hence, t f (t) < λt for 0 < t < δ and, hence, Z

0

t

f ≤λ

t2 , for 0 < t < δ, 2

2 + which gives C1 (f ; 0, t) ≤ λ for 0 < t < δ. Hence, C1 D f (0) ≤ λ. The rest is t clear. Corollary 1.8.5 Let f be Cr−1 P -integrable, r ≥ 1. If f is Cr -continuous at x, then f is Cr+1 -continuous at x and, if Cr Df (x) exists, then Cr+1 Df (x) exists and equals Cr Df (x). Corollary 1.8.6 Let f be C0 P -integrable. If f is continuous at x, then f is C1 -continuous at x and, if the ordinary derivative f ′ (x) exists, then C1 Df (x) exists and equals f ′ (x). Theorem 1.8.7 Let f be Cr−1 P -integrable, r ≥ 1 and let Φ be its indefinite Cr−1 P -integral. Then the upper and lower Cr−1 -derivates of Φ at x are, respectively, the upper and lower Cr -limits of f at x.

38

Higher Order Derivatives Integrating by parts, we have

r hr

Z

x+h

r−1

(x + h − t)

x

and so from (1.8.1)

Z t x+h 1 r r−1 (x + h − t) f f (t) dt = h hr−1 t=x x Z x+h Z t r−1 + r−1 (x + h − t)r−2 f (ξ) dξ dt h x x Z r r − 1 x+h r−2 = Φ(t) − Φ(x) dt (x + h − t) h hr−1 x i rh = Cr−1 (Φ; x, x + h) − Φ(x) , h

Cr (f ; x, x + h) = which proves the theorem.

i rh Cr−1 (Φ; x, x + h) − Φ(x) , h

Corollary 1.8.8 A function f is Cr -continuous at x if and only if f (x) is the Cr−1 -derivate of its indefinite Cr−1 P -integral.

1.8.2

Ces` aro Boundedness

A function f is said to be Cesàro bounded of order r, r ≥ 1, briefly Cr bounded at x if Cr (f ; x, x + h) − f (x) = O(h) as h → 0. Theorem 1.8.9 A function f is Cr -bounded at x if and only if Cr−1 Df (x) and Cr Df (x) are finite.

This follows from the definition.

Theorem 1.8.10 If f is Cr -bounded at x, then f is Cr+1 -bounded at x. By Theorem 1.8.9, Cr Df (x) and Cr Df (x) are finite and, hence, by Theorem 1.8.3, Cr+1 Df (x) and Cr+1 Df (x) are finite, and so by Theorem 1.8.9, f is Cr+1 -bounded at x. We now exhibit a function that will show that in Theorems 8.3 and 8.4 strict inequality is possible and will complete the discussion in Corollary 1.8.5. Example. Let f (x) =

(

2x sin 0,

2 1 1 − cos 2 , if x 6= 0, 2 x x x if x = 0.


39

Then C0 Df (0) = −∞, C0 Df (0) = ∞, and remember that these derivates are the ordinary first order lower and upper derivates of f at 0. Since f is the first order derivative of F , where ( 1 x2 sin 2 , if x 6= 0, F (x) = x 0, if x = 0, f is special Denjoy integrable in every neighbourhood of 0 and F is its indefinite integral. So, Z 1 h 1 F (h) C1 (f ; 0, h) = = h sin 2 , f= h 0 h h and, hence, f is C1 -continuous at 0 and C1 Df (0) = lim inf h→0

2 1 C1 (f ; 0, h) − f (0) = lim inf 2 sin 2 = −2. h→0 h h

Similarly, C1 Df (0) = 2. Also, Z h Z h i h 2 2h C2 (f ; 0, h) = 2 (h − t)f (t) dt = 2 (h − t)F (t) + F h 0 h t=0 0 Z h Z h 2 1 2 F = 2 t2 sin 2 dt, = 2 h 0 h 0 t and, hence, 3 3! C2 Df (0) = lim C2 (f ; 0, h) − f (0) = lim 3 h→0 h h→0 h

Now

Z

h

t2 sin

0

1 dt. t2

(1.8.6)

d 5 1 1 1 t cos 2 = 5t4 cos 2 + 2t2 sin 2 dt t t t

and so Z Z h 1 5 h 4 1 1 1 t cos 2 dt = o(h4 ) as h → 0. t2 sin 2 dt = h5 cos 2 − t 2 h 2 t 0 0

(1.8.7)

From (1.8.6) and (1.8.7), C2 Df (0) = 0. This shows that in Theorem 1.8.4 (ii) the inequalities may be strict and that for the last part of Corollary 1.8.5 the converse is not true. Example. Consider further the function ( 2 1 4 1 1 2 sin 2 − 2 cos 2 − 4 sin 2 , if x 6= 0, g(x) = x x x x x 0, if x = 0. Then, with f as in the previous example, f ′ (x) = g(x) for all x 6= 0. Since f is

40


continuous everywhere except x = 0 where it is C1 -continuous and C1 Df (0) and C1 Df (0) are finite, f is both a C1 P -major and a C1 P -minor function of g and, hence, g is C1 P -integrable and f is its indefinite C1 P -integral. Integrating by parts, Z h h Z h i 2h 2 f (h − t)g(t) dt = 2 (h − t)f (t) + C2 (g; 0, h) = 2 h 0 h 0 0 1 2 = 2 F (h) = 2 sin 2 . h h Hence, C2 Dg(0) = lim sup h→0

3! 1 3 C2 (g; 0, h) − g(0) = lim sup sin 2 = ∞. h h h h→0

Similarly, C2 Dg(0) = −∞. Since g is C1 P -integrable, it is C2 P -integrable and f is its indefinite C2 P -integral. So, Z h Z h h i 3 3h (h − t)f (t) dt C3 (g; 0, h) = 3 (h − t)2 g(t) dt = 3 (h − t)2 f (t) + 2 h 0 h 0 0 Z Z h h h i 3! 1 3! h F = 3 t2 sin 2 dt, = 3 (h − t)F (t) + h h 0 t 0 0

and so by (1.8.7)

4 4! C3 D(g; 0, h) = lim [C3 (g; 0, h) − g(0) = lim 4 h→0 h h→0 h

Z

0

h

t2 sin

1 dt = 0. t2

There is no difficulty in considering one-sided limits and one-sided derivates and so the inequalities in Theorems 1.8.3 and 1.8.4 may be strict inequalities.

1.9

Symmetric Ces` aro Derivatives

As for the Ces` aro derivative, the symmetric Cesàro derivative of order r is defined with the help of the Cesàro–Perron integral of order r − 1. Let f be Cr−1 P -integrable in [a, b] and let the Cesàro mean Cr (f ; x, x + h) of f of order r on the interval with endpoints x, x + h ∈ [a, b] be defined by (1.8.1).

1.9.1 Let

Symmetric Ces` aro Continuity and Symmetric Ces` aro Derivative + SC r f (x) = lim sup Cr (f ; x, x + h) − Cr (f ; x, x − h) , h→0+


41

−

− the other limits SC r f (x), SC + r f (x) and SC r f (x) being defined analogously by using the appropriate limits of Cr (f ; x, x + h) − Cr (f ; x, x − h) . If +

−

− SC r f (x) = SC r f (x) = SC + r f (x) = SC r f (x), then the common value, written SCr f (x), is called the symmetric Cr -limit, briefly the SCr -limit of f at x. If SCr f (x) = 0, then f is said to be symmetrically Cr -continuous, or SCr -continuous, at x. The upper and lower SCr derivates of f at x are defined by

r+1 [Cr (f ; x, x + h) − Cr (f ; x, x − h) , 2h h→0 r+1 SCr Df (x) = lim inf [Cr (f ; x, x + h) − Cr (f ; x, x − h) , h→0 2h

SCr Df (x) = lim sup

respectively. If SCr Df (x) = SCr Df (x), the common value, denoted by SCr Df (x), is called the SCr -derivative of f at x. Note that for the concept of SCr D there is no effect whether h → 0+ or h → 0−, but that it matters in the case of SCr above. Theorem 1.9.1 If r ≥ 1 and f is Cr−1 P -integrable in [a, b], then for a < x 0 such that Cr (f ; 0, ξ) − Cr (f ; 0, −ξ) < λ

2ξ for 0 < ξ < δ. r+1

So, r ξr

Z

ξ

(ξ − t)r−1 f (t) dt −

0

which gives Z hZ ξ r (ξ − t)r−1 f (t) dt + 0

0

Integrating, r

hZ

0

hZ ξ 0

r (−ξ)r

−ξ

Z

−ξ

(−ξ − t)r−1 f (t) dt < λ

0

2ξ , r+1

i 2ξ r+1 for 0 < ξ < δ. (ξ + t)r−1 f (t) dt < λ r+1

(ξ − t)r−1 f (t) dt dξ +

Z

0

h

Z

−ξ

0

2hr+2 , for 0 < h < δ. ≤λ (r + 1)(r + 2)

(ξ + t)r−1 f (t) dt dξ

i (1.9.1)

42


By Lemma 1.8.2 Z h

(h − t)r f (t) dt = r

Z

Z

h

0

0

ξ

(ξ − t)r−1 f (t) dt dξ.

(1.9.2)

0

Since Lemma 1.8.2 is also true for negative values of h, Z −h Z ξ Z −h (ξ − t)r−1 f (t) dt dξ (−h − t)r f (t) dt = r 0

0

=− r

Z

h

0

and, hence, Z −h

r

(h + t) f (t) dt = r

0

Z

0

h Z −ξ

Z

0 −ξ

0

(ξ + t)r−1 f (t) dt dξ.

(1.9.3)

0

From (1.9.1), (1.9.2) and (1.9.3) Z −h Z h (h + t)r f (t) dt ≤ λ (h − t)r f (t) dt + 0

0

(−ξ − t)r−1 f (t) dt dξ

2hr+2 , for 0 < h < δ. (r + 1)(r + 2)

Hence, Z Z −h r+1 r+2 r+1 h r r (h − t) f (t) dt − (−h − t) f (t) dt ≤ λ, 2h hr+1 0 (−h)r+1 0 for 0 < h < δ, which gives r+2 Cr+1 (f ; 0, h) − Cr+1 (f ; 0, −h) ≤ λ, for 0 < h < δ. 2h Hence, SCr+1 Df (0) ≤ λ. Since λ is arbitrary, SCr+1 Df (0) ≤ SCr Df (0), proving the last inequality of (ii). The proof of the first inequality of (ii) is similar. Theorem 1.9.2 If r ≥ 1 and f is Cr−1 P -integrable in [a, b], then for a < x 0 such that f (t) − f (−t) < 2λt for 0 < t < δ. Integrating we have Z Z −h i 2 h1 h 1 1 C1 (f ; 0, h) − C1 (f ; 0, −h) = f− f h 2h h 0 −h 0 Z Z i 2 h1 h 1 h = f (t) dt − f (−t) dt 2h h 0 h 0 ≤ λ for 0 < h < δ, and so SC1 Df (0) ≤ λ. Since λ is arbitrary, the right-hand inequality follows. The left-hand inequality has a similar proof. Corollary 1.9.4 If f is C0 P -integrable in [a, b], then for a < x < b (i) if f is SC0 -continuous at x, then f is SC1 -continuous at x; (ii) if SC0 Df (x) exists, then SC1 Df (x) exists and SC0 Df (x) = SC1 Df (x). Remark. The SCr -continuity of f at x is analogous to the smoothness of order r + 1 at x of the function that is obtained by taking the r times repeated integral of f . This will be discussed in Section 5 of Chapter II.

1.9.2

Symmetric Ces` aro Boundedness

A function f is said to be symmetric Cesàro bounded of order r at x, or simply SCr -bounded at x if Cr (f ; x, x + h) − Cr (f ; x, x − h) = O(h) as h → 0. Theorem 1.9.5 A function f is SCr -bounded at x if and only if SCr Df (x) and SCr Df (x) are finite. If f is SCr -bounded at x, then f is SCr+1 -bounded at x. The first part is clear and the second part follows from Theorem 1.9.1 and the first part of the theorem.

44


1.10

Borel Derivatives

1.10.1

Bilateral Borel Derivatives

Let f be special Denjoy integrable and let r be a fixed positive integer. If there is a polynomial Q(t) = Qx (t) of degree at most r such that Z h f (x + t) − Q(t) dt = o(h) as h → 0, (1.10.1) tr 0 then f is said to have a Borel derivative of order r at x, and if ar /r! is the coefficient of tr in Q(t), then ar is called the Borel derivative of f at x of order r and is denoted by BDr f (x). The integral in (1.10.1) is assumed to be convergent at t = 0. Integrating by parts we have Z h Z h f (x + t) − Q(t) tr f (x + t) − Q(t) dt = dt tr 0 0 Z h f (x + t) − Q(t) r dt = h tr 0 Z h Z t f (x + ξ) − Q(ξ) − r tr−1 dξ dt ξr 0 0 = o(hr+1 ) as h → 0. (1.10.2) This, as we now see, shows that the Borel derivative, if it exists, is unique. If Q1 (t) is another polynomial of degree at most r satisfying (1.10.1), then it also satisfies (1.10.2)and so Z h Z h Z h f (x + t) − Q1 (t) dt = o(hr+1 ) f (x + t) − Q(t) dt − (Q − Q1 ) = 0

0

0

as h → 0.

Rh

Since 0 (Q − Q1 ) is a polynomial in h of degree at most r + 1, it vanishes identically and so Q(t) = Q1 (t) for all t. Now we show that if r ≥ 2 and if BDr f (x) exists, then BDi f (x) exists, 0 < i < r. P For if aj /j! is the coefficient of tj in Q(t), 0 ≤ j ≤ r, and if Q(t) = r Q(t) − j=i+1 (aj /j!)tj , then using (1.10.1) and integrating by parts Z

0

h

Z h Z h X r aj j−i f (x + t) − Q(t) r−i f (x + t) − Q(t) dt = dt + t t dt ti tr 0 0 j=i+1 j! Z h f (x + t) − Q(t) r−i dt =h tr 0

Higher Order Derivatives Z h Z t f (x + ξ) − Q(ξ) dξ dt + o(h) tr−i−1 −(r − i) ξr 0 0 = o(hr−i+1 ) + o(h) = o(h) as h → 0.

45

So (1.10.1) is satisfied if Q(t) and r are replaced by Q(t) and i, respectively. Since Q(t) is a polynomial of degree at most i, BDi f (x) exists and BDi f (x) = ai . If a0 is the constant term in Q(t), then writing BD0 f (x) = a0 we have from (1.10.2) Z

0

h

f (x + t) −

r X ti i=0

i

BDi f (x) dt = o(hr+1 ) as h → 0.

and so 1 BD0 f (x) = lim h→0 h

Z

h

f (x + t) dt,

0

which is f (x) if f is continuous at x. So we make the following definition. Definition. If limh→0 continuous at x.

1 h

Rh 0

f (x + t) dt = f (x), then f is said to be Borel

Thus, Borel continuity is the same as Cesàro continuity of order 1, C1 continuity. We shall see in Chapter II that a Borel derivative is not the same as a Ces` aro derivative. To define upper and lower Borel derivates of order k at x, we suppose the existence of BDk−1 f (x) and define Q(t) =

k−1 X i i=0

t BDi f (x). i!

(1.10.3)

Then for each h > 0 and 0 < ǫ < h, 1/t is of bounded variation in [ǫ, h] and so the function k! f (x + t) − Q(t) tk is special Denjoy integrable in [ǫ, h] as a function of t. The right upper and lower Borel derivates of f at x of order k are defined by Z h 1 h f (x + t) − Q(t) i + dt , BDk f (x) = k! lim sup lim sup tk ǫ→0+ h ǫ h→0+ Z h 1 h f (x + t) − Q(t) i BD+ dt , (1.10.4) f (x) = k! lim inf lim inf k h→0+ ǫ→0+ h ǫ tk +

respectively. If BDk f (x) = BD + k f (x), the common value, possibly infinite, is called the right Borel derivative of f at x of order k and is denoted by − − − BD+ k f (x). The definitions of BD k f (x), BD k f (x) and BD k f (x) are obtained by considering [−h, −ǫ] and taking the integral from −h to −ǫ, where h > 0 + − − and 0 < ǫ < h. If BDk f (x) = BD+ k f (x) = BD k f (x) = BD k f (x), then

46


the common value, possibly infinite, is the Borel derivative of f at x of order k, written BDk f (x). We now show that if BDk f (x) exists finitely, then this definition agrees with the previous one. Writing Z h f (x + t) − Q(t) F (x, ǫ, h) = dt, (1.10.5) tk ǫ +

where Q is defined in (1.10.3), observe that if BDk f (x) and BD+ k f (x) are finite, then lim supǫ→0+ F (x, ǫ, h) and lim inf ǫ→0+ F (x, ǫ, h) are finite. We first show that they are equal. Let Φ(x, h) = lim sup F (x, ǫ, h), ǫ→0+

φ(x, h) = lim inf F (x, ǫ, h).

(1.10.6)

ǫ→0+

Since +

BD k f (x) = lim sup h→0+

Φ(x, h) h

and BD + k f (x) = lim inf h→0+

φ(x, h) , h

and since they are finite, we have lim sup Φ(x, h) = 0 = lim inf φ(x, h), h→0+

h→0+

and, therefore, since φ(x, h) ≤ Φ(x, h), lim Φ(x, h) − φ(x, h) = 0.

h→0+

Let η > 0 be small, then

(1.10.7)

Φ(x, h + η) − Φ(x, h) (1.10.8) Z h Z h+η f (x + t) − Q(t) f (x + t) − Q(t) dt − lim sup dt = lim sup k t tk ǫ→0+ ǫ→0+ ǫ ǫ Z h hZ h Z h+η i f (x + t) − Q(t) f (x + t) − Q(t) dt − lim sup dt = lim sup + k t tk ǫ→0+ ǫ→0+ ǫ ǫ h Z h+η f (x + t) − Q(t) = dt. tk h Similarly, φ(x, h + η) − φ(x, h) =

Z

h+η

h

f (x + t) − Q(t) dt. tk

From (1.10.8) and (1.10.9), Φ(x, h + η) − Φ(x, h) = φ(x, h + η) − φ(x, h)

(1.10.9)


47

or Φ(x, h + η) − φ(x, h + η) = Φ(x, h) − φ(x, h). This shows that Φ(x, h) − φ(x, h) is independent of h and so from (1.10.7), Φ(x, h) = φ(x, h) for all h. Hence, from (1.10.5) and (1.10.6), Rh + dt exists and so from (1.10.4), if BDk f (x) and limǫ→0 ǫ f (x+t)−Q(t) tk BD+ k f (x) are finite, then Z 1 h f (x + t) − Q(t) + dt, BD k f (x) = k! lim sup tk h→0+ h 0 Z 1 h f (x + t) − Q(t) BD + dt. (1.10.10) f (x) = k! lim inf k h→0+ h 0 tk Now suppose that BD k f (x) exists finitely in this latter sense. Then from (1.10.10) and its analogue Z 1 1 h f (x + t) − Q(t) dt = BD k f (x) + o(1) as h → 0, h 0 tk k! and so 1 h

Z

h

0

f (x + t) − Q(t) − tk

tk k! BD k f (x)

dt = o(1) as h → 0,

and this gives by (1.10.3) 1 h

Z

0

h

P f (x + t) − Q(t) − ki=0 tk

ti i! BD k f (x)

dt = o(1) as h → 0,

showing that BDk f (x) is also the Borel derivative in the former sense.

1.10.2

Unilateral Borel Derivatives

If in (1.10.1) a one-sided limit is taken, the definition of the one-sided Borel − derivatives BD + r f (x) and BD r f (x) are obtained and if in (1.10.3) we take Q(t) =

k−1 X i i=0

t BD+ i f (x), i!

(1.10.3*)

the definition of the unilateral Borel derivates are obtained. Note that the bilateral Borel derivates and the unilateral Borel derivates are different since (1.10.3)* and (1.10.3) are different.

1.10.3

Borel Boundedness

If in (1.10.1) the condition o(1) is replaced by O(1) as t → 0, then f is said to be Borel bounded of order r at x.

48


Theorem 1.10.1 If f is Borel bounded of order r at x, then f has a finite Borel derivative of order r − 1 and BD r f (x) and BD r f (x) are finite. Conversely, if BDr f (x) and BDr f (x) are finite, then f is Borel bounded of order r at x. We have from the definition of Borel boundedness that Z 1 h f (x + t) − Q(t) dt = O(1) as h → 0, (1.10.11) h 0 tr Pr where Q(t) is a polynomial of degree at most r. Let Q(t) = i=0 bi ti . Writing Pr−1 i Q(t) = i=0 bi t , we have from (1.10.11)

Z

h

0

f (x + t) − Q(t) dt = tr−1

Z

h

t

0

Z

f (x + t) − Q(t) dt + tr

h

Z

h

br t dt

0

f (x + t) − Q(t) dt tr 0 Z hZ t f (x + ξ) − Q(ξ) − dξ dt + O(h2 ) ξr 0 0 Z h = hO(h) − O(t) dt + O(h2 ) = O(h2 ) as h → 0. = h

0

Hence, 1 h

Z

h

0

f (x + t) − Q(t) dt = O(h) as h → 0. tr−1

(1.10.12)

Since O(h) implies o(h) as h → 0, (1.10.12) shows that BDr−1 f (x) exists Pr−1 finitely and that Q(t) = i=0 (ti /i!)BDi f (x). Since Q(t) = Q(t) + br tr , from (1.10.11), 1 h

Z

h

f (x + t) −

0

=

1 h

Z

Pr−1

i=0 (t tr

i

/i!)BDi f (x)

dt

h

br dt + O(1) = O(1) as h → 0.

(1.10.13)

0

Hence, BDr f (x) and BD r f (x) are finite. For the converse, since BD r f (x) and BDr f (x) are finite by definition, (1.10.13) holds. Let C be any finite constant and let Q(t) = Pr−1 i r i=0 (t /i!)BDi f (x) + Ct . Then from (1.10.13) 1 h

Z

0

h

1 f (x + t) − Q(t) dt = O(1) − r t h

Z

h

C dt = O(1) as h → 0.

0

Since Q(t) is a polynomial of degree of most r, f is Borel bounded of order r at x.


1.11 1.11.1

49

Symmetric Borel Derivatives Symmetric Borel Derivatives and Symmetric Borel Continuity

Let f be special Denjoy integrable in some neighbourhood of x and let r be a positive integer. If there is a polynomial P (t) = Px (t) of degree at most r such that Z h 1 r 2 f (x + t) + (−1) f (x − t) − P (t) dt = o(h), as h → 0, (1.11.1) tr 0 then f is said to have a symmetric Borel derivative at x of order r and if ar /r! is the coefficient of tr in P (t), then ar is called the symmetric Borel derivative of f at x of order r, and is denoted by SBDr f (x). The integral in (1.11.1) is assumed to be convergent at t = 0. The derivative, if it exists, is unique. Integrating by parts and using (1.11.1), we have Z hh 0

i f (x + t) + (−1)r f (x − t) − P (t) dt Z h 1 r r 2 f (x + t) + (−1) f (x − t) − P (t) t = dt tr 0 Z h 1 r 2 f (x + t) + (−1) f (x − t) − P (t) = hr dt tr 0 Z h Z t 1 r r−1 2 f (x + ξ) + (−1) f (x − ξ) − P (ξ) t −r dξ dt ξr 0 0 = o(hr+1 ), as h → 0. (1.11.2) 1 2

So, if P1 (t) is another polynomial of degree at most r satisfying (1.11.1), then it also satisfies (1.11.2) and so Z

0

h

P1 (t) − P (t) dt =

Z hh 0

1 2

−

i f (x + t) + (−1)r f (x − t) − P (t) dt

Z hh 0

= o(h

r+1

1 2

i f (x + t) + (−1)r f (x − t) − P1 (t) dt

), as h → 0.

(1.11.3)

Since P1 (t) − P (t) is a polynomial of degree at most r, the left-hand side of ( 11.3) is a polynomial in h of degree at most r + 1 and so from (1.11.3), P1 (t) − P (t) vanishes identically. Hence, P1 (t) = P (t) for all t. Therefore, if SBDr f (x) exists, it is unique.

50


We now show that the polynomial P (t) in (1.11.1) has only even or only odd powers of t according as r is even or odd. For, since (1.11.2) is true for all h, changing h to −h in (1.11.2) we have Z 0 h i r r+1 1 ) as h → 0. 2 f (x + t) + (−1) f (x − t) − P (t) dt = o(h −h

Again, changing the variable t to −t and multiplying by (−1)r Z hh i r r 1 f (x + t) + (−1) f (x − t) − (−1) P (−t) dt = o(hr+1 ) as h → 0. 2 0

(1.11.4)

From (1.11.2) and (1.11.4), Z h P (t) − (−1)r P (−t) dt = o(hr+1 ) as h → 0.

(1.11.5)

0

Since P (t)−(−1)r P (−t) is a polynomial of degree at most r, the left-hand side of (1.11.5) is a polynomial in h of degree at most r+1 and so P (t)−(−1)r P (−t) vanishes identically. Hence, P (t) has only even or odd powers of t according as r is even or odd. If r ≥ 2 and SBDr fP (x) exists, then SBDr−2 f (x) exists. r For if P (t) = i=0 ai ti , then writing S(t) = P (t) − ar tr and, since P (t) has only odd or even powers, we conclude that S(t) is of degree at most r − 2. Also from (1.11.1) integration by parts gives Z h 1 r 2 f (x + t) + (−1) f (x − t) − S(t) dt tr−2 0 Z h 1 r r 2 f (x + t) + (−1) f (x − t) − P (t) + ar t = dt tr−2 0 Z h 1 f (x + t) + (−1)r f (x − t) − P (t) h3 dt + a = t2 2 r tr 3 0 Z h 1 r f (x + t) + (−1) f (x − t) − P (t) 2 = h2 dt tr 0 Z h Z t 1 r h3 2 f (x + ξ) + (−1) f (x − ξ) − P (ξ) −2 t dξ dt + ar r ξ 3 0 0 Z h 3 h = o(h2 ) as h → 0. = h2 o(h) − 2 t o(t) dt + ar 3 0 Since o(h2 ) implies o(h), as h → 0, SBDr−2 f (x) exists. When r is even, we have to consider SBD0 f (x), which is given by Z 1 h1 SBD0 f (x) = lim f (x + t) + f (x − t) dt, h→0 h 0 2

which is f (x) if f is continuous at x. So, we make the following definitions.


51

Definitions. If 1 h→0 h lim

Z

h 1 2

0

f (x + t) + f (x − t) dt = f (x),

then f is said to be symmetrically Borel continuous at x of even order. If 1 h→0 h lim

Z

0

h 1 2

f (x + t) − f (x − t) dt = 0,

then f is said to be symmetrically Borel continuous at x of odd order. Clearly symmetric Borel continuity of odd order is equivalent to SC1 continuity. Also, it is clear that symmetric continuity of even, (odd) order (see Section 6.1 for the definitions), implies symmetric Borel continuity of even, (odd) order. To define upper and lower symmetric Borel derivates of f at x of order k, k ≥ 2, we suppose that SBDk−2 f (x) exists and define ̟k (f ; x, t) as follows: t̟1 (f ; x, t) = 21 f (x + t) − f (x − t) , (1.11.6) k t ̟k (f ; x, t) = 12 f (x + t) + (−1)k−2 f (x − t) − P (t) for k ≥ 2, k!

where P (t) is the polynomial of degree at most k − 2 given by  (k−2)/2   X t2i   SBD2i f (x), if k is even,    i=0 (2i)! P (t) =  (k−3)/2  X  t2i+1    SBD2i+1 f (x), if k is odd.  (2i + 1)! i=0

(1.11.7)

For small h > 0 and 0 < ǫ < h, the function 1/t is of bounded variation in [ǫ, h] and so ̟1 (f ; x, t), as a function of t, is special Denjoy integrable in [ǫ, h]. The upper and lower symmetric Borel derivates of f at x of order k are defined by 1 SBDk f (x) = lim sup lim sup h h→0+ ǫ→0+ SBDk f (x) = lim inf lim inf h→0+

ǫ→0+

1 h

Z

h

̟k (f ; x, t) dt,

ǫ

Z

h

̟k (f ; x, t) dt,

ǫ

respectively. If SBDk f (x) = SBDk f (x), then the common value, denoted by SBDk f (x), is called the symmetric Borel derivative, possibly infinite, of

52


f at x of order k. It can be shown, as in the case of the unsymmetric Borel derivative, that if SBDk f (x) and SBDk f (x) are finite, then Z 1 h SBDk f (x) = lim sup ̟k (f ; x, t) dt, h→0+ h 0 (1.11.8) Z h 1 SBDk f (x) = lim inf ̟k (f ; x, t) dt, h→0+ h 0 and that if SBDk f (x) exists finitely in this latter sense, then it also exists in the former sense and they are equal. Note that h → 0+ in (1.11.8) can be replaced by h → 0.

1.11.2

Borel Smoothness

Let a function f be defined in some neighbourhood of x and let f be special Denjoy integrable in that neighbourhood. Let SBDk−2 f (x) exist finitely. Define Z 1 h t̟k (f ; x, t) dt, BSk f (x) = lim sup h→0 h 0 (1.11.9) Z h 1 t̟k (f ; x, t) dt. BSk f (x) = lim inf h→0 h 0 Then BSk f (x) and BSk f (x) are, respectively, called the index of upper and lower Borel smoothness of f of order k at x. If BSk f (x) = BSk f (x), then the common value, denoted by BSk f (x), is called the index of Borel smoothness of f of order k at x. If BSk f (x) = 0, then f is said to be Borel smooth of order k at x. Theorem 1.11.1 If SBDk f (x) and SBDk f (x) are finite, then f is Borel smooth of order k at x. Since SBDk f (x) and SBDk f (x) are finite, it follows from (1.11.8) that Rh limh→0 0 ̟k (f ; x, t) dt = 0. So, integrating by parts, Z

0

h

t̟k (f ; x, t) dt = h

Z

h

̟k (f ; x.t) dt −

0

Z

0

h

Z

t

̟k (f ; x, ξ) dξ dt

0

and hence, Z h Z Z Z 1 h t 1 h ̟k (f ; x.t) dt − ̟k (f ; x, ξ) dξ dt, t̟k (f ; x, t) dt = h 0 h 0 0 0 and so the left-hand side tends to 0 as h → 0.


53

Corollary If f is Borel smooth of order k at x, then f is Borel smooth of order k − 2 at x. If f is Borel smooth of order k at x, then by definition SBDk−2 f (x) exists finitely and so the result follows from the above theorem.

1.11.3

Symmetric Borel Boundedness

If, in (1.11.1), o(h) is replaced by O(h) as h → 0, then f is said to be symmetric Borel bounded of order r at x. Theorem 1.11.2 If f is symmetric Borel bounded of order r at x, then f has a finite symmetric Borel derivative of order r − 2 at x and SBDr f (x) and SBDr f (x) are finite. Conversely, if SBDr f (x) and SBDr f (x) are finite, then f is symmetric Borel bounded of order r at x. We have from the definition of symmetric Borel boundedness Z 1 h 21 f (x + t) + (−1)r f (x − t) − P (t) dt = O(1), as h → 0, (1.11.12) h 0 tr P where P (t) is a polynomial of degree at most r. Let P (t) = ri=0 bi ti . Then Pr−2 writing P (t) = i=0 bi ti we have from (1.11.12) Z h 1 r−2 f (x − t) − P (t) 2 f (x + t) + (−1) dt tr−2 0 Z h 1 Z h f (x + t) + (−1)r f (x − t) − P (t) = t2 2 dt + (br−1 t + br t2 ) dt r t 0 0 Z h 1 r f (x + t) + (−1) f (x − t) − P (t) 2 = h2 dt tr 0 Z h Z t 1 r 2 f (x + ξ) + (−1) f (x − ξ) − P (ξ) dξ dt + O(h2 ) −2 t ξr 0 0 Z h = h2 O(h) − 2 tO(t) dt + O(h2 ) = O(h2 ) as h → 0.

0

Hence, 1 h

Z

0

h 1 2

f (x + t) + (−1)r−2 f (x − t) − P (t) dt = O(h) as h → 0. (1.11.13) tr−2

Since O(h) implies o(1) as h → 0 and P (t) is a polynomial of degree at most r − 2, (1.11.13) shows that SBDr−2 f (x) exists and that SBDr−2 f (x) = (r − 2)!br−2 .

54

Higher Order Derivatives We now show that br−1 = 0. Using (1.11.12)

Z

0

h

1 f (x + t) + (−1)r f (x − t) − P (t) dt 2 Z h 1 r r 2 f (x + t) + (−1) f (x − t) − P (t) = t dt tr 0 Z h 1 r 2 f (x + t) + (−1) f (x − t) − P (t) = hr dt (1.11.14) r t 0 Z h Z t 1 r r−1 2 f (x + ξ) + (−1) f (x − ξ) − P (ξ) dξ dt −r t ξr 0 0 Z h = hr O(h) − r tr−1 O(t) dt = O(hr+1 ) as h → 0. 0

Since (1.11.14) holds for all h, Z −h 1 r f (x + t) + (−1) f (x − t) − P (t) dt = O(hr+1 ) as h → 0. 2 0

Changing the variable t to −t and multiplying by −(−1)r , we get Z h 1 f (x + t) + (−1)r f (x − t) − (−1)r P (−t) dt = O(hr+1 ) as h → 0. 2 0 (1.11.15) Subtracting (1.11.14) from (1.11.15), Z h P (t) − (−1)r P (−t) dt = O(hr+1 ) as h → 0. (1.11.16) 0

Rh Since P (t) −(−1)r P (−t) is a polynomial of degree at most r, then 0 P (t) − (−1)r P (−t) dt is a polynomial in h of degree at most r + 1 and so from (1.11.16), P (t) − (−1)r P (−t) = Ctr for some constant C. Hence, if r is even, P (t) does not contain any odd powers of t and, if r is odd, then P (t) does not contain any even powers of t. So, P (t) does not contain the term br−1 tr−1 ; that is, br−1 = 0 and P (t) = P (t) + br tr . Let r be even, r = 2m, say. Then from (1.11.13), P (t) = Pm−1 t2i Pm−1 t2i 2m and, i=0 (2i)! SBD2i f (x) and so P (t) = i=0 (2i)! SBD2i f (x) + b2m t hence, from (1.11.12), Pm−1 t2i Z 1 1 h 2 f (x + t) + f (x − t) − i=0 (2i)! SBD2i f (x) dt h 0 t2m Z 1 h = b2m dt + O(1) = O(1) as h → 0, (1.11.17) h 0 which shows that SBD2m f (x) and SBD2m f (x) are finite.


55

If r is odd, r = 2m + 1, say, then from (1.11.13), P (t) = Pm t2i−1 Pm t2i−1 i=1 (2i−1)! SBD2i−1 f (x) + i=1 (2i−1)! SBD2i−1 f (x) and so P (t) = 2m+1 b2m+1 t and, hence, from (1.11.12) 1 h

Z

1 2 f (x

h

0

1 h

=

Z

h

Pm

t2i−1 i=1 (2i−1)! SBD2i−1 f (x) t2m+1

+ t) − f (x − t) −

b2m+1 dt + O(1) = O(1) as h → 0,

dt (1.11.18)

0

which shows that SBD2m+1 f (x) and SBD2m+1 f (x) are finite. For the converse part, since SBDr f (x) and SBDr f (x) are finite, by definition (1.11.17) or (1.11.18) holds according as r = 2m or r = 2m + 1. Let C be any finite real constant and let  m−1 X t2i    SBD2i f (x) + Ct2m , if r = 2m,    i=0 (2i)! P (t) =  m  X  t2i−1   SBD2i−1 f (x) + Ct2m+1 , if r = 2m + 1.  (2i − 1)! i=1 Then from (1.11.17) or (1.11.18) 1 h

Z

h 1 2

0

Z f (x + t) + (−1)r f (x − t) − P (t) 1 h dt = O(1) − C dt = O(1), tr h 0

as h → 0, which shows that f is symmetric Borel bounded of order r at x.

1.12 1.12.1

Lp -Derivatives Lp -Derivatives and Lp -Continuity

Let f ∈ Lp in some neighbourhood of a point x for some p, 1 ≤ p < ∞. If there is a polynomial P (t) = Px (t) of degree at most k for which h1 Z h

0

h

i1/p f (x + t) − P (t) p dt = o(hk ) as h → 0,

(1.12.1)

then f is said to have a kth Lp -derivative at x of order k and, if ak /k! is the coefficient of xk in P (t), then ak is called the kth Lp -derivative of f at x denoted by f(k),p (x). If f(k),p (x) exists, then it is unique.

56


Let P1 (t) be another polynomial of degree at most k that satisfies (1.12.1). Then by Minkowski’s inequality h1 Z h

h 0

i1/p i1/p h1 Z h P1 (t) − P (t) p dt f (x + t) − P (t) p dt ≤ h 0 h1 Z h i1/p f (x + t) − P1 (t) p dt + h 0 = o(hk ) as h → 0.

Taking h > 0 we have 1 h

Z

0

h

P1 (t) − P (t) p dt = o(hkp ) as h → 0.

(1.12.2)

p Pk Let Q(t) = P1 (t) − P (t) = i=0 ci ti . Since Q(t) is a continuous function p p R h of t, we have from (1.12.2), Q(0) = limh→0+ h1 0 Q(t) dt = 0, and so c0 = 0. Suppose that we have proved c0 = c1 = · · · = cj = 0, 0 ≤ j < k. Then Pk Q(t) = i=j+1 ci ti and so integrating by parts, Z

0

h

Q(t) p dt =

Z

h

0

k p X ci ti−j−1 dt tp(j+1) i=j+1

Z h X k p ci ti−j−1 dt = hp(j+1) 0

i=j+1

Z −p(j + 1)

h

t

p(j+1)−1

0

0

Hence, 1 hpj+p+1

Z

0

h

Q(t) p dt = 1 h −

Z t X k p ci ξ i−j−1 dξ dt. i=j+1

Z h X k p ci ti−j−1 dt 0

i=j+1

p(j + 1) hpj+p+1

Z

0

h

tp(j+1)−1

Z t X k p ci ξ i−j−1 dξ dt. 0

i=j+1

(1.12.3)

pj+p Now as h → 0, the right-hand side of (1.12.3) tends to cj+1 − pj+p+1 cj+1 , that is, to cj+1 (pj + p + 1). But, by (1.12.2), the left-hand side of (1.12.3) tends to 0 and so cj+1 = 0. Hence, by induction, cj = 0, 0 ≤ j ≤ k and so P1 (t) = P (t). If h < 0, the proof is similar. If f(k),p (x) exists and k ≥ 2, then f(k−1),p (x) also exists.


