An Introduction to Ultrametric Summability Theory [2ed.] 8132225589, 978-81-322-2558-4, 978-81-322-2559-1, 8132225597

Table of contents :
Front Matter....Pages i-xiii
Introduction and Preliminaries....Pages 1-14
Some Arithmetic and Analysis in \({\mathbb Q}_p\) ; Derivatives in Ultrametric Analysis....Pages 15-22
Ultrametric Functional Analysis....Pages 23-28
Ultrametric Summability Theory....Pages 29-61
The Nörlund and The Weighted Mean Methods....Pages 63-88
The Euler and The Taylor Methods....Pages 89-102
Tauberian Theorems....Pages 103-114
Silverman-Toeplitz Theorem for Double Sequences and Double Series....Pages 115-131
The Nörlund Method and The Weighted Mean Method for Double Sequences....Pages 133-155
Back Matter....Pages 157-159

Citation preview

Forum for Interdisciplinary Mathematics 2

P.N. Natarajan

An Introduction to Ultrametric Summability Theory Second Edition

Forum for Interdisciplinary Mathematics Volume 2

Editor-in-chief P.V. Subrahmanyam, Indian Institute of Technology Madras, Chennai, India Editorial Board Bhu Dev Sharma, Forum for Interdisciplinary Mathematics, Meerut, India Janusz Matkowski, University of Zielona Góra, Zielona Góra, Poland Mathieu Dutour Sikirić, Institute Rudjer Boúsković, Zagreb, Croatia Thiruvenkatachari Parthasarathy, Chennai Mathematical Institute, Kelambakkam, India Yogendra Prasad Chaubey, Concordia University, Montréal, Canada

The Forum for Interdisciplinary Mathematics (FIM) series publishes high-quality monographs and lecture notes in mathematics and interdisciplinary areas where mathematics has a fundamental role, such as statistics, operations research, computer science, financial mathematics, industrial mathematics, and biomathematics. It reflects the increasing demand of researchers working at the interface between mathematics and other scientific disciplines.

More information about this series at http://www.springer.com/series/13386

P.N. Natarajan

An Introduction to Ultrametric Summability Theory Second Edition

123

P.N. Natarajan Formerly of the Department of Mathematics Ramakrishna Mission Vivekananda College Chennai, Tamil Nadu India

ISSN 2364-6748 ISSN 2364-6756 (electronic) Forum for Interdisciplinary Mathematics ISBN 978-81-322-2558-4 ISBN 978-81-322-2559-1 (eBook) DOI 10.1007/978-81-322-2559-1 Library of Congress Control Number: 2015945119 Springer New Delhi Heidelberg New York Dordrecht London © Springer India 2014, 2015 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper Springer (India) Pvt. Ltd. is part of Springer Science+Business Media (www.springer.com)

Dedicated to my parents S. Thailambal and P.S. Narayanasubramanian

Preface to the Second Edition

Most of the material discussed in the present revised, enlarged edition has appeared in the first edition of the book, An Introduction to Ultrametric Summability Theory (Springer, 2013). The first three chapters of the first edition have been retained in the second edition. In Chaps. 4–9 of the present edition, we present a survey of the literature on “ultrametric summability theory”. We have supplemented substantially to the survey in the current edition. Our survey starts with a paper by Andree and Petersen of 1956 (it is the earliest known paper on the topic). In Chap. 4, the Silverman–Toeplitz theorem is proved by using the “sliding hump method”. Schur’s theorem and Steinhaus theorem also find a mention here. It is proved that certain Steinhaus-type theorems fail to hold. An interesting characterization of infinite matrices in ð‘α ; ‘α Þ, α [ 0 is proved. There is, as such, no classical analogue for this result. The core of a sequence and Knopp’s core theorem are discussed. Towards the end of the chapter, an important result on Cauchy multiplication of series is proved. In Chap. 5, we introduce the Nörlund and Weighted mean methods in the ultrametric set-up, and their properties are elaborately discussed. We also show that the Mazur–Orlicz theorem and Brudno’s theorem fail to hold in the ultrametric case. In Chap. 6, we introduce the Euler and Taylor methods and discuss their properties extensively. In Chap. 7, we prove Tauberian theorems for the Nörlund, the weighted mean and the Euler methods. In Chap. 8, we introduce double sequences and double series in ultrametric analysis. We prove Silverman–Toeplitz theorem for four-dimensional infinite matrices. We also prove Schur’s and Steinhaus theorems for four-dimensional infinite matrices. Towards the end of the chapter, we obtain some interesting characterizations of two-dimensional Schur matrices. Finally, in Chap. 9, we introduce the Nörlund and the weighted mean methods for double sequences and make a detailed study of their properties. The author thanks E. Boopal for typing the manuscript.

vii

viii

Preface to the Second Edition

Errors in the first edition, both typographical and conceptual, have been corrected in the present revised, enlarged edition. Chennai, India