57

To see this, note that if h > 0, then h1 Z h i1/p h 1 Z h i1/p h hkp i1/p = O(hk ) as h → 0, |tk |p dt tkp dt = = h 0 h 0 kp + 1 (1.12.4) and, if h < 0, then h1 Z h i1/p i1/p h 1 Z −h i1/p h 1 Z 0 k p k p |t | dt |t | dt |tk |p dt = = h 0 −h h −h 0 h 1 Z −h i1/p h (−h)kp i1/p = = = O(hk ) as h → 0, ξ kp dξ −h 0 kp + 1 (1.12.5) and so in any case h 1 1 f(k),p (x) k! h

Z

0

h

|tk |p dt

i1/p

= o(hk−1 ) as h → 0.

(1.12.6)

Now, by Minkowski’s inequality h1 Z h i1/p k f (x + t) − P (t) + t f(k),p (x) p dt h 0 k! Z h1 Z h h 1 h k p i1/p p i1/p 1 f(k),p (x) | |t | dt f (x + t) − P (t) dt + ≤ h 0 k! h 0

and so by (1.12.1) and (1.12.6) h1 Z h i1/p k f (x + t) − P (t) + t f(k),p (x) p dt = o(hk−1 ) as h → 0. (1.12.7) h 0 k! k

Since P (t) − tk! f(k),p (x) is a polynomial of degree at most k − 1, f has a (k − 1)th Lp derivative at x. Thus, if f(k),p (x) exists, then all the previous derivatives f(i),p (x), 1 ≤ i ≤ k − 1 exist. If a0 is the constant term in P (t), then writing f(0),p (x) = a0 we get, as in (1.12.7) h 1 Z h i1/p f (x + t) − f(0),p (x) p dt = o(1) as h → 0. h 0

Definition. A function f is said to be Lp -continuous at x if i1/p h 1 Z h f (x + t) − f (x) p dt = o(1) as h → 0. h 0

If f is continuous at x, then f is Lp -continuous at x and, if f is Lp -continuous at x, then f(0),p (x) exists and f(0),p (x) = f (x). In particular, if p = 1, then f is Lp -continuous at x if and only if x is a Lebesgue point of f . It can be shown that if f ∈ Lp , p ≥ 1, then f is Lp -continuous almost everywhere [196; I, p. 65].

58


Lemma 1.12.1 If 0 < r < p < ∞ and f ∈ Lp in [a, b], then f ∈ Lr in [a, b]. If E1 and E2 are the sets of points x in [a, b] where f (x) ≤ 1 and where f (x) > 1, respectively, then Z

a

b

|f |r =

Z

E1

|f |r +

Z

|f |r ≤ (b − a) +

Z

b

|f |p < ∞,

a

E2

and so the result follows.

Lemma 1.12.2 If 0 < r < p < ∞ and f ∈ Lp in some neighbourhood of x, then for small h h1 Z h i1/r h 1 Z h i1/p f (x + t) − P (t) r dt f (x + t) − P (t) p dt ≤ (1.12.8) h 0 h 0 for any polynomial P (t).

By hypothesis, f ∈ Lp and so f (x + t) − P (t) ∈ Lp . As a result, by Lemma 1.12.1, f (x + t) − P (t) ∈ Lr in some neighbourhood of x. Hence, by H¨ older’s inequality, assuming h > 0, Z

0

h

f (x + t) − P (t) r dt ≤ h(p−r)/p

Z

h

0

r/p f (x + t) − P (t) p dt ,

which is (1.12.8). If h < 0, taking the integrals from h to 0, we again get (1.12.8). Theorem 1.12.3 If 0 < r < p < ∞ and if f has a kth Lp -derivative at x, then f has a kth Lr -derivative at x and they are equal.

If f has a kth Lp -derivative at x, then k 1 Z h X p 1/p ti f (x + t) − f(i),p (x) dt = o(hk ) as h → 0, h 0 i! i=0

and so by Lemma 1.12.2

k 1 Z h X r 1/r ti f (x + t) − f(i),p (x) dt = o(hk ) as h → 0. h 0 i! i=0

Hence, f(k),r (x) exists and f(i),r (x) = f(i),p (x) for i = 0, 1, . . . , k.


59

Lp -Boundedness

1.12.2

If in (1.12.1) o(hk ) is replaced by O(hk ) as h → 0, then f is said to be L -bounded of order k at x. p

Theorem 1.12.4 If f is Lp -bounded of order k at x, then f(k−1),p (x) exists. From the definition of Lp -boundedness, we have

1 Z h 1/p f (x + t) − Q(t) p dt = O(hk ) as h → 0, h 0

where Q(t) is a polynomial of degree at most k. Let Q(t) = Pk−1 Q(t) = i=0 bi ti . We have seen in (1.12.4) and (1.12.5) that

(1.12.9) Pk

i=0 bi t

Z h 1/p 1 |t|k dt = O(hk ) as h → 0. h 0

i

and

(1.12.10)

Applying Minkowski’s inequality, we get from (1.12.9) and (1.12.10) 1 h

Z

0

1/p Z h f (x + t) − Q(t) p dt = 1 f (x + t) − Q(t) + bk tk p dt h 0 1 Z h 1/p f (x + t) − Q(t) p dt ≤ h 0 1/p Z h 1 |t|k dt + |bk | h 0

h

= O(hk ) + O(hk ) = O(hk ) as h → 0. (1.12.11)

Since O(hk ) implies o(hk−1 ) as h → 0, (1.12.11) shows that f(k−1),p (x) exists. Pk−1 ti The argument also shows that Q(t) = i=0 i! f(i),p (x) and so, from (1.12.11), we get: Corollary 1.12.5 If f is Lp -bounded of order k at x, then f(k−1),p (x) exists and k−1 1 Z h X ti p 1/p f (x + t) − f(i),p (x) dt = O(hk ) as h → 0. h 0 i! i=0

60

1.13 1.13.1


Symmetric Lp -Derivatives Symmetric Lp -Derivatives and Symmetric Lp -Continuity

Assume that f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x. If there is a polynomial Q(t) = Qx (t) of degree at most k for which Z h p 1/p 1 f (x + t) + (−1)k f (x − t) −Q(t) dt = o(hk ) as h → 0, (1.13.1) h 0 2 then f is said to have a kth symmetric Lp -derivative at x and, if ak /k! is the coefficient of tk in Q(t), then ak is called the kth symmetric Lp -derivative of (s) f at x and is denoted by f(k),p (x). It can be shown as in the case of the unsymmetric Lp -derivative that if (s) f(k),p (x) exists it is unique. (s)

We show that if f(k),p (x) exists, then Q(t) has only even or odd powers of t according as k is even or odd. Let k = 2m. Then for h > 0 we have from (1.13.1) Z h p 1/p 1 f (x + t) + f (x − t) − Q(t) dt = o(h2m ) as h → 0. (1.13.2) h 0 2

This being true for all h, changing h to −h, Z −h p 1/p 1 f (x + t) + f (x − t) − Q(t) = o(h2m ) as h → 0. dt −h 0 2

Changing t to −t this becomes Z h p 1/p 1 f (x − t) + f (x + t) − Q(−t) dt = o(h2m ) as h → 0. (1.13.3) h 0 2

Therefore, applying Minkowski’s inequality, we have from (1.13.2) and (1.13.3) 1/p 1 Z h Q(t) − Q(−t) p dt = o(h2m ) as h → 0. (1.13.4) h 0 P2m Pm If we put Q(t) = i=0 bi ti , then Q(t) − Q(−t) = 2 i=1 b2i−1 t2i−1 and so Q(t) − Q(−t) Q(t) − Q(−t) limt→0 = 2b1 . Therefore, if b1 6= 0, limt→0 >0 t t Q(t) − Q(−t) and so there are ǫ > 0 and h > 0 such that > ǫ for 0 < t < h, t which gives 1 Z h 1/p h Q(t) − Q(−t) p dt >ǫ , h 0 (p + 1)1/p


61

which contradicts (1.13.4). So, b1 = 0. Suppose that b1 = b3 = · · · = b2r−1 = 0, where r < m. If b2r+1 6= 0, then Q(t) − Q(−t) limt→0 > 0 and so, as above, there is an ǫ > 0 and an h > 0 t2r+1 such that 1 Z h 1/p h2r+1 Q(t) − Q(−t) p dt >ǫ 1/p , h 0 p(2r + 1) + 1

which contradicts (1.13.4) since 2r + 1 < 2m. Hence, b2r+1 = 0 and so by induction b2i−1 = 0 for i = 1, 2, . . . , m. k is odd is similar. In this case the integrand in (1.13.4) is The case when Q(t) + Q(−t) p and the right-hand side of (1.13.4) is o(h2m+1 ) from which we get ultimately that Q(t) = −Q(−t). (s)

(s)

Also, if f(k),p (x) exists, then f(k−2),p (x) exists. For, as in (1.12.4), Z 1/p 1 h kp 1 (s) |t | dt f(k),p (x) = o(hk−1 ) as h → 0, k! h 0

(1.13.5)

and, applying Minkowski’s inequality, we have from (1.13.1 and (1.13.5) Z h p 1/p tk (s) 1 f (x + t) + (−1)k f (x − t) − Q(t) + f(k),p (x) dt h 0 2 k! = o(hk−2 ) as h → 0. k

(s)

(s)

Since Q(t)− tk! f(k),p (x) is a polynomial in t of degree at most k −2, f(k−2),p (x) exists. (s) (s) (s) Thus, if f(k),p (x) exists, then f(k−2),p (x), f(k−4),p (x), . . . , exists. If k is even, (s)

(s)

the existence of f(0),p (x) means that f(0),p (x) = a0 , where a0 is the constant term in the polynomial Q(t) in (1.13.1) and so Z h p 1/p 1 f (x + t) + (−1)k f (x − t) (s) − f(0),p (x) dt = o(1) as h → 0. h 0 2 Definition. A function f is said to be symmetric Lp -continuous at x of even order if Z h p 1/p 1 f (x + t) + f (x − t) − f (x) = o(1) as h → 0, dt h 0 2 and f is said to be symmetric Lp -continuous at x of odd order if Z h p 1/p 1 = o(1) as h → 0. f (x + t) − f (x − t) dt h 0

Clearly, if f is symmetric continuous at x of even, (odd), order, then f is symmetric Lp -continuous at x of even, (odd), order.

62


1.13.2

Lp -Smoothness (s)

Let f ∈ Lp , 1 ≤ p < ∞, on some neighbourhood of x and let f(k−2),p (x) exist, where k ≥ 2. Let  (k/2)−1 X t2i (s)    (x), if k is even, f   (2i)! (2i),p i=0 P (t) = (k−1)/2 (1.13.6)  2i−1 X  t (s)   (x), if k is odd. f  (2i − 1)! (2i−1),p i=0 If

Z h p 1/p 1 f (x + t) + (−1)k f (x − t) − P (t) dt = o(hk−1 ) as h → 0, h 0 2 (1.13.7) then f is said to be Lp -smooth of order k at x. (s)

Theorem 1.13.1 If f(k),p (x) exists, then f is Lp -smooth of order k at x.

(s)

Let f(k),p (x) exist, then Z h p 1/p 1 tk (s) f (x + t) + (−1)k f (x − t) − P (t) − f(k),p (x) dt h 0 2 k! = o(hk ) as h → 0,

(1.13.8)

where P (t) is as in (1.13.6). Applying Minkowski’s inequality, we have from (1.13.8) and (1.12.10) Z h p 1/p 1 f (x + t) + (−1)k f (x − t) − P (t) dt h 0 2 Z h p 1/p 1 tk (s) tk (s) f (x + t) + (−1)k f (x − t) = − P (t) − f(k),p (x) + f(k),p (x) dt h 0 2 k! k! Z h p 1/p tk (s) 1 f (x + t) + (−1)k f (x − t) − P (t) − f(k),p (x) dt ≤ h 0 2 k! Z p i1/p 1 h h tk (s) f(k),p (x) dt + h 0 k! = o(hk ) + O(hk ) as h → 0.

(1.13.9)

From (1.13.9), we conclude that (1.13.7) is satisfied and so the result follows. Corollary 1.13.2 If f is Lp -smooth of order k at x, then f is Lp -smooth of order k − 2 at x. (s)

If f is Lp -smooth of order k at x, then by definition f(k−2),p (x) exists and so the result follows from Theorem 1.13.1.


63

Symmetric Lp -Boundedness

1.13.3

If in (1.13.1), o(hk ) is replaced by O(hk ) as h → 0, then f is said to be symmetric Lp -bounded at x of order k. Theorem 1.13.3 If f is symmetric Lp -bounded at x of order k, then (s) f(k−2),p (x) exists. Since f is symmetric Lp -bounded at x of order k,

Z h p 1/p 1 f (x + t) + (−1)k f (x − t) − Q(t) dt = O(hk ) as h → 0, h 0 2 (1.13.10) Pk where Q(t) is a polynomial of degree at most k. Let Q(t) = i=0 bi ti and Pk−2 Q(t) = i=0 bi ti . As in (1.12.10), h1 Z h

h

0

|tk |p dt

i1/p

= O(hk ),

h1 Z h

0

h

|tk−1 |p dt

i1/p

= O(hk−1 ) as h → 0.

(1.13.11) Applying Minkowski’s inequality, we have from (1.13.10) and (1.13.11) Z h p 1/p 1 f (x + t) + (−1)k−2 f (x − t) − Q(t) dt h 0 2 Z h p 1/p 1 f (x + t) + (−1)k−2 f (x − t) − Q(t) dt ≤ h 0 2 i1/p i1/p h1 Z h h1 Z h |tk−1 |p dt |tk |p dt +|bk−1| + |bk | h 0 h 0 = O(hk ) + O(hk−1 ) + O(hk ) as h → 0.

(1.13.12)

Since the right-hand side of (1.13.12) implies o(hk−2 ) as h → 0, it follows that (s) f(k−2),p (x) exists.

1.14 1.14.1

Abel Derivatives Abel Summability

The concept of the Abel derivative of a function f at a point x is closely related to the concept of Abel summability at x of the differentiated series of the Fourier series of f . So, we first discuss Abel summability and deduce some results needed P∞ P∞ in the sequel. n Let n=0 un be a series of constant terms n=0 un r converges P∞such that n for each r, 0 < r < 1, and let φ(r) = u r . Then lim sup r→1 φ(r) n=0 n

64


and lim inf r→1 φ(r) are respectively called the upper P∞ and lower Abel sums of P ∞ upper and lower Abel sums of n=0 un are equal and finite, n=0 un . If theP then the series ∞ is called Abel summable and the common value is n=0 unP P∞ ∞ called the Abel sum of n=0 un . It is clear that if n=0 un is convergent, then P∞ it is Abel summable and the Abel sum is equal to the sum of the series n=0 un . P∞ Lemma 1.14.1 P∞ If themupper and lower Abel sums of n=0 un are finite, then the series n=1 un /n is Abel summable for all integers m ≥ 1. P∞ Let P∞Abel sums, P∞m = 1.n Since n=0 un has finite upper and lower the series n=0 un r converges for each r, 0 < rP< 1, and n=0 un rn ren mains bounded as r → 1−. So, the power series ∞ n=0 un r has a radius of convergence of at least 1. Let φ(r) =

∞ X

un r n ,

Φ(r) =

n=0

Then Z

r

0

φ(t) − u0 dt = t

Z

∞ rX

un tn−1 dt =

0 n=1

Since φ(r) is bounded as r → 1− , Z Φ(r) − Φ(r′ ) =

∞ X un n r . n n=1

∞ X un n r = Φ(r), 0 < r < 1. n n=1

φ(t) − u0 is bounded as t → 1−. Hence, t

φ(t) − u0 dt → 0 as r, r′ → 1 − . t r P∞ So, limr→1− Φ(r) exists finitely, which shows that n=1 un /n is Abel summable. If m ≥ 1, then applying this result for m = 1 repeatedly for m − 1 times completes the proof.

1.14.2

r′

Abel Derivatives

Let f be a 2π-periodic Lebesgue integrable function and let Z Z 1 π 1 π f (t) cos nt dt, bn = f (t) sin nt dt an = π −π π −π

(1.14.1)

be the Fourier coefficients of f . By the Riemann–Lebesgue theorem an → 0 and bn → 0 as n → ∞ and, so, there is an M > 0 such that |an | ≤ M, |bn | ≤ M for all n. Hence, (an cos nx + bn sin nx)rn ≤ 2M rn for all n and x, 0 < r < 1. (1.14.2)

Higher Order Derivatives 65 P∞ Thus, the series 12 a0 + n=1 (an cos nx + bn sin nx)rn converges uniformly and absolutely for fixed r, 0 < r < 1, and let f (r, x) be its sum. Then, using (1.14.1), f (r, x) = =

= = =

∞ X 1 a0 + (an cos nx + bn sin nx)rn 2 n=1 Z π 1 f 2π −π Z ∞ 1Z π X 1 π n + r f (t) cos nt dt cos nx + f (t) sin nt dt sin nx π −π π −π n=1 Z ∞ 1 π 1 X n + r cos n(x − t) dt f (t) π −π 2 n=1 Z ∞ 1 X n 1 π + r cos nt dt f (x + t) π −π 2 n=1 Z π 1 f (x + t)P (r, t) dt, (1.14.3) π −π

where

∞

1 X n r cos nt. P (r, t) = + 2 n=1

(1.14.4)

Clearly the interchange of the order of summation and integration and the change of variable are justified. Now consider 1 11+z + z + z2 + z3 · · · = , 2 21−z

z = reit , 0 < r < 1.

Equating its real part, we have ∞ i 1 − r2 1h 1 X n + r cos nt = 2 n=1 2 1 − 2r cos t + r2

and so P (r, t) = So, we have

i i 1h 1h 1 − r2 1 − r2 . = 2 1 − 2r cos t + r2 2 (1 − r)2 + 4r sin2 t/2

(1.14.5)

P (r, t) ≥ 0; P (r, −t) = P (r, t); (1.14.6) Z 1 π P (r, t) dt = 1; (1.14.7) π −π 1 P (r, t) < ; (1.14.8) 1−r 1−r 1 P (r, t) ≤ C 2 if ≤ r < 1 and 0 < |t| ≤ π, C being a constant.(1.14.9) t 2

66


In fact, (1.14.6) follows from (1.14.5) and (1.14.7) follows from (1.14.4) using term-by-term integration. From (1.14.5), P (r, t) ≤

11−r 1 1 1 − r2 = < , 2 (1 − r)2 21+r 1−r

proving (1.14.8). Finally, (t cos t − sin t)′ < 0 if 0 < t < π and, so, t cos t − sin t is decreasing on that interval. Thus, if 0 < t < π, t cos t − sin t < 0. Hence, (sin t/t)′ = (t cos t−sin t)/t2 < 0 if 0 < t < π showing that sin t/t is decreasing if 0 < t < π. Hence, sin t ≥ 2t/π if 0 ≤ t ≤ π/2 and, so, from (1.14.5) if 1 2 ≤ r < 1 and |t| ≤ π, P (r, t) ≤

1 1 − r2 1−r π2 1 − r , ≤ ≤ 2 4r sin2 t/2 2 t2 2 sin2 t/2

proving (1.14.9).

Lemma 1.14.2 If f is a 2π-periodic Lebesgue integrable function and if at a point x, limt→0+ f (x + t) + f (x − t) − 2f (x) = 0, then the Fourier series of f is Abel summable at x to f (x). Let ǫ > 0 be arbitrary. Then there is a δ > 0, 0 < δ < π, such that f (x + t) + f (x − t) − 2f (x) < ǫ whenever 0 < t < δ. So, from (1.14.6) and (1.14.7)

Z 1 Z δ ǫ δ f (x + t) + f (x − t) − 2f (x) P (r, t) dt ≤ P (r, t) dt π 0 π 0 Z ǫ π ≤ P (r, t) dt = ǫ. π −π (1.14.10)

Also, from (1.14.9), taking 1 π

and, thus,

Z

1 2

≤ r < 1, we have

f (x + t) + f (x − t) − 2f (x) P (r, t) dt δ Z 1 − r π f (x + t) + f (x − t) − 2f (x) ≤C dt π t2 Zδ π 1−r f (x + t) + f (x − t) − 2f (x) dt, ≤C 2 δ π δ π

1 Z π f (x + t) + f (x − t) − 2f (x) P (r, t) dt = 0. lim r→1− π δ

(1.14.11)


67

From (1.14.3), (1.14.7) and (1.14.6) Z 1 π f (x + t) − f (x) P (r, t) dt f (r, x) − f (x) = π −π Z 0 Z π 1 f (x + t) − f (x) P (r, t) dt = + π −π 0 Z 1 π = f (x + t) + f (x − t) − 2f (x) P (r, t) dt π 0 Z δ Z π 1 = f (x + t) + f (x − t) − 2f (x) P (r, t) dt + π δ 0 (1.14.12) and so from (1.14.10), (1.14.11) and (1.14.12), lim supr→1− f (r, x)−f (x) ≤ ǫ. Since ǫ was arbitrary, it follows that limr→1− f (r, x) = f (x), completing the proof. Let k be a positive integer. If k is even, then from (1.14.2) (−1)k/2 (an nk cos nx + bn nk sin nx)rn ≤ 2M nk rn for all x and n, 0 < r < 1. P∞ k n 1/n k n Since limn→∞ P∞(n r ) k/2 = r k< 1, the seriesk n=1 nnr is convergent and so the series n=1 (−1) (an n cos nx + bn n sin nx)r is uniformly and absolutely convergent for fixed r, 0 < r < 1. Hence, the series f (r, x) =

∞ X 1 a0 + (an cos nx + bn sin nx)rn 2 n=1

(1.14.13)

can be differentiated term-by-term k times with respect to x. If k is odd, then as in (1.14.2) (an sin nx − bn cos nx)rn ≤ 2M rn for all n and x, 0 < r < 1,

and so

(−1)(k+1)/2 (an nk sin nx − bn nk cos nx)rn ≤ 2M nk rn for all x and n, 0 < r < 1,

and, as above, we can differentiate the series (1.14.13) term-by-term k times with respect to x. So differentiating (1.14.13) k times with respect to x, we have ∂kf (r, x) = (1.14.14) ∂xk P ∞ (−1)k/2 n=1 (an nk cos nx + bn nk sin nx)rn , if k is even, P∞ (−1)(k+1)/2 n=1 (an nk sin nx − bn nk cos nx)rn , if k is odd.

68

Higher Order Derivatives The upper and lower Abel derivates of f at x of order k are defined by ADk f (x) = lim sup r→1−

ADk f (x) = lim inf r→1−

∂kf (r, x), ∂xk

∂kf (r, x), ∂xk

respectively. If ADk f (x) = ADk f (x), the common value is called the Abel derivative of f at x of order k and is denoted by ADk f (x). Theorem 1.14.3 If AD k f (x) and ADk f (x) are finite and k ≥ 2, then ADk−2 f (x) exists finitely.

The proof follows from Lemma 1.14.1.

While the finiteness of ADk f (x) and ADk f (x) implies the existence of ADk−2 f (x) finitely, nothing can be said about the existence of ADk−1 f (x) since Abel summability at x does not ensure the Abel summability of the once integrated series at the point x. P∞ Example. The series n=1 sin nx is Abel summable at x = 0, but the series P ∞ n=1 (cos nx)/n is not Abel summable at x = 0.

Because of this the Abel derivative is considered as a symmetric derivative. From Theorem 1.14.3, it follows that if AD k f (x) exists finitely, then ADk−2 f (x) exists finitely and so AD0 f (x) or AD1 f (x) exists finitely according as k is even or odd, where AD0 f (x) = limr→1− f (r, x). Thus, the Abel derivative at x of order 0 is just the Abel sum at x of the Fourier series of f . From Lemma 1.14.2, if f is continuous at x, then AD0 f (x) = f (x).

1.14.3

Abel Continuity

Let f be a 2π-periodic Lebesgue integrable function and let f (r, x) be as in (1.14.3). Then lim supr→1− f (r, x) and lim inf r→1− f (r, x) are respectively called the upper and lower Abel limits of f at x. If lim supr→1− f (r, x) = lim inf r→1− f (r, x), then this common value is called the Abel limit if f at x. If the Abel limit exists finitely with value f (x), then f is said to be Abel continuous at x. The upper and lower Abel sums of the Fourier series of f at x are the upper and lower Abel limits of f at x, respectively. Also the Fourier series of f is Abel summable to S at x if and only if the Abel limit of f at x exists and is equal to S, and the Fourier series of f is Abel summable to f (x) at x if and only if f is Abel continuous at x. In the following, we shall see that if f is continuous at x, then f is Abel continuous at x. The converse is obviously not true.


69

Theorem 1.14.4 Let f be 2π-periodic and Lebesgue integrable. Then for any point x lim inf t→0

f (x+t)+f (x−t) 2

≤ lim inf f (r, x) ≤ lim sup f (r, x) r→1−

≤ lim sup t→0

r→1−

f (x +t)+f (x−t) . 2

We prove the right-hand side inequality, the proof for left-hand side inequality being similar. f (x + t) + f (x − t) . Then there Let ǫ > 0 be arbitrary and L = lim supt→0 2 f (x + t) + f (x − t) < L + ǫ for 0 < t < δ. So, exists δ, 0 < δ < π, such that 2 from (1.14.6) and (1.14.7), 1 π

Z

Z (2L + 2ǫ) δ f (x + t) + f (x − t) P (r, t) dt ≤ P (r, t) dt π 0 0 Z Z 2(L + ǫ) π (L + ǫ) π ≤ P (r, t) dt = P (r, t) dt = L + ǫ. (1.14.15) π π 0 −π δ

From (1.14.9), taking 1/2 < r < 1, Z Z 1 π 1 π f (x + t) + f (x − t) P (r, t) dt ≤ f (x + t) + f (x − t) P (r, t) dt π δ π δ Z 1−r 1 π ≤ f (x + t) + f (x − t) C 2 dt π δ t Z π C(1 − r) f (x + t) + f (x − t) dt ≤ πδ 2 Zδ C(1 − r) π ≤ f (x + t) + f (x − t) dt. 2 πδ 0 (1.14.16) From (1.14.15) and (1.14.16), Z Z C(1 − r) π 1 π f (x+t)+f (x−t) P (r, t) dt ≤ L+ǫ+ f (x+t)+f (x−t) dt. 2 π 0 πδ 0

Letting r → 1−, we have Z 1 π f (x + t) + f (x − t) P (r, t) dt ≤ L + ǫ. lim sup r→1− π 0

(1.14.17)

70


Also, from (1.14.3) and (1.14.6), we have Z Z 1 π 1 π f (r, x) = f (x + t)P (r, t) dt = f (x − t)P (r, t) dt π −π π −π Z π 1 = f (x + t) + f (x − t) P (r, t) dt 2π −π Z 1 π f (x + t) + f (x − t) P (r, t) dt. (1.14.18) = π 0 Letting r → 1− in (1.14.18), we have from (1.14.17) that lim supr→1− f (r, x) ≤ L + ǫ. Since ǫ is arbitrary, the proof is complete. Corollary 1.14.5 If f is continuous at x then f is Abel continuous at x.

1.14.4

Abel Smoothness

Let f be a 2π-periodic Lebesgue integrable function and let k be a fixed positive integer. Define AS k f (x) = lim sup(1 − r) r→1−

AS k f (x) = lim inf (1 − r) r→1−

∂k f (r, x) ∂xk

∂k f (r, x) ∂xk

∂kf where (r, x) is as in (1.14.14). Then AS k f (x) and AS k f (x) are called the ∂xk upper and lower indices of Abel smoothness of f of order k at x, respectively. If AS k f (x) = AS k f (x), the common value, denoted by AS k f (x), is called the index of Abel smoothness of f of order k at x. If AS k f (x) = AS k f (x) = 0, then f is said to be Abel smooth of order k at x. Theorem 1.14.6 If AD k f (x) and ADk f (x) are finite, then f is Abel smooth of order k at x. ∂kf (r, x) remains bounded as ∂xk r → 1− and, so, AS k f (x) = AS k f (x) = 0, completing the proof.

Since AD k f (x) and ADk f (x) are finite,

1.15 1.15.1

Laplace Derivatives Laplace Derivatives and Laplace Continuity

Let f be special Denjoy integrable in some neighbourhood N (x) of x and let r be a fixed positive integer. If there exists a polynomial Q(t) = Qx (t) of


71

degree at most r such that Z δ sr+1 e−st f (x + t) − Q(t) dt = o(1) as s → ∞

(1.15.1)

0

for some δ > 0 such that x + δ ∈ N (x), then f is said to have a right Laplace derivative at x of order r and if ar /r! is the coefficient of tr in Q(t), then ar is called the right Laplace derivative of f at x of order r and is denoted by LDr+ f (x). Note that if δ1 > 0, δ2 > 0, δ1 6= δ2 , x + δ1 , x + δ2 ∈ N (x), then integrating by parts, Z δ2 Z δ2 h f (x + t) − Q(t) dt e−st f (x + t) − Q(t) dt = sr+1 e−sδ2 sr+1 δ1

+s

Z

δ1

δ2

e−st

δ1

Z

δ1

= o(1) as s → ∞,

i f (x + ξ) − Q(ξ) dξ dt

t

and, so, if (1.15.1) is true for δ = δ1 , then it is also true for δ = δ2 and, therefore, (1.15.1) does not depend on δ. If (1.15.1) is replaced by Z 0 sr+1 (1.15.1′) est f (x + t) − Q(t) dt = o(1) as s → ∞, −δ

or equivalently

sr+1

Z

δ 0

e−st f (x − t) − Q(−t) dt = o(1) as s → ∞,

then f is said to have a left Laplace derivative at x of order r and if ar /r! is the coefficient of tr in Q(t), then ar is called the left Laplace derivative of f at x of order r and is denoted by LDr− f (x). We shall consider LDr+ f (x); the case for LDr− f (x) is similar. We need the following lemma. Lemma 1.15.1 If δ > 0 and p, q are positive integers, then Z δ sq e−st tp dt = p!sq−p−1 + o(1) as s → ∞. 0

Putting τ = st, Z δ Z sδ q −st p q−p−1 s e t dt = s e−τ τ p dτ 0 0 Z ∞ Z ∞ q−p−1 −τ p q−p−1 = s e τ dτ − s e−τ τ p dτ 0 sδ Z ∞ = p!sq−p−1 − sq−p−1 e−τ τ p dτ. sδ

72


Let k be a positive integer such that q − p + k ≥ 2. Then, since

τ q+k < eτ , (q + k)!

we have Z

∞

Z

∞

(q + k)! dτ q−p+k sδ sδ τ (q + k)! = = o(1) as s → ∞, (q − p + k − 1)(sδ)q−p+k−1 e

−τ p

τ dτ
0 be arbitrary. Then there is a δ > 0 such that f (x + t) − f (x) < ǫ for 0 < t < δ and, therefore, Z s

δ

e

−st

0

Z f (x + t) − f (x) dt ≤ ǫs

δ

e−st dt → ǫ as s → ∞.

So, from (1.15.5) and (1.15.6), LD+ f (x) − f (x) 0 Z h Z δ = lim s e−st f (x + t) − f (x) dt + s s→∞ 0

≤ ǫ.

(1.15.6)

0

0

δ

e

−st

f (x) dt − f (x) i

Since ǫ is arbitrary the assertion follows. To define the right upper and lower Laplace derivates of f at x + of order k, we suppose that LDk−1 f (x) exists finitely and let Q(t) = Pk−1 i + (t /i!)LD f (x). Define i i=0

74


+

LDk f (x) = lim sup sk+1 s→∞

LD+ k f (x)

= lim inf s s→∞

k+1

Z

Z

δ

0 δ

0

+

e−st f (x + t) − Q(t) dt;

e−st f (x + t) − Q(t) dt.

(1.15.7) (1.15.8)

If LDk f (x) = LD+ k f (x), the common value, possibly infinite, is the derivative LDk+ f (x). This definition agrees with the former definition, as we now show. Let LDk+ f (x) exist in this sense and be finite. Then since, by Rδ Lemma 1.15.1, lims→∞ sk+1 0 e−st (tk /k!) dt = 1 and since, from (1.15.7) and (1.15.8), lim s

k+1

s→∞

Z

0

δ

k−1 X (ti /i!)LDi+ f (x) dt = LDk+ f (x), e−st f (x + t) − i=0

we have lim s

s→∞

k+1

Z

0

δ

k X (ti /i!)LDi+ f (x) dt = 0, e−st f (x + t) − i=0

showing that LDk+ f (x) exists in the former sense with the same value. The converse is clear.

Definition A special Denjoy integrable function f is said to be Laplace right-continuous at x if lim s

s→∞

Z

δ

e−st f (x + t) dt = f (x)

0

for an arbitrary δ > 0, and f is said to be Laplace left-continuous at x if lim s

s→∞

Z

δ

e−st f (x − t) dt = f (x)

0

for an arbitrary δ > 0. If f is both Laplace left- and right-continuous at x, then f is said to be Laplace continuous at x. We now show that if f is right-continuous, left-continuous, at x, then it is Laplace right-continuous, Laplace left-continuous, at x. Suppose that f is right-continuous at x and let ǫ > 0 be arbitrary. Then there is a δ > 0 such that f (x + t) − f (x) < ǫ for 0 < t < δ. So, Z s

0

δ

Z e−st f (x + t) − f (x) dt ≤ ǫs

0

δ

e−st dt → ǫ as s → ∞.

(1.15.9)


75

Therefore, from (1.15.6), lim s s→∞

Z

δ

e

−st

0

h Z δ f (x + t) dt − f (x) = lim s e−st f (x + t) − f (x) dt s→∞ 0 Z δ i +s e−st f (x) dt − f (x) 0

≤ ǫ.

Since ǫ was arbitrary, the result follows.

1.15.2

Laplace Boundedness

If in (1.15.1) the condition o(1) is replaced by O(1) as s → ∞, then f is said to be right Laplace bounded, or Laplace bounded on the right, at x of order r. Theorem 1.15.3 If f is Laplace bounded on the right at x of order r, then f + has a right Laplace derivative of order r − 1 at x and LDr f (x) and LD+ r f (x) + + are finite. Conversely, if LDr f (x) and LDr f (x) are finite, then f is Laplace bounded on the right at x of order r.

Let f be Laplace bounded on the right at x of order r. Then Z δ sr+1 e−st f (x + t) − Q(t) dt = O(1) as s → ∞, (1.15.10) 0

P where Q(t) is a polynomial of degree at most r. Let Q(t) = ri=0 bi ti and Pr−1 define Q(t) = i=0 bi ti . From (1.15.9) and Lemma 1.15.1, we have sr

Z

δ

0

e−st f (x + t) − Q(t) dt

= sr

hZ

0

δ

e−st f (x + t) − Q(t) dt +

Z

δ

e−st br tr dt

0

i

1 = O(1) + br r!sr−r−1 + o(1) = o(1) as s → ∞, s Pr−1 + and, hence, LDr−1 f (x) exists finitely and Q(t) = i=0 (ti /i!)LDi+ f (x). Since Q(t) = Q(t) + br tr , from (1.15.10) and Lemma 1.15.1, sr+1

Z

0

δ

r−1 X (ti /i!)LDi+ f (x) dt = O(1) + sr+1 e−st f (x + t) − i=0

= O(1) + O(1) = O(1) as s → ∞, +

and, so, LDr f (x) and LD+ r f (x) are finite.

Z

δ

e−st br tr dt

0

(1.15.11)

76

Higher Order Derivatives +

Conversely, suppose that LDr f (x) and LD+ are finite. Then (1.15.11) r f (x) Pr−1 holds. Let c be any finite constant and let Q(t) = i=0 (ti /i!)LDi+ f (x) + ctr . Then, from (1.15.11) Z δ Z δ sr+1 e−st f (x + t) − Q(t) dt = O(1) + sr+1 e−st ctr dt 0

0

= O(1) + O(1) = O(1) as s → ∞.

Since Q(t) is a polynomial of degree at most r, f is Laplace bounded on the right of x of order r.

1.15.3

Bilateral Laplace Derivatives

The left upper and the left lower Laplace derivates of f at x of order k, say, − written LDk f (x) and LD− k f (x), respectively, are defined as follows. Suppose − that the left Laplace derivative of order r − 1, LDr−1 f (x), exists finitely and Pk−1 i − let Q(t) = i=0 (t /i!)LDi f (x). Define Z 0 − k k+1 LDk f (x) = lim sup(−1) s est f (x + t) − Q(t) dt, s→∞

−δ

k k+1 LD− k f (x) = lim inf (−1) s s→∞

Z

0

−δ

−

est f (x + t) − Q(t) dt,

respectively. If LDk f (x) = LD− k f (x), the common value is the left Laplace derivative of f at x of order k denoted by LDk− f (x). This definition of LDk− f (x) is equivalent to the previous definition if it is finite, as we now show. Suppose that LDk− f (x) exists in the sense of the previous definition. − Then LDi f (x) exists for i = 0, 1, · · · , k − 1 and sk+1

Z

0

−δ

k X (ti /i!)LDi− f (x) dt = o(1) as s → ∞. est f (x + t) − i=0

So, using Lemma 1.15.1, sk+1

Z

0

−δ

k−1 X (ti /i!)LDi− f (x) dt est f (x + t) − i=0

= sk+1

Z

0

−δ

est (tk /k!)LDk− f (x) dt + o(1)

1 = sk+1 LDk− f (x) k!

Z

δ

e−st (−t)k dt + o(1)

0

= (−1)k LDk− f (x) + o(1).


77

Hence, lim (−1)k sk+1

s→∞

Z

0

−δ

k−1 X (ti /i!)LDi− f (x) dt = LDk− f (x), est f (x + t) − i=0

showing that LDk− f (x) is the derivative Conversely, if LDk− f (x) exists and

in the sense of the present definition. is finite in the sense of the present

definition, then

k k+1

lim (−1) s

s→∞

Z

0

−δ

k−1 X (ti /i!)LDi− f (x) dt = LDk− f (x), est f (x + t) − i=0

and, so, Z 0 k−1 X (ti /i!)LDi− f (x) dt = (−1)k LDk− f (x)+o(1) as s → ∞, sk+1 est f (x+t)− −δ

i=0

and using Lemma 1.15.1 Z 0 k X (ti /i!)LDi− f (x) dt = o(1) as s → ∞, sk+1 est f (x + t) − −δ

i=0

LDk− f (x)

showing that is the derivative in the sense of the previous definition. Now, if LDr+ f (x) and LDr− f (x) exist and LDi+ f (x) = LDi− f (x), i = 1, . . . , r, then we say that f has a Laplace derivative at x of order r and this common value is denoted by LDr f (x), the (bilateral) Laplace derivative at x of order r. The upper and lower (bilateral) Laplace derivates of f at x of order r are defined by +

−

LDr f (x) = max{LDr f (x), LDr f (x)}, − LDr f (x) = min{LD+ r f (x), LD r f (x)}, respectively, and f is said to be Laplace bounded of order r at x if f is both left and right Laplace bounded of order r at x. Analogues of Theorems 15.2 and 15.3 hold.

1.16 1.16.1

Symmetric Laplace Derivatives and Laplace Smoothness Symmetric Laplace Derivatives and Symmetric Laplace Continuity

Let f be special Denjoy integrable in some neighbourhood N (x) of x and let r be a fixed positive integer. If there exists a polynomial P (t) = Px (t) of

78


degree at most r such that s

r+1

Z

0

δ

e−st

h f (x + t) + (−1)r f (x − t) 2

= o(1) as s → ∞

−

P (t) + (−1)r P (−t) i dt 2 (1.16.1)

for some δ > 0 such that x ± δ ∈ N (x), then f is said to have a symmetric Laplace derivative at x of order r, and, if ar /r! is the coefficient of tr in P (t), then ar is called the symmetric Laplace derivative of f at x of order r and is denoted by SLDr f (x). Now, if 0 < δ1 < δ2 , then, as in the case of the Laplace derivatives, it can be shown that Z δ2 h f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) i e−st sr+1 dt − 2 2 δ1 = o(1) as s → ∞ and, so, if (1.16.1) is true for δ = δ1 , it is also true for δ = δ2 and, so (1.16.1) does not depend on δ. Theorem 1.16.1 If the derivative SLDr f (x) exists, it is unique. P P Let P1 (t) = ri=0 ci ti and P2 (t) = ri=0 di ti be two polynomials for which (1.16.1) is true. Then, by Lemma 1.15.1 sr+1

Z

0

δ

Z r X (ci − di )sr+1 e−st P1 (t) − P2 (t) dt = =

i=0

e−st ti dt

0

i=0

r X

δ

(ci − di )i!sr−i + o(1) as s → ∞,

(1.16.2)

and sr+1

Z

0

δ

Z r X (ci − di )sr+1 (−1)i e−st P1 (−t) − P2 (−t) dt = =

i=0 r X i=0

δ

e−st ti dt

0

(ci − di )(−1)i i!sr−i + o(1)

as s → ∞. (1.16.3)


79

Also, by (1.16.1) sr+1 2

Z

0

δ

h i e−st P1 (t) − P2 (t) + (−1)r P1 (−t) − P2 (−t) dt Z

P2 (t) + (−1)r P2 (−t) i dt 2 2 0 Z δ h f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) i 1 1 −sr+1 e−st − dt 2 2 0 = o(1) − o(1) = o(1) as s → ∞. (1.16.4) = sr+1

δ

e−st

h f (x + t) + (−1)r f (x − t)

−

From (1.16.2), (1.16.3) and (1.16.4), r h i X 1 + (−1)r−i (ci − di ) i!sr−i = o(1) as s → ∞. 2 i=0

(1.16.5)

If r is even, r = 2m say, then (1.16.5) is (c0 − d0 )s2m + (c2 − d2 )2!s2m−2 + · · · · · · + (c2m−2 − d2m−2 )(2m − 2)!s2 + (c2m − d2m )(2m)! = o(1) as s → ∞. (1.16.6) and so dividing by s2m and then letting s → ∞, we have c0 = d0 and (1.16.6) becomes (c2 − d2 )2!s2m−2 + · · · + (c2m−2 − d2m−2 )(2m − 2)!s2 + (c2m − d2m )(2m)! = o(1) as s → ∞. Dividing by s2m−2 and then letting s → ∞, we have c2 = d2 . Continuing this process we get c2m−2 = d2m−2 and (c2m − d2m )(2m)! = o(1) as s → ∞, and so finally letting s → ∞, c2m = d2m . If r is odd, r = 2m + 1 say, then (1.16.5) is (c1 −d1 )s2m +(c3 −d3 )3!s2m−2 +· · ·+(c2m+1 −d2m+1 )(2m+1)! = o(1) as s → ∞. and, proceeding as above, we have c1 = d1 , c3 = d3 , . . . , c2m+1 = d2m+1 . This shows the uniqueness of SLDr f (x).

Theorem 1.16.2 If r ≥ 2 and SLDr f (x) exists, then SLDr−2 f (x) exists. Suppose that SLDr f (x) exists. Then there is a polynomial P (t) of degree at most r such that (1.16.1) holds. Let aj /j! be the coefficient of tj in ar−1 r−1 ar r P (t) and let P (t) = P (t) − t − t . Then (r − 1)! r! P (t) + (−1)r−2 P (−t) = P (t) + (−1)r P (−t) − 2

ar r t r!

80


and so, by (1.16.1) and Lemma 1.15.1, Z δ h f (x + t) + (−1)r−2 f (x − t) P (t) + (−1)r−2 P (−t) i r−1 s e−st dt − 2 2 0 Z δ h f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) i dt − = sr−1 e−st 2 2 0 Z δ ar + sr−1 e−st tr dt r! 0 ar r−1−r−1 1 + o(1) = o(1) as s → ∞. (1.16.7) = 2 o(1) + r!s s r! Since P (t) is a polynomial of degree at most r − 2, SLDr−2 f (x) exists and with value ar−2 . If r is even, then by the above SLD2 f (x) exists and, as in (1.16.7), Z δ i h f (x + t) + f (x − t) − a0 dt = o(1) as s → ∞. s e−st 2 0 Rδ Since s 0 e−st dt → 1 as s → ∞, SLD0 f (x) exists and SLD0 f (x) = a0 = lim s s→∞

Z

δ

0

e−st

f (x + t) + f (x − t) dt. 2

Similarly, if r is odd, SLD1 f (x) exists. Theorem 1.16.3 (a) If limt→0 12 f (x+t)+f (x−t) exists finitely, with value ℓ0 , say, then SLD0 f (x) exists and SLD0 f (x) = ℓ0 . f (x + t) − f (x − t) (b) If limt→0 exists finitely, with value ℓ1 , say, then 2t SLD1 f (x) exists and SLD1 f (x) = ℓ1 . (a) Let ℓ0 be as defined in (a) above. Then for arbitrary ǫ > 0, δ > 0 f (x + t) + f (x − t) − ℓ0 < ǫ for 0 < t < σ, there is a σ, 0 < σ < δ, such that 2 and so Z σ Z σ h i −st f (x + t) + f (x − t) s e e−st dt. − ℓ0 dt < ǫs 2 0 0

Hence,

s

Z

0

δ

e−st

h f (x + t) + f (x − t)

i − ℓ0 dt

2 i h f (x + t) + f (x − t) − ℓ0 dt e−st +s 2 0 σ Z σ Z δ h f (x + t) + f (x − t) i ≤ ǫs e−st dt + s e−st − ℓ0 dt. 2 0 σ

Z = s

σ

Z

δ


81

Letting s → ∞, Z

lim sup s s→∞

δ

e−st

0

Similarly, Z

lim inf s s→∞

δ

e−st

0

h f (x + t) + f (x − t) 2

h f (x + t) + f (x − t)

i − ℓ0 dt ≤ ǫ.

i − ℓ0 dt ≥ −ǫ.

2

Since ǫ > 0 is arbitrary, Z δ i h f (x + t) + f (x − t) − ℓ0 dt ≤ ǫ, lim s e−st s→∞ 2 0

showing that SLD0 f (x) = ℓ0 . (b) Let ℓ1 be as defined in (b) above. Then for arbitrary ǫ > 0, δ > 0 there f (x + t) − f (x − t) is a σ, 0 < σ < δ, such that − ℓ1 < ǫ for 0 < t < σ, 2t f (x + t) − f (x − t) − ℓ1 t < ǫt for 0 < t < σ. So, by Lemma or equivalently 2 1.15.1, Z σ Z σ h f (x + t) − f (x − t) i s e−st − ℓ1 t dt ≤ ǫs e−st t dt 2 0 0 ǫ = + o(1) as s → ∞. s So, Z δ i h f (x + t) − f (x − t) − ℓ1 t dt e−st s 2 0 Z σ Z δ h f (x + t) − f (x − t) i = s +s e−st − ℓ1 t dt 2 o σ Z δ i h f (x + t) − f (x − t) ǫ e−st − ℓ1 t dt + o(1). ≤ +s s 2 σ Hence,

lim sup s s→∞

lim inf s and, so, lim s

s→∞

δ

h f (x + t) − f (x − t)

i − ℓ1 t dt ≤ 0.

h f (x + t) − f (x − t)

i − ℓ1 t dt ≥ 0,

h f (x + t) − f (x − t)

i − ℓ1 t dt = 0

e−st

0

Similarly, s→∞

Z Z

δ

e−st

0

Z

δ

e−st

0

and, hence, SLD1 f (x) = ℓ1 .

2

2

2

82


Definition. A special Denjoy integrable function f is said to be symmetric Laplace continuous at x of even order if Z δ f (x + t) + f (x − t) lim s e−st dt = f (x), s→∞ 2 0 and f is said to be symmetric Laplace continuous at x of odd order if Z δ e−st f (x + t) − f (x − t) dt = 0. lim s s→∞

0

It can be verified that if f is special Denjoy integrable in some neighbourhood of x, then f is symmetric Laplace continuous at x of even order if f is symmetric continuous of even order at x, and f is symmetric Laplace continuous at x of odd order if f is symmetric continuous of odd order at x. To define the upper and lower symmetric Laplace derivates of f at x of order k we suppose that SLDk−2 f (x) exists. If k is even, k = 2m say, define SLD2m f (x) = lim sup s

2m+1

s→∞

Z

δ

e−st

0

h f (x + t) + f (x − t) 2

−

m−1 X i=0

i t2i SLD2i f (x) dt, (2i)!

with a similar definition, but taking lim inf, for SLD2m f (x). If k is odd, k = 2m + 1, say, define Z δ f (x + t) − f (x − t) dt, if m = 0; SLD1 f (x) = lim sup s2 e−st 2 s→∞ 0 and if m ≥ 1, SLD2m+1 f (x) = Z lim sup s2m+2 s→∞

0

δ

e−st

h f (x + t) − f (x − t) 2

−

m−1 X i=0

i t2i+1 SLD2i+1 f (x) dt, (2i + 1)!

with a similar definition, but taking lim inf, for SLD2m+1 f (x). If SLDk f (x) = SLDk f (x), then the common value is the symmetric Laplace derivative of f at x of order k, possibly infinite, SLDk f (x). Clearly this definition agrees with the previous one if SLDk f (x) is finite.

1.16.2

Symmetric Laplace Boundedness

If, in (1.16.1), o(1) is replaced by O(1) as s → ∞, then f is said to be symmetric Laplace bounded at x of order r. Theorem 1.16.4 If f is symmetric Laplace bounded at x of order r, then f has a finite symmetric Laplace derivative at x of order r − 2 and SLDr f (x) and SLDr f (x) are finite. Conversely, if SLDr f (x) and SLDr f (x) are finite, then f is symmetric Laplace bounded at x of order r


83

Let f be symmetric Laplace bounded at x of order r. Then there is a polynomial P (t) of degree at most r such that sr+1

Z

sr−1

Z

P (t) + (−1)r P (−t) i dt = O(1) 2 2 0 as s → ∞. Pr Pr−2 Let P (t) = i=0 bi ti . Then writing P (t) = i=0 bi ti we have δ

e−st

h f (x + t) + (−1)r f (x − t)

−

P (t) + (−1)r−2 P (−t) i dt 2 2 0 Z δ h f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) i r−1 = s e−st − dt 2 2 0 Z δ +sr−1 e−st br tr dt δ

e−st

h f (x + t) + (−1)r−2 f (x − t)

−

0

1 = 2 O(1) + br r!sr−1−r−1 + o(1) = o(1) as s → ∞. s

(1.16.8)

Since P (t) is a polynomial of degree at most r−2, SLDr−2 f (x) exists. Further, P m−1 t2i  if r = 2m,  i=0 (2i)! SLD2i f (x), P (t) + (−1)r−2 P (−t) =   Pm−1 t2i+1 SLD2i+1 f (x), if r = 2m + 1. i=0 (2i+1)!

So, from (1.16.9), s

Z

h f (x + t) + (−1)r−2 f (x − t)

P (t) + (−1)r−2 P (−t) i dt 2 2 0 Z δ h f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) i = sr+1 e−st dt − 2 2 0 Z δ + sr+1 e−st br tr dt

r+1

δ

e−st

−

0

= O(1) + br r!sr+1−r−1 + o(1) as s → ∞.

(1.16.9)

Hence, taking lim sup and lim inf in (1.16.10), SLDr f (x) and SLDr f (x) are finite. Conversely, let SLDr f (x) and SLDr f (x) be finite. Then from the definition it follows that SLDr−2 f (x) exists finitely and so for a polynomial P (t) of degree at most r − 2 Z

h f (x + t) + (−1)r f (x − t) P (t) + (−1)r−2 P (−t) i dt e−st − 2 2 0 = O(1) as s → ∞. (1.16.10)

sr+1

δ

84


Let P1 (t) = P (t) + ctr−1 + dtr where c and d are constants. Then, from (1.16.11) and by Lemma 1.15.1, Z δ h f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) i 1 1 sr+1 e−st dt − 2 2 0 Z δ h f (x + t) + (−1)r f (x − t) P (t) + (−1)r−2 P (−t) i dt − = sr+1 e−st 2 2 0 Z δ −sr+1 e−st dtr dt 0

= O(1) − dr!sr+1−r−1 − o(1) = O(1) as s → ∞,

and so f is symmetric Laplace bounded at x of order r.

1.16.3

Laplace Smoothness

Let f be special Denjoy integrable in some neighbourhood of x and let SLDr−2 f (x) exist finitely. Let X tr−2i T (t) = SLDr−2i f (x), (1.16.11) (r − 2i)! i=1

where the summation in (1.16.12) extends to i = r/2 if r is even and to i = (r − 1)/2 if r is odd. Let Z δ i h f (x + t) + (−1)r f (x − t) − T (t) dt, LS r f (x) = lim sup sr e−st 2 s→∞ 0 LS r f (x) = lim inf s s→∞

r

Z

0

δ

e−st

h f (x + t) + (−1)r f (x − t) 2

i − T (t) dt.

Then LS r f (x) and LS r f (x) are called the upper and lower index of Laplace smoothness of f of order r at x. If LS r f (x) = LS r f (x) = LSr f (x) = 0, then f is said to be Laplace smooth of order r at x. Theorem 1.16.5 If SLDr f (x) and SLDr f (x) are finite, then f is Laplace smooth of order r at x. Since SLDr f (x) and SLDr f (x) are finite, by definition SLDr−2 f (x) exists and Z δ i h f (x + t) + (−1)r f (x − t) r+1 − T (t) dt = O(1) as s → ∞, s e−st 2 0 where T (t) is as defined in (1.16.12). So, Z δ i h f (x + t) + (−1)r f (x − t) r − T (t) dt = o(1) as s → ∞, s e−st 2 0

completing the proof.

Chapter 2 Relations between Derivatives

In this chapter, we shall study relations between the various derivatives considered in Chapter I where all the appropriate notations are defined. All the kth order derivatives considered are generalizations of the kth order ordinary derivative f (k) . We first show that this is the case for the kth order Peano derivative f(k) .

2.1

Ordinary and Peano Derivatives, f (k) and f(k)

The following theorem sharpens Theorem 1.4.1 of Chapter I. Theorem 2.1.1 If f (k) (x) exists finitely so does f(k) (x) and f(k) (x) = f (k) (x). Moreover, if f (k) (x) exists finitely +

(x) ≤ f (k+1) (x) ≤ D+ f (k) (x), D+ f (k) (x) ≤ f + (k+1) where D+ f (k) = D+ (f (k) ) and D+ f (k) = D+ (f (k) ) are, respectively, the lower and upper Dini right derivates of f (k) . Similarly, −

D− f (k) (x) ≤ f − (x) ≤ f (k+1) (x) ≤ D− f (k) (x). (k+1) The proof of the first part is in Theorem 1.4.1 of Chapter I. For the second part, suppose that f (k) (x) exists finitely. Then so does f(k) (x) and so let k i X ti (k + 1)! h f(i) (x) . γk+1 (f ; x, t) = k+1 f (x + t) − t i! i=0 Then, as in the relation (1.4.5) of Chapter I, since f(i) (x) = f (i) (x) for i = 0, 1, . . . , k, γk+1 (f ; x, t) =

f (k) (x + ξt ) − f (k) (x) , 0 < ξt < t. ξt

85

86


Letting t → 0+, +

f (k+1) (x) = lim sup γk+1 (f ; x, t) = lim sup t→0+

t→0+

f (k) (x + ξt ) − f (k) (x) ξt

f (k) (x + t) − f (k) (x) ≤ lim sup = D+ f (k) (x). t t→0+ This proves the right inequality in the first set of inequalities and the rest of the inequalities can be proved in a similar manner. Since + − (x), f − (x) f (k+1) (x) = max f (k+1) (x), f (k+1) (x) , f (k+1) (x) = min f + (k+1) (k+1) and

+ − Df (k) (x) = max D f (k) (x), D f (k) (x) , Df (k) (x) = min D+ f (k) (x), D− f (k) (x) ,

we have from the above

Df (k) (x) ≤ f (k+1) (x) ≤ f (k+1) (x) ≤ Df (k) (x). The converse of Theorem 2.1.1 is not true; see the example following Theorem 1.4.1 of Chapter I.

2.2

∗ Riemann∗ and Peano Derivatives, f(k) and f(k)

∗ Theorem 2.2.1 If k ≥ 2 and if f(k−1) (x0 ) exists finitely, then f(k−1) (x0 ) also ∗ exists finitely and f(i) (x0 ) = f(i) (x0 ) for i = 1, 2, . . . , k − 1. Moreover,

lim

xk−1 →x0

· · · lim k!Qk (f ; x0 , . . . , xk ) = γk (f ; x0 , xk − x0 ) x1 →x0

(2.2.1)

∗ and, hence, f(k) (x0 ) exists, possibly infinitely, if and only if f(k) (x0 ) exists; in ∗ either case, f(k) (x0 ) = f(k) (x0 ). ∗ ∗ Since f(k−1) (x0 ) exists finitely, by Theorem 1.5.2 of Chapter I, f(i) (x0 ) exists finitely for i = 1, 2, . . . , k − 1 and so ∗ lim · · · lim i!Qi (f ; x0 , . . . , xi ) = f(i) (x0 ), i = 1, 2, . . . , k − 1.

xi →x0

x1 →x0

(2.2.2)

By Theorem 1.5.1 of Chapter I, f is continuous at x0 and so limx1 →x0 Q1 f : x1 , x2 = Q1 (f ; x0 , x2 ). Also, since by (2.2.2) we have limx1 →x0 Q1 f :x0 , x1 =

Relations between Derivatives

87

∗ f(1) (x0 ) = f ′ (x0 ), it follows that

Q1 (f ; x0 , x1 ) − Q1 (f ; x1 , x2 ) x0 − x2 f (x0 ) − Q1 (f ; x0 , x2 ) = x0 − x2 f (x2 ) − f (x0 ) − (x2 − x0 )f ′ (x0 ) = (x2 − x0 )2 1 = γ2 (f ; x0 , x2 − x0 ), 2!

lim Q2 (f ; x0 , x1 , x2 ) = lim

x1 →x0

x1 →x0 ′

proving the result in the case k = 2. ∗ Suppose that the result is true for k = r and that f(r) (x0 ) exists finitely. ∗ Then by Theorem 1.5.2 of Chapter I, f(i) (x0 ) exists finitely for i = 1, 2, . . . , r, and (2.2.2) is true for i = 1, 2, . . . , r. Since the result is assumed for k = r, ∗ f(r−1) (x0 ) exists finitely and f(i) (x0 ) = f(i) (x0 ) for i = 1, 2, . . . , r − 1, and by (2.2.1) lim

xr−1 →x0

· · · lim r!Qr (f ; x0 , . . . , xr ) = γr (f ; x0 , xr − x0 ). x1 →x0

(2.2.3)

∗ Since f(r) (x0 ) exists, taking the limit in (2.2.3) as xr → x0 ,

lim · · · lim r!Qr (f ; x0 , x1 , . . . , xr ) = f(r) (x0 ).

xr →x0

x1 →x0

(2.2.4)

Since f is continuous at x0 , lim Qr (f ; x1 , x2 , . . . , xr+1 ) = Qr (f ; x0 , x2 , . . . , xr+1 ).

x1 →x0

(2.2.5)

So, from (2.2.5) and (2.2.3), lim · · · lim r!Qr (f ; x1 , x2 . . . , xr+1 )

xr →x0

=

x1 →x0

lim · · · lim r!Qr (f ; x0 , x2 . . . , xr+1 )

xr →x0

x2 →x0

= γr (f ; x0 , xr+1 − x0 ).

(2.2.6)

From (2.2.4) and (2.2.6), lim · · · lim (r + 1)!Qr+1 (f ; x0 , x1 . . . , xr+1 )

xr →x0

x1 →x0

r+1 lim · · · = x0 − xr+1 xr →x0 × lim r!Qr (f ; x0 , x1 , . . . , xr ) − r!Qr (f ; x1 , x2 . . . , xr+1 ) x1 →x0

=

r+1 f(r) (x0 ) − γr (f ; x0 , xr+1 − x0 ) = γr+1 (f ; x0 , xr+1 − x0 ). x0 − xr+1 (2.2.7)

88


The relation (2.2.7) shows that (2.2.1) is true for k = r + 1. So, the result is proved by induction. Theorem 2.2.2 If k ≥ 2, f is continuous at x0 and f(k−1) (x0 ) exists finitely, ∗ ∗ then f(k−1) (x0 ) exists finitely and f(i) (x0 ) = f(i) (x0 ) for i = 1, 2, . . . , k − 1. Moreover the relation (2.2.1) holds. Hence, f(k) (x0 ) exists, possibly infinitely, ∗ ∗ if and only if f(k) (x0 ) exists and in either case f(k) (x0 ) = f(k) (x0 ). Let k = 2; f is continuous at x0 so limx1 →x0 Q1 (f ; x1 , x2 ) = Q1 (f ; x0 , x2 ); also, limx1 →x0 Q1 (f ; x0 , x1 ) = f ′ (x0 ) = f(1) (x0 ). Hence, as in Theorem 2.2.1, lim 2!Q2 (f ; x0 , x1 , x2 ) = γ2 (f ; x0 , x2 − x0 )

x1 →x0

completing the proof for the case k = 2. Suppose that the result is true for k = r and that f(r) (x0 ) exists finitely. ∗ Then f(r−1) (x0 ) exists finitely and since the result is true for k = r, f(i) (x0 ) ∗ exists finitely and f(i) (x0 ) = f(i) (x0 ) for i = 1, 2, . . . , r − 1, and the relation (2.2.3) holds. Since f(r) (x0 ) exists taking the limit in (2.2.3) as xr → 0, (2.2.4) ∗ ∗ holds and f(r) (x0 ) exists with f(r) (x0 ) = f(r) (x0 ). Since f is continuous at x0 and the result is true for k = r, (2.2.6) holds. Hence, as in the previous theorem, we obtain (2.2.7) showing that (2.2.1) is also true for k = r + 1 and completing the proof. ∗ Remark. It follows from Theorems 1.1 and 1.2 that f(k) and f(k) are the same in the sense that if one of them exists, so does the other with equal value and, further, the relation (2.2.1) shows that the upper, lower and unilateral derivates are the same.

2.3

Symmetric Riemann∗ and Symmetric de la Vall´ ee ∗(s) (s) Poussin Derivatives, f(k) and f(k)

Theorem 2.3.1 Let k ≥ 2. (s) (a) If f(2k−2) (x0 ) exists finitely, then lim . . . lim (2k)!Q2k (f ; x0 − hk , . . . , x0 − h1 , x0 , x0 + h1 , . . . , x0 + hk )

hk−1 →0

h1 →0

= (s)

∗(s)

̟2k (f ; x0 , hk )

(2.3.1)

and so f(2k) (x0 ) exists if and only if f(2k) (x0 ) exists and in either case they are equal;


89

(s)

(b) if f(2k−3) (x0 ) exists finitely, then lim . . . lim (2k − 1)!Q2k−1 (f ; x0 − hk , . . . , x0 − h1 , x0 + h1 , . . . , x0 + hk )

hk−1 →0

h1 →0

=

̟2k−1 (f ; x0 , hk ),

(2.3.2)

∗(s)

(s)

and so f(2k−1) (x0 ) exists if and only if f(2k−1) (x0 ) exists and, in either case, they are equal. We prove (a) and discuss briefly the proof of (b). We may suppose without loss in generality that x0 = 0 and, by adding a suitable constant if necessary, that f (x0 ) = 0. Applying (1.1.1) of Chapter I,

Q2k (f ; −hk , . . . , −h1 , 0, h1 , . . . , hk ) = Q2k−1 (f ; −hk , . . . , −h1 , 0, h1 , . . . , hk−1 )−Q2k−1 (f ; −hk−1 , . . . , −h1 , 0, h1 , . . . , hk ) −2hk 1 Q2k−2 (f ; −hk , . . . , −h1 , 0, h1 , . . . , hk−2 ) = 2hk (hk−1 + hk ) −Q2k−2 (f ; −hk−1 , . . . − h1 , 0, h1 , . . . , hk−1 ) −Q2k−2 (f ;−hk−1 , . . . ,−h1 , 0, h1 , . . . , hk−1 )

=

1 2hk (hk−1 + hk )

+Q2k−2 (f ; −hk−2 , . . . , −h1 , 0, h1 , . . . , hk )

Q2k−2 (f ; −hk , . . . , −h1 , 0, h1 , . . . , hk−2 ) +Q2k−2 (f ; −hk−2 , . . . , −h1 , 0, h1 , . . . , hk )

−2Q2k−2 (f ; −hk−1 , . . . − h1 , 0, h1 , . . . , hk−1 ) .

(2.3.3)

Assume now that k = 2. Then (2.3.3) reduces to Q4 (f ; −h2 , −h1 , 0, h1 , h2 ) Q2 (f ; −h2 , −h1 , 0) + Q2 (f ; 0, h2 , h1 ) − 2Q2 (f ; −h1 , 0, h1 ) = . (2.3.4) 2h2 (h1 + h2 ) Since f (0) = 0, we have the following: f (−h2 ) f (−h1 ) − , h2 (h2 − h1 ) h1 (h2 − h1 ) f (h1 ) f (h2 ) − , Q2 (f ; 0, h2 , h1 ) = h2 (h2 − h1 ) h1 (h2 − h1 ) f (−h1 ) f (h1 ) − . Q2 (f ; −h1 , 0, h1 ) = 2h21 2h21

Q2 (f ; −h2 , −h1 , 0) =

(2.3.5) (2.3.6) (2.3.7)

90


From (2.3.4), (2.3.5), (2.3.6) and (2.3.7), f (h2 ) + f (−h2 ) f (h1 ) + f (−h1 ) f (h1 ) + f (−h1 ) − − h2 (h2 − h1 ) h1 (h2 − h1 ) h21 Q4 (f ; −h2 , −h1 , 0, h1 , h2 ) = 2h2 (h1 + h2 ) and, hence, lim 4!Q4 (f ; −h2 , −h1 , 0, h1 , h2 )

h1 →0

f (h2 ) + f (−h2 ) f (h1 ) + f (−h1 ) h1 + 1 − h2 (h2 − h1 ) h21 h2 − h1 = lim 4! h1 →0 2h2 (h1 + h2 ) i 4! h f (h2 ) + f (−h2 ) (s) (0) − f = 2 (2) 2h2 h22 h 2 4! f (h2 ) + f (−h2 ) h2 (s) i − f(2) (0) = ̟4 (f ; 0, h2 ), = 4 h2 2 2!

proving the result for k = 2. We now suppose that the result is true for k = r ≥ 2, and prove it for (s) k = r + 1. Let k = r + 1 and assume that f(2r) (0) exists finitely. Then, from (2.3.3), we have Q2r+2 (f ; −hr+1 , . . . , −h1 , 0, h1 , . . . , hr+1 ) = 1 1 Q2r (f ; −hr+1 , . . . , −h1 , 0, h1 , . . . , hr−1 ) = hk (hk−1 + hk ) 2

+Q2r (f ; −hr−1 , . . . , −h1 , 0, h1 , . . . , hr+1 ) −Q2r (f ; −hr , . . . − h1 , 0, h1 , . . . , hr ) . (2.3.8)

Since f (0) = 0, Q− 2r = Q2r (f ; −hr+1 , . . . , −h1 , 0, h1 , . . . , hr−1 ) =

r+1 X j=1

hj

+

(2.3.9)

f (−hj ) Qr−1 i=1 (hj − hi ) i=1 (hj + hi )

Qr+1 i6=j

r−1 X j=1

hj

f (hj ) , Qr+1 i=1 (hj − hi ) i=1 (hj + hi )

Qr−1 i6=j

Q+ 2r = Q2r (f ; −hr−1 , . . . , −h1 , 0, h1 , . . . , hr+1 ) =

r−1 X

j=1 hj

+

f (−hj ) Qr+1

Qr−1

r+1 X

i=1 (hj − hi )

i6=j

j=1 hj

Qr+1

i=1 (hj

f (hj ) Qr−1

i=1 (hj − hi )

i6=j

+ hi )

i=1 (hj

+ hi )

.

(2.3.10)


91

From (2.3.9) and (2.3.10), + Q− 2r + Q2r 2

(2.3.11)

f (hj ) + f (−hj ) 2Q = Qr+1 r−1 (h − h ) i=1 (hj + h1 ) h j i j i=1 j=1 r+1 X

i6=j

f (hj ) + f (−hj ) 2Q + . Qr−1 r+1 i=1 (hj − hi ) i=1 (hj + h1 ) j=1 hj r−1 X

Since

i6=j

r h2r+2 f (hj ) + f (−hj ) X h2i j j s (0) + = f(2i) ̟2r+2 (f ; 0, hj ), 2 (2i)! (2r + 2)! i=1

(2.3.12)

we get from (2.3.11) + Q− 2r + Q2r = I1 + I2 + I3 + I4 2

(2.3.13)

where Pr

h2i j s i=1 (2i)! f(2i) (0) Qr+1 Qr−1 i=1 (hj − hi ) i=1 (hj j=1 hj i6=j

I1 =

r+1 X

I2 =

r+1 X

I3 =

r−1 X

I4 =

r−1 X

+ h1 )

h2r+2 j (2r+2)! ̟2r+2 (f ; 0, hj ) Qr+1 Qr−1 i=1 (hj − hi ) i=1 (hj + j=1 hj i6=j

Pr

h2i j s i=1 (2i)! f(2i) (0) Qr−1 Qr+1 i=1 (hj − hi ) i=1 (hj j=1 hj i6=j

h1 )

+ h1 )

h2r+2 j (2r+2)! ̟2r+2 (f ; 0, hj ) Qr−1 Qr+1 i=1 (hj − hi ) i=1 (hj + j=1 hj i6=j

h1 )

,

,

,

.

Pr x2i (s) Let P (x) = i=1 (2i)! f(2i) (0). Then since P (0) = 0 and P (x) = P (−x), I1 +I3 is the divided difference of P (x) at the points −hr+1 , . . . , −h1 , 0, h1 , . . . , hr−1 , and also at the points −hr−1 , . . . , −h1 , 0, h1 , . . . , hr+1 . Since P (x) is a polynomial of degree at most 2r, this divided difference is equal to the coefficient of x2r . Hence, 1 (s) I1 + I3 = (2.3.14) f (0). (2r)! (2r)

92

Higher Order Derivatives (s)

To estimate I2 , note that since f(2r) (0) exists r f (hj ) + f (−hj ) X h2i j (s) = f(2i) (0) + o(h2r j ), 2 (2i)! i=1

and so, from (2.3.12) and (2.3.15),

(2.3.15)

h2r+2 j ̟2r+2 (f ; 0, hj ) = o(h2r j ). Using (2r + 2)!

this we have from (2.3.13) I2 =

r−1 X j=1

+

=

h2r+2 j (2r+2)! ̟2r+2 (f ; 0, hj ) Qr+1 Qr−1 i=1 (hj − hi ) i=1 (hj + j=r hj i6=j

j=1

Hence, lim . . . lim I2 =

hr−1 →0

h1 →0

i6=j

r+1 X

r−1 X

+

o(h2r j ) Qr+1 Q hj i=1 (hj − hi ) r−1 i=1 (hj + hi ) hi )

o(hj2r−1 ) Qr−1 i=1 (hj − hi ) i=1 (hj + hi )

Qr+1 i6=j

h2r+1 j (2r+2)! ̟2r+2 (f ; 0, hj ) Qr−1 Qr+1 i=1 (hj − hi ) i=1 (hj + j=r i6=j

r+1 X

lim . . . lim

hr−1 →0

h1 →0

hi )

.

h2r+1 ̟2r+2 (f ; 0, hr ) r Qr−1 Qr+1 (2r + 2)! i=1 (hr − hi ) i=1 (hr + hi )

→ i6=r 2r+1 hr+1 ̟2r+2 (f ; 0, hr+1 ) + lim . . . lim Qr Qr−1 hr−1 →0 h1 →0 (2r + 2)! i=1 (hr+1 − hi ) i=1 (hr+1 + hi ) " # 2r+1 2r+1 hr+1 ̟2r+2 (f ; 0, hr+1 ) hr ̟2r+2 (f ; 0, hr ) 1 = + r−1 (2r + 2)! hr−1 (hr − hr+1 )hr−1 hr+1 (hr+1 − hr )hr−1 r r r+1

" # h3r ̟2r+2 (f ; 0, hr ) h3r+1 ̟2r+2 (f ; 0, hr+1 ) 1 . + = (2r + 2)! (hr − hr+1 ) (hr+1 − hr )

Similarly, I4 =

r−1 X j=1

=

r−1 X

hj

o(h2r j ) Qr−1 i=1 (hj − hi ) i=1 (hj + hi )

Qr+1

i6=j

o(hj2r−1 ) Qr+1 Qr−1 i=1 (hj i=1 (hj + hi ) j=1 i6=j

− hi )

,

(2.3.16)


93

and so lim . . . lim I4 = 0.

hr−1 →0

(2.3.17)

h1 →0

From (2.3.13), (2.3.14), (2.3.16) and (2.3.17), + Q− 2r + Q2r = (2.3.18) hr−1 →0 h1 →0 2 " # 1 (s) h3r ̟2r+2 (f ; 0, hr ) h3r+1 ̟2r+2 (f ; 0, hr+1 ) 1 . f (0) + + (2r)! (2r) (2r + 2)! (hr − hr+1 ) (hr+1 − hr )

lim . . . lim

Since the result is true for k = r, 1 ̟2r (f ; 0, hr ). (2r)! (2.3.19) from (2.3.9) and (2.3.10) we get from

lim . . . lim Q2r (f ; −hr , . . . , −h1 , 0, h1 , . . . , hr ) =

hr−1 →0

h1 →0

+ Restoring the values of Q− 2r and Q2r (2.3.8), (2.3.18) and (2.3.19)

lim . . . lim Q2r+2 (f ; −hr+1 , . . . , −h1 , 0, h1 , . . . , hr+1 ) h1 →0 " h h3 ̟ 1 (s) 1 1 r 2r+2 (f ; 0, hr ) f(2r) (0) + = hr+1 (hr + hr+1 ) (2r)! (2r + 2)! (hr − hr+1 ) # h3 ̟2r+2 (f ; 0, hr+1 ) i 1 + r+1 − ̟2r (f ; 0, hr ) . (2.3.20) (hr+1 − hr ) (2r)!

hr−1 →0

Since, (s) hr ̟2r+2 (f ; 0, hr ) = (2r + 1)(2r + 2) ̟2r (f ; 0, hr ) − f(2r) (0) (s)

and since ̟2r (f ; 0, hr ) → f(2r) (0) as hr → 0, we have, taking the limit as hr → 0 in (2.3.20), lim . . . lim Q2r+2 (f ; −hr+1 , . . . , −h1 , 0, h1 , . . . , hr+1 ) =

hr →0

h1 →0

completing the proof of (a) by induction.

̟2r+2 (f ; 0, hr+1 ) , (2r + 2)!

94

Higher Order Derivatives To prove (b), note, as in (2.3.3), that 1

Q3 (f ; −h2 , −h1 , h1 , h2 ) =

Q1 (f ; −h2 , −h1 ) + Q1 (f ; h1 , h2 ) −2Q1 (f ; −h1 , h1 ) h f (−h ) − f (−h ) f (h ) − f (h ) 1 2 1 2 1 = + 2h2 (h1 + h2 ) h1 − h2 h2 − h1 f (h1 ) − f (−h1 ) i −2 2h1 h f (h ) − f (−h ) 1 2 2 = − 2h2 (h1 + h2 ) h2 − h1 f (h1 ) − f (−h1 ) i f (h1 ) − f (−h1 ) 2h1 , −2 − 2h1 h2 − h1 2h1 2h2 (h1 + h2 )

and so i 1 h f (h2 ) − f (−h2 ) (s) (0) − h f 2 (1) h1 →0 h32 2 1 = ̟3 (f ; 0, h2 ), 3! proving (b) for k = 2. Suppose that it is true for k = r ≥ 2 and let k = r + 1. Then, as in (2.3.8), lim Q3 (f ; −h2 , −h1 , h1 , h2 ) =

Then

Q2r+1 (f ; −hr+1 , . . . , −h1 , h1 , . . . , hr+1 ) − 1 Q2r−1 + Q+ = 2r−1 − 2Q2r−1 . 2hr+1 (hr + hr+1 ) Q− 2r−1 = Q2r−1 (f ; −hr+1 , . . . , −h1 , h1 , . . . , hr−1 ) =

r+1 X j=1

+

−f (−hj ) Qr−1 i=1 (hj − hi ) i=1 (hj + hi )

Qr+1 i6=j

r−1 X j=1

and

f (hj ) , Qr+1 i=1 (hj − hi ) i=1 (hj + hi )

Qr−1 i6=j

Q+ 2r−1 = Q2r−1 (f ; −hr−1 , . . . , −h1 , h1 , . . . , hr+1 ) =

r+1 X j=1

+

Qr+1

r−1 X j=1

f (hj ) Qr−1

i=1 (hj − hi )

i6=j

Qr−1

−f (−hj ) Qr+1

i=1 (hj − hi )

i6=j

i=1 (hj

+ hi )

i=1 (hj

+ hi )

,


95

and so + Q− 2r−1 + Q2r−1 2 r+1 1 X 2 f (hj ) − f (−hj ) = Qr+1 Qr−1 i=1 (hj − hi ) i=1 (hj + hi ) j=1 i6=j

+

r−1 X j=1

Using the relation

f (hj ) − f (−hj ) . Qr+1 Qr−1 i=1 (hj − hi ) i=1 (hj + hi ) 1 2

i6=j

r−1 X h2i+1 (s) h2r+1 1 f (hj ) − f (−hj ) = f(2i+1) (0) + ̟2r+1 (f ; 0, hj ) 2 (2i + 1)! (2r + 1)! i=0

and, proceeding as in (a), the proof of (b) for k = r + 1 can be completed. Theorem 2.3.2 Let k ≥ 2. (a) If lim . . . lim Q2k (f ; x0 − hk , . . . , x0 − h1 , x0 , x0 + h1 , . . . , x0 + hk ) (2.3.21)

hk−1 →0

h1 →0

(s)

exists finitely, then f(2k−2) (x0 ) exists and (2.3.21) equals (b) if

1 (2k)! ̟2k (f ; x0 , hk );

lim . . . lim Q2k−1 (f ; x0 − hk , . . . , x0 − h1 , x0 + h1 , . . . , x0 + hk ) (2.3.22)

hk−1 →0

h1 →0

(s)

exists finitely, then f(2k−3) (x0 ) exists and (2.3.22) equals

1 (2k−1)! ̟2k−1 (f ; x0 , hk ).

We prove (a). As in Theorem 2.3.1, we may suppose that x0 = 0 = f (x0 ). Since f (0) = 0, we have by (1.1.3) of Chapter I Q2k (f ; −hk , . . . , −h1 , 0, h1 , . . . , hk ) =

+

=

k X

j=1 hj

k X

j=1 hj

f (−hj ) Qk i=1 (hj − hi ) i=1 (hj + hi )

Qk

i6=j

f (hj ) Qk i=1 (hj − hi ) i=1 (hj + hi )

Qk

i6=j

k

X f (hj ) + f (−hj ) f (h1 ) + f (−h1 ) + . Q Qk k 2h21 i=2 (h21 − h2i ) j=2 2h2j i=1 (h2j − h2i ) i6=j

(2.3.23)

96


Now, by hypothesis, the left-hand side of (2.3.23) tends to a finite limit as h1 → 0. Also, the summation on the right-hand side of (2.3.23) tends to a finite limit as h1 → 0. Hence, the first term on the right-hand side of (2.3.23) f (h1 ) + f (−h1 ) tends to a finite limit as h1 → 0. So tends to a finite limit as h21 (s) h1 → 0 and this shows, since f (0) = 0, that f(2) (0) exists finitely. In addition, if k = 2, then from (2.3.23) 1 (s) f (h2 ) + f (−h2 ) f (0) + 2h22 (2) 2h42 1 1 h f (h2 ) + f (−h2 ) h22 (s) i − f(2) (0) = ̟4 (f ; 0, h2 ), = 4 h2 2 2 4!

lim Q4 (f ; −h2 , −h1 , 0, h1 , h2 ) = −

h1 →0

completing the proof in the case k = 2. (s) So, let k > 2 and suppose that f(2r) (0) exists finitely, 1 ≤ r < k − 1. Then r

f (h) + f (−h) X h2i (s) h2r+2 = f(2i) (0) + ̟2r+2 (f ; 0, h). 2 (2i)! (2r + 2)! i=1

(2.3.24)

From (2.3.23) and (2.3.24), we have

Q2k = Q2k (f ; −hk , . . . , −h1 , 0, h1 , . . . , hk ) =

=

Pr

h2i (s) i=1 (2i)! f(2i) (0) Qk 2 2 2 i=1 (hj − hi ) j=1 hj i6=j

k X

k X f (hj ) + f (−hj ) Qk 2 2 2 i=1 (hj − hi ) j=1 2hj i6=j

+

1 (2r + 2)!

k X

h2r+2 ̟2r+2 (f ; 0, hj ) j . Qk 2 2 2 i=1 (hj − hi ) j=1 hj i6=j

(2.3.25) (s) Let P (x) = j=1 x2i /(2i)! f(2i) (0). Then P (0) = 0 and P (x) = P (−x) so that the first summation on the right-hand side of (2.3.25) is half the divided difference of P (x) at the points −hk , . . . , −h1 , 0, h1 , . . . , hk . Since P (x) is a polynomial of degree at most 2r < 2k, this divided difference is zero. So, from (2.3.25), k X h2r 1 j ̟2r+2 (f ; 0, hj ) Q2k = . (2.3.26) Qk 2 2 (2r + 2)! j=1 i=1 (hj − hi ) Pr

i6=j

Since

r

we have

h2r+2 f (h) + f (−h) X h2i (s) ̟2r+2 (f ; 0, h) = − f (0), (2r + 2)! 2 (2i)! (2i) i=1 r i (2r)! h f (h) + f (−h) X h2i (s) (0) − f (2i) h2r 2 (2i)! i=1 (s) = (2r + 2)(2r + 1) ̟2r (f ; 0, h) − f(2i) (0) ,

h2 ̟2r+2 (f ; 0, h) = (2r + 2)(2r + 1)


97

and so h2 ̟2r+2 (f ; 0, h) → 0 as h → 0. Hence, from (2.3.26), k

X h2r 1 j ̟2r+2 (f ; 0, hj ) Q (2r + 2)! j=2 h2j ki=2 (h2j − h2i )

lim Q2k =

h1 →0

i6=j

k X hj2r−2 ̟2r+2 (f ; 0, hj ) 1 = , Qk 2 2 (2r + 2)! j=2 i=2 (hj − hi ) i6=j

and so lim lim Q2k

h2 →0 h1 →0

=

k k X X hj2r−2 ̟2r+2 (f ; 0, hj ) hj2r−4 ̟2r+2 (f ; 0, hj ) 1 1 = . Q Qk 2 2 (2r + 2)! j=3 h2j ki=3 (h2j − h2i ) (2r + 2)! j=3 i=3 (hj − hi ) i6=j

i6=j

Continuing this process, we get lim . . . lim Q2k =

hr →0

h1 →0

"

k X ̟2r+2 (f ; 0, hj ) 1 Qk (2r + 2)! j=r+1 i=r+1 (h2j − h2i ) i6=j

# k X ̟2r+2 (f ; 0, hr+1 ) ̟2r+2 (f ; 0, hj ) 1 = . (2.3.27) + Qk Qk 2 2 2 2 (2r + 2)! i=r+1 (hj − hi ) i=r+2 (hr+1 − hi ) j=r+2 i6=j

Since r < k − 1, r + 1 ≤ k − 1 and, so, by the hypothesis limhr+1 →0 . . . Pk ̟ (f ;0,h ) limh1 →0 Q2k exists finitely. Also, limhr+1 →0 j=r+2 Qk2r+2 (h2 −hj 2 ) exists i=r+1 i6=j

j

i

finitely. From (2.3.27), limhr+1→0 ̟2r+2 (f ; 0, hr+1 ) exists finitely and so (s) f(2r+2) (0) exists finitely. (s)

Hence, by induction on r, we conclude that f(2k−2) (0) exists finitely, completing the proof of the first part of (a). To prove the second part of (a), note that we have proved (2.3.27) assuming (s) (s) the existence of f(2r) (0) and, therefore, since f(2k−2) (0) exists finitely, (2.3.27) is true if r = k − 1 and hence, from (2.3.27) with r = k − 1, we have lim . . . lim Q2k =

hk−1 →0

completing the proof of (a).

h1 →0

1 ̟2k (f ; 0, hk ), 2k!

98

2.4 2.4.1


Ces` aro and Peano Derivatives Ces` aro and Peano derivatives, Ck Df and f(k)

Theorem 2.4.1 Let f be Cr−1 P -integrable in [a, b]. If x ∈ [a, b], let  Rx  (Cr−1 P ) a f, for k = 1, Fk (x) = (2.4.1) Rx  (C r−k P ) a Fk−1 , for 2 ≤ k ≤ r and write Φ = Fr . If f is Cr -continuous at x then: (a) Φ(r) (x) exists and equals f (x); (b) Φ(k) (x) = Fr−k (x) if 0 < k < r; (c) Cr (f ; x, x + h) = γr (Φ; x, h) for x, x + h ∈ [a, b].

Since r Cr (f ; x, x + h) = r (Cr−1 P ) h

Z

x+h

(x + h − t)r−1 f (t) dt,

x

integrating by parts successively we have by Theorem 1.8.1 of Chapter I and using (2.4.1), Z x+h 1 hr Cr (f ; x, x + h) = (Cr−1 P ) (x + h − t)r−1 f (t) dt r! (r − 1)! x x+h 1 = (x + h − t)r−1 F1 (t) x (r − 1)! Z x+h 1 (x + h − t)r−2 F1 (t) dt (Cr−2 P ) + (r − 2)! x x+h hr−1 1 = − F1 (x) + (x + h − t)r−2 F2 (t) x (r − 1)! (r − 2)! Z x+h 1 (Cr−3 P ) (x + h − t)r−3 F2 (t) dt + (r − 3)! x hr−2 hr−1 F1 (x) − F2 (x) + · · · = − (r − 1)! (r − 2)! Z x+h r−1 i X h = − Fr−i (x) + (C0 P ) Fr−1 i! x i=1 = Φ(x + h) − Φ(x) −

r−1 i X h i=1

i!

Fr−i (x).

(2.4.2)


99

Since f is Cr -continuous, Cr (f ; x, x + h) → f (x) as h → 0 and, so, r−1 i X r! h Fr−i (x) → f (x) as h → 0. Φ(x + h) − Φ(x) − hr i! i=1

This shows that Φ(x + h) − Φ(x) −

r−1 i X h i=1

i!

Fr−i (x) −

hr f (x) = o(hr ), as h → 0, r!

proving (a) and (b). Also, by (2.4.2), Cr (f ; x, x + h) = γr (Φ; x, h), which completes the proof. Theorem 2.4.2 Let f be Cr -continuous in [a, b] and let Φ be the function of Theorem 2.4.1. Then, for x ∈ [a, b], f (x) = Φ(r) (x) and Cr Df (x) = Φ(r+1) (x), etc.; further, Cr Df (x) exists if and only if Φ(r+1) (x) exits and in either case they are equal. The first part of this theorem follows from Theorem 2.4.1 (c), and using parts (c) and (a) of that theorem, r + 1 r + 1 Cr (f ; x, x + h) − f (x) = γr (Φ; x, h) − Φ(r) (x) = γr+1 (Φ; x, h), h h

the result follows.

Theorem 2.4.3 Let Φ be the function of Theorem 2.4.1. If Φ(r) exists finitely in [a, b], then for 1 ≤ k ≤ r (a) Φ(k) is the Ck−1 -derivative of Φ(k−1) in [a, b]; (b) Φ(k) is Ck -continuous in [a, b]; (c) Φ(k−1) is an indefinite Ck−1 P integral of Φ(k) in [a, b]; (d) Ck (f ; x, x + h) = γk (Φ; x, h) for x, x + h ∈ [a, b]. If r = 1, then Φ(1) , being the ordinary first derivative of Φ, is the C0 -derivative of Φ, Φ(1) is C0 P -integrable and Φ(0) = Φ is its indefinite C0 P integral; further, Z x+h Φ(x + h) − Φ(x) 1 Φ(1) = = γ1 (Φ; x, x + h). C1 (Φ(1) ; x, x + h) = (C0 P )− h h x Hence, C1 (Φ(1) ; x, x + h) → Φ(1) (x) as h → 0, proving the result in the case r = 1. Suppose now that the result is true for r = 1, 2, . . . , r0 , we prove the result for r = r0 + 1. Let Φ(r0 +1) exist in [a, b]. Then Φ(r0 ) exists in [a, b]. By

100


the induction hypothesis, Φ(r0 ) is the Cr0 −1 -derivative of Φ(r0 −1) , Φ(r0 ) is Cr0 continuous, Φ(r0 −1) is the indefinite Cr0 −1 P -integral of Φ(r0 ) and Cr0 (f ; x, x+ h) = γr0 (Φ; x, h). Hence, r0 + 1 r0 + 1 Cr0 (Φ(r0 ) ; x, x + h) − Φ(r0 ) (x) = γr0 (Φ; x, h) − Φ(r0 ) (x) h h = γr0 +1 (Φ; x, h), and, since Φ(r0 +1) (x) exists, γr0 +1 (Φ; x, h) → Φ(r0 +1) (x) as h → 0 and so, r0 + 1 Cr0 (Φ(r0 ) ; x, x + h) − Φ(r0 ) (x) = Φ(r0 +1) (x). h→0 h lim

Thus, Cr0 DΦ(r0 ) (x) = Φ(r0 +1) (x) and, since this holds for all x ∈ [a, b], it follows, from the definition of the Cr P -integral, that Φ(r0 +1) is Cr0 P -integrable and Φ(r0 ) is its indefinite Cr0 P -integral. So, integrating by parts successively as in (2.4.2), hr0 +1 Cr +1 (Φ(r0 +1) ; x, x + h) = (r0 + 1)! 0 =

R x+h 1 (x r0 ! (Cr0 P )− x

+ h − t)r0 Φ(r0 +1) (t) dt

Φ(x + h) − Φ(x) −

Pr0

hi i=1 i! Φ(i) (x).

Hence, Cr0 +1 (Φ(r0 +1) ; x, x + h) = γr0 +1 (Φ; x, h), thus Cr0 +1 (Φ(r0 +1) ; x, x + h) → Φ(r0 +1) (x), proving that Φ(r0 +1) is Cr0 +1 -continuous at x. So, the result is true for r = r0 + 1 and the proof is now completed by induction. Theorem 2.4.4 Let Φ be the function of Theorem 2.4.1 and let Φ(r) exist finitely in [a, b] and let x ∈ [a, b]. Then Φ(r+1) (x) exists if and only if Cr DΦr (x) exists and in either case they are equal.

From (d) of Theorem 2.4.3, we have

r + 1 r+ 1 Cr (Φ(r) ; x, x+h)−Φ(r) (x) = γr (Φ; x, h)−Φ(r) (x) = γr+1 (Φ; x, h), h h

and the result follows.

Remark. Remark. If Φ(r) exists at a single point, say ξ, then it may be the case that the previous derivatives Φ(k) , 1 ≤ k ≤ r−1, do not exist at any point except ξ and it may even be the case that the function Φ is not measurable and, so, the existence of the Cesàro derivative at ξ is not guaranteed. To see this, let A ⊂ I = [− 21 , 12 ] be a nonmeasurable set such that its outer measure and that of its complement in I are both equal to 1. It follows that there is a point ξ ∈ I such that every neighbourhood N (ξ) of ξ contains a portion of A and its complement. Define (x − ξ)r+1 if x ∈ A, Φ(x) = (x − ξ)r+2 if x ∈ / A.

Relations between Derivatives Then Φ(ξ + h) =

hr+1 hr+2

101

if ξ + h ∈ A, if ξ + h ∈ / A.

So, Φ(ξ + h) = o(hr ) as h → 0, and hence, Φ(r) (ξ) exists. However, there is no neighbourhood of ξ in which Φ(k) exists, 1 ≤ k ≤ r, except at ξ. Clearly Φ is not measurable.

2.4.2

Ces` aro and absolute Peano derivatives, Ck Df and f ∗

Theorem 2.4.5 Let f be defined in a neighbourhood N (x0 ) of x0 in R. If f ∗ (x0 ) exists, then Cn Df (x0 ) exists for some n and Cn Df (x0 ) = f ∗ (x0 ). The converse is not true. Suppose that f ∗ (x0 ) exists. Then there is a function g and a nonnegative integer n such that g(n) (x) = f (x), x ∈ N (x0 ), g(n+1) (x0 ) = f ∗ (x0 ).

(2.4.3) (2.4.4)

By Theorem 2.4.4, g(n+1) (x0 ) is the Cesàro derivative of g(n) at x0 of order n. Hence, by (2.4.3) and (2.4.4), f ∗ (x0 ) is the Cesàro derivative Cn Df (x0 ). This proves the first part. For the second nq part, consider the following o example. 2 Let E = {0} ∪ and define (4n+1)π ; n = 1, 2, 3, . . .

0, if x ∈ E, sin(1/x2 ), if x ∈ R \ E. Rx Then f is Riemann integrable and set F (x) = 0 f . Changing the variable and applying the second mean value theorem, Z 1 Z ∞ sin u |x3 | Z ξ x F (x) = du = sin(1/t2 ) dt = sin u du ≤ |x3 |. 2 x−2 u3/2 2 0 x−2 f (x) =

This implies that F(1) (0) = F(2) (0) = 0, and F(1) (x) = sin(1/x2 ), x 6= 0. Hence, R 1 h C1 (f ; 0, h) − f (0) h o f − f (0) = lim lim 1 1 h→0 h→0 2h 2h 2 = lim 2 F (h) − F (0) − F(1) (0) = F(2) (0). h→0 h

So, C1 Df (0) exists with value 0. However, f ∗ (0) does not exist. For, if possible, suppose the contrary, that f ∗ (0) does exist. Let δ > 0 be arbitrary and choose q q q 1 2 4 n such that < δ. Then putting x = , x = 1 2 2nπ (4n+1)π (8n+1)π , we

102


have from the definition of f that f (x1 ) = 0, f (x2 ) = √12 and f (x) > √12 for x1 < x < x2 . Hence, f does not satisfy the Darboux property on the interval I = (−δ, δ). Therefore, there is no function g such that for some n, g(n) (x) = f (x) in I since a Peano derivative has the Darboux property. Since δ was arbitrary, f ∗ (0) cannot exist. Remark. Remark Since every Peano derivative is a generalized Peano derivative, in view of the Remark in 3.1, if f has generalized Peano derivative at x0 , the Ces` aro derivative of f at x0 may not exist.

2.5

Peano and Symmetric de la Vall´ ee Poussin Deriva(s) tives, f(k) and f(k) , and Smoothness of Order k (s)

Theorem 2.5.1 If f(k) (x) exists finitely, then f(k) (x) exists finitely and the two derivatives are equal. Moreover, γk+1 (f ; x, h) + γk+1 (f ; x, −h) = ̟k+1 (f ; x, h); (a) 2 γk+1 (f ; x, h) − γk+1 (f ; x, −h) h (b) = ̟k+2 (f ; x, h). 2 k+2 Let f(k) (x) exist finitely and suppose that k is even, k = 2m, say. Then, since k X hi f(i) (x) + o(hk ) as h → 0, f (x + h) = i! i=0 we have

2m

f (x + h) + f (x − h) 1 X hi + (−h)i = f(i) (x) + o(h2m ) 2 2 i=0 i! = (s)

m X h2j f(2j) (x) + o(h2m ), (2j)! j=0 (s)

which shows that f(2m) (x) exists and f(2j) (x) = f(2j) (x), j = 1, 2, . . . , m. Also, since γk+1 (f ; x, h) =

k X (k + 1)! hi f (x + h) − f(i) (x) , k+1 h i! i=0


103

we have that γk+1 (f ; x, h) + γk+1 (f ; x, −h) (2m + 1)! h f (x + h) − f (x − h) = 2 h2m+1 2 2m i i i X h − (−h) 1 f(i) (x) − 2 i=0 i! =

m−1 i (2m + 1)! h f (x + h) − f (x − h) X h2j+1 − f (x) (2j+1) h2m+1 2 (2j + 1)! j=0

= ̟2m+1 (f ; x, h)

and that γk+1 (f ; x, h) − γk+1 (f ; x, −h) (2m + 1)! h f (x + h) + f (x − h) = 2 h2m+1 2 2m i i i 1 X h + (−h) f(i) (x) − 2 i=0 i! i (2m + 1)! h f (x + h) + f (x − h) X h2j − f(2j) (x) 2m+1 h 2 (2j)! j=0 m

= =

h ̟2m+2 (f ; x, h). 2m + 2

This proves the result for k even. Let now k be odd, k = 2m + 1, say. Then, as above, 2m+1 1 X hi − (−h)i f (x + h) − f (x − h) = f(i) (x) + o(h2m+1 ) 2 2 i=0 i! m X h2j+1 = f(2j+1) (x) + o(h2m+1 ), (2j + 1)! j=0 (s)

(s)

showing that f(2m+1) (x) exists and f(2j+1) (x) = f(2j+1) (x), j = 0, 1, . . . , m. Also, γk+1 (f ; x, h) + γk+1 (f ; x, −h) (2m + 2)! h f (x + h) + f (x − h) = 2 h2m+2 2 2m+1 i 1 X hi + (−h)i f(i) (x) − 2 i=0 i!

i (2m + 2)! h f (x + h) + f (x − h) X h2j − f(2j) (x) 2m+2 h 2 (2j)! j=0 m

=

= ̟2m+2 (f ; x, h)

104


and γk+1 (f ; x, h) − γk+1 (f ; x, −h) (2m + 2)! h f (x + h) − f (x − h) = 2 h2m+2 2 2m+1 i i X h − (−h)i 1 f(i) (x) − 2 i=0 i!

i (2m + 2)! h f (x + h) − f (x − h) X h2j+1 − f (x) (2j+1) h2m+2 2 (2j + 1)! j=0 m

=

=

h ̟2m+3 (f ; x, h), 2m + 3

showing that the result is also true when k is odd.

(s)

Corollary 2.5.2 If f(k) (x) exists, possibly infinitely, then f(k) (x) exists with the same value.

The proof follows from Theorem 2.5.1(a). The converse of Theorem 2.5.1 is not true. Suppose that k is even and sin 1/x, if x 6= 0, f (x) = 0, if x = 0. (s)

Then f(k) (0) = 0, but f(k) (0) does not exist since f(1) (0) does not exist. If k is odd, replace the sine by cosine in the above example to get a similar result. However, we have the following result. Theorem 2.5.3 If f is smooth at x of orders k + 1 and k + 2, then f(k) (x) exists finitely. If f(k) (x) exists finitely, then f is smooth at x of order k+1. More +

−

(x), f − (x) are generally, if the Peano derivates f (k+1) (x), f (k+1) (x), f + (k+1) (k+1) all finite, then +

lim sup h̟k+2 (f ; x, h) ≤ (k + 2)

f (k+1) (x) − f − (x) (k+1) 2

h→0+

,

−

lim inf h̟k+2 (f ; x, h) ≥ (k + 2)

(x) − f (k+1) (x) f+ (k+1) 2

h→0+

.

(s)

Since f is smooth of order k + 1 at x, f(k−1) (x) exists finitely and h̟k+1 (f ; x, h) = o(1) as h → 0. That is, if k is even, k = 2m, say, m−1 i (2m + 1)! h f (x + h) − f (x − h) X h2i+1 (s) (x) = o(1) as h → 0, − f h2m 2 (2i + 1)! (2i+1) i=0


105

which gives m−1 X h2i+1 (s) f (x + h) − f (x − h) = f(2i+1) (x) + o(h2m ) as h → 0. (2.5.1) 2 (2i + 1)! i=0

Similarly, smoothness of order k + 2 = 2m + 2 gives m

f (x + h) + f (x − h) X h2i (s) = f(2i) (x) + o(h2m+1 ) as h → 0. 2 (2i)! i=0

(2.5.2)

From (2.5.1) and (2.5.2), f (x + h) =

m m−1 X X h2i+1 (s) h2i (s) f(2i) (x) + f(2i+1) (x) + o(h2m ) as h → 0, (2i)! (2i + 1)! i=0 i=0

which shows that f(2m) (x) exists. If k is odd, the argument is similar. Now suppose that f(k) (x) exists finitely. Then hγk+1 (f ; x, h) = (k + 1) γk (f ; x, h) − f(k) (x) (2.5.3)

and since the right-hand side of (2.5.3) tends to 0 as h → 0, hγk+1 (f ; x, h) = o(1); and so, from Theorem 2.5.1(a), h̟k+1 (f ; x, h) = o(1), showing that f is smooth of order k + 1 at x. The last two inequalities follow from Theorem 2.5.1(b). Theorem 2.5.4 If f(k) (x) exists finitely and if f is smooth of order k + 2 at x, then + − (x) = f − (x). f (k) (x) = f (k) (x) and f + (k) (k)

From Theorem 2.5.1(b), γk+1 (f ; x, h) =

2h ̟k+2 (f ; x, h) + γk+1 (f ; x, −h), k+2

and so the result follows by letting h → 0+.

Theorem 2.5.5 If f is Peano bounded of order r at x, then f is d.l.V.P. bounded of order r at x. By Theorem 1.4.4 of Chapter I, f(r−1) (x) exists finitely and so, by Theorem 2.5.1(a), the result follows.

106


2.6

Symmetric Ces` aro and Symmetric de la Vall´ ee (s) Poussin Derivatives, SCk Df and f(k)

Theorem 2.6.1 Let f be Cr−1 P − integrable in [a, b] and for x ∈ [a, b] let ( Rx (Cr−1 P ) a f, if k = 1, Fk (x) = Rx (Cr−k P ) a Fk−1 , if 2 ≤ k ≤ r,

and write Φ = Fr . Then the Peano derivative Φ(r−1) exists in [a, b] and for x ∈ [a, b] i r + 1h Cr (f ; x, x + h) − Cr (f ; x, x − h) = ̟r+1 (Φ; x, h). 2h

As in (2.4.2), we have

r−1 i X h hr Cr (f ; x, x + h) = Φ(x + h) − Φ(x) − Fr−i (x), r! i! i=1

and so Cr (f ; x, x + h) − Cr (f ; x, x − h) = r−1 i i X h r! h Φ(x + h) − Φ(x) − F (x) r−i hr i! i=1

r−1 i X (−h)i r! h Φ(x − h) − Φ(x) − F (x) r−i (−h)r i! i=1 r! = r Φ(x + h) − (−1)r Φ(x − h) − Φ(x) + (−1)r Φ(x) h r−1 i X h − 1 − (−1)r−i Fr−i (x) . i! i=1

−

Hence, r + 1 Cr (f ; x, x + h) − Cr (f ; x, x − h) = 2h (r + 1)! h Φ(x + h) − (−1)r Φ(x − h) 1 − (−1)r Φ(x) hr+1 2 2 r−1 i X 1 − (−1)r−i hi Fr−i (x) . − 2 i! i=1

(2.6.1)


107

So, if r is even, r = 2m say, then r + 1 Cr (f ; x, x + h) − Cr (f ; x, x − h) = 2h 2m−1 i X 1 − (−1)2m−i hi (2m + 1)! h Φ(x + h) − Φ(x − h) − F (x) 2m−i h2m+1 2 2 i! i=1 =

m−1 i (2m + 1)! h Φ(x + h) − Φ(x − h) X h2j+1 − F2m−2j−1 (x) . (2.6.3) 2m+1 h 2 (2j + 1)! j=0

Since a Cr−k P -integral is Cr−k−1 P -integrable and Cr−k -continuous, and since Fr−i is a Ci P -integral, Fr−i is Ci−1 P -integrable and Ci -continuous, and so, in particular, F1 is Cr−2 P -integrable and Cr−1 -continuous in [a, b]. Hence, by Theorem 2.4.1, Φ(r−1) exists and Φ(k) = Fr−k in [a, b] for 1 ≤ k ≤ r − 1. Therefore, Φ(2j+1) = F2m−2j−1 for 0 ≤ j ≤ m − 1, and so from (2.6.3) r + 1 Cr (f ; x, x + h) − Cr (f ; x, x − h) = 2h m−1 i (2m + 1)! h Φ(x + h) − Φ(x − h) X h2j+1 − Φ(2j+1) (x) . (2.6.4) 2m+1 h 2 (2j + 1)! j=0 Similarly, if r is odd, r = 2m + 1 say, then from (2.6.2) r + 1 Cr (f ; x, x + h) − Cr (f ; x, x − h) = 2h 2m i (2m + 2)! h Φ(x + h) + Φ(x − h) X 1 − (−1)2m−i+1 hi − F (x) , 2m−i+1 h2m+2 2 2 i! i=0

(2.6.5)

and so, as above, Φ(2j) = F2m−2j+1 and hence, from (2.6.5) r + 1 Cr (f ; x, x + h) − Cr (f ; x, x − h) = 2h m i (2m + 2)! h Φ(x + h) + Φ(x − h) X h2j − Φ (x) . (2j) h2m+2 2 (2j)! j=0

(2.6.6)

Hence, from (2.6.4) and (2.6.6) and from the definition of ̟r+1 (Φ; x, h), we obtain (2.6.1). Corollary 2.6.2 Let the hypotheses of the above theorem hold. Then, f is SCr -continuous at x if and only if Φ is smooth of order r + 1 at x. Also, (s) (s) SCDr f (x) = Φ(r+1) (x), SCDr f (x) = Φ(r+1) (x) and SCr Df (x) exists if and (s)

only if Φ(r+1) (x) exists.

The proof follows from (2.6.1).

108


Borel and Peano Derivatives, BDk f and f(k)

2.7

Theorem 2.7.1 Let f be special Denjoy integrable in some neighbourhood of x. If f(r) (x) exists, then BDr f (x) exists and the two derivatives are equal, but the converse is not true. More generally, if f(r−1) (x) exists finitely, then f (r) (x) ≤ BD r f (x) ≤ BDr f (x) ≤ f (r) (x). If f(r) (x) exists finitely, then

f (x + t) =

r X ti i=0

and so 1 h

Z

0

h

i!

f(i) (x) + o(tr ) as t → 0,

Pr f (x + t) − i=0 tr

ti i! f(i) (x)

dt = o(1) as h → 0.

Hence, BDr f (x) exists and BDr f (x) = f(r) (x). If f(r) (x) is infinite, then f(i) (x) exists finitely, 0 ≤ i ≤ r − 1, and so, by the above argument, BDi f (x) exists and is equal to f(i) (x), 0 ≤ i ≤ r − 1. Suppose that f(r) (x) = ∞. Then for any M > 0 there is a δ > 0 such that

Pr−1 f (x + t) − i=0 tr /r!

ti i! f(i) (x)

> M for 0 < |t| < δ.

Since f(i) (x) = BDi f (x), 0 ≤ i ≤ r − 1, this gives r! h

Z

ǫ

h

P f (x + t) − r−1 i=0 tr

ti i! BDi f (x)

dt > M

h−ǫ for 0 < ǫ < h < δ. h

Letting ǫ → 0+ first and then h → 0+, we have that BD+ r f (x) ≥ M . f (x) ≥ M . f (x) ≥ M and so BD Similarly, BD− r r M being arbitrary this shows that BD r f (x) = ∞. Hence, BD r f (x) exists and equals f(r) (x). If f(r) (x) = −∞, the proof is similar. Regarding the converse, let r be a fixed positive integer and consider sin x, if x is rational, f (x) = xr sin(1/xr−1 ), if x is irrational. Then f ′ (0) does not exist and so f(r) (0) cannot exist. To see that BDr (0) exists first note that Z Z 1 h 1 h f (t) dt = sin(1/tr−1 ) dt. (2.7.1) h 0 tr h 0


109

Now if r ≥ 2, ′ tr cos(1/tr−1 ) = rtr−1 cos(1/tr−1 ) + (r − 1) sin(1/tr−1 ).

(2.7.2)

The right-hand side of (2.7.2) being Riemann integrable, (r − 1)

Z

h

sin(1/tr−1 ) dt = hr cos(1/hr−1 ) − r

0

Z

h

tr−1 cos(1/tr−1 ) dt

0

and hence, Z Z 1 h r 1 h r−1 hr−1 cos(1/hr−1 ) − sin(1/tr−1 ) dt = t cos(1/tr−1 ) dt. h 0 r−1 r−1h 0 (2.7.3) The right-hand side of (2.7.3) tends to 0 as h → 0 and so 1 h

Z

h

sin(1/tr−1 ) dt = o(1) as h → 0.

(2.7.4)

0

From (2.7.1) and (2.7.4) and the definition of BDr f , it follows that BDi (0) exists 0 ≤ i ≤ r. To prove the last part, we consider the last inequality, the consideration of the first being similar. We may then assume that f (r) (x) < ∞, f (r) (x) < M , say. Then there is a δ > 0 such that γr (f ; x, t) < M for 0 < |t| < δ, where γr (f ; x, t) is as defined in (1.4.3) of Chapter I. Hence, 1 h

Z

ǫ

h

γr (f ; x, t) dt ≤ M

h−ǫ for 0 < ǫ < h < δ. h

(2.7.5)

As f(r−1) (x) exists finitely, we have by the first part that f(r−1) (x) = BDr−1 f (x), and a fortiorti f(i) (x) = BDi f (x), 0 ≤ i ≤ r − 1. So from (2.7.5) +

by letting ǫ → 0+ and then h → 0+, we have that BDr f (x) ≤ M , Simi− larly, BDr f (x) ≤ M showing that BDr f (x) ≤ M . Since M is arbitrary, the inequality is proved. Theorem 2.7.2 Let f be special Denjoy integrable in some neighbourhood of x with F its indefinite integral in that neighbourhood. If F(r+1) (x) exists, then BDr f (x) exists with BDr f (x) = F(r+1) (x) and F (r+2) (x) ≤ BDr+1 f (x) ≤ BDr+1 f (x) ≤ F (r+2) (x).

(2.7.6)

To prove this theorem, we need the following: Rh Lemma 2.7.3 If limt→0 φ(x + t) = ℓ, then limh→0 h1 0 φ(x + t) dt = ℓ, where −∞ ≤ ℓ ≤ ∞. More generally, if lim suph→0 φ(x + t) = M, −∞ < M ≤ ∞, Rh then lim suph→0 h1 0 φ(x + t) dt ≤ M . There is a similar result for lim inf.

110


We first prove the second part of the lemma. We may without loss in generality suppose that M < ∞. If then M > M , there is a δ > 0 such that h φ(x + t) < M if |t| < δ. Hence, if 0 < |h| < δ, then h1 0 φ(x + t) dt < M and, h hence, lim suph→0 h1 0 φ(x+ t) dt ≤ M . Since M is arbitrary, this proves this part of the lemma. The proof of the lim inf case is similar. Now if is finite, the proof of the the first statement is similar to the above proofs. Suppose then = ∞ and let G > 0. Then there is a δ > 0 such that h φ(x + t) > G if |t| < δ. So, if 0 < |h| < δ, then h1 0 φ(x + t) dt > G and, since G was arbitrary, this proves that limt→0 φ(x + t) dt = ∞. The case when = −∞ is similar. Proof of Theorem 2.7.2: Let r = 1. Since F(2) (x) exists, F(1) (x) exists finitely and F (x + h) − F (x) 1 = lim h→0 h→0 h h

F(1) (x) = lim

h

f (x + t) dt = BD0 f (x)

0

by definition. Let 0 < < h. Then integrating by parts, h f (x + t) − BD0 f (x) dt (2.7.7) t h h F (x + t) − F (x) − tBD0 (x) F (x + t) − F (x) − tBD0 f (x) = + dt. t t2

If F(2) (x) is finite, the right-hand side of (2.7.7) tends to a limit as → 0. So, letting → 0 and dividing by h, we have from (2.7.7) 1 h

h

f (x + t) − BD0 f (x) dt (2.7.8) t 0 F (x + h) − F (x) − hF(1) (x) 1 h F (x + t) − F (x) − tF(1) (x) = + dt. h2 h 0 t2

Letting h → 0, we have by the lemma 1 1 1 h f (x + t) − BD0 f (x) dt = F(2) (x)+ F(2) (x) = F(2) (x). BD1 f (x) = lim h→0 h 0 t 2 2 If F(2) (x) = ∞, then for any G > 0 there is a δ > 0 such that 2 t2 F (x + t) − F (x) − tF(1) (x) > G for |t| < δ and, so if |h| < δ, then 1 h F (x+t)−F (x)−tF(1) (x) dt > G, and so, letting h → 0 in (2.7.8), h 0 t2 /2 lim inf h→0

1 h

0

h

f (x + t) − BD0 f (x) 1 dt ≥ ∞ + G = ∞, t 2


111

showing that BD1 f (x) = ∞. If F(2) (x) = −∞, the proof is similar. To prove the inequality (2.7.6) in this case we consider only the last inequality and we may suppose that F (3) (x) < ∞, F (3) (x) < M , say. Since F(2) (x) is finite, we have by the above that BD1 f (x) is also finite. So, for 0 < ǫ < h, integrating by parts, Z

h

ǫ

f (x + t) − BD0 f (x) − tBD1 f (x) dt t2 /2

h F (x + t) − F (x) − tBD0 f (x) − (t2 /2)BD1 f (x) t2 /2 ǫ Z h 4 F (x + t) − F (x) − tBD0 f (x) − (t2 /2)BD1 f (x) dt. (2.7.9) + 3! ǫ t3 /3!

=

Since F (3) (x) < M , there is a δ > 0 such that the integrand in the last term on the right-hand side of (2.7.9) is less than M if |t| < δ and, so, if 0 < ǫ < h < δ, the integral in the last term on the right-hand side of (2.7.9) does not exceed M (h − ǫ). Hence, from (2.7.9) Z

1 h

h

ǫ

f (x + t) − BD0 f (x) − tBD1 f (x) dt t2 /2

h 1 F (x + t) − F (x) − tBD0 f (x) − (t2 /2)BD1 f (x) 2 M (h − ǫ) ≤ . +3 h t2 /2 h ǫ +

Letting ǫ → 0 and then h → 0+ we have BD2 f (x) ≤ M . − In a similar manner BD2 f (x) ≤ M and so BD2 f (x) ≤ M . Since M was arbitrary, this shows that BD 2 f (x) ≤ F (3) (x), completing the proof of the theorem in the case r = 1. Now assume the result to be known for r = m and let r = m + 1. Then, since F(m+2) (x) exists, F(m+1) (x) exists finitely and by the induction hypothesis, BDm f (x) exists and BDi f (x) = F(i+1) (x), 0 ≤ i ≤ m. Then P ti writing Q(t) = m i=0 i! BDi f (x) as in (1.10.3) of Chapter I we have, integrating by parts, Z

ǫ

h

Pm ti+1 F (x + t) − F (x) − i=0 (i+1)! F(i+1) (x) h f (x + t) − Q(t) dt = tm+1 tm+1 ǫ P i+1 Z h F (x + t) − F (x) − m t F (x) i=0 (i+1)! (i+1) +(m + 1) dt. (2.7.10) m+2 t ǫ

112


If F(m+2) (x) is finite then the right-hand side of (2.7.10) tends to a limit as ǫ → 0, and so, letting ǫ → 0 and then dividing by h, we get from (2.7.10) Pm+1 i Z F (x + h) − F (x) − i=1 hi! F(i) (x) 1 h f (x + t) − Q(t) dt = h 0 tm+1 hm+2 Pm+1 ti Z h F (x + t) − F (x) − i=1 i! F(i) (x) m+1 + dt, (2.7.11) h tm+2 0 and so letting h → 0, we have by of Lemma 2.7.3 that BDm+1 f (x) =

1 m+1 F(m+2) (x) + F(m+2) (x) = F(m+2) (x). m+2 m+2

If F(m+2) (x) = ∞, then for any G > 0 there is a δ > 0 such that m+1 i X ti (m + 2)! h F (x + t) − F (x) − F(i) (x) > G for |t| < δ, m+2 t i! i=1

and so 1 h

Z

0

h

m+1 i X ti (m + 2)! h F (x + t) − F (x) − F (x) dt > G for |h| < δ, (i) tm+2 i! i=1

and hence, letting h → 0 in (2.7.11), Z m+1 1 h f (x + t) − Q(t) dt ≥ ∞ + G = ∞. lim inf m+1 h→0 h 0 t (m + 2)! This shows that BDm+1 f (x) = ∞. If F(m+2) (x) = −∞, the proof is similar. This completes the proof of the first part of the theorem. For the second part, we can suppose that F (m+3) (x) < M < ∞. Then F(m+2) (x) exists finitely and so by the first part BDm+1 f (x) exists and BDi f (x) = F(i+1) (x), 0 ≤ i ≤ m + 1. So, for 0 < ǫ < h, integrating by parts, Z

h

ǫ

f (x + t) −

Pm+1

ti i=0 i! BDi f (x) tm+2

F (x + t) − F (x)−

Pm+1

i=0 tm+2

+(m + 2)

Z

h F (x

dt =

(2.7.12)

h ti+1 (i+1)! F(i+1) (x)

+ t) − F (x) −

ǫ

ǫ Pm+1

i=0 tm+3

ti+1 (i+1)! F(i+1) (x)

Since F (m+3) (x) < M , there is a δ > 0 such that Pm+2 F (x + t) − F (x) − i=1 tm+3 /(m + 3)!

ti i! F(i) (x)

< M for |t| < δ.

dt.


113

So, if 0 < ǫ < h < δ, then Z

h

ǫ

Pm+2 F (x + t) − F (x) − i=1 tm+3 /(m + 3)!

ti i! F(i) (x)

dt < M (h − ǫ)

and, hence, from (2.7.12) 1 h

Z

h

f (x + t) −

ǫ

Pm+1

ti i=0 i! BDi f (x) tm+2

" Pm+1 1 F (x + t) − F (x) − i=0 ≤ h tm+2

dt #

ti+1 h (i+1)! F(i+1) (x)

Pm+2 F (x + ǫ) − F (x) − i=1 M − < (m + 3)! hǫm+2

ǫ ǫi F (x) i! (i)

+

m + 2 M (h − ǫ) (m + 3)! h

+

m + 2 M (h − ǫ) . (m + 3)! h

Letting ǫ → 0+ first and then h → 0+, we have +

BDm+2 f (x) ≤

(m + 2)M M + = M. m+3 m+3

−

Similarly, BDm+2 f (x) ≤ M and so BDm+2 f (x) ≤ M . Since M is arbitrary, this proves that BD m+2 f (x) ≤ F (m+3) (x). Theorem 2.7.4 Let f be special Denjoy integrable in some neighbourhood of x and let F be its indefinite integral in that neighbourhood. If BDr f (x) exists finitely, then F(r+1) (x) exists and equals BDr f (x). If, moreover, Z

h

f (x + t) −

0

Pr

ti i=0 i! BDi f (x) tr+1

dt = lim

ǫ→0

Z

h

f (x + t) −

ǫ

Pr

ti i=0 i! BDi f (x) tr+1

dt

exists, then (r + 2)BDr+1 f (x) − (r + 1)BDr+1 f (x) ≤ F (r+2) (x) ≤ F (r+2) (x) ≤ (r + 2)BDr+1 f (x) − (r + 1)BDr+1 f (x),

(2.7.13)

provided the two extreme members on the left and right of (2.7.13) are well defined. The last condition excludes the possibility that BDr+1 f (x) exists and is infinite. Further note that (2.7.13) is obvious if either BDr+1 f (x) = ∞ and BDr+1 f (x) < ∞ or if BDr+1 f (x) = −∞ and BDr+1 f (x) > −∞. First, we prove the result when r = 1, that is, if BD1 f (x) exists finitely. Let Z h f (x + t) − BD0 f (x) φ(h) = dt, t 0 the integral existing from the definition of BD1 f (x).

114

Higher Order Derivatives Since φ is a special Denjoy integral, φ′ (t) = 1t f (x + t) − BD0 f (x) almost everywhere, and so tφ′ (t) = f (x + t) − BD0 f (x) almost everywhere. Hence, Z

h

′

tφ (t) dt =

0

Z

0

h

f (x + t) − BD0 f (x) dt = F (x + h) − F (x) − hBD0 f (x).

Also, since φ is a special Denjoy integral, from (2.7.14), on integrating by parts,

Rh 0

(2.7.14) φ′ = φ(h) − φ(0) = φ(h), and so Z

F (x + h) − F (x) − hBD0 f (x) = hφ(h) −

h

φ.

0

Hence, φ(h) 2 F (x + h) − F (x) − hBD0 f (x) =2 − 2 h2 /2 h h

Z

h

φ.

(2.7.15)

0

Rh φ(h) → BD1 f (x) and h22 0 φ → BD1 f (x) as h h → 0, so from (2.7.15), letting h → 0, F(2) (x) exists with F(1) (x) = BD0 f (x) and F(2) (x) = 2BD1 f (x) − BD1 f (x) = BD1 f (x). This gives the first part in this case. For the second part, let

Since BD1 f (x) exists finitely,

ψ(h) =

Z

0

h

f (x + t) − BD0 f (x) − tBD1 f (x) dt t2

exist. Then, as above, t2 ψ ′ (t) = f (x + t) − BD0 f (x) − tBD1 f (x) almost everywhere and, so, on integrating, Z

h

t2 ψ ′ (t) dt = F (x + h) − F (x) − hBD0 f (x) −

0

h2 BD1 f (x). 2

Hence, integrating by parts and noting that F(1) (x) = BD0 f (x) and F(2) (x) = BD1 f (x), h2 F (x + h) − F (x) − hF(1) (x) − F(2) (x) = h2 ψ(h) − 2 2

Z

h

tψ(t) dt.

0

Applying the mean value theorem, F (x + h) − F (x) − hF(1) (x) − h3 /3!

h2 2 F(2) (x)

Z ψ(h) 3!2 h − 3 tψ(t) dt h h 0 ψ(h) 3!2 ξψ(ξ) − = 3! h 3 ξ2

= 3!


115

for some ξ, 0 < ξ < h. Letting h → 0, ψ(h) ψ(ξ) − 2!2 lim inf h→0 h ξ h→0 ψ(h) ≤ 3BD2 f (x) − 2!2 lim inf h→0 h = 3BD2 f (x) − 2BD2 f (x).

F (3) (x) ≤ 3! lim sup

In a similar manner, we can obtain F (3) (x) ≥ 3BD2 f (x) − 2BD2 f (x), completing the proof of the theorem for this case, r = 1. Now, assume the result to be known for r = m and let r = m + 1. Let BDm+1 f (x) exist finitely. Then BDi f (x) exists, 0 ≤ i ≤ m. Let Pm i Z h f (x + t) − i=0 ti! BDi f (x) φ(h) = dt, tm+1 0 the integral existing by the definition of BDm+1 f (x). Since φ is a Denjoy P ti f (x + t) − m i=0 i! BDi f (x) ′ almost everywhere, which gives integral, φ (t) = tm+1 Z h m X hi+1 tm+1 φ′ (t) dt = F (x + h) − F (x) − BDi f (x). (2.7.16) (i + 1)! 0 i=0 By the induction hypothesis, F(i+1) (x) exists and is equal to BDi f (x), 0 ≤ i ≤ m, and, so, from (2.7.16) on integrating by parts, Z h m+1 X hi F(i) (x) = hm+1 φ(h) − (m + 1) tm φ(t) dt. F (x + h) − F (x) − i! 0 i=1

Hence,

Pm+1 i h φ(h) m + 1 Z h i F (x + h) − F (x) − i=1 hi! F(i) (x) m = (m + 2)! − t φ(t) dt . hm+2 /(m + 2)! h hm+2 0 (2.7.17) Since (m + 1)! φ(h)/h → BDm+1 f (x) as h → 0, we have from (2.7.17),

letting h → 0,

F(m+2) (x) = (m + 2)BDm+1 f (x) − (m + 1)BDm+1 f (x) = BDm+1 f (x). (2.7.18) For the second part, let P Z h ti f (x + t) − m+1 i=0 i! BDi f (x) dt ψ(h) = tm+2 0 P ti exist. Then, as above, tm+2 ψ ′ (t) = f (x + t) − m+1 i=0 i! BDi f (x) almost everywhere and so Z h m+1 X hi+1 BDi f (x). (2.7.19) tm+2 ψ ′ (t) dt = F (x + h) − F (x) − (i + 1)! 0 i=0

116


As BDm+1 f (x) exists finitely, by the induction hypothesis, BDi f (x) = F(i+1) (x), 0 ≤ i ≤ m. Also, from (2.7.18), F(m+2) (x) = BDm+1 f (x) and so, from (2.7.19) on integrating by parts, Z h m+2 X hi F (x + h) − F (x) − F(i) (x) = hm+2 ψ(h) − (m + 2) tm+1 ψ(t) dt. i! 0 i=1 Hence, applying the mean value theorem, Pm+2 i F (x + h) − F (x) − i=1 hi! F(i) (x) hm+3 /(m + 3)! h ψ(h) (m + 2) Z h i = (m + 3)! − m+3 tm+1 ψ(t) dt h h 0 ξ m+1 ψ(ξ) ψ(h) , − (m + 2)!(m + 2) = (m + 3)! h ξ m+2

for some ξ, 0 < ξ < h. Letting h → 0, F (m+3) (x) ≤ (m + 3)! lim sup h→0

ψ(h) ψ(ξ) − (m + 2)!(m + 2) lim inf h→0 h ξ

≤ (m + 3)BDm+2 f (x) − (m + 2)BDm+2 f (x). In a similar manner F (m+3) (x) ≥ (m + 3)BDm+2 f (x) − (m + 2)BDm+2 f (x), which completes the proof. Remark. All the results concerning upper and lower derivates proved above are also true for the corresponding unilateral upper and lower derivates.

2.8

Symmetric Borel and Symmetric de la Vall´ ee (s) Poussin Derivatives, SBDk f and f(k)

Theorem 2.8.1 Let f be special Denjoy integrable in some neighbourhood (s) of x. If f(r) (x) exists, then SBDr f (x) exists and the two derivatives are equal; (s)

the converse is not true. More generally, if f(r−2) (x) exists finitely, then (s)

f (s) (x) ≤ SBDr f (x) ≤ SBDr f (x) ≤ f (r) (x). (r) f (x + t) + (−1)r f (x − t) (s) = P (t) + o(tr ) If f(r) (x) exists finitely, then 2 where  2i   Pr/2 t f (s) (x), if r is even,  i=0 (2i)! (2i) P (t) = P(r−1)/2 t2i+1  (s)  (x) if r is odd.  i=0 f (2i + 1)! (2i+1)


117

Hence, 1 h

Z

0

h

1 f (x + t) + (−1)r f (x − t) − P (t) dt = o(1) as h → 0. tr 2 (s)

So, SBDr f (x) exists and SBDr f (x) = f(r) (x). (s)

(s)

If f(r) (x) is infinite, then by definition f(i) (x) exists finitely, i = r − 2, (s)

r −4, · · · , and so, by the above argument, SBDi f (x) exists and equals f(i) (x), i = r − 2, r − 4, · · · . (s) Let f(r) (x) = ∞. Then for any M > 0 there is a δ > 0 such that ̟r (f ; x, t) > M for 0 < |t| < δ, where ̟r (f ; x, t) is defined in (2.7.5) of Chapter I. So, Z 1 h h−ǫ , for 0 < ǫ < h < δ. (2.8.1) ̟r (f ; x, t) dt > M h ǫ h (s)

Since by the above argument, SBDi f (x) exists and equals f(i) (x), i = r − 2, r − 4, · · · , ̟r (f ; x, t) = ̟r (f ; x, t), where ̟ r (f ; x, t) is defined as in (1.11.6) of Chapter I, and, hence, from (2.8.1) we have 1 h

Z

h

̟r (f ; x, t) dt > M

ǫ

h−ǫ , for 0 < ǫ < h < δ. h

Letting ǫ → 0+ first and then h → 0+, we get that SBDr f (x) ≥ M and, so, (s) since M is arbitrary, SBDr f (x) = ∞; this proves that SBDr f (x) = f(r) (x). (s)

If f(r) (x) = −∞, the argument is similar. Regarding the converse, let r be a fixed positive integer and consider the functions: cos x, sin x, if x is rational, f (x) = g(x) = xr sin(1/xr−1 ); xr sin(1/xr−1 ), if x is irrational. (s)

(s)

Since f(2) (0) does not exist, then f(r) (0) does not exist if r is even, and since (s)

(s)

g(1) (0) does not exist, then g(r) (0) does not exist if r is odd. However, for any r, Z Z 1 h f (t) + (−1)r f (−t) 1 h g(t) + (−1)r g(−t) dt = dt = I, say. h 0 2tr h 0 2tr Further, I=

0, 1 h

Rh 0

sin(1/t

r−1

if r is even, ) dt, if r is odd.

Hence, proceeding as in Theorem 2.7.1, I = o(1) as h → 0, see (2.7.4) , and so whether r is even or odd both SBDr f (0) and SBDr g(0) exist.

118


To prove the converse, take f or g according as r is even or odd. We now consider the last part of the theorem and consider the last inequality; the first can be treated in a similar manner. We may suppose that (s) (s) f (r) (x) < ∞, and let f (r) (x) < M < ∞. Then there is a δ > 0 such that ̟r (f ; x, t) < M for 0 < |t| < δ. Hence, Z h−ǫ 1 h , for 0 < ǫ < h < δ. (2.8.2) ̟r (f ; x, t) dt ≤ M h ǫ h (s)

Since f(r−2) (x) exists finitely, then by the first part SBDr−2 f (x) exists finitely (s)

(s)

and is equal to f(r−2) (x) and thus f(i) (x) = SBDi f (x), i = r − 2, r − 4, · · · , from which if follows that ̟r (f ; x.t) = ̟r (f ; x.t) and, so, letting ǫ → 0 and then h → 0 in (2.8.2), we have SBDr f (x) ≤ M . Since M was arbitrary, this (s)

completes the proof that SBDr f (x) ≤ f (r) (x).

Theorem 2.8.2 Let f be special Denjoy integrable in some neighbourhood of x and let F be its indefinite integral in that neighbourhood. If SBDr f (x) (s) exists finitely, then F(r+1) (x) exists with value SBDr f (x). Moreover, if Rh 0 ̟ r+2 (f ; x.t) dt exists, then (s)

(s)

(r + 3)SBDr+2 f (x) − (r + 2)SBDr+2 f (x) ≤ F (r+3) (x) ≤ F (r+3) (x)

≤ (r + 3)SBDr+2 f (x) − (r + 2)SBDr+2 f (x),(2.8.3) provided the two extreme members in the left and right of the inequality are properly defined. If either SBDr+2 f (x) = ∞ and SBDr+2 f (x) < ∞ or if SBDr+2 f (x) = −∞ and SBDr+2 f (x) > −∞, then (2.8.3) is obvious. Also, if SBDr+2 f (x) exists, it must be finite; for otherwise the extreme members become undefined. We first prove the case r = 1. Let SBD1 f (x) exist finitely and write Z h f (x + t) − f (x − t) φ(h) = dt; 2t 0 this integral exists by the definition of SBD1 f (x). Since φ is a special Denjoy Rh f (x + t) − f (x − t) integral, φ′ (t) = almost everywhere and 0 φ′ = φ(h) − 2t φ(0) = φ(h). Since 2tφ′ (t) = f (x + t) − f (x − t) almost everywhere, Z h Z h ′ f (x + t) − f (x − t) dt = F (x + h) − 2F (x) + F (x + h). 2 tφ (t) dt = 0

0

On integrating by parts and dividing by h2 , this gives Z h Z h φ(h) 2 2 F (x + h) − 2F (x) + F (x + h) φ = 2 = hφ(h) − − φ. h2 h2 h h2 0 0 (2.8.4)

Relations between Derivatives 119 Z h 2 φ(h) and 2 Since SBD1 f (x) exists φ → SBD1 f (x) as h → 0. Hence, h h 0 (s) from (2.8.4) on letting h → 0, we obtain F(2) (x) = SBD1 f (x), giving the first part of the theorem when r = 1. For the second part of the theorem in this case, let Z h ψ(h) = ̟3 (f ; x, t) dt 0

exist. Then, as above, ψ ′ (t) = ̟3 (f ; x, t) almost everywhere and so f (x + t) − f (x − t) t3 ′ ψ (t) = − tSBD1 f (x) 3! 2

almost everywhere and, hence, Z Z h 1 h 3 ′ f (x + t) − f (x − t) − tSBD1 f (x) dt t ψ (t) dt = 3! 0 2 0 F (x + h) + F (x − h) h2 = − F (x) − SBD1 f (x) 2 2 h2 (s) F (x + h) + F (x − h) − F (x) − F(2) (x). = 2 2 Integrating by parts and dividing by h4 /4! this becomes 4! F (x + h) + F (x − h) h2 (s) − F (x) − 2 F(2) (x) h4 2 Rh 2 Rh 2 ψ(h) 3 4 3 = h4 h ψ(h) − 3 0 t ψ(t) dt = 4 h − h4 0 t ψ(t) dt .

(2.8.5)

By the mean value theorem there is a ξ, 0 < ξ < h, such that 1 h4

Z

0

h

t2 ψ(t) dt =

1 ψ(ξ) 1 2 ξ ψ(ξ) = , 4ξ 3 4 ξ

(2.8.6)

and so letting h → 0, we have from (2.8.5) and (2.8.6) (s)

F (4) (x) ≤ 4 lim sup h→0

ψ(h) ψ(h) − 3 lim inf ≤ 4SBD3 f (x) − 3SBD3 f (x), h→0 h h

and (s)

F (4) (x) ≥ 4 lim inf h→0

ψ(h) ψ(h) − 3 lim sup ≥ 4SBD3 f (x) − 3SBD3 f (x). h h h→0

This completes the discussion of the case r = 1. Let us now suppose that the result is true for r = 2m − 1 and prove it for r = 2m + 1.

120


Assume that SBD2m+1 f (x) exists. Then SBD2i+1 f (x), 1 = 0, 1, · · · , m − 1 exists. Let Z h

φ(h) =

̟2m+1 (f ; x, t) dt,

0

the integral existing by the definition of SBD2m+1 f (x). Then, as above, Z h 1 t2m+1 φ′ (t) dt (2m + 1)! 0 Z h m−1 f (x + t) − f (x − t) X t2i+1 − SBD2i+1 f (x) dt = 2 (2i + 1)! 0 i=0

(2.8.7)

m−1 X h2i+2 F (x + h) + F (x − h) − F (x) − SBD2i+1 f (x). = 2 (2i + 2)! i=0

Since SBD2i+1 f (x) exists finitely, i = 0, 1, · · · , m − 1, by the induction hy(s) pothesis, F(2i+2) (x) exists and is equal to SBD2i+1 f (x), i = 0, 1, · · · , m − 1 and so (2.8.7) becomes m X F (x + h) + F (x − h) h2i (s) − F (x) − F (x) 2 (2i)! (2i) i=1 Z h 1 t2m+1 φ′ (t) dt = (2m + 1)! 0 Z h 1 h2m+1 φ(h) − (2m + 1) t2m φ(t) dt . = (2m + 1)! 0

So, (2m + 2)! h2m+2

F (x+h)+F (x−h) 2

− F (x) −

= (2m + 2) φ(h) h −

(s) h2i i=1 (2i)! F(2i) (x)

(2m+1)(2m+2) h2m+2

φ(h) 2m + 2 → SBD2m+1 f (x) and 2m+2 h h h → 0, letting h → 0 in (2.8.8), we obtain Since

Pm

Z

h

Rh 0

t2m φ(t) dt.

(2.8.8)

t2m φ(t) dt → SBD2m+1 f (x) as

0

(s)

F(2m+2) (x) = (2m+2)SBD2m+1 f (x)−(2m+1)SBD2m+1 f (x) = SBD2m+1 f (x), proving the first part of of the theorem for r = 2m + 1. For the second part, let ψ(h) =

Z

0

h

̟2m+3 (f ; x, t) d


121

exist. Then as above, Z h 1 t2m+3 ψ ′ (t) dt (2m + 3)! 0 Z h m f (x + t) − f (x − t) X t2i+1 = − SBD2i+1 f (x) dt (2.8.9) 2 (2i + 1)! 0 i=0 =

m X F (x + h) + F (x − h) h2i+2 − F (x) − SBD2i+1 f (x). 2 (2i + 2)! i=0 (s)

By the induction hypothesis SBD2i+1 f (x) = F(2i+2) (x), i = 0, 1, · · · , m − 1, and, so, from (2.8.9) 1 (2m + 3)! =

Z

h

t2m+3 ψ ′ (t) dt

0

m+1 X h2i (s) F (x + h) + F (x − h) − F (x) − F (x). 2 (2i)! (2i) i=1

(2.8.10)

Also, integrating by parts and applying the mean value theorem, 1 h2m+4

Z

0

h

t

2m+3

Z ψ(h) 2m + 3 h 2m+2 − 2m+4 ψ (t) dt = t ψ(t) dt h h 0 ψ(h) 2m + 3 ψ(ξ) − , (2.8.11) = h 2m + 4 ξ ′

where 0 < ξ < h. So, multiplying both sides of (2.8.10) by (2m + 4)!/h2m+4 and using (2.8.11), we have ̟2m+4 (F ; x, t) = (2m + 4)

ψ(h) ψ(ξ) − (2m + 3) . h ξ

Letting h → 0, we then have (s)

F (2m+4) (x) ≤ (2m + 4) lim sup h→0

ψ(ξ) ψ(h) − (2m + 3) lim inf h→0 h ξ

≤ (2m + 4)SBD2m+3 f (x) − (2m + 3)SBD2m+3 f (x), and similarly (s)

F (2m+4) (x) ≥ (2m + 4)SBD2m+3 f (x) − (2m + 3)SBD2m+3 f (x), completing the proof for r = 2m + 1 and so the result is proved by induction for all odd values of r. The proof for even values of r is similar.

122


Theorem 2.8.3 Let f be special Denjoy integrable in some neighbourhood of x. If f is smooth of order r at x, then it is Borel smooth of order r at x. (s) More generally, if f(r−2) (x) exists finitely, then S r f (x) ≤ BS r f (x) ≤ BS r f (x) ≤ S r f (x),

(2.8.12)

where S r f (x), [S r f (x)], and BS r f (x), [BS r f (x)], are, respectively, the index of lower, [upper], smoothness and Borel smoothness of f of order r at x. [For the various definitions, see §6.2 and §11.2 of Chapter I.] (s) We prove (2.8.12), which implies the first part. Since f(r−2) (x) exists finitely, then by the first part of Theorem 2.8.1 SBDi f (x) exists and is equal (s) to f(i) (x), i = r − 2, r − 4, · · · . Hence, ̟r (f ; x, t) = ̟r (f ; x, t). To prove the last inequality of (2.8.12), we may suppose that S r f (x) < ∞ and then choose an M such that S r f (x) < M . Since S r f (x) = lim supt→0 t̟r (f ; x, t), there is a δ > 0 such that t̟r (f ; x, t) < M for 0 < t < δ and, so, since ̟r (f ; x, t) = ̟r (f ; x, t), we have 1 h

Z

h

t̟r (f ; x, t) dt ≤ M

ǫ

h−ǫ , for 0 < ǫ < h < δ. h

Z 1 h Letting ǫ → 0 and then h → 0, we get that lim sup t̟r (f ; x, t) dt ≤ M . h→0 h ǫ Since M was arbitrary, the last inequality of (2.8.12) is proved. The proof of the first inequality of (2.8.12) is similar. The converse of the first part of Theorem 2.8.3 is not true and to see this we first give the following lemma that will be of use in the sequel. Lemma 2.8.4 If φ is special Denjoy integrable in some neighbourhood of 0 Rh Rh and if 0 φ = o(hα ) as h → 0, then 0 tφ(t) dt = o(hα+1 ) as h → 0. Integrating by parts,

Z

0

h

tφ(t) dt = h

Z

0

h

φ−

Z hZ 0

t

φ(ξ) dξ dt = o(hα+1 ) as h → 0.

0

Now consider the functions f and g in Theorem 2.8.1 according as r is even or odd and apply Lemma 2.8.4 above to show that as h → 0, 1 h

Z

0

h

1 1 f (t) + (−1)r f (−t) dt = r−1 t 2 h

Z

0

h

1 g(t) + (−1)r g(−t) dt = o(1). tr−1 2

This proves that f or g is Borel smooth at 0 of order r. However, these functions (s) (s) are not smooth of order r at 0 since f(2) (0), g(2) (0) do not exist.


123

Theorem 2.8.5 Let f be special Denjoy integrable in some neighbourhood of x and let F be its indefinite integral in that neighbourhood. If f is Borel smooth at x of order r, r ≥ 2, then F is smooth of order r + 1 at x. More Rh generally, if SBDr−2 f (x) exists finitely and if 0 t̟r (f ; x, t) dt exists, then r2 − 1 BS r f (x) ≤ S r+1 F (x) r r2 − 1 BS r f (x). ≤ S r+1 F (x) ≤ (r + 1)BS r f (x) − r

(r + 1)BS r f (x) −

(2.8.13)

We prove (2.8.13) from which the first part of the theorem follows. We first consider the case r = 2. Let Z h

t̟2 (f ; x, t) dt

ψ(h) =

(2.8.14)

0

exist. Then, as in Theorem 2.8.2, ψ ′ (t) = t̟2 (f ; x, t) almost everywhere and so Z h i R hh (x−t) − SBD f (x) dt tψ ′ (t) dt = 2 0 f (x+t)+f 0 2 0

= F (x + h) − F (x − h) − 2hSBD0 f (x).

On integrating by parts, this gives F (x + h) − F (x − h) − 2hSBD0 f (x) = hψ(h) −

Z

h

ψ.

(2.8.15)

0

Rh By hypothesis ψ(h) → 0 as h → 0 and so h1 0 ψ → 0 as h → 0. Hence, dividing both sides of (2.8.15) by 2h and then letting h → 0, we conclude that (s) F(1) (x) exists with value SBD0 f (x). So, from (2.8.15), we have by the mean value theorem that Z h 3ψ(h) 3 3! F (x + h) − F (x − h) (s) − hF(1) (x) = − 2 ψ h2 2 h h 0 3ψ(h) 3ψ(ξ) − , (2.8.16) = h 2ξ where 0 < ξ < h. Z ψ(h) 1 h Since = t̟2 (f ; x, t) dt, we have from (2.8.16), on letting h → 0, h h 0 that 3 3 S 3 F (x) ≤ 3BS 2 f (x) − BS 2 f (x), and S 3 F (x) ≥ 3BS 2 f (x) − BS 2 f (x), 2 2 proving the result in the case r = 2.

124


Suppose now that r > 2. We prove the case of even r; the case of odd r can be handled in a similar manner. Let r = 2m and suppose SBD2m−2 f (x) exists finitely and that ψ(h) =

Z

h

t̟2m (f ; x, t) dt

(2.8.17)

0

exists. Then, as above, (2.8.17) gives Z

h

t

0

2m−1

Z hh m−1 i f (x + t) + f (x − t) X t2i ψ (t) dt = (2m)! − SBD2i f (x) dt 2 (2i)! 0 i=0 m−1 F (x + h) − F (x − h) X h2i+1 − SBD2i f (x) . = (2m)! 2 (2i + 1)! i=0 ′

(2.8.18)

(s)

By Theorem 2.8.2, F(2i−1) (x) exists and equals SBD2i−2 f (x), i = 1, 2, · · · , m, and so from (2.8.18) on integrating by parts m−1 X h2i+1 F (x + h) − F (x − h) (s) (x) − F (2m)! 2 (2i + 1)! (2i+1) i=0 Z h 2m−1 t2m−2 ψ(t) dt, = h ψ(h) − (2m − 1) 0

and, so, by the mean value theorem m−1 (2m + 1)! F (x + h) − F (x − h) X h2i+1 (s) (x) − F h2m 2 (2i + 1)! (2i+1) i=0 Z ψ(h) (4m2 − 1) h 2m−2 − t ψ(t) dt = (2m + 1) h h2m 0 ψ(h) (4m2 − 1) ψ(ξ) = (2m + 1) − , (2.8.19) h 2m ξ where 0 < ξ < h. So, letting h → 0, we have from (2.8.17) and (2.8.19) that S 2m+1 F (x) ≤ (2m + 1)BS 2m f (x) −

4m2 − 1 BS 2m f (x) 2m

and S 2m+1 F (x) ≥ (2m + 1)BS r f (x) − which completes the proof.

4m2 − 1 BS 2m f (x), 2m

(2.8.20)


125

Theorem 2.8.6 Let f be special Denjoy integrable in some neighbourhood of (s) x and let F be its indefinite integral in that neighbourhood. If F(k) (x) exists, (s)

k ≥ 2, then SBDk−1 f (x) exists and SBDk−1 f (x) = F(k) (x). More generally, if F is smooth at x of order k, then (s)

(s)

F (k) (x) ≤ SBDk−1 f (x) ≤ SBDk−1 f (x) ≤ F (k) (x).

(2.8.21)

We prove the theorem when k is even, the proof when k is odd being similar. Further, we only prove the right-hand inequality in (2.8.20), the left being similar and the inequality implying the first part of the theorem. First, consider the case of k = 2. (s) (s) We may suppose that F (2) (x) < ∞ and then choose M such that F (2) (x) < M . Then there is a δ1 > 0 such that F (x + t) + F (x − t) − 2F (x) < M, for 0 < t < δ1 . t2

(2.8.22)

Since F is an indefinite integral of f in some neighbourhood of x, then F ′ = f almost everywhere in that neighbourhood. So, there is a δ2 > 0 such that for almost t ∈ (0, δ2 ) ′ f (x + t) − f (x − t) F (x + t) + F (x − t) − 2F (x) = t t F (x + t) + F (x − t) − 2F (x) − . t2 (2.8.23) Let 0 < p < q < min{δ1 , δ2 }. Integrating (2.8.22)between p and q and then dividing by q, we have F (x + q) + F (x − q) − 2F (x) F (x + p) + F (x − p) − 2F (x) − q2 pq Z q Z 1 q F (x + t) + F (x − t) − 2F (x) 1 f (x + t) − f (x − t) dt, dt − = q p t q p t2 and so Z Z 1 q p 1 q f (x + t) − f (x − t) dt = ̟2 (F ; x, q)+ ̟2 (F ; x, t) dt− ̟2 (F ; x, p). q p t q p q Hence, by (2.8.21), Z 1 q f (x + t) − f (x − t) 1 p dt < M + M (q − p) − ̟2 (F ; x, p). q p t q q Since F is smooth of order 2 at x, p̟2 (F ; x, p) → 0 as p → 0. So letting p → 0, Z 1 q f (x + t) − f (x − t) lim sup dt ≤ 2M. t p→0 q p

126


Now, letting q → 0, we get that SBD1 f (x) ≤ M , which, since M was arbitrary, proves the right-hand inequality in this case. Suppose now that the right-hand inequality holds for k = 2, 4, · · · , 2m − 2. Then we now deduce it holds for k = 2m. This will complete the proof by induction. (s) As before, choose M such that F (2m) (x) < M < ∞. Then there is δ1 > 0 such that ̟2m (F ; x, t) < M, for 0 < t < δ1 . (2.8.24) Since F is an indefinite integral of f in some neighbourhood of x, then F ′ = f almost everywhere in that neighbourhood. So, there is a δ2 > 0 such that for almost t ∈ (0, δ2 ) m−1 t ′ (2m − 1)! f (x + t) − f (x − t) X t2i−1 (s) ̟2m (F ; x, t) = − F (x) 2m t2m−1 2 (2i − 1)! (2i) i=1 m−1 F (x + t) + F (x − t) X t2i (s) 2m − 1 (2m)! − . (2.8.25) − F(2i) (x) 2 (2i)! 2m t2m i=0 (s)

Since F is smooth of order 2m at x, we have by definition that F(2i) (x) exists finitely with value SBD2i−1 f (x) for i = 1, 2, · · · , m−1, and, so, from (2.8.24), that for almost all t ∈ (0, δ2 ) ′ t 2m − 1 ̟2m (F ; x, t) = ̟ 2m−1 (f ; x, t) − ̟2m (F ; x, t). (2.8.26) 2m 2m Let 0 < p < q < min{δ1 , δ2 }. Integrating (2.8.25) between p and q and then multiplying by 2m/q, we have p ̟2m (F ; x, p) q Z Z 2m q 2m − 1 q = ̟ 2m−1 (f ; x, t) dt − ̟2m (F ; x, t) dt, q p q p

̟2m (F ; x, q) −

or 2m q

Z

q

̟2m−1 (f ; x, t) dt

p

2m − 1 = ̟2m (F ; x, q) + q

Z

p

q

p ̟2m (F ; x, t) dt − ̟2m (F ; x, p). q

Hence, by (2.8.23), Z 2m − 1 p 2m q ̟2m−1 (f ; x, t) dt < M + M (q − p) − ̟2m (F ; x, p). q p q q Since F is smooth of order 2m at x, letting p → 0, we have Z q 2m ̟2m−1 (f ; x, t) dt ≤ 2mM. lim sup q p→0 p


127

Now, letting q → 0, 2mSBD2m−1 f (x) ≤ 2mM, that is, SBD2m−1 f (x) ≤ M . (s)

Since M is arbitrary, this proves that SBD2m−1 f (x) ≤ F (2m) (x).

2.9

Borel and Symmetric Borel Derivatives, BDk f and SBDk f , and Borel Smoothness of Order k

Theorem 2.9.1 If BDr f (x)exists finitely, then SBDr f (x) exists with the same value. Moreover for small h and 0 < ǫ < h Z h Z 1 h γ (f ; x, t) + γ r+1 (f ; x, −t) dt = ̟r+1 (f ; x, t) dt; (i) 2 ǫ r+1 ǫ 1 (ii) 2 1 (iii) 2

Z

ǫ

Z

ǫ

where

h

t γ r+1 (f ; x, t) + γ r+1 (f ; x, −t) dt =

h

γ r+1 (f ; x, t) − γ r+1 (f ; x, −t) dt = γ r+1 (f ; x, t) =

Z

h

t̟r+1 (f ; x, t) dt;

ǫ

1 r+2

Z

h

t̟r+2 (f ; x, t) dt;

ǫ

r i X ti (r + 1)! h f (x + t) − BD f (x) i tr+1 i! i=0

(2.9.1)

and ̟k (f ; x, t) is as defined in (1.11.6) of Chapter I.

Let BDr f (x) exist finitely. Then from the definition of BDr f (x), 1 h

Z

h

f (x + t) −

0

Pr

ti i=0 i! BDi f (x) tr

dt = o(1), as h → 0.

Let r be even, r = 2m, say. Then from (2.9.2) 1 h

f (x + t) + f (x − t) Pm t2i − i=0 (2i)! BD2i f (x) 2 dt t2m 0 P2m i Z h f (x + t) − i=0 ti! BDi f (x) 1 = dt 2h 0 t2m P Z h (−t)i f (x − t) − 2m 1 i=0 i! BDi f (x) dt + 2h 0 (−t)2m = o(1), as h → 0.

Z

h

(2.9.2)

128


So, SBD2m f (x) exists and SBD2i f (x) = BD2i f (x) for i = 0, 1, · · · , m. Also, from (2.9.1) and the definition of ̟k (f ; x, t) 1 2

Z

ǫ

h

γ 2m+1 (f ; x, t) + γ 2m+1 (f ; x, −t) dt

1 = 2

+

Z

h

ǫ

2m X i=0

1 = 2 Z =

Z

ǫ

h

ǫ

=

Z

h

h

2m i X t (2m + 1)! h f (x + t) − BDi f (x) − f (x − t) t2m+1 i! i=0

i (−t)i BDi f (x) dt i!

2m i i X t 1 − (−1)i (2m + 1)! h f (x + t) − f (x − t) − BD f (x) dt i t2m+1 i! i=0

m−1 i (2m + 1)! h f (x + t) − f (x − t) X t2i+1 − BD f (x) dt 2i+1 t2m+1 2 (2i + 1)! i=0

̟2m+1 (f ; x, t) dt.

ǫ

Similarly, 1 2

Z

h

ǫ

t γ 2m+1 (f ; x, t) + γ 2m+1 (f ; x, −t) dt =

Z

h

t̟(f ; x, t) dt.

ǫ

Also Z 1 h γ 2m+1 (f ; x, t) − γ 2m+1 (f ; x, −t) dt 2 ǫ Z 2m i X t 1 h (2m + 1)! h f (x + t) − BDi f (x) + f (x − t) = 2m+1 2 ǫ t i! i=0 −

2m X (−t)i i=0

= =

1 2 Z

Z

ǫ

h

ǫ

=

Z

ǫ

h

h

i!

i BDi f (x) dt

2m i i X t 1 + (−1)i (2m + 1)! h f (x + t) + f (x − t) − BDi f (x) dt 2m+1 t i! i=0

m i (2m + 1)! h f (x + t) + f (x − t) X t2i − BD f (x) dt 2i t2m+1 2 (2i)! i=0

t ̟2m+2 (f ; x, t) dt. 2m + 2

This proves (i), (ii) and (iii) when r is even. The proof when r is odd is similar.


129

Corollary 2.9.2 If BDr f (x) exists, possibly infinite, then SBDr f (x) exists + with the same value. If f is Borel smooth of order r + 1 at x, then BDr f (x) = − − BDr f (x) and BD+ r f (x) = BD r f (x). The proof of the first part follows from Theorem 2.9.1 (i) and that of the second part follows from Theorem 2.9.1 (iii). Theorem 2.9.3 If f is Borel smooth of orders r + 1 and r + 2 at x, then BDr f (x) exists finitely. If BDr f (x) exists finitely, then f is Borel smooth of order r + 1 at x. We prove the theorem for even r; the proof is similar when r is odd. Let r = 2m. Then, since f is Borel smooth of order 2m + 1 at x, SBD2m−1 f (x) exists finitely and as h → 0+, Z

m−1 i (2m + 1)! h f (x + t) − f (x − t) X t2i+1 − SBD f (x) dt = o(h). 2i+1 t2m 2 (2i + 1)! 0 i=0 (2.9.3) Again, since f is Borel smooth of order 2m+2 at x, SBD2m f (x) exists finitely and as h → 0+, h

Z

h

0

m i (2m + 2)! h f (x + t) + f (x − t) X t2i − SBD f (x) dt = o(h), 2i t2m+1 2 (2i)! i=0

and, so, by Lemma 2.8.4 as h → 0+, Z

m i (2m + 2)! h f (x + t) + f (x − t) X t2i − SBD f (x) dt = o(h2 ). 2i t2m 2 (2i)! 0 i=0 (2.9.4) From (2.9.3) and (2.9.4), we have h

Z

0

h

2m i i X t 1 h f (x + t) − SBDi f (x) dt = o(h), 2m t i! i=0

which shows that BD2m f (x) exists with value SBD2m f (x). R h Now, suppose that BDr f (x) exists finitely. Then 0 γ r (f (; x, t) − BDr f (x) dt = o(h) as h → 0. Hence, since tγ r+1 (f (; x, t) = (r + Rh 1) γ r (f (; x, t)−BDr f (x) we have that 0 tγ r+1 (f (; x, t) dt = o(h) as h → 0+, Rh and so, applying Theorem 2.9.1(ii), we have 0 tω r+1 (f (; x, t) dt = o(h) as h → 0+, showing that f is Borel smooth of order r + 1 at x.

130

2.10


Peano and Lp -Derivatives, f(k) and f(k),p

Theorem 2.10.1 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x. If f(r) (x) exists finitely, then f(r),p (x) exists with the same value, but the converse is false.

Since f(r) (x) exists finitely, we have from (1.4.2) of Chapter I f (x + t) −

r X ti i=0

i!

f(i) (x) = o(tr ) as t → 0.

Let ǫ > 0 be arbitrary. Then there is a δ > 0 such that P i f (x + t) − ri=0 ti! f(i) (x) < ǫ for 0 < |t| < δ. tr

(2.10.1)

Let 0 < h < δ. Then from (2.10.1) p r X ti f (x + t) − f(i) (x) < (ǫtr )p for 0 < t < h, i! i=0

and

p r X (−ξ)i f (x − ξ) − f(i) (x) < (ǫξ r )p for 0 < ξ < h. i! i=0

Hence, 1/p Z h r p 1/p rp X hr 1 ti p h =ǫ f(i) (x) dt ≤ ǫ , f (x + t) − h 0 i! rp + 1 (rp + 1)1/p i=0 (2.10.2) and Z h r p 1/p X 1 (−ξ)i hr . (2.10.3) f(i) (x) dξ ≤ǫ f (x − ξ) − h 0 i! (rp + 1)1/p i=0 Changing variables in (2.10.3) we have Z −h r p 1/p X 1 ti hr f(i) (x) dt ≤ǫ . f (x + t) − −h 0 i! (rp + 1)1/p i=0

From (2.10.2) and (2.10.4) Z h r p 1/p X 1 ti f (x) f (x + t) − = o(hr ), as h → 0. dt (i) h 0 i! i=0

(2.10.4)


131

Hence, f(r),p (x) exists and f(r),p (x) = f(r) (x). For the converse, consider the function r x if x = 0 or x is irrational, f (x) = 1 if x is rational and x 6= 0 . Z 1 h Then f (t) − tr dt = 0 so f(r),p (0) exists with value r!, while h 0 f(i),p (0) = 0, i = 1, 2, · · · , r − 1. However, f(1) (0) does not exist and so f(r) (0) cannot exist. Theorem 2.10.2 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x and let F be an indefinite integral of f in that neighbourhood. If f(r),p (x) exists finitely, then F(r+1) (x) exists with value f(r),p (x). By H¨ older’s inequality, Z h Z h r r p 1/p X X ti ti f(i),p (x) dt ≤ f(i),p (x) dt h1−1/p , f (x+t)− f (x+t)− i! i! 0 0 i=0 i=0 and so, Z h Z r r p 1/p X X ti ti 1 1 h f(i),p (x) dt ≤ f(i),p (x) dt . f (x + t) − f (x + t) − h 0 i! h i! 0 i=0 i=0

When h < 0, put k = −h to obtain Z k Z r r p 1/p X X ti ti 1 1 k f(i),p (x) dt ≤ f(i),p (x) dt , f (x + t) − f (x + t) − k 0 i! k i! 0 i=0 i=0 and, so, for any small |h| > 0 Z hh Z r r i X X 1 ti ti 1 h f (x) dt f (x) ≤ f (x + t) − (x + t) − dt f (i),p (i),p h i! h i! 0 0 i=0 i=0 Z h r p 1/p X ti 1 = o(hr ), as h → 0. f(i),p (x) dt ≤ f (x + t) − h 0 i! i=0 Hence,

which gives

Z hh r i X ti f (x + t) − f(i),p (x) dt = o(hr+1 ), as h → 0, i! 0 i=0

F (x + h) − F (x) −

r X hi+1 f(i),p (x) = o(hr+1 ), as h → 0, (i + 1)! i=0

which shows that F(r+1) (x) exists and equals f(r),p (x).

132


Theorem 2.10.3 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x and let F be an indefinite integral of f in that neighbourhood. If f is Lp bounded of order r at x, then F(r) (x) exists finitely and both F (r+1) (x) and F (r+1) (x) are finite.

By Corollary 1.12.5 of Chapter I, f(r−1),p (x) exists and Z h r−1 i p 1/p X t 1 f (x) (x + t) − = O(hr ), as h → 0. dt f (i),p h 0 i! i=0

So, proceeding as in Theorem 2.10.2, we have

Z hh r−1 i i X t f (x + t) − f(i),p (x) dt = O(hr+1 ), as h → 0, i! 0 i=0 which gives F (x + h) − F (x) −

r−1 X hi+1 f(i),p (x) = O(hr+1 ), as h → 0. (i + 1)! i=0

So, F is Peano bounded of order r + 1. The result then follows by Theorem 1.4.4 of Chapter I. Theorem 2.10.4 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x. If f is Peano bounded of order r at x, then f is Lp bounded of order r at x. The converse is not true. Since f is Peano bounded of order r at x by Theorem 1.4.4 of Chapter I, f(r−1) (x) exits finitely and f (x + t) −

r−1 i X t i=0

i!

f(i) (x) = O(tr ), as t → 0.

So, there is an M > 0 and a δ > 0 such that h r−1 i i X 1 t < M, for 0 < |t| < δ. f (x) f (x + t) − (i) tr i! i=0

(2.10.5)

This relation is analogous to (2.10.1). Proceeding as in the proof of Theorem 2.10.1, we get from (2.10.6) Z h r−1 i p 1/p X 1 t f(i) (x) dt = O(hr ), as h → 0, f (x + t) − h 0 i! i=0

which shows that f is Lp -bounded of order r at x. The converse part is similar to that of Theorem 2.10.1.


2.11

133

(s)

Lp - and Symmetric Lp -Derivatives, f(k),p and f(k),p (s)

Theorem 2.11.1 If f(r),p (x) exists, then f(r),p (x) exists with the same value, but the converse is false.

Since f(r),p (x) exists, then r h 1 Z h X p i1/p ti f (x + t) − = o(hr ), as h → 0. f(i),p (x) dt h 0 i! i=0

(2.11.1)

Let r be even, r = 2m, say. Then

m f (x + t) + f (x − t) X t2i − f(2i),p (x) 2 (2i)! i=0 P2m i P2m f (x + t) − i=0 ti! f(i),p (x) f (x − t) − i=0 = + 2 2

(−t)i i! f(i),p (x)

.

Hence, by Minkowski’s inequality and by (2.11.1), Z h m p 1/p 1 f (x + t) + f (x − t) X t2i − f(2i),p (x) dt h 0 2 (2i)! i=0 " Z #1/p P2m ti h f (x + t) − i=0 i! f(i),p (x) p 1 ≤ dt h 0 2 " Z P2m (−t)i p #1/p 1 h f (x − t) − i=0 i! f(i),p (x) + = o(hr ), as h → 0, dt h 0 2 (s)

and so, f(2m),p (x) exists and is equal to f(2m),p (x). If r is odd, r = 2m + 1, say, then

m f (x + t) − f (x − t) X t2i+1 − f(2i+1),p (x) 2 (2i + 1)! i=0 P2m+1 P2m+1 i f (x + t) − i=0 ti! f(i),p (x) f (x − t) − i=0 − = 2 2

(−t)i i! f(i),p (x)

,

134


and hence, as above, Z h m p 1/p 1 f (x + t) − f (x − t) X t2i+1 − f(2i+1),p (x) dt h 0 2 (2i + 1)! i=0 " Z P2m+1 ti p #1/p h f (x) f (x + t) − 1 (i),p i=0 i! dt ≤ h 0 2 " Z P2m+1 (−t)i p #1/p 1 h f (x − t) − i=0 i! f(i),p (x) + = o(hr ), as h → 0, dt h 0 2 (s)

and so f(2m+1),p (x) exists and is equal to f(2m+1),p (x). Regarding the converse, let 1, if x 6= 0, f (x) = 0, if x = 0 .

If f(1),1 (0) exist, with value λ say, then Z

1 h

0

h

f (t) − tλ dt = o(h), as h → 0.

Let h > 0 be so small that 1 − λt > 1 h 1 0 be arbitrary. Then there is a δ > 0 such that h i 1 f (x + t) + (−1)r f (x − t) − P (t) < ǫ for 0 < |t| < δ. tr 2 p Let 0 < h < δ. Then 21 f (x + t) + (−1)r f (x − t) − P (t) < (ǫtr )p for 0 < t < h. Hence, Z h p 1/p ǫp hrp 1/p 1 1 . (2.12.2) ≤ f (x + t) + (−1)r f (x − t) − P (t) dt h 0 2 rp + 1 Similarly, if −δ < h < 0, Z h p 1/p ǫp (−h)rp 1/p 1 1 r . (2.12.3) ≤ f (x+t)+(−1) f (x−t) −P (t) dt h 0 2 rp + 1

From (2.12.2) and (2.12.3), Z h p 1/p 1 1 r = o(hr ), as h → 0. f (x + t) + (−1) f (x − t) − P (t) dt h 0 2 (s)

(s)

Hence, f(r),p (x) exists, and from (2.12.1) and from the definition of f(r),p (x), (s)

(s)

it follows that f(r),p (x) = f(r) (x). For the converse, consider ( 1, if x > 0 and rational, f (x) = 0, otherwise. (s)

(s)

(s)

Then f(1) (0) and f(2) (0) do not exist and, so, f(r) (0) cannot exist for any r, (s)

(s)

odd or even, but f(r),p (0) exists with f(r),p (0) = 0 for any r.

Theorem 2.12.2 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x and (s) let F be an indefinite integral of f in that neighbourhood. If f(r),p (x) exists, (s)

(s)

then F(r+1) (x) exists with value f(r),p (x). By H¨ older’s inequality, we have, when h > 0, that Z h 1 f (x + t) + (−1)r f (x − t) − P (t) dt 2 0 Z h p 1/p 1 h1−1/p , ≤ f (x + t) + (−1)r f (x − t) − P (t) dt 0 2

where P (t) is given by (see relation (1.13.6) of Chapter I)  r/2  2i   X t f (s) (x), if r is even, (2i)! (2i),p P (t) = i=0    P(r−1)/2 t2i+1 f (s) if r is odd, i=0 (2i+1)! (2i+1),p (x),

138


which can be written P (t) =

tr (s) tr−2 (s) (x) + · · · . f(r),p (x) + f r! (r − 2)! (r−2),p

(2.12.4)

So, 1 h

Z h 1 f (x + t) + (−1)r f (x − t) − P (t) dt 2 0 Z h p 1/p 1 1 ≤ f (x + t) + (−1)r f (x − t) − P (t) dt h 0 2 = o(hr ), as h → 0+.

A similar result holds when h → 0− and, so, Z h 1 f (x + t) + (−1)r f (x − t) − P (t) dt = o(hr+1 ), as h → 0. 2 0

So, using (2.12.4), we have

1 hr+1 (s) F (x + h) + (−1)r+1 F (x − h) − (x) − · · · = o(hr+1 ), as h → 0. f 2 (r + 1)! (r),p (s)

(s)

(s)

Hence, F(r+1) (x) exists and F(r+1) (x) = f(r),p (x).

Theorem 2.12.3 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x and let F be an indefinite integral of f in that neighbourhood. If f is symmetric (s) (s) (s) Lp -bounded of order r at x, then F(r−1) (x) exists and F (r+1) (x) and F (r+1) (x) are finite.

(s)

By Theorem 1.13.3 of Chapter I, f(r−2),p (x) exists and so, by Theorem (s)

2.12.2, F(r−1) (x) exists and further from relation (1.13.4) of Chapter I, Z h 1 f (x + t) + (−1)r f (x − t) h 0 2

p 1/p tr−4 (s) tr−2 (s) f(r−2),p (x) − f(r−4),p (x) − · · · dt (r − 2)! (r − 4)! = O(hr ), as h → 0.

−

Hence, as in Theorem 2.12.2, Z hh tr−2 (s) f (x + t) + (−1)r f (x − t) (x) − f 2 (r − 2)! (r−2),p 0 i tr−4 (s) f(r−4),p (x) − · · · dt − (r − 4)! = O(hr+1 ), as h → 0,


139

and so 1 hr−1 (s) F (x+h)+(−1)r+1 F (x−h) − (x)−· · · = O(hr+1 ), as h → 0, f 2 (r − 1)! (r−2),p (s)

(s)

(s)

which shows that F(r−1) (x) exists and F (r+1) (x) and F (r+1) (x) are finite. Theorem 2.12.4 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x. If f is smooth of order r at x, then f is Lp -smooth of order r at x, but not conversely. (s)

Let f be smooth of order r at x. Then f(r−2) (x) exists finitely and h̟r (f ; x, h) = o(1) as h → 0, where ̟r (f ; x, h) is defined in relation (1.6.5) of Chapter I. Hence, 1 f (x + t) + (−1)r f (x − t) − P (t) = o(hr−1 ) as h → 0, 2

where P (t) is given by P (t) =

tr−2 (s) tr−4 (s) (x) + · · · f(r−2) (x) + f (r − 2)! (r − 4)! (r−4)

(see relation (1.6.5) of Chapter I). So, proceeding as in Theorem 2.12.1, we have Z h p 1/p 1 1 r = o(hr−1 ), as h → 0. f (x + t) + (−1) f (x − t) − P (t) dt h 0 2 Hence, f is Lp -smooth of order r at x. For the converse, the function in Theorem 2.12.1 will suffice.

Theorem 2.12.5 Let f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x. (s) (s) If f(r) (x) exists finitely and if f (r+2) (x) and f (s) (x) are finite, then f is (r+2) symmetric Lp -bounded of order r + 2 at x.

(s)

(x) are finite, we have Since f (r+2) (x) and f (s) (r+2) 1 f (x + t) + (−1)r+2 f (x − t) − P (t) = O(tr+2 ), as t → 0, 2

where P (t) is defined in (2.12.1). So, there are M > 0 and δ > 0 such that i 1 h f (x + t) + (−1)r+2 f (x − t) < M, for 0 < |t| < δ. − P (t) tr+2 2

Proceeding as in the proof of Theorem 2.12.1, replacing ǫ and r by M and r + 2, we have for 0 < |h| < δ Z h p 1/p M p h(r+2)p 1/p 1 1 , ≤ f (x + t) + (−1)r+2 f (x − t) − P (t) dt h 0 2 (r + 2)p + 1

140


and, hence, Z h p 1/p 1 1 = O(hr+2 ), , as h → 0. f (x + t) + (−1)r+2 f (x − t) − P (t) dt h 0 2

So, f is symmetric Lp -bounded of order r + 2 at x.

Corollary 2.12.6 If f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x and if f is d.l.V.P bounded of order r at x, then f is symmetric Lp -bounded of order r at x.

Borel and Lp -Derivatives, BD(k) f and f(k),p

2.13

Theorem 2.13.1 If f(r),p (x) exists, so does BDr f (x) with the same value; the converse is not true. Let f(r),p (x) exist. Then

r h 1 Z h X p i1/p ti f (x + t) − f(i),p (x) dt = o(hr ), as h → 0. h 0 i! i=0

So, applying H¨ older’s inequality, we get as in (2.10.5) Z

0

h

f (x + t) −

r X ti i=0

i!

f(i),p (x) dt = o(hr+1 ), as h → 0.

(2.13.1)

Hence, there is δ > 0 such that r 1 Z t X ξi f(i),p (x) dξ < 1, for 0 < |t| < δ, f (x + ξ) − r+1 t i! 0 i=0

(2.13.2)

and so

r 1 Z h 1 Z t X ξi f (x) dξ dt f (x + ξ) − ≤ 1, for 0 < |h| < δ. (i),p h 0 tr+1 0 i! i=0

Hence, Z

0

h

1 tr+1

Z

0

t

f (x + ξ) −

r X ξi i=0

i!

f(i),p (x) dξ dt ≤ |h|, for 0 < |h| < δ. (2.13.3)


141

Let 0 < h < δ and choose ǫ, 0 < ǫ < h. Then, on integrating by parts, Z h Z h r r X X ti ti 1 1 f (x + t) − f (x) dt = f(i),p (x) dt f (x + t) − (i),p r r i! h ǫ i! ǫ t i=0 i=0 Z h Z t r X 1 ξi +r f(i),p (x) dξ dt. (2.13.4) f (x + ξ) − r+1 t i! ǫ ǫ i=0

Now, Z

h

0

1 tr+1 =

Z t

f (x + ξ) −

0

Z

ǫ

+

Z h

r X ξi i=0

1

f(i),p (x) dξ dt

i! Z t

r X ξi

f (x + ξ) − f(i),p (x) dξ dt tr+1 0 i! i=0 Z ǫ Z t Z h r X ξi 1 f(i),p (x) dξ dt + f (x + ξ) − = r+1 i! 0 ǫ ǫ t i=0 Z ǫ Z t r X ξi 1 + f (x + ξ) − f (x) dξ dt. (2.13.5) (i),p r+1 i! 0 t 0 i=0 0

ǫ

Applying (2.13.2), we have r Z h ǫr+1 Z h 1 Z ǫ X ξi ǫr+1 1 1 , dt = − f (x) dξ dt ≤ f (x+ξ)− (i),p r+1 r+1 i! r ǫ r hr ǫ t 0 ǫ t i=0 which tends to zero as ǫ → 0 and so, Z h Z ǫ r X ξi 1 lim f(i),p (x) dξ dt = 0. f (x + ξ) − ǫ→0 ǫ tr+1 0 i! i=0

(2.13.6)

Applying (2.13.3), r Z ǫ 1 Z t X ξi f (x + ξ) − f (x) dξ dt ≤ ǫ, (i),p r+1 t i! 0 0 i=0

and so,

lim

ǫ→0

Z

0

ǫ

1 tr+1

Z

0

t

f (x + ξ) −

r X ξi i=0

i!

f(i),p (x) dξ dt = 0.

From (2.13.5), (2.13.6) and (2.13.7), Z h Z t r X ξi 1 lim f(i),p (x) dξ dt f (x + ξ) − r+1 ǫ→0 ǫ t i! ǫ i=0 Z h Z t r X ξi 1 = f (x + ξ) − f(i),p (x) dξ dt. r+1 i! 0 t 0 i=0

(2.13.7)

(2.13.8)

142


Letting ǫ → 0 in (2.13.4), we have (using (2.13.8)) Z

h

0

Z h r r X X ti ti 1 1 f (x + t) − f (x) dt = f(i),p (x) dt f (x + t) − (i),p r r t i! h 0 i! i=0 i=0 Z h Z t r X ξi 1 +r f(i),p (x) dξ dt. (2.13.9) f (x + ξ) − r+1 i! 0 t 0 i=0

If 0 < −h < δ, (2.13.9) remains valid. So applying (2.13.1) in the right-hand side of (2.13.9), we have that as h → 0 Z

Z h r X ti 1 1 1 r+1 f (x + t) − o(h ) + r o(tr+1) dt = o(h). f (x) dt = (i),p r r r+1 t i! h t 0 0 i=0 (2.13.10) This shows that BDr f (x) exists and BDr f (x) = f(r),p (x). To prove that the converse is not true, we exhibit a function f such that BD1 f (0) exists, but f(1),p (0) does not exist. Let f, φ, ψ and F be functions defined for all x, 0 ≤ x ≤ 1, with f (0) = φ(0) = F (0) = 0 and if 0 < x ≤ 1, h

1 cos(1/x3 ), x

f (x) =

φ(x) = x2 sin(1/x3 ),

F (x) = x3 sin(1/x3 );

further, let ψ(x) = 3 φ(x) − f (x) . Since F ′ (x) = ψ(x), for all x, ψ is special Denjoy integrable in [0, 1]. Since φ is continuous, it follows that f is special Denjoy integrable. However, f is not Lebesgue integrable. For if it were, then since φ is continuous, the function ψ also would be Lebesgue integrable. This implies that F is absolutely continuous and qthis we now show isqnot the case. 2 1 Let cn = 3 (4n+1)π and dn = 3 2nπ , n = 1, 2, . . .. Then (cn , dn ) is a P∞ sequence of disjoint Hence, the series 1 (dn − cn ) P∞ intervals lying in 2[0,P1]. ∞ 1 diverges. This implies that converges, but 1 F (dn ) − F (cn ) = π 1 4n+1 F is not absolutely continuous. Hence f is not Lebesgue integrable, it follows that for all p, 1 ≤ p < ∞, f∈ / Lp (0, 1) and this implies f(1),p (0) does not exist. ′ Now for x 6= 0, x2 sin(1/x3 ) = 2x sin(1/x3 ) − (3/x2 ) cos(1/x3 ) and so for 0 < ǫ < h, Z h Z h 1 1 1 1 1 3 x sin 3 dx − h2 sin 3 + ǫ2 sin 3 . cos dx = 2 (2.13.11) 2 3 x x x h ǫ ǫ ǫ Since the right-hand side of (2.13.11) tends to a limit as ǫ → 0, we can let ǫ → 0 in (2.13.11) to obtain 3

Z

0

h

1 1 cos 3 dx = 2 x2 x

Z

0

h

x sin

1 1 dx − h2 sin 3 , x3 h

(2.13.12)


143

and so, from the definition of f and from (2.13.12), Z h Z h 1 1 f (x) dx = 2 x sin 3 dx − h2 sin 3 = o(h), as h → 0. 3 x x h 0 0 Hence, BD1 f (0) exists with value 0.

Theorem 2.13.2 If f is Lp -bounded of order r at x, then f is Borel bounded of order r at x. The converse is not true. Since f is Lp -bounded of order r at x, by Corollary 1.12.5 of Chapter I, f(p−1),p (x) exists and r−1 i h 1 Z h X p i1/p t f (x + t) − f(i),p (x) dt = O(hr ), as h → 0. h 0 i! i=0

(2.13.13)

From (2.13.13), we get as in (2.13.1) Z

h

f (x + t) −

0

r−1 i X t i=0

i!

f(i),p (x) dt = O(hr+1 ), as h → 0.

(2.13.14)

So, there are M > 0 and δ > 0 such that r−1 i 1 Z t X ξ f(i),p (x) dξ < M, for 0 < |t| < δ, f (x + ξ) − r+1 t i! 0 i=0

and hence, Z

h

0

1 tr+1

Z

0

t

f (x + ξ) −

r−1 i X ξ i=0

i!

f(i),p (x) dξ dt ≤ M |h|, for 0 < |h| < δ.

(2.13.15) The relation (2.13.15) is analogous to (2.13.3). Applying arguments similar to those used to deduce (2.13.9) from (2.13.3) and then using (2.13.14) instead of (2.13.1), we get the following result, which is analogous to (2.13.10): Z

0

h

r−1 X t 1 f (x + t) − f(i),p (x) dt = O(h), as h → 0. r t i! i=0

Hence, f is Borel bounded of order r at x. The converse is as in Theorem 2.13.1.

(2.13.16)

Corollary 2.13.3 If f is Lp -bounded of order r at x, then the Borel derivative BDr−1 f (x) exists finitely and BDr f (x) and BDr f (x) are finite. I.

This follows from Theorem 2.13.2 above and Theorem 1.10.1 of Chapter

144


Theorem 2.13.4 Let f be special Denjoy integrable in some neighbourhood of x and let F be an indefinite integral of f in that neighbourhood. If BDr f (x) exists finitely, then F(r+1),p (x) exists with value BDr f (x). By Theorem 2.7.4, F(r+1) (x) exists with value BDr f (x). Since F is an indefinite integral of f , it is continuous and so is in Lp in some neighbourhood of x for all p, 1 ≤ p < ∞. So, by Theorem 2.10.1, the result follows.

2.14

Symmetric Borel and Symmetric Lp -Derivatives, (s) SBD(k) f and f(k),p (s)

Theorem 2.14.1 If the symmetric Lp -derivative f(r),p (x) exists, then the symmetric Borel derivative SBDr f (x) exists with the same value, but the converse is not true.

(s)

Let f(r),p (x) exist. Then as h → 0,

Z h m p 1/p X tr−2i 1 1 (s) f(r−2i),p (x) dt = o(hr ), f (x+t)+(−1)r f (x−t) − h 0 2 (r − 2i)! i=0

where m = r/2 or (r − 1)/2 according as r is even or odd. Applying H¨ older’s inequality, we have as in (2.13.1) Z hh m i X tr−2i 1 (s) f (x+t)+(−1)r f (x−t) − f(r−2i),p (x) dt = o(hr+1 ), as h → 0. (r − 2i)! 0 2 i=0 So, there is a δ > 0 such that if 0 < |t| < δ, Z th m i 1 X ξ r−2i (s) 1 r f (x + ξ) + (−1) f (x − ξ) − f(r−2i),p (x) dξ < 1. tr+1 (r − 2i)! 0 2 i=0

Hence, for 0 < |h| < δ, Z h Z th m i X ξ r−2i (s) 1 1 r f (x + ξ) + (−1) f (x − ξ) − f(r−2i),p (x) dξ dt r+1 t 2 (r − 2i)! 0 0 i=0 ≤ |h|.

(2.14.1)

This is similar to that in (2.13.3) and applying arguments similar to those used to deduce (2.13.10) from (2.13.3), we have that as h → 0 Z h h m i X tr−2i 1 1 (s) r f (x + t) + (−1) f (x − t) − (x) dt = o(h). f r 2 (r − 2i)! (r−2i),p 0 t i=0 (2.14.2)


145 (s)

This shows that SBDr f (x) exists and SBDr f (x) = f(r),p (x). For the converse, consider the function ( 1 cos(1/x3 ), if −1 < x ≤ 1, x 6= 0, f (x) = x 0, if x = 0. (s)

Then, we have seen in Theorem 2.13.1 that f ∈ / Lp (−1, 1) and so, f(2),p (0) 1 cannot exist. However, f is odd and, so, 2 f (x) + f (−x) = 0 for all x, which implies that SBDr f (0) = 0 for all even values of r. Theorem 2.14.2 If f is symmetric Lp -bounded of order r at x, then f is symmetric Borel bounded of order r at x. The converse is not true. Since f is symmetric Lp -bounded of order r at x, by the relation (2.14.4) (s) and Theorem 1.13.3 of Chapter I, f(r−2),p (x) exists and as h → 0 Z h m p 1/p X 1 tr−2i 1 (s) r f(r−2i),p (x) dt = O(hr ), f (x+t)+(−1) f (x−t) − h 0 2 (r − 2i)! i=1 where m = r/2 or (r − 1)/2 according as r is even or odd. Then, applying H¨ older’s inequality, we have as in (2.13.1)

Z hh m i X tr−2i 1 (s) f (x+t)+(−1)r f (x−t) − f(r−2i),p (x) dt = O(hr+1 ), as h → 0. (r − 2i)! 0 2 i=1 So, there is an M > 0 and a δ > 0 such that if 0 < |t| < δ, Z th m i 1 X ξ r−2i (s) 1 r f (x + ξ) + (−1) f (x − ξ) − f(r−2i),p (x) dξ < M. tr+1 (r − 2i)! 0 2 i=1

Hence, for 0 < |h| < δ, Z h Z th m i X ξ r−2i (s) 1 1 r f (x + ξ) + (−1) f (x − ξ) − f(r−2i),p (x) dξ dt r+1 t 2 (r − 2i)! 0 0 i=1 ≤ M |h|.

This is similar to (2.14.1) and applying arguments similar to those used to deduce (2.13.10) from (2.13.3), we have that as h → 0 Z

0

h

m i X tr−2i 1 h1 (s) r f (x + t) + (−1) f (x − t) − f(r−2i),p (x) dt = O(h). r t 2 (r − 2i)! i=1

Hence, f is symmetric Borel bounded of order r at x. The converse is as in Theorem 2.14.1.

146


Theorem 2.14.3 If f is Lp -smooth of order r at x, then f is Borel smooth of order r at x. (s)

Since f is Lp -smooth of order r at x, f(r−2),p (x) exists and as h → 0

Z h m p 1/p X 1 tr−2i 1 (s) r f (x+t)+(−1) f (x−t) − (x) f = o(hr−1 ), dt (r−2i),p h 0 2 (r − 2i)! i=1

where m = r/2 or (r − 1)/2 according as r is even or odd. Then, proceeding as in Theorem 2.14.1, we get as in (2.14.2) Z h m i X tr−2i 1 h1 (s) r f (x + t) + (−1) f (x − t) − (x) dt = o(h). f (r−2i),p r−1 2 (r − 2i)! 0 t i=1 (2.14.3) Now, by Theorem 2.14.1, SBDr−2 f (x) exists and SBDr−2i f (x) = (s) f(r−2i),p (x), i = 1, 2, . . . , m, and so, from (2.14.3) and from the definition of ̟r (f ; x, t), (see relation (1.11.6) of Chapter I), we get Z h t̟r (f ; x, t) dt = o(h), as h → 0. 0

Hence, f is Borel smooth of order r at x.

Theorem 2.14.4 If f ∈ Lp , 1 ≤ p < ∞, in some neighbourhood of x let F be an indefinite integral of f in that neighbourhood. If SBDr f (x) exists finitely, (s) then F(r+1),p (x) exists and equals SBDr f (x). Moreover, if SBDr+2 f (x) and SBDr+2 f (x) are finite, then F is Lp -bounded of order r + 3 at x. (s)

By Theorem 2.8.2, F(r+1) (x) exists with value SBDr f (x). So, by The-

(s)

(s)

orem 2.12.1, F(r+1),p (x) exists and equals F(r+1) (x), which gives the first part of the theorem. (s) For the second part, we have by Theorem 2.8.2 that F (r+3) (x) and (s)

F (r+3) (x) are finite and, so, by Theorem 2.12.5, F is Lp -bounded of order r + 3 at x.

2.15

Ces` aro and Borel Derivatives, Ck Df and BDk f

Theorem 2.15.1 Let f be Cr−1 P -integrable in some neighbourhood of x and let Φ be its rth indefinite integral in that neighbourhood. If f is Cr -continuous at x, then Cr Df (x) ≤ BDr+1 Φ(x) ≤ BDr+1 Φ(x) ≤ Cr Df (x)

(2.15.1)


147

and (r + 2)BDr+1 Φ(x) − (r + 1)BDr+1 Φ(x) ≤ Cr+1 Df (x) ≤ Cr+1 Df (x) ≤ (r + 2)BDr+1 Φ(x) − (r + 1)BDr+1 Φ(x),

(2.15.2)

provided that the two extremes of (2.15.2) are defined.

By Theorem 2.4.2, Φ(r) (x) = f (x) and Cr Df (x) = Φ(r+1) (x), Cr Df (x) = Φ(r+1) (x).

(2.15.3)

Also, Φ is a C0 P -integral (see Theorem 2.4.1) and, hence, is continuous. Since Φ(r) (x) exists finitely, by the last part of Theorem 2.7.1, Φ(r+1) (x) ≤ BDr+1 Φ(x) ≤ BDr+1 Φ(x) ≤ Φ(r+1) (x).

(2.15.4)

The proof of (2.15.1) follows from (2.15.3) and (2.15.4). To prove (2.15.2), we note that since Φ(r) (x) exists, BDr Φ(x) exists by Theorem 2.7.1. So, by Theorem 2.7.4, (r + 2)BDr+1 Φ(x) − (r + 1)BDr+1 Φ(x) ≤ Ψr+2 (x) ≤ Ψr+2 (x) ≤ (r + 2)BDr+1 Φ(x) − (r + 1)BDr+1 Φ(x),

(2.15.5)

where Ψ is an indefinite integral of Φ in some neighbourhood of x, and provided the extreme terms in (2.15.5) are defined. Since f is Cr−1 P -integrable, it is Cr P -integrable and, so, Z x+h hr+1 1 (x + h − t)r f (t) dt Cr+1 (f ; x, x + h) = Cr P − (r + 1)! r! x Z x+h 1 Cr−1 P − (x + h − t)r f (t) dt. (2.15.6) = r! x Now, using the relation (2.4.1) of Theorem 2.4.1 and integrating by parts successively, we have Z 1 x+h (x + h − t)r f (t) dt (2.15.7) r! x Z x+h x+h 1 1 (x + h − t)r−1 F1 (t) dt + = (x + h − t)r F1 (t) r! (r − 1)! x x x+h hr 1 = − F1 (x)+ (x + h − t)r−1 F2 (t) r! (r − 1)! x Z x+h 1 (x + h − t)r−2 F2 (t) dt + (r − 2)! x = ···························································· r−2 x+h Z x+h X hr−i Fi+1 (x) + (x + h − t)Φ(t) + Φ(t) dt =− (r − i)! x x i=0 =−

r−2 X hr−i Fi+1 (x) − hΦ(x) + Ψ(x + h) − Ψ(x), (r − i)! i=0

148


where F1 , F2 , . . . are as in (2.4.1) of Theorem 2.4.1. Since, by Theorem 2.4.1, Φ(k) (x) = Fr−k (x), 0 < k < r, from (2.15.6) and (2.15.7) r−1 X hr−i hr+1 Cr+1 (f ; x, x + h) = Ψ(x + h) − Ψ(x) − Φ(r−i−1) (x), (r + 1)! (r − i)! i=0

or Cr+1 (f ; x, x + h) =

r i X hi (r + 1)! h Ψ(x + h) − Ψ(x) − Φ (x) . (2.15.8) (i−1) hr+1 i! i=1

Since Φ is a C0 P -integral, it is continuous and, hence, by Theorem 2.4.1 r i X hi (r + 1)! h Φ (x) Ψ(x + h) − Ψ(x) − (i−1) h→0 hr+1 i! i=1

lim

r−1 i i X h r! h Φi (x) = lim γr (Φ; x, h) Φ(x + h) − r h→0 h h→0 i! i=0

= lim

= lim Cr (f ; x, x + h) = f (x), h→0

since f is Cr -continuous at x. So, Ψ(x + h) − Ψ(x) −

r X hi i=1

i!

Φ(i−1) (x) −

hr+1 f (x) = o(hr+1 ), as h → 0, (r + 1)!

which shows that Ψ(i) (x) = Φ(i−1) (x), i = 1, 2, . . . , r, and Ψ(r+1) (x) = f (x). Therefore, from (2.15.8), Cr+1 (f ; x, x + h) = γr+1 (Ψ; x, h). So, r + 2 r + 2 Cr+1 (f ; x, x + h) − f (x) = γr+1 (Ψ; x, h) − f (x) = γr+2 (Ψ; x, h), h h

which shows that

Cr+1 Df (x) = Ψ(r+2) (x),

Cr+1 Df (x) = Ψ(r+2) (x).

From (2.15.5) and (2.15.9), we get (2.15.2).

(2.15.9)

Corollary 2.15.2 Let f be Cr−1 P -integrable in some neighbourhood of x and let Φ be its rth indefinite integral in that neighbourhood. If f is Cr continuous at x, then (i) if Cr Df (x) exists, BDr+1 Φ(x) exists with value Cr Df (x); (ii) if BDr+1 Φ(x) exists finitely, then Cr+1 Df (x) exists and equals BDr+1 Φ(x).


149

We now exhibit a function that gives strict inequality in the extreme terms of (2.15.1). Let ( 2 2 x sin 1/x2 , if x 6= 0, 2x sin 1/x2 − cos 1/x2 , if x 6= 0, f (x) = F (x) = x 0, if x = 0. 0, if x = 0; Then, F ′ = f so f is special Denjoy integrable in every neighbourhood of 0 and F is its indefinite integral. Hence, Z F (h) 1 h = h sin 1/h2 , f= C1 (f ; 0, h) = h o h showing that f is C1 -continuous at 0. Further, C1 Df (0) = lim inf h→0

2 C1 (f ; 0, h) − f (0) = −2, h

and similarly C1 Df (0) = 2. Also, (x3 cos 1/x2 )′ = 3x2 cos 1/x2 + 2 sin 1/x2 , x 6= 0, and so Z Z h 1 h 3 h2 cos 1/h2 − sin 1/x2 dx = x2 cos 1/x2 dx → 0 as h → 0. h 0 2 2h 0 Hence, 1 h

Z

0

h

1 F (x) dx = 2 x h

that is, BD2 F (0) = 0.

2.16

Z

h

sin 1/x2 dx → 0 as h → 0;

0

Symmetric Ces` aro and Symmetric Borel Derivatives, SCk Df and SBDk f

Theorem 2.16.1 Let f be Cr−1 P -integrable in some neighbourhood of x and let Φ be its rth indefinite integral in that neighbourhood. Then, SCr Df (x) ≤ SBDr+1 Φ(x) ≤ SBDr+1 Φ(x) ≤ SCr Df (x),

(2.16.1)

and (r + 2)SBDr+1 Φ(x) − (r + 1)SBDr+1 Φ(x) ≤ SCr+1 Df (x) ≤ SCr+1 Df (x) ≤ (r + 2)SBDr+1 Φ(x) − (r + 1)SBDr+1 , (2.16.2) provided the extreme terms in (2.16.2) are defined.

150


We first observe that the relation Φ(k) (x) = Fr−k (x), 0 < k < r, still holds even if f is not Cr -continuous at x. For F1 , being a Cr−1 P -integral is Cr−1 -continuous and Cr−1 P -integrable and, so, applying Theorem 2.4.1 with f and r replaced by F1 and r −1, respectively, we have that Φ(r−1) (x) = F1 (x) and Φ(k) (x) = Fr−k (x), 0 < k < r. The hypotheses of Theorem 2.6.1 imply the existence of Φ(r−1) (x) in some neighbourhood of x and by Corollary 2.6.2 (s)

SCr Df (x) = Φ(r+1) (x),

(s)

SCr Df (x) = Φ(r+1) (x).

(2.16.3)

Also Φ, being a C0 P -integral, is continuous and so, by the last part of Theorem 2.8.1, (s)

(s)

Φ(r+1) (x) ≤ SBDr+1 Φ(x) ≤ SBDr+1 Φ(x) ≤ Φ(r+1) (x).

(2.16.4)

Hence, we have (2.16.1) from (2.16.3) and (2.16.4). As for (2.16.2), since Φ(r−1) (x) exists, BDr−1 Φ(x) exists by Theorem 2.7.1 and so SBDr−1 Φ(x) exists by Theorem 2.9.1. Hence, by the last part of Theorem 2.8.2, (s)

(r + 2)SBDr+1 Φ(x) − (r + 1)SBDr+1 Φ(x) ≤ Ψ(r+2) (x)

(2.16.5)

(s)

≤ Ψ(r+2) (x) ≤ (r + 2)SBDr+1 Φ(x) − (r + 1)SBDr+1 , where Ψ is an indefinite integral of Φ in some neighbourhood of x and provided that the extreme terms in (2.16.5) are defined. Since f is Cr−1 P -integrable, f is Cr P -integrable and, so, by the observation made at the beginning of this proof, and by using the argument applied to deduce (2.15.8) from (2.15.6), r

Cr+1 (f ; x, x + h) =

X hi (r + 1)! Ψ(x + h) − Ψ(x) − Φ(i−1) (x) . (2.16.6) hr+1 i! i=1

Since, by the observation, Φ(r−1) (x) exists, Φ(x + h) − Φ(x) −

r−1 i X h i=1

i!

Φ(i) (x) = o(hr−1 ), as h → 0,

and so, Φ being continuous, we have

r

X hi r! Ψ(x + h) − Ψ(x) − Φ(i−1) (x) r h→0 h i! i=1 lim

r−1 i X h (r − 1)! Φ(x + h) − Φ(i) (x) = 0. r−1 h→0 h i! i=0

= lim This gives

Ψ(x + h) − Ψ(x) −

r X hi i=1

i!

Φ(i−1) (x) = o(hr ), as h → 0.


151

Therefore, Ψ(i) (x) = Φ(i−1) (x), 1 ≤ i ≤ r. So, from (2.16.6), Cr+1 (f ; x, x + h) =

r X (r + 1)! hi Ψ(i) (x) = γr+1 (Ψ; x, h). Ψ(x + h) − Ψ(x) − r+1 h i! i=1

So, by Theorem 2.5.1 (ii) r + 2 Cr+1 (f ; x, x + h) − Cr+1 (f ; x, x − h) = 2h r + 2 γr+1 (Ψ; x, h) − γr+1 (Ψ; x, −h) = ̟(Ψ; x, h), 2h

which shows that

(s)

SCr+1 Df (x) = Ψ(r+2) (x),

(s)

SCr+1 Df (x) = Ψ(r+2) (x).

The relation (2.16.5) and(2.16.7) imply (2.16.2).

(2.16.7)

Corollary 2.16.2 Let f be Cr−1 P -integrable in some neighbourhood of x and let Φ be its rth indefinite integral in that neighbourhood. Then, (i) if SCr Df (x) exists, SBDr+1 Φ(x) exists with value SCr Df (x); (ii) if SBDr+1 Φ(x) exists finitely, then SCr+1 Df (x) exists and SCr+1 Df (x) = SBDr+1 Φ(x). Theorem 2.16.3 Let f be Cr−1 P -integrable in some neighbourhood of x and let Φ be its rth indefinite integral in that neighbourhood. If f is SCr continuous at x then Φ is Borel smooth of order r + 1 at x, but the converse is not true.

2.17

The proof follows from Corollary 2.6.2 and Theorem 2.8.3.

Abel and Symmetric de la Vall´ ee Poussin (s) Derivatives, ADk f and f(k)

The Abel derivative has a different form and is rather complicated; the arguments used in this section are greatly influenced by Zygmund [196] and Verblunsky [177]. Theorem 2.17.1 Let f be 2π-periodic and Lebesgue integrable. Then for all x, (s) f (s) (x) ≤ AD1 f (x) ≤ AD1 f (x) ≤ f (1) (x). (1)

152


We prove the right-hand inequality with x = x0 ; the proof of the lefthand inequality is similar. (s) There is no loss in generality in assuming f (1) (x0 ) < ∞. Let ∞ X 1 a0 + (an cos nx + bn sin nx) 2 n=1

(2.17.1)

be the Fourier series of f and write f (r, x) =

∞ X 1 (an cos nx + bn sin nx)rn , 0 < r < 1. a0 + 2 n=1

Then, by (1.14.3) of Chapter I, Z Z 1 π 1 π f (x + t)P (r, t) dt = f (t)P (r, t − x) dt, f (r, x) = π −π π −π where P (r, t) =

∞ X 1 1 1 − r2 a0 + rn cos nt = , 2 2 ∆ n=1

(2.17.2)

(2.17.3)

(2.17.4)

where ∆ = ∆(t) = ∆(r, t) = 1 − 2r cos t + r2 . Hence, writing P ′ (r, x) = (∂/∂x)P (r, x), P ′ (r, x) = −

∞ X

n=1

nrn sin nx =

−r(1 − r2 ) sin x . ∆2 (x)

(2.17.5)

In this section, we will use the prime notation exclusively for differentiation with respect to the second variable in P . In what follows, we shall write ∆ to mean ∆(t) unless otherwise stated. From (2.17.3) and (2.17.5), Z π 1 ∂ 1 f (r, x) = − f (t)P ′ (r, t − x0 ) dt r ∂x πr −π x=x0 Z π 1 = − f (x0 + t)P ′ (r, t) dt πr −π Z π 1 f (x0 − t)P ′ (r, t) dt (2.17.6) = πr −π Z π f (x0 + t) − f (x0 − t) ′ 1 P (r, t) dt = − πr −π 2 Z 1 π g(t)K(r, t) dt, = π −π where g(t) =

f (x0 + t) − f (x0 − t) P ′ (r, t) sin t (1 − r2 ) sin2 t . and K(r, t) = − = 2 sin t r ∆2


153

Since g(t) = g(−t) and K(r, t) = K(r, −t), we have from (2.17.6) that Z 2 π 1 ∂ f (r, x) = g(t)K(r, t) dt. (2.17.7) r ∂x π 0 x=x0 (s)

Since lim supt→0 g(t) = f (1) (x0 ), given an ǫ > 0 there is a δ > 0 such that (s)

g(t) < f (1) (x0 ) + ǫ if 0 < t < δ. Hence, noting that K(r, t) ≥ 0, 0 ≤ t ≤ π, 2 π

Z

0

δ

Z δ 2 (s) f (x0 ) + ǫ g(t)K(r, t) dt ≤ K(r, t) dt π (1) Z0 π 2 (s) f (1) (x0 ) + ǫ K(r, t) dt. < π 0

(2.17.8)

The relation (2.17.7) holds for all 2π-periodic Lebesgue integrable functions and for all choices of x0 and, hence, we can put f (x) R π = sin x and x0 = 0 when g(t) = 1 for all t and, so, from (2.17.7), 1 = π2 0 K(r, t) dt. Using this in (2.17.8), Z 2 δ (s) (2.17.9) g(t)K(r, t) dt < f (1) (x0 ) + ǫ. π 0 Also,

2 π

Z

δ

π

Z π 2 g(t) dt. sup K(r, t) g(t)K(r, t) dt ≤ π δ≤t≤π δ

(2.17.10)

−1 For δ ≤ t ≤ π, ∆(t) ≤ (1 − 2r cos δ + r2 )−1 and, so,

2 ′ r(1 − r2 ) P (r, t) = −r(1 − r ) sin t ≤ . ∆2 (t) (1 − 2r cos δ + r2 )2

Hence, if δ ≤ t ≤ π,

P ′ (r, t) sin t lim sup K(r, t) = lim sup r→1 δ≤t≤π r→1 δ≤t≤π r ≤ lim sup

r→1 δ≤t≤π

(1 − r2 ) = 0. (1 − 2r cos δ + r2 )2

Hence, from (2.17.10), 2 r→1 π lim

Z

π

g(t)K(r, t) dt = 0.

δ

From (2.17.7), (2.17.9) and (2.17.11), lim sup r→1

1 ∂ (s) f (r, x) ≤ f (1) (x0 ) + ǫ, r ∂x x=x0

(2.17.11)

154


and since is arbitrary, this proves that lim sup r→1

1 ∂ (s) f (r, x) ≤ f (1) (x0 ), r ∂x x=x0 (s)

which implies that AD 1 f (x0 ) ≤ f (1) (x0 ) as had to be proved.

Corollary 2.17.2 Let f be 2π-periodic and Lebesgue integrable. Then if (s) f(1) (x) exists so does AD1 f (x), with the same value. This result has not been extended to higher order Abel derivatives because inequalities analogous to (2.17.1) are not known. However, a weaker result for the second order Abel derivative can be given. The following result is known as Rajchman’s Lemma; see [196] [p. 353] and [177] [p. 445]. Theorem 2.17.3 Let f be 2π-periodic and Lebesgue integrable with a Fourier series Abel summable at x0 to f (x0 ). Then, (s)

(s) AD2 f (x0 ) ≥ f (2) (x0 ) and AD2 f (x0 ) ≤ f (2) (x0 ).

(2.17.12)

We prove the first inequality and the second inequality follows by applying the first inequality to the function −f . Let (2.17.1) be the Fourier series of f and the function f (r, x), the Abel means of f , be as in (2.17.2). Then the hypothesis of the theorem says that f (x0 ) = limr→1 f (r, x0 ). Further, without loss in generality, we may suppose that x0 = 0, f (x) = f (−x) and f (0) = 0. (s) (0) and let k be such that Suppose, if possible, that AD 2 f (0) < f (2) (s) AD2 f (0) < k < f (2) (0). Put F (x) = f (x) − k(1 − cos x) when F (0) = 0 and F (x) = F (−x). The Abel mean of the Fourier series of F , as in (2.17.2), is ∞ 1 a0 − k + kr cos x + (an cos nx + bn sin nx)rn , 0 < r < 1. 2 n=1 (2.17.13) Further, since f (x0 ) = limr→1 f (r, x0 ), we have that F (0) = limr→1 F (r, 0). (s) (s) Also, AD 2 f (0) − k = AD2 F (0) and f (2) (0) − k = F (2) (0) and, so, by our

F (r, x) =

(s)

supposition, AD 2 F (0) < 0 < F (2) (0). Let φ(t) = (s)

F (t) + F (−t) − 2F (0) 2F (t) = , 2 sin t sin2 t

then since F (2) (0) > 0, lim inf t→0 φ(t) > 0. So, choosing h, 0 < h < lim inf t→0 φ(t), we find an η, 0 < η < π/3, such that φ(t) > h if 0 < t < η. If

Relations between Derivatives 0 < t < η, then cos t > −

Z

η

1 2

155

and, by (2.17.5) P ′ (r, t) < 0, we have

′

Z

η

sin tP ′ (r, t) dt Z η η −h sin tP (r, t) 0 − cos tP (r, t) dt Z0 η h P (r, t) dt −h sin ηP (r, η) + 2 0 Z π Z h h π −h sin ηP (r, η) + P (r, t) dt − P (r, t) dt 2 0 2 η Z hπ h π −h sin ηP (r, η) + − P (r, t) dt. 4 2 η

φ(t) sin tP (r, t) dt ≥ −h

0

0

= ≥ = =

So, using the relations (1.14.5) and (1.14.9) of Chapter I, Z η hπ ′ . φ(t) sin tP (r, t) dt ≥ lim inf − r→1 4 0 Further, by (2.17.5), lim inf r→1

and so lim inf r→1

−

−

Z

π

′

φ(t) sin tP (r, t) dt

η

Z

π ′

φ(t) sin tP (r, t) dt 0

= 0,

≥

hπ . 4

(2.17.14)

Writing G(r) = F (r, 0) and applying (2.17.3) and (2.17.4), Z Z 1 π 1 π 1 − r2 G(r) = dt. F (t)P (r, t) dt = F (t) π −π π 0 ∆ So, differentiating with respect to r and using (2.17.5), Z Z d rG(r) d 1 π 1 π r 1 − r2 = dt dt = F (t) F (t) dr 1 − r2 dr π 0 ∆ π 0 ∆2 Z π 1 − r2 1 dt φ(t) sin2 t = 2π 0 ∆2 Z π 1 = − φ(t) sin tP ′ (r, t) dt, 2πr 0 hence, from (2.17.14), d lim inf r→1 dr

rG(r) 1 − r2

≥

h . 8

(2.17.15)

156


Now, d 1 − r2 rG(r) dr r log r 1 − r2 1 − r2 d rG(r) d 1 − r2 rG(r) = + . (2.17.16) r log r dr 1 − r2 dr r log r 1 − r2 d 1 − r2 1 − r2 = O(1 − r) as r → 1 and noting that → −2 and Since r log r dr r log r limr→1 G(r) = F (0) = 0, we get from (2.17.15) and (2.17.16) that d G(r) lim sup < 0. log r r→1 dr d G(r) So, there is an r0 , 0 < r0 < 1, such that < 0, r0 < r < 1. This dr log r G(r) is strictly decreasing on that interval; that is, implies that log r d dr

G(r) log r

=

G(r) G(p) > , if r0 < r < p < 1. log r log p

(2.17.17)

Now, from (2.17.13), ∞ X ∂2 F (r, x) = −krx cos x − (an cos nx + bn sin nx)n2 rn ∂x2 n=1 ∂ ∂ r F (r, x) . (2.17.18) = −r ∂r ∂r ∂2 Since, by definition, AD2 F (0) = lim supr→1 2 F (r, x) < 0, ∂x x=0 ∂ ∂ r F (r, x) (2.17.18) implies that lim inf r→1 r > 0 and so, ∂r ∂r x=0 d d lim inf r→1 r r G(r) > 0. So, there is an r1 , 0 < r1 < 1, such that dr dr d d r G(r) > 0, r1 < r < 1; and so rG′ (r) is strictly increasing in (r1 , 1). dr dr Since limr→1 G(r) = 0, for each r, r1 < r < 1, there is, by the mean value G(r) d theorem, a p, r < p < 1, such that = p G(r) . Hence, for each log r dr r=p r, r1 < r < 1, there are p and σ such that d d G(r) G(p) − = p G(r) −σ G(r) < 0, r < p < σ < 1. (2.17.19) log r log p dr dr r=p r=σ

Relations (2.17.19) and (2.17.17) are contradictory and show that the assumption made is false and thus, the theorem is proved.


157

Theorem 2.17.4 Let f be 2π-periodic and Lebesgue integrable with a Fourier series Abel summable at x0 to f (x0 ). If Z h

f (x0 + h) + f (x0 − h) − 2f (x0 ) 0 − f (x0 + u) + f (x0 − u) − 2f (x0 ) cos u du, 3 2h3

I(h) =

then

(2.17.20)

lim inf I(h) ≤ AD2 f (x0 ) ≤ AD2 f (x0 ) ≤ lim sup I(h). h→0

h→0

As before, we may suppose that x0 = 0, f (0) = 0 andf (x) = f (−x). Then (2.17.20) becomes I(h) = =

3 2h3 3 h3

= 3

Z

0

h

2f (h) − 2f (u) cos u du

f (h) sin h −

φ(h) , where h3

Z

0

h

f (u) cos u du

φ(h) = f (h) sin h −

Z

h

f (u) cos u du. (2.17.21)

0

Thus, we have to prove lim inf 3 t→0

φ(t) φ(t) ≤ AD2 f (0) ≤ AD2 f (0) ≤ lim sup 3 3 . 3 t t t→0

(2.17.22)

We prove the left-hand inequality, as the proof of the right-hand inequality then follows by applying the left-hand inequality to the function −f . Further, φ(t) we may (without loss in generality) suppose that lim inf t→0 3 > −∞ and t φ(t) choose a k such that lim inf t→0 3 > k > −∞. Then there is a δ > 0, 0 < t φ(t) δ < π, such that 3 > k if 0 < t < δ. So, using notation introduced above, t Z δ Z δ 3 Z δ 3 sin t t sin t φ(t) t sin t φ(t) dt = dt ≥ k dt. 3 3 3 ∆ ∆ t ∆3 0 0 0 Therefore, Z

0

π

Z δ Z π sin t sin t φ(t) dt = φ(t) dt + ∆3 ∆3 0 δ Z δ 3 Z π t sin t sin t ≥k dt + φ(t) dt 3 3 ∆ 0 δ ∆ Z π Z π Z π t3 sin t sin t dt + φ(t) dt. (2.17.23) = k − 3 3 ∆ δ ∆ 0 δ

158


Since, Z (1 − r2 )

Z π 3 t sin t t sin t 2 dt ≤ (1 − r ) dt → 0 as r → 1 3 (r, δ) ∆3 ∆ δ

π 3

δ

and

Z (1 − r ) 2

δ

π

Z π sin t 1 − r2 φ(t) dt → 0 as r → 1, φ(t) dt ≤ 3 3 ∆ ∆ (r, δ) δ

we have from (2.17.23) Z π 3 Z π t sin t sin t 2 φ(t) dt ≥ lim inf (1 − r )k dt. lim inf (1 − r2 ) 3 r→1 r→1 ∆ ∆3 0 0

(2.17.24)

We now evaluate the right-hand side of (2.17.24). Integrating by parts and using (2.17.23), Z π t=π Z π π(1 − r2 ) π −πr ′ tP (r, t) dt = tP (r, t) P (r, t) dt = − − = , 2 2(1 + r) 2 1 +r t=0 0 0

and so, using (2.17.5) Z π 2 Z π 2 −t −r(1 − r2 ) sin t t dt = dt r(1 − r2 ) 2 sin t ∆2 0 0 ∆ Z π Z π t2 ′ tP ′ (r, t) dt (2.17.25) P (r, t) dt − t− = sin t 0 0 Z π πr t2 = g(t)P ′ (r, t) dt + , where g(t) = t − . 1+r sin t 0

2t sin t − t2 cos t t t 2− cos t . So g(t) → 0 = 1− 2 sin t sin t sin t g ′ (t) → 0 as t → 0. Let ǫ > 0 be arbitrary and choose δ > 0 such that and g ′ (t) < ǫ if 0 < t < δ. So, integrating by parts, Then g ′ (t) = 1 −

Z δ Z δ t=δ ′ g ′ (t)P (r, t) dt g(t)P (r, t) dt = g(t)P (r, t) t=0 − 0 0 Z δ ′ ≤ g(δ)P (r, δ) + g (t)P (r, t) dt 0 Z π ≤ g(δ) P (r, δ) + ǫ P (r, t) dt

Hence,

π = g(δ) P (r, δ) + ǫ . 2

0

Z δ π π lim g(t)P ′ (r, t) dt ≤ g(δ) lim P (r, δ) + ǫ = ǫ . r→1 0 r→1 2 2

(2.17.26)


159

Also, lim

r→1

Z

π

Z π t2 −r(1 − r2 ) sin t t− dt r→1 δ sin t ∆2 Z π 2 t − t sin t dt = 0. (2.17.27) = lim r(1 − r2 ) r→1 ∆2 δ

g(t)P ′ (r, t) dt = lim

δ

From (2.17.26) and (2.17.27), Z π π lim g(t)P ′ (r, t) dt ≤ ǫ r→1 0 2 and, since ǫ is arbitrary, lim

r→1

Z

π

g(t)P ′ (r, t) dt = 0.

(2.17.28)

0

Now integrating by parts, Z π Z π 2 2r sin t 8r (1 − r2 )t3 sin t 2 dt = 4r(1 − r ) t3 dt 3 ∆ ∆3 0 0 Z t Z π Z t t=π 2r sin ξ 2r sin ξ 2 dξ dξ dt t − 3 = 4r(1 − r2 ) t3 3 3 t=0 0 0 ∆ (r, ξ) 0 ∆ (r, ξ) Z π −1 t=π 3 t2 + = 4r(1 − r2 ) t3 dt 2 2 2∆ (r, t) t=0 2 0 ∆ (r, t) Z π t2 −2r(1 − r2 )π 3 2 + 6r(1 − r ) dt. = 4 2 (1 + r) 0 ∆ (r, t) Letting r → 1 and using (2.17.25) and (2.17.28), Z π 2 8r (1 − r2 )t3 sin t rπ lim dt = lim 6 = 3π. r→1 0 r→1 1 + r ∆3 Using (2.17.29), we get from (2.17.24), Z π 3πk sin t dt ≥ lim inf (1 − r2 ) . 3 r→1 8 0 ∆ It can be verified by differentiating (2.17.2) and(2.17.4) that Z 1 π ∂2 f (r, x) = f (t)P ′′ (r, t) dt. ∂x2 π −π Now, from (2.17.4), −r(1 − r2 ) cos x 2r(1 − r2 ) sin x 2r sin x + ∆(x)2 ∆3 (x) h 4r sin2 x cos x i , − 2 = r(1 − r2 ) 3 ∆(x) ∆ (x)

P ′′ (r, x) =

(2.17.29)

(2.17.30)

(2.17.31)

160


and, so, from (2.17.31), ∂2 f (r, x) (2.17.32) ∂x2 x=x0 Z 4r sin2 t cos t 1 π dt − f (t) 2r(1 − r2 ) = π 0 ∆3 ∆2 Z Z 1 π 8r2 (1 − r2 ) sin t 1 π 2r(1 − r2 ) = f (t) sin t dt − f (t) cos t dt. π 0 ∆3 π 0 ∆2 Integrating by parts, Z π 1 f (t) cos t dt (2.17.33) 2 ∆ 0 Z Z t Z t=π t π 1 2.2r sin t f (ξ) cos ξ dξ dt f (ξ) cos ξ dξ = 2 + ∆ 0 ∆3 t=0 0 0 Z t Z π Z π 1 4r sin t f (ξ) cos ξ dξ dt. = f (ξ) cos ξ dξ + (1 + r)2 0 ∆3 0 0 From (2.17.32) and (2.17.33), ∂2 f (r, x) = 2 ∂x x=x0 Z π 2 Z π 1 2r(1 − r2 ) 8r (1 − r2 ) sin t 1 f (t) sin t dt − f (t) cos t dt π 0 ∆3 π (1 + r)2 0 Z t Z π 4r sin t f (ξ) cos ξ dξ dt + ∆3 0 0 Z Z 2r(1 − r2 ) π 8r2 (1 − r2 ) π sin t f (t) sin t =− f (t) cos t dt + 3 π(1 + r)2 0 π 0 ∆ Z t − f (ξ) cos ξ dξ dt 0 Z Z 2r(1 − r2 ) π 8r2 (1 − r2 ) π sin t =− φ(t) dt, f (t) cos t dt + 3 π(1 + r)2 0 π 0 ∆ where φ(t) is defined in (2.17.21). So, using (2.17.30), lim inf r→1

∂2 f (r, x) ≥ 3k. ∂x2 x=x0

Since k was arbitrary, we can conclude

AD2 f (0) ≥ 3 lim inf t→0

proving the left-hand inequality in (2.17.22).

φ(t) , t3


161

Theorem 2.17.5 Let f be 2π-periodic and Lebesgue integrable, with f (x0 + t) + f (x0 − t) limt→0 = f (x0 ). If 2 Z h 3 J(h) = f (x0 + h) + f (x0 − h) − 2f (x0 ) 3 2h 0 (2.17.34) − f (x0 + u) + f (x0 − u) − 2f (x0 ) du,

then

lim inf J(h) ≤ AD2 f (x0 ) ≤ AD2 f (x0 ) ≤ lim sup J(h). h→0

h→0

We show that limh→0 J(h)−I(h) = 0, where I(h) is the integral defined in (2.17.20) and the result follows from Theorem 2.17.4 because by Lemma 2.15.2 of Chapter I, the Fourier series of f is Abel summable at x0 to f (x0 ). As before, we may suppose that x0 = 0, f (0) = 0 and f (x) = f (−x) when Z h J(h) − I(h) = 3 f (h) − f (u) (1 − cos u) du . 2h3 0

2 Let ǫ > 0 be arbitrary, then since f (t) → f (0) = 0 and (1 − cos t)/t → 1/2 as t → 0, there is a δ > 0 such that f (t) < ǫ and |1 − cos t| < (ǫ + 1/2)t2 if 0 < t < δ. Then, from the above, Z h f (h) − f (u) (1 − cos u) du J(h) − I(h) ≤ 3 3 2h 0 Z Z h i 3 h h f (u) (1 − cos u) du ≤ f (h) (1 − cos u) du + 2h3 0 3 03 ǫ 1 h 1 h 3 = (1 + 2ǫ). ǫ +ǫ +ǫ +ǫ ≤ 2h3 2 3 2 3 2 Hence, limh→0 J(h) − I(h) = 0.

Theorem 2.17.6 Let f be 2π-periodic and Lebesgue integrable and let f (x0 + t) + f (x0 − t) = f (x0 ). Then, limt→0 2

3 (s) 3 (s) 1 1 (s) f (x0 ) − f (2) (x0 ) ≤ AD2 f (x0 ) ≤ AD2 f (x0 ) ≤ f (2) (x0 ) − f (s) (x0 ), 2 (2) 2 2 2 (2) (2.17.35) provided the extreme terms of these inequalities are defined. We will only consider the the left-hand inequality as once this is proved the right-hand inequality is obtained by applying the left-hand inequality to the function −f . (s) Note that, in particular, the extreme terms are not defined if f(2) (x0 ) = (s)

(x ) < ∞ ±∞. So, we may assume that f (2) (x0 ) < ∞ for if it is ∞, then f (s) (2) 0

162

Higher Order Derivatives (s)

and the left-hand side is −∞ and the result is obvious. If f (2) (x0 ) < ∞ and f (s) (x ) = −∞, the result is again obvious, thus, we may assume that both (2) 0 derivates are finite. In addition, we may assume without loss in generality that x0 = 0, f (0) = 0 and f (x) = f (−x). Choose then k1 , k2 such that k2 < (s)

(0) ≤ f (2) (0) < k1 . Then there exists δ > 0 such that f (s) (2) k2
3k2 . h3

h

3 2f (h) − 2f (u) du ≥ lim inf 3 h→0 h 0 Z h 3 f (u) du − lim sup 3 h→0 h 0 3 1 ≥ k2 − k1 . 2 2

Z

h

f (h) du

0

(2.17.36)

Thus, by Theorem 2.17.5, AD2 f (0) ≥ 23 k2 − 12 k1 and, since k1 and k2 are arbitrary, this gives the left-hand inequality in (2.17.35). (s)

Corollary 2.17.7 Let f be 2π-periodic and Lebesgue integrable. If f(2) (x0 ) (s)

exists finitely, then AD2 f (x0 ) exists with value f(2) (x0 ). Remark. Another proof of this corollary is in Lemma 2.9.2 of [196] [356]. Analogues of Theorem 2.17.3 and Corollary 2.17.7 are not known for higher order Abel and d.l.V.P. derivates. However, a slightly less general result is the following: Theorem 2.17.8 Let f be 2π-periodic and Lebesgue integrable. If the upper (s) (x ) are finite, then there is a and lower d.l.V.P. derivates f (r) (x0 ) and f (s) (r) 0 K > 0 such that −Kλ ≤ AD r f (x0 ) ≤ ADr f (x0 ) ≤ Kλ, (s) (s) (x ) }. Moreover, if f(r) (x0 ) exists finitely, where λ = max{ f (r) (x0 ) , f (s) (r) 0 (s)

then ADr f (x0 ) exists and equals f(r) (x0 ).


163

This result is proved in [121, Theorem 2.1 and Corollary] when r is even, and [121, Theorem 2.2 ] when r is odd. Theorem 2.17.9 Let f be 2π-periodic and Lebesgue integrable with a Fourier series Abel summable at x0 to f (x0 ). Then, S 2 f (x0 ) ≤ πAS 2 f (x0 ),

S 2 f (x0 ) ≥ πAS 2 f (x0 ).

We prove the first inequality by showing that if for any k, S 2 f (x0 ) > k, then πAS 2 f (x0 ) ≥ k; the proof of the second is similar. We may suppose that x0 = 0, f (x0 ) = 0 and f (x) = f (−x). Define the 2π-periodic function g(t) by letting g(t) = 21 k|t|, −π < t ≤ π, where k is as above. Then, as in (2.17.31), Z Z 1 π 2 π t ′′ ∂2 ′′ g(r, x) = g(t)P (r, t) dt = k P (r, t) dt ∂x2 π −π π 0 2 x=0 t=π Z π k = tP ′ (r, t) P ′ (r, t) dt − π t=0 0 t=π k = − P (r, t) , by (2.17.5), π t=0 1 − r2 k 1 1 − r2 , by (2.17.4), − = π 2 (1 − r)2 (1 + r)2 k 2r = . π(1 − r) 1 + r ∂2 = k. This shows that the theorem Hence, limr→1− π(1 − r) 2 g(r, x) ∂x x=0 holds for the function g and x0 = 0. As a result, we may subtract g fromf and it is sufficient to prove that if S 2 f (0) > 0, then πAS 2 f (0) ≥ 0, or just that AS 2 f (0) ≥ 0. Suppose then that S 2 f (0) > 0 when limh→0+ f (h)/h > 0. Hence, here is a δ > 0 such that f (t) > 0 for 0 < t < δ. We may assume that f > 0 everywhere. For, since Z ∂2 1 π f (r, x) = f (t)P ′′ (r, t) dt ∂x2 π 0 x=0 Z δ Z π 1 + f (t)P ′′ (r, t) dt = π 0 δ and lim (1 − r)

r→1−

Z

π ′′

f (t)P (r, t) dt = lim (1 − r) δ

r→1−

Z

π δ

1 − r2 ∂ 2 f (t) 2 ∂t2

1 ∆

dt = 0,

we have πAS 2 f (0) = lim sup π(1−r) r→1−

Z δ ∂2 f (r, x) = lim sup (1−r) f (t)P ′′ (r, t) dt. ∂x2 x=0 r→1− 0

164


Therefore, the value of AS 2 f (0) does not depend on the values of f outside (0, δ). So, Z Z 1 π 1 − r2 1 π dt > 0, for 0 < r < 1. f (t)P (r, t) dt = f (t) f (r, 0) = π −π π 0 ∆ (2.17.37) P n Since f (r, x) is given by (2.17.2), f (r, 0) = a0 /2 + ∞ a r and, thus, n=1 n ∞ X d d ∂2 −r r f (r, 0) = − n2 an rn = 2 f (r, x) . dr dr ∂r x=0 n=1

Now, suppose if possible that AS 2 f (0) < 0. Then lim sup(1 − r) ∂ 2 f (r, x) ∂x2 r→−1

x=0

(2.17.38)

0 and a σ, 0 < σ < 1, such that −c ∂2 f (r, x) , for 1 − σ < r < 1. < 2 ∂x 1−r x=0

Hence, from (2.17.39), d d c r f (r, 0) > , for 1 − σ < r < 1. dr dr 1−r

(2.17.39)

Let 1 − σ < r1 < r2 < 1. Then, from (2.17.40), Z r2 d 1 d f (r, 0) dr − r f (r, 0) ≥c dr dr 1 − r r=r2 r=r1 r1 1 − r1 → ∞ as r2 → 1−. = c log 1 − r2 d Hence, limr2 →1− r f (r, 0) = ∞ and, so, f (r, 0) is strictly increasing in dr r=r2 some left neighbourhood of 1. Since the Fourier series of f is Abel summable at 0 to f (0) = 0, f (r, 0) → 0 as r → −1. It follows that f (r, 0) is negative in some left neighbourhood of 1. However, this contradicts (2.17.38) and so our assumption that AS 2 f (0) < 0 is false and, so, AS 2 f (0) ≥ 0, as had to be proved. r

Just as Theorem 2.17.9 is the analogue of Theorem 2.17.3 for 2-smoothness, a theorem is proved in [122, Theorem 2.1 and Theorem 2.2] and is an analogue of Theorem 2.17.8 for r-smoothness. We now give two examples. The first will show that the converse of Corollary 2.17.7 does not hold; the second will show that the use of the Lebesgue integral in this theory is necessary.


165

Example 1. Let f be 2π-periodic and  π−x  , if 0 < x ≤ 2π and irrational,  f (x) = 1, 2 if 0 < x ≤ 2π and rational,   0, if x = 0. P∞ Then, n=1 sin nx/n is the Fourier series of f and, thus, f (r, x) = P∞ n n=1 (r sin nx)/n. Hence, ∞ X −r(1 − r2 ) sin x ∂2 n ′ f (r, x) = − nr sin nx = P (r, x) = . ∂x2 (1 − 2r cos x + r2 )2 n=1 (s)

Therefore, AD2 f (x) = 0 for all x; but f(2) (x) does not exist for any x. This same example shows that AD2 f (x) = 0 for all x does not imply that f is a polynomial of degree at most, unless other conditions are satisfied 2. Example 2. If F (x) =

x2 cos 1/x2 , if x 6= 0, 0, if x = 0,

then F is differentiable. Define f as a 2π-periodic function with ( F ′ (x), if 0 ≤ x ≤ π, f (x) = f (−x), if −π ≤ x ≤ 0. Since F ′ is not Lebesgue integrable on [0, π], f is not Lebesgue integrable on [−π, π] and, thus, the existence of ADk f (x) cannot arise for any k and any x. Note that although F ′ is special Denjoy integrable and, hence, f is also special Denjoy integrable, the Fourier Denjoy coefficients of f need not tend to zero and, therefore, the definition of ADk f fails. However, f (k) (x) exists for all k and for all x, x 6= 0. Remark. Theorem 2.17.3 is due to Rajchman and Zygmund, see [177] [p. 445] and [196] [p. 353]. Theorems 2.17.4 and 2.17.5 are due to Verblunsky; see [177] [pp. 446–447]. Theorem 2.17.9 is proved in [177] [p. 443] and also in [196] [p. 357]. The proofs given in these references are related to Fourier series. We have modified and simplified these proofs to serve our purpose.

2.18

Laplace, Peano and Generalized Peano Derivatives, LDk f , f(k) and f[k]

Lemma 2.18.1 Let f be special Denjoy integrable in I = [a, b] and C1 continuous at x0 ∈ I. If ℓ ∈ N, write f (−ℓ) for the ℓ-th indefinite integral of f

166


on I. Then, for each k ∈ N (f (−k) )(r) (x0 ) = f (−k+r) (x0 ), for r = 1, 2, . . . , k.

(2.18.1)

Since f (−1) is an indefinite Denjoy integral of f on I and since f is C1 continuous at x0 , f (−1) (x0 + t) − f (−1) (x0 ) 1 x0 +t = f → f (x0 ) as t → 0. (2.18.2) t t x0 Since f (−1) is continuous in I, (f (−k) )(r) (x0 ) = f (−k+r) (x0 ), for r = 1, 2, . . . , k − 1,

(2.18.3)

and, from (2.18.2), (f (−k) )(k) (x0 ) = (f (−1) )(1) (x0 ) = f (x0 ). This together with (2.18.3) gives (2.18.1). Theorem 2.18.2 Let f be special Denjoy integrable and C1 -continuous in I = [a, b]. If n ∈ N, then for all k ∈ N, LDn+k f (−k) (x) exists if and only if LDn f (x) exists, in which case they are equal. More generally, for all k +

+

LDn+k f (−k) (x) = LDn f (x),

(−k) LD+ (x) = LD+ n f (x), n+k f

(2.18.4)

with similar relations for the left-hand Laplace derivates. Let n and k be fixed. By Lemma 2.18.1, the kth derivative of f (−k) at x exists. Let αj = (f (−k) )(j) (x), j = 0, 1, 2, . . . , k. If n > 1, choose arbitrarily αk+1 , . . . , αn+k−1 ; if n = 1, there is nothing to choose. Integrating by parts, n+k−1 ti αi dt e−st f (−k) (x + t) − i! 0 i=0

−st n+k−1 ti t=δ n+k+1 e (−k) f αi =s (x + t) − −s i! t=0 i=0 δ n+k−1 ti−1 n+k −st (−k+1) αi dt. f +s e (x + t) − (i − 1)! 0 i=1

sn+k+1

δ

(2.18.5)

The first term on the right-hand side of (2.18.5 ) is o(1) as s → ∞ and so, as s → ∞, n+k−1 ti αi dt e−st f (−k) (x + t) − i! 0 i=0 δ n+k−1 ti−1 αi dt. (2.18.6) = o(1) + sn+k e−st f (−k+1) (x + t) − (i − 1)! 0 i=1

sn+k+1

δ


167

In a similar manner we can also prove n+k−1 X ti−1 (2.18.7) αi dt e−st f (−k+1) (x + t) − (i − 1)! 0 i=1 Z δ n+k−1 X ti−2 = o(1) + sn+k−1 e−st f (−k+2) (x + t) − αi dt. (i − 2)! 0 i=2

sn+k

Z

δ

From (2.18.6) and (2.18.7), we have n+k−1 X ti (−k) (2.18.8) αi dt f (x + t) − s e i! 0 i=0 Z δ n+k−1 X ti−2 n+k−1 (−k+2) −st = o(1) + s f (x + t) − e αi dt. (i − 2)! 0 i=2 n+k+1

Z

δ

−st

As we get (2.18.8) from (2.18.6) using (2.18.7), we get, continuing this process, n+k−1 X ti αi dt (2.18.9) e−st f (−k) (x + t) − i! 0 i=0 Z δ n+k−1 X ti−k = o(1) + sn+1 e−st f (x + t) − αi dt (i − k)! 0 i=k Z δ n−1 X ti n+1 −st αi+k dt. f (x + t) − = o(1) + s e i! 0 i=0

sn+k+1

Z

δ

Now, suppose that LDn+k f (−k) (x) exists finitely. Then taking αi = LDi+ f (−k) (x), i = k + 1, k + 2, . . . , n + k − 1, the left-hand side of (2.18.9) + tends to LDn+k f (−k) (x) as s → ∞ and, so, the right-hand side of (2.18.9) + tends to LDn f (x) with equal value. If LDn+ f (x) exists finitely, then taking αi+k = LDi+ f (x), i = 1, 2, . . . , n − 1, the right-hand side of (2.18.9) tends to LDn+ f (x) as s → ∞ and, so, the left-hand side of (2.18.9) tends + to LDn+k f (−k) (x). +

To prove the last part: It follows from the definitions of LDn+k f (−k) (x) and +

+ + LDn f (x) that LDn+k−1 f (−k) (x) and LDn−1 f (x) exist finitely and, so, from the first part, are equal. So, writing αi = LDi+ f (−k) (x), i = 0, 1, . . . , n + k − 1, in (2.18.9) and letting s → ∞, we get (2.18.4).

Theorem 2.18.3 Let f be special Denjoy integrable in I = [a, b] and x ∈ I. If f(n−1) (x) exists finitely, then +

+

f+ (x) ≤ LD+ n f (x) ≤ LDn f (x) ≤ f (n) (x). (n)

(2.18.10)

168


We prove the right-hand inequality only as the proof of the left-hand inequality is similar. + If f (n) (x) = ∞, there is nothing to prove; so, we may suppose that +

f (n) (x) = M < ∞. Choose an M1 < ∞ such that M < M1 . Then there is a δ > 0 such that f (x + t) −

n−1 X i i=0

t tn f(i) (x) < M1 , for 0 < t < δ. i! n!

Therefore, using Lemma 1.15.1 of Chapter I, s

n+1

Z

δ

e

−st

f (x + t) −

0

n−1 X i i=0

=

M1 t f(i) (x) dt ≤ sn+1 i! n!

Z

δ

e−st tn dt

0

M1 n!sn+1−n−1 + o(1) = M1 + o(1), as s → ∞. n!

+

+

So, letting s → ∞, LDn f (x) ≤ M1 and, thus, LDn f (x) ≤ M as M1 was arbitrary. Theorem 2.18.4 Let f be special Denjoy integrable and C1 -continuous in + I = [a, b]. If f[n−1] (x) exists finitely, x ∈ I, then +

+

f+ (x) ≤ LD+ n f (x) ≤ LDn f (x) ≤ f [n] (x). [n]

+ Since f[n−1] (x) exists finitely, there is a k ∈ N such that the kth in+

+

definite integral of f in I, f (−k) , satisfies (f (−k) )(n+k) (x) = f [n] (x) and, by +

+

Theorem 2.18.3, LDn+k (f (−k) )(x) ≤ (f (−k) )(n+k) (x). So, applying Theorem +

2.18.2, we get that LDn f (x) ≤ f [n] (x). The other inequality follows similarly. From Theorems 17.3 and 17.4 we get:

Theorem 2.18.5 Let f be special Denjoy integrable in I = [a, b]. Then, (i) if f(n) (x) exists, possibly infinite, x ∈ I, then LDn f (x) exists with the same value; + (ii) if f is C1 -continuous in I and if f[n] (x) exists, possibly infinite, x ∈ I, then LDn f (x) exists with the same value.

The following example shows that the converse is not true. Example. Let f (x) =

(

xn , if either x is irrational or x = 0, 1,

if x is rational, x 6= 0.


169

Then, f ′ (0) does not exist. However, for any r, 1 ≤ r ≤ n, Z δ Z δ r+1 −st r+1 s e f (t) dt = s e−st tn dt 0

0

= n!s So, lim s

s→∞

r+1

Z

r+1−n−1

δ

e

−st

+ o(1) = n!sr−n + o(1), as s → ∞.

f (t) dt =

0

(

n!, if r = n, 0,

if r < n.

Hence, LDn f (0) exists with value n!.

2.19

Laplace and Borel Derivatives, LDk f and BDk f

Theorem 2.19.1 Let f be special Denjoy integrable in some neighbourhood of x. If BDr+ f (x) exists finitely, so does LDr+ f (x) with the same value. Moreover, +

+ (r + 2)BD+ r+1 f (x) − (r + 1)BD r+1 f (x) ≤ LD r+1 f (x) +

+

≤ LDr+1 f (x) ≤ (r + 2)BDr+1 f (x) − (r + 1)BD+ r+1 f (x), (2.19.1) provided the extreme terms have meaning. Similar relations hold for the other derivates. Let BDr+ f (x) exist finitely and let ǫ > 0, σ > 0 be arbitrary. Then, there is a δ > 0 such that Z h f (x + t) − Pr ti BD+ f (x) i i=0 i! dt < ǫh, for 0 < h < δ. (2.19.2) tr 0 So, if 0 < δ1 < min{δ, σ}, integrating by parts and using (2.19.2), r Z δ1 X ti BDi+ f (x) dt e−st f (x + t) − i! 0 i=0 Pr i Z δ1 f (x + t) − i=0 ti! BDi+ f (x) = e−st tr dt tr 0 Pr i Z t f (x + ξ) − i=0 ξi! BDi+ f (x) t=δ1 −st r ≤ e t dξ ξr 0 t=0 Pr i Z t Z δ1 f (x + ξ) − i=0 ξi! BDi+ f (x) −st r −st r−1 dξ dt + (se t − re t ) r ξ 0 0

170

Higher Order Derivatives Z δ1 Z δ1 e−st tr−1 ǫt dt e−st tr ǫt dt + r < e−sδ1 δ1r ǫδ1 + s 0

0

= e−sδ1 δ1r+1 ǫ + sǫ

Z

δ1

e−st tr+1 dt + rǫ

0

Z

δ1

e−st tr dt.

0

So, by Lemma 1.15.1 of Chapter I, r+1 s

Z

δ1

e

−st

0

r X ti + f (x + t) − BDi f (x) dt i! i=0

≤ sr+1 e−sδ1 δ1r+1 ǫ + ǫ(r + 1)!sr+2−r−1−1 + rǫr!sr+1−r−1 + o(1), as s → ∞.

Also, it can be shown using integration by parts that Z σ r X ti r+1 + −st BD f (x) dt = o(1) as s → ∞. f (x + t) − e s i i! δ1 i=0

Thus, letting s → ∞, since ǫ is arbitrary, lim sr+1

s→∞

Z

0

σ

r X ti BDi+ f (x) dt = 0. e−st f (x + t) − i! i=0

Hence, LDr+ f (x) exists finitely and LDr+ f (x) = BDr+ f (x), which proves the first part of the theorem. For the second part, we prove the right-hand inequality; the proof of the + left-hand inequality is similar. If BDr+1 f (x) = ∞, then since the righthand side of (2.19.1) is well defined, BD+ r+1 f (x) 6= ∞, giving the right-hand side of (2.19.1) the value ∞ and the inequality is obvious. So, we may sup+ pose that BDr+1 f (x) < ∞. Similarly, if BD + r+1 f (x) = −∞, we must have +

BDr+1 f (x) > −∞ and again the inequality is obvious. So, we need only con+

sider the case −∞ < BD+ r+1 f (x) ≤ BD r+1 f (x) < ∞. Then, choose m, M such that +

−∞ < m < BD+ r+1 f (x) ≤ BD r+1 f (x) < M < ∞.

(2.19.3)

Then, there is δ > 0 such that m
0 arbitrary and let 0 < δ1 < min{δ, σ}. By the first part, LDi+ f (x) exists and equals BDi+ f (x), i = 0, 1, . . . , r, and, so, we get, integrating by


171

parts and applying (2.19.4), Z δ1 r X ti + −st f (x + t) − e BDi f (x) dt i! 0 i=0 Pr i Z δ1 f (x + t) − i=0 ti! BDi+ f (x) e−st tr+1 = dt tr+1 0 Pr i Z t f (x + ξ) − i=0 ξi! BDi+ f (x) t=δ1 = e−st tr+1 dξ ξ r+1 0 t=0 Pr Z t Z δ1 f (x + ξ) − i=0 se−st tr+1 − (r + 1)e−st tr + r+1 ξ 0 0 =

e−sδ1 δ1r+1 +s

Z

Z

δ1

f (x + ξ) −

0

δ1

e

−st r+1

t

0

−(r + 1)

Z

Z

t

Pr

ξi + i=0 i! BDi f (x) ξ r+1

f (x + ξ) −

0

δ1

e−st tr

0

sM e−sδ1 δ1r+2 M + < (r + 1)! (r + 1)!

Z

Z

t

Pr

e

f (x + ξ) −

0

−st r+2

t

ξi + i! BDi f (x)

dξ dt

dξ


0

δ1

(2.19.5)

Pr

dξ dt


m dt − r!

Z

δ1

dξ dt

e−st tr+1 dt.

0

From (2.19.5) and Lemma 1.15.1 of Chapter I, sr+2

Z

δ1

e−st f (x + t) −

0

< sr+2

r X ti i=0

i!

BDi+ f (x) dt

e−sδ1 δ1r+2 M + (r + 2)M − (r + 1)m + o(1), as s → ∞. (r + 1)!

Also, sr+2

r X ti e−st f (x + t) − BDi+ f (x) dt = o(1) as s → ∞. i! δ1 i=0

Z

σ

+

So, letting s → ∞, LDr+1 f (x) ≤ (r + 2)M − (r + 1)m. Since M and m are arbitrary, this gives the right-hand inequality in (2.19.1). Theorem 2.19.2 If f is special Denjoy integrable in some neighbourhood of x and if f is Borel bounded of order r at x, then f is Laplace bounded of order r at x. By Theorem 1.10.1 of Chapter I, the Borel derivative of order r − 1 at x exists finitely and the Borel derivates of order r at x are finite. So, by Theorem

172


2.19.1, the Laplace derivative of order r − 1 at x exists finitely and the Laplace derivates of order r at x are finite. So, by Theorem 1.15.3 of Chapter I, f is Laplace bounded of order r at x.

2.20

Symmetric Laplace and Symmetric de la Vall´ ee (s) Poussin Derivatives, SLDk f and f(k)

Theorem 2.20.1 Let f be special Denjoy integrable in some neighbourhood (s) of x. If the d.l.V.P. derivative of f at x of order r, f(r) (x), exists finitely, then the symmetric Laplace derivative of f at x of order r, SLDr f (x), exists and (s) is equal to f(r) (x). Moreover, (s)

(x) ≤ SLDr+2 f (x) ≤ SLDr+2 f (x) ≤ f (r+2) (x). f (s) (r+2)

(2.20.1)

(s)

(s)

Let f(r) (x) exist finitely. Then the derivative f(k) (x) exists, k = r, r − 2, . . . , 0 or 1, according as r is even or odd. Write τ (t) =

m X i=0

tr−2i (s) (x), f (r − 2i)! (r−2i)

(2.20.2)

where m = r/2 or (r − 1)/2 according as r is even or odd. Let ǫ > 0, σ > 0 be arbitrary. Then there is a δ > 0 such that r f (x + t) + (−1)r f (x − t) < ǫ t , for 0 ≤ t < δ. − τ (t) 2 r!

Let 0 < δ1 < min{δ, σ}. Then, by Lemma 1.15.1 of Chapter I, Z Z r+1 δ1 −st f (x + t) + (−1)r f (x − t) ǫ r+1 δ1 −st r s − τ (t) dt s e t dt ≤ e 2 r! 0 0 = ǫ + o(1), as s → ∞. Also, s

r+1

Z

σ

δ1

e

−st

f (x + t) + (−1)r f (x − t) 2

− τ (t) dt = o(1), as s → ∞.

So, letting s → ∞, Z σ f (x + t) + (−1)r f (x − t) lim sr+1 e−st − τ (t) dt ≤ ǫ. s→∞ 2 0


173

Hence, since ǫ was arbitrary, Z σ r r+1 −st f (x + t) + (−1) f (x − t) s e − τ (t) dt = o(1), as s → ∞. 2 0 Further, τ (t) = τ (t) + (−1)r τ (−t) /2 and, so, SLDr f (x) exists with value (s) f(r) (x). To prove the second part of the theorem, it is sufficient to consider the (s) right-hand inequality and to assume that f (r+2) (x) < ∞. (s)

So, choose an M such that f (r+2) (x) < M < ∞ and σ > 0. Then, there is a δ > 0 such that f (x + t) + (−1)r f (x − t) tr+2 − τ (t) ≤ M , for 0 ≤ t < δ. 2 (r + 2)! Then, if 0 < δ1 < min{δ, σ}, sr+3

Z

0

δ1

e−st

f (x + t) + (−1)r f (x − t) Z

2

δ1

tr+2 dt (r + 2)! 0 = M + o(1), as s → ∞.

≤ sr+3

− τ (t) dt

e−st M

And, so, as before, Z σ f (x + t) + (−1)r f (x − t) − τ (t) dt = M + o(1), as s → ∞. sr+3 e−st 2 0 Therefore, letting s → ∞, SLDr+2 f (x) ≤ M and since M was arbitrary, (s)

this proves that SLDr+2 f (x) ≤ f (r+2) (x).

Theorem 2.20.2 Let f be special Denjoy integrable in some neighbourhood (s) of x. If the d.l.V.P. derivative of f order r, f(r) (x), exists finitely, then 1 1 S r+2 f (x). f (x) ≤ SLr+2 f (x) ≤ SLr+2 f (x) ≤ S r + 2 r+2 r+2

(2.20.3)

(The definitions of the indices of smoothness in (2.20.3) can be found in Sections 6.2 and 16.3 of Chapter I.) We consider the right-hand inequality in (2.20.3) and assume without loss in generality that S r+2 f (x) < M < ∞. Then, there is a δ > 0 such that f (x + t) + (−1)r f (x − t) tr+1 − τ (t) < M , for 0 ≤ t < δ, 2 (r + 2)!

174


where τ (t) is as defined in (2.20.2). Let σ > 0 be arbitrary. Then for 0 < δ1 < min{δ, σ}, Z δ1 r −st f (x + t) + (−1) f (x − t) r+2 e s − τ (t) dt 2 0 Z δ1 tr+1 e−st M ≤ sr+2 dt (r + 2)! 0 M = + o(1), as s → ∞ r+2 and, hence, Z σ r M r+2 −st f (x + t) + (−1) f (x − t) − τ (t) dt ≤ + o(1), as s → ∞. s e 2 r +2 0 So, letting s → ∞, and since by Theorem 2.20.2, SLDk f (x) exists with value (s) f(k) (x), k = r, r − 2, . . ., we have SLr+2 f (x) ≤ M/(r + 2). Since, then, M is arbitrary, the proof is complete. Corollary 2.20.3 Under the hypotheses of Theorem 2.20.2, if f is d.l.V P. smooth of order r + 2 at x, then f is Laplace smooth of order r + 2 at x. Theorem 2.20.4 If f is d.l.V.P. bounded of order r+2 at x, then f is Laplace bounded of order r + 2 at x. (s)

Since f is d.l.V.P. bounded of order r+2 at x, then f(r) (x) exists finitely

(s)

(x) are finite. Hence, and, by Theorem 1.6.3 of Chapter I, f (r+2) (x) and f (s) (r+2) there is an M and a δ > 0 such that f (x + t) + (−1)r f (x − t) tr+2 − τ (t) < M , for 0 ≤ t < δ, 2 (r + 2)! where τ (t) is as in (2.20.2). Let σ > 0 be fixed. Then for 0 < δ1 < min{δ, σ}, Z δ1 f (x + t) + (−1)r f (x − t) r+3 e−st − τ (t) dt s 2 0 Z δ1 tr+2 r+3 ≤s e−st M dt (r + 2)! 0 = M + o(1), as s → ∞. So, s

r+3

Z

0

σ

e

−st

f (x + t) + (−1)r f (x − t)

which completes the proof.

2

− τ (t) dt = M + o(1) as s → ∞,


2.21

175

Laplace and Symmetric Laplace Derivatives, LDk f and SLDk f

Theorem 2.21.1 Let f be special Denjoy integrable in some neighbourhood of x. If the Laplace derivative of f of order r at x, LDr f (x), exists finitely, then the symmetric Laplace derivative of order r at x, SLDr f (x), exists with the same value. Moreover, 1 − LD+ r+1 f (x) + LD r+1 f (x) ≤ SLDr+1 f (x) 2 1 + − ≤ SLDr+1 f (x) ≤ LDr+1 f (x) + LDr+1 f (x) , 2

(2.21.1)

provided the extreme terms are well defined. Let LDr f (x) exist finitely and define

Q(t) =

r X ti i=0

i!

LDi f (x).

(2.21.2)

Then, s

r+1

Z

δ

0

and s

r+1

Z

δ

0

e−st f (x + t) − Q(t) dt = o(1), as s → ∞

(2.21.3)

e−st f (x − t) − Q(−t) dt = o(1), as s → ∞.

(2.21.4)

From (2.21.3) and (2.21.4), as s → ∞ Z δ f (x + t) + (−1)r f (x − t) Q(t) + (−1)r Q(−t) − dt = o(1). sr+1 e−st 2 2 0 Hence, SLDr f (x) exists with value LDr f (x). + To prove (2.21.1), we may suppose that LDr+1 f (x) − LDr+1 f (x) < ∞. Choose M1 and M2 − and LDr+1 f (x) < M2 < ∞. Then, there

s

r+2

Z

δ

0

and (−1)r+1 sr+2

0 such that

e−st f (x + t) − Q(t) dt < M1 , for s > N

Z

0

δ

∞ and

+ LDr+1 f (x)

e−st f (x − t) − Q(−t) dt < M2 , for s > N.

(2.21.5)

(2.21.6)

176


From (2.21.5) and(2.21.6), if s > N , Z δ f (x + t) + (−1)r+1 f (x − t) Q(t) + (−1)r+1 Q(−t) dt − sr+2 e−st 2 2 0 M1 + M2 < . 2 Letting s → ∞ SLDr+1 f (x) ≤ (M1 + M2 )/2 and then letting M1 → + − LDr+1 f (x) and M2 → LDr+1 f (x), the right-hand side of (2.21.1) follows. The left-hand side is proved similarly. Theorem 2.21.2 If LDr+1 f (x) exists finitely, then f is Laplace smooth at x of order r + 2. Let LDr+1 f (x) exist finitely and let ǫ > 0 be arbitrary. Then there exists an N > 0 such that Z δ r+2 e−st f (x + t) − P (t) dt < ǫ, for s > N (2.21.7) s 0

and

Z δ r+2 e−st f (x − t) − P (−t) dt < ǫ, for s > N , s

(2.21.8)

0

where

P (t) =

r+1 i X t i=0

i!

LDi f (x).

From (2.21.7) and (2.21.8), if s > N ,

Z δ r+2 −st f (x + t) + (−1)r f (x − t) P (t) + (−1)r P (−t) e dt < ǫ. − s 2 2 0

Hence, f is Laplace smooth of order r + 2.

2.22

Peano and Unsymmetric Riemann Derivatives, f(k) and RDk f

Theorem 2.22.1 If f(k) (x) exists finitely, then RDk f (x) exists and is equal to f(k) (x). Further, if f (k+1) (x) and f (k+1) (x) are finite, then RDk+1 f (x) and RDk+1 f (x) are also finite, but not conversely.


The following relation will be used in the proof. ( k X 0, if s = 0, 1, . . . , k − 1, s k−i k i = (−1) i k!, if s = k. i=0

177

(2.22.1)

Since f(k) (x) exists finitely, f (x + t) =

k X ti

i!

i=0

f(i) (x) + o(tk ), as t → 0.

(2.22.2)

From (2.22.1) and (2.22.2) (see (2.3.16) and (2.3.21) of Chapter I), k f (x + it) (−1) ∆k (f, x, t) = i i=0   X k k j X k (it)  (−1)k−i = f(j) (x) + o(tk ) i j! j=0 i=0 ! k k X X tj k j i + o(tk ) (−1)k−i f(j) (x) = j! i i=0 j=0 k X

k−i

= tk f(k) (x) + o(tk ), as t → 0,

and, thus, ∆k (f, x, t) = f(k) (x) + o(1), as t → 0. tk Hence, RDk f (x) exists with value f(k) (x). For the second part, suppose that f (k+1) (x) and f (k+1) (x) are finite. Then f (x + t) =

k X ti i=0

i!

f(i) (x) + O(tk+1 ), as t → 0.

(2.22.3)

From (2.22.1) and (2.22.3), k+1 f (x + it) (−1) ∆k+1 (f, x, t) = i i=0   X k k+1 j X k+1  (it) (−1)k+1−i = f(j) (x) + O(tk+1 ) i j! j=0 i=0 ! k k+1 X X tj k + 1 ij + O(tk+1 ) (−1)k+1−i = f(j) (x) j! i j=0 i=0 k+1 X

k+1−i

= O(tk+1 ), as t → 0.

178


Hence, RDk+1 f (x) and RDk+1 kf (x) are finite. For the converse, consider f (x) = |x|. Then, since f(1) (0) does not exist, it follows that f(2) (0) does not exist. However, limt→0± ∆2 (f, 0, t) t2 = 0 and, so, RD2 f (0) exists with value 0.

2.23

Symmetric de la Vall´ ee Poussin and Symmetric (s) (s) Riemann Derivatives, f(k) and RDk f (s)

(s)

Theorem 2.23.1 If f(k) (x) exists finitely, then RDk f (x) exists and is equal (s)

(s)

(s)

(x) are finite, then RDk+2 f (x) and to f(k) (x). Further, if f (k+2) (x) and f (s) (k+2) (s)

RDk+2 f (x) also are finite. We prove the case when k is even, k = 2m, say; then, the case of odd k (s) is similar. If f(2m) (x) exists finitely, we have m

f (x + t) + f (x − t) X t2i (s) = f(2i) (x) + o(t2m ), as t → 0, 2 (2i)! i=0

(2.23.1)

and, so (see (1.2.20) and (1.2.22) of Chapter I), using (2.23.1) 2m X 2m f (x + 2it − 2mt) (−1)2m−i ∆s2m (f, x, 2t) = i i=0 2m X 2m−i 2m f (x − 2it + 2mt) (−1) = i i=0 2m X 2m−i 2m f (x + 2it − 2mt) + f (x − 2it + 2mt) (−1) = i 2 i=0   X m 2m 2j X 2m (2it − 2mt) (s)  (−1)2m−i f(2j) (x) + o(t2m ) = i (2j)! j=0 i=0

! 2m X t2j (s) 2j 2m 2m−i 2m 2 (i − m) + o(t ) (−1) = f (x) i (2j)! (2j) j=0 i=0 m X

= 22m

Hence,

2j

t2m (s) (x)(2m)! + o(t2m ), as t → 0, using (2.22.1). f (2m)! (2m) (2.23.2)

∆s2m (f, x, 2t) (s) = f(2m) (x) + o(1) as t → 0, (2t)2m

Relations between Derivatives (s)

179

(s)

and, so, RD2m f (x) exists and is equal to f(2m) (x). Next, suppose that (s)

f (2m+2) (x) and f (s) (x) are finite. Then, (2m+2) m

f (x + t) + f (x − t) X t2i (s) = f(2i) (x) + O(t2m+2 ), as t → 0. 2 (2i)! i=0

(2.23.3)

So, as above, applying (2.23.3) we get, ∆s2m+2 (f, x, 2t)

=

2m+2 X

2m+2−i

(−1)

i=0

2m + 2 f (x + 2it − 2mt − 2t) i

2m + 2 f (x + 2it − 2mt − 2t) + f (x − 2it + 2mt + 2t) i 2 i=0   X m 2m+2 2j X 2m + 2  (2i − 2m − 2) 2j (s) (−1)2m+2−i t f(2j) (x) + O(t2m+2 ) = i (2j)! j=0 i=0 ! m 2m+2 X X (2t)2j (s) 2j 2m+2 2m+2−i 2m + 2 = (i − m − 1) + O(t ) (−1) f (x) (2j)! (2j) i j=0 i=0 =

2m+2 X

(−1)2m+2−i

= O(t2m+2 ), as t → 0, using (2.22.1). So,

∆s2m+2 (f, x, 2t) = O(1), as t → 0. (2t)2m+2

Hence, lim supt→0

∆s2m+2 (f, x, 2t) ∆s2m+2 (f, x, 2t) and lim sup are both t→0 (2t)2m+2 (2t)2m+2

(s)

(s)

finite; that is, RD2m+2 f (x) and RD2m+2 f (x) are finite. The converse of the above theorem is known only for k = 1, 2, 3, 4. From (s) (s) (s) the definition, RDk f (x) = f (k) (x) and RDk f (x) = f (s) (x) when k = 1, 2. (k) If k ≥ 3, the definitions of the two derivatives differ; we prove the converse when k = 3, 4. Theorem 2.23.2 Let RD1s f (x) exist finitely. If RD3s f (x) exists finitely, then s (s) f(3) (x) exists and is equal to RD3s f (x). Moreover, if RDs3 f (x) and RD3 f (x) (s)

(x) and f (3) (x) are also finite. are finite, then f (s) (3) Without loss in generality, we may assume that x = 0 and first we suppose that RD1s f (0) = RD3s f (0) = 0. (2.23.4) From (2.23.4), we have f (t) − f (−t) = o(t), as t → 0

(2.23.5)

180 and

Higher Order Derivatives f (3t) − f (−3t) − 3 f (t) − f (−t) = o(t3 ), as t → 0.

Replacing t by t/3

n+1

(2.23.6)

in (2.23.6), we get

f (t/3n ) − f (−t/3n ) − 3 f (t/3n+1 ) − f (−t/3n+1) = o(t3 /33n+3 ) as t → 0, (2.23.7) or, since o(t3 /33n+3 ) as t → 0 implies 3−(3n+3) o(t3 ) as t → 0, 3n f (t/3n ) − f (−t/3n) − 3n+1 f (t/3n+1 ) − f (−t/3n+1 ) 1 = 2n+3 o(t3 ) as t → 0. (2.23.8) 3 Putting n = 0, 1, . . . , N and adding in (2.23.8), N X N +1 N +1 N +1 3 f (t/3 ) − f (−t/3 ) = o(t ) f (t) − f (−t) − 3

1 . (2.23.9) 2n+3 3 n=0

By (2.23.5), for fixed t lim 3N +1 f (t/3N +1 ) − f (−t/3N +1 ) = 0.

(2.23.10)

N →∞

Therefore, letting N → ∞, we have from (2.23.9) that as t → 0 ∞ 1X 1 f (t) − f (−t) = o(t3 ), = o(t3 ) 2 2 n=0 32n+3

(2.23.11)

(s)

which shows that f(3) (0) exists with value 0.

s

For the second part let RDs3 f (0) and RD3 f (0) be finite. Then, instead of (2.23.4), we now consider RD1s f (0) = 0. Relation (2.23.5) remains true and on the right-hand sides of (2.23.6),(2.23.7), (2.23.8) and (2.23.9) the little-oh terms are replaced by similar big-oh terms; (2.23.10) remains the same as it only uses (2.23.5). So, finally (2.23.11) becomes ∞ 1X 1 f (t) − f (−t) = O(t3 ). = O(t3 ) 2 2 n=0 32n+3

(2.23.12)

(s)

(0) and f (3) (0) are finite, completing the It follows from (2.23.12) that f (s) (3) proof of the special case. For the general case, we begin by considering the first part of the theorem and define x3 g(x) = f (x) − xRD1s f (0) − RD3s f (0). 3! (s)

Then, g satisfied the conditions of the special case and, so g(3) (0) exists and (s)

(s)

(s)

g(3) (0) = 0. So, f(3) (0) exists and f(3) (0) = RD3s f (0).


181

For the second part, put g(x) = f (x) − xRD1s f (0). (s)

(s)

Then, RD1s g(0) = 0 and using the special case g (3) (0) and g (3) (0) are finite (s)

(s) (0) are finite. and, hence, f (3) (0) and f (3)

Theorem 2.23.3 Let RD2s f (x) exists finitely. If RD4s f (x) exists finitely, then s (s) f(4) (x) exists and is equal to RD4s f (x). If RD4 f (x) and RDs4 f (x) are finite, (s)

(s) (x) are also finite. then f (4) (x) and f (4)

Without loss in generality, we may suppose that x = 0 and first consider the special case f (0) = RD2s f (0) = RD4s f (0) = 0. (2.23.13) From (2.23.13), f (t) + f (−t) = o(t2 ), as t → 0,

f (2t) + f (−2t) − 4 f (t) + f (−t) = o(t4 ), as t → 0.

(2.23.14) (2.23.15)

In (2.23.15), replace t by t/2n+1 to get

f (t/2n ) + f (−t/2n) − 4 f (t/2n+1 ) + f (−t/2n+1) = o(t4 /24n+4 ),

and, as in (2.23.8),

22n f (t/2n ) + f (−t/2n ) − 22n+2 f (t/2n+1 ) + f (−t/2n+1) 1 (2.23.16) = 2n+4 o(t4 ), as t → 0. 2 Putting n = 0, 1, . . . , N in (2.23.16) and adding gives N

f (t) + f (−t) − 22N +2 f (t/2N +1 ) + f (−t/2N +1) = o(t4 )

n=0

Now, by (2.23.14), we have for fixed t

lim 22N +2 f (t/2N +1 ) + f (−t/2N +1) = 0. N →∞

1 22n+4

. (2.23.17)

(2.23.18)

So, letting N → ∞ in (2.23.17), we get from(2.23.18) that ∞ 1 1 f (t) + f (−t) = o(t4 ) = o(t4 ), as t → 0. 2 2 n=0 22n+4 (s)

This shows that f(4) (0) exists with value 0.

(2.23.19)

182


For the second part of the special case, replace (2.23.13) by f (0) = RD2s f (0) = 0. Then (2.23.14) is true and the right-hand sides of (2.23.15),(2.23.16), and (2.23.17) the little-oh terms are replaced by similar big-oh terms; (2.23.18) remains the same as it only uses (2.23.14). So, finally (2.23.19) becomes ∞ f (t) + f (−t) 1X 1 = O(t4 ), as t → 0. = O(t4 ) 2 2 n=0 22n+4

(2.23.20)

(s)

From (2.23.20), f (4) (0) and f (s) (0) are finite, completing the proof of the (4) special case. For the general case, we begin by considering the first part of the theorem and define g(x) = f (x) − f (0) −

x2 x4 RD2s f (0) − RD4s f (0). 2 4!

(2.23.21) (s)

Then, g satisfied the conditions of the special case and, so, g(4) (0) exists and (s)

(s)

(s)

g(4) (0) = 0. So, f(4) (0) exists and, from (2.21), f(4) (0) = RD4s f (0). For the second part, put g(x) = f (x) − f (0) −

x2 RD2s f (0). 2

(2.23.22)

Then, g satisfies the conditions of the second part of the special case and (s) (s) using the special case g(4) (0) and g (4) (0) are finite and, hence, from (2.23.22), (s)

f (4) (0) and f (s) (0) are finite. (4)

2.24

Generalized Riemann GRDk f and f(k)

and

Peano

Derivatives,

Since the existence of the Peano derivative, f(n) (x), implies the existence of (s) the d.l.V.P. derivative, f(n) (x) (see Theorem 2.5.1), it follows from Theorem 2.23.1 that the existence of f(n) (x) implies the existence of the symmetric Riemann derivative, RDns f (x), and their values will be the same. Also, a particular choice of the system S = {a0 , a1 , . . . , an+ℓ ; A0 , A1 , . . . , An+ℓ ; L} in the definition of the generalized Riemann derivative, GDRn f (x), (see Section 3 of Chapter I), gives the unsymmetric Riemann derivative, RDn f (x), and the symmetric Riemann derivative, RDns f (x). However the general system S may give a derivative that differs from both RDn f (x) and RDns f (x), so we prove the following.


183

Theorem 2.24.1 If f(n) (x) exists finitely, then GRDn f (x, S) exists and equals f(n) (x) for any system S. Further, if f (n) (x) and f (n) (x) are finite, then GRDn f (x, S) and GRDn f (x, S) are also finite.

Let f(n) (x) exist finitely. Then f (x + h) =

n X hi i=0

i!

f(i) (x) + o(hn ), as h → 0.

(2.24.1)

From (2.24.1) and relation (1.3.1) of Chapter I,   n n+ℓ n+ℓ n! X n! X X (ai h)j Ai f (x + ai h) = Ai f(j) (x) + o(hn ) Lhn i=0 Lhn i=0 j! j=0 ! n n+ℓ X n! X hj j Ai ai + o(1) f(j) (x) = Lhn j=0 j! i=0 =

n! hn f(n) (x)L + o(1) = f(n) (x) + o(1), as h → 0. Lhn n!

So, letting h → 0 n+ℓ n! X Ai f (x + ai h) = f(n) (x). h→0 Lhn i=0

GRDn f (x) = lim

For the second part, (2.24.1) is replaced by f (x + h) =

n−1 X i=0

hi f(i) (x) + O(hn ), as h → 0. i!

(2.24.2)

From (2.24.2)and relation (1.3.1) of Chapter I,   n−1 n+ℓ n+ℓ n! X  X (ai h)j n! X Ai f (x + ai h) = Ai f(j) (x) + O(hn ) Lhn i=0 Lhn i=0 j! j=0 ! n−1 n+ℓ X n! X hj j Ai ai + O(1) (2.24.3) = f(j) (x) Lhn j=0 j! i=0 = O(1), as h → 0.

So, letting h → 0, GRDn f (x, S) and GRDn f (x, S), respectively, the upper and lower limits of the left-hand side of (2.24.3) are finite.

184

2.25


e k f and f(k) MZ- and Peano Derivatives, D

e k f (x), is defined in Section 3 of The MZ-derivative of order k at x, D Chapter I; it is a special generalized Riemann derivative, namely one with S = {0, 1, 2, 22, . . . , 2k−1 ; A0 , A1 , . . . , Ak ; L};

L =

k X

2(i−1)k Ai ;

ℓ = 0;

(2.25.1)

i=1

Ai , 0 ≤ i ≤ k, satisfying the equations (1.3.3) of Section 3 of Chapter I. So, e k f (x) = lim λk ∆ e k (f, x, t) D t→0 tk k X λk Ai f (x + 2i−1 t) , = lim k A0 f (x) + t→0 t i=1

(2.25.2)

provided the limit exists, where λk = Pk

i=1

k! 2(i−1)k Ai

.

(2.25.3)

For details of the relevant calculations, see Section 3 of Chapter I. If the limit does not exist, then the MZ-derivates of f at x of order k are defined to be e k f (x) = lim sup λk ∆ e k (f, x, t), D tk t→0 e k (f, x, t). e f (x) = lim inf λk ∆ D k t→0 tk

(2.25.4)

e 1 f (x), D e 2 f (x), . . . D e k f (x) exist finitely, then f(k) (x) Theorem 2.25.1 If D e i f (x), 1 = 1, 2, . . . , k and exists and f(i) (x) = D e e D k+1 f (x) ≤ f (k+1) (x) ≤ f (k+1) (x) ≤ D k+1 f (x).

(2.25.5)

e k+1 f (x) exists, possibly infinite, then f(k+1) (x) exists with In particular, if D the same value.

We may suppose without loss in generality that x = 0. Let g(t) = f (t) − f (0) −

k X ti i=1

i!

e i f (0). D

(2.25.6)


185

Then, from the definition of the MZ-derivative, it follows that e 1 g(0) = · · · = D e k g(0) = 0. g(0) = D

(2.25.7)

Let ǫ > 0 be arbitrary. Then, using the last equality in (2.25.7), there is a δ > 0 such that ∆ e k (g, 0, t) < ǫ|t|k , for |t| < δ. (2.25.8)

e k (g, 0, t) = ∆ e k−1 (g, 0, 2t) − 2k−1 ∆ e k−1 (g, 0, t) (see relation (1.3.7) of Since ∆ Chapter I), we have from (2.25.8) ∆ e k−1 (g, 0, 2t) − 2k−1 ∆ e k−1 (g, 0, t) < ǫ|t|k , for |t| < δ. (2.25.9) Keeping t fixed in (2.25.9)and replacing t by t/2n+1 , we get ∆ e k−1 (g, 0, t/2n ) − 2k−1 ∆ e k−1 (g, 0, t/2n+1 )
0 be g(i) (0) = 0 i = 1, 2, . . . , k, and D arbitrary, then there exists a δ > 0 such that e k+1 (g, 0, t) < ǫtk+1 , for 0 < t < δ. ∆

(2.25.18)

The relation (2.25.18) is analogous to (2.25.8) and, so, applying arguments similar to those that deduced (2.25.16) from (2.25.8), we deduce from (2.25.18) that g(t) < ǫtk+1 , for 0 < t < δ. (2.25.19) + Hence, g + (k+1) (0) ≤ (k + 1)!ǫ, and since ǫ is arbitrary, this gives g (k+1) (0) ≤ 0 +

or, equivalently, f (k+1) (0) ≤ M . For the negative side, note that when k + 1 is odd, the relation (2.25.18) becomes e k+1 (g, 0, t) > ǫtk+1 , for −δ < t < 0. ∆ From this, we get g(t) > ǫtk+1 , for −δ < t < 0, which implies g − (k+1) (0) ≤ 0 −

or f (k+1) (0) ≤ M . Thus, f (k+1) (0) ≤ M , which completes the proof.

e k f (x) does not imply the existence of D e k−1 f (x). Remark. The existence of D To see this, consider ( x, if x ≥ 0, f (x) = 0, if x < 0. e 1 f (0) does not exists, but D e 2 f (0) does, D e 2 f (0) = 0. Here, D

e k f (x) exists with value Theorem 2.25.2 If f(k) (x) exists finitely, then D e k+1 f (x) and (x) are finite, then D f(k) (x). Moreover, if f (k+1) (x) and f (k+1)

e D k+1 f (x) are also finite.

Since the MZ- derivatives and derivates are special cases of the generalized Riemann derivatives and derivates considered in the previous section, this result follows from Theorem 2.24.1. The argument used in that proof can be given for this special case, but the argument is not really any simpler.

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