P.N. Natarajan

Preface to the First Edition

The purpose of the present monograph is to discuss briefly what summability theory is like when the underlying field is not ℝ (the field of real numbers) or ℂ (the field of complex numbers) but a field K with a non-Archimedean or ultrametric valuation, i.e., a mapping jj: K ! R satisfying the ultrametric inequality jx þ yj  maxðj xj; j yjÞ instead of the usual triangle inequality jx þ yj  j xj þ j yj; x; y 2 K. To make the monograph really useful to those who wish to take up the study of ultrametric summability theory and do some original work therein, some knowledge of real and complex analysis, functional analysis and summability theory over ℝ or ℂ is assumed. Some of the basic properties of ultrametric fields—their topological structure and geometry—are discussed in Chap. 1. In this chapter, we introduce the p-adic valuation, p being prime and prove that any valuation of ℚ (the field of rational numbers) is either the trivial valuation, a p-adic valuation or a power of the usual absolute value jj1 on ℝ, i.e., jjα1 , where 0\α  1. We discuss equivalent valuations too. In Chap. 2, we discuss some arithmetic and analysis in ℚp, the p-adic field for a prime p. In Chap. 2, we also introduce the concepts of differentiability and derivatives in ultrametric analysis and very briefly indicate how ultrametric calculus is different from our usual calculus. In Chap. 3, we speak of ultrametric Banach space, and also mention the many results of the classical Banach space theory, viz., the closed graph, the open mapping and the Banach-Steinhaus theorems carry over in the ultrametric set-up. However, the Hahn-Banach theorem fails to hold. To salvage the Hahn-Banach theorem, the concept of a ‘‘spherically complete field’’ is introduced and Ingleton’s version of the Hahn-Banach theorem is proved. The lack of ordering in an ultrametric field K makes it quite difficult to find a substitute for classical ‘‘convexity’’. However, classical convexity is replaced, in the ultrametric setting, by a notion called ‘‘K-convexity’’, which is briefly discussed towards the end of the chapter. In the main Chap. 4, our survey of the literature on ‘‘Ultrametric Summability Theory’’, starts with the paper of Andree and Petersen of 1956 (it was the earliest known paper on the topic) to the present. As far as the author of the present

ix

x

Preface to the First Edition

monograph knows, most of the material discussed in the survey has not appeared in book form earlier. Almost all of Chap. 4 consists entirely of the work of the author of the present monograph. Suitable references have been provided at appropriate places indicating where further developments may be found. The author takes this opportunity to thank Prof. P. V. Subrahmanyam of the Department of Mathematics, Indian Institute of Technology (Madras), Chennai for encouraging him to write the monograph during the author’s short stay at the Institute (July 8–August 5, 2011) as a Visiting Faculty. The author thanks the Forum for Inter-disciplinary Mathematics for being instrumental in getting the monograph published. Chennai, India

P.N. Natarajan

Contents

1

Introduction and Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 14

2

Some Arithmetic and Analysis in Qp ; Derivatives in Ultrametric Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 22

3

Ultrametric Functional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23 28

4

Ultrametric Summability Theory. . . . . . . . . . . . . . . 4.1 Classes of Matrix Transformations . . . . . . . . . . . 4.2 Steinhaus-Type Theorems . . . . . . . . . . . . . . . . . 4.3 Core of a Sequence and Knopp’s Core Theorem . 4.4 A Characterization of Regular and Schur Matrices 4.5 Cauchy Multiplication of Series . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

The Nörlund and The Weighted 5.1 The Nörlund Method . . . . . 5.2 Some More Properties of the 5.3 The Weighted Mean Method References . . . . . . . . . . . . . . . .

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29 29 35 48 52 59 60

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63 63 70 73 88

6

The Euler and The Taylor Methods . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89 102

7

Tauberian Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

103 114

Mean Methods . ............. Nörlund Method . ............. .............

. . . . .

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xi

xii

8

Contents

Silverman-Toeplitz Theorem for Double Sequences and Double Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Double Sequences and Double Series . . . . . . . . . . . . . . 8.2 Silverman-Toeplitz Theorem . . . . . . . . . . . . . . . . . . . . 8.3 Schur’s and Steinhaus Theorems . . . . . . . . . . . . . . . . . 8.4 Some Characterizations of 2-Dimensional Schur Matrices References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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115 115 116 123 128 131

The Nörlund Method and The Weighted Mean Method for Double Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 The Nörlund Method for Double Sequences. . . . . . . . 9.2 Cauchy Multiplication of Double Series . . . . . . . . . . 9.3 The Weighted Mean Method for Double Sequences . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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133 133 135 139 154

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

157

9

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. . . . .

About the Author

P.N. Natarajan is former professor and head, Department of Mathematics, Ramakrishna Mission Vivekananda College, Chennai. He did his Ph.D. from the University of Madras, under Prof. M.S. Rangachari, former director and head, The Ramanujan Institute for Advanced Study in Mathematics, University of Madras. An active researcher, Prof. Natarajan has over 100 research papers to his credit published in several international journals like Proceedings of the American Mathematical Society, Bulletin of the London Mathematical Society, Indagationes Mathematicae, Annales Mathematiques Blaise Pascal, and Commentationes Mathematicae (Prace Matematyczne). His research interest includes summability theory and functional analysis (both classical and ultrametric). Professor Natarajan was honoured with the Dr. Radhakrishnan Award for the Best Teacher in Mathematics for the year 1990–1991 by the Government of Tamil Nadu. Besides visiting several institutes of repute in Canada, France, Holland and Greece on invitation, Prof. Natarajan has participated in several international conferences and has chaired sessions.

xiii

Chapter 1

Introduction and Preliminaries

Abstract Some basic properties of ultrametric fields—their topological structure and geometry are discussed in this chapter. We introduce the p-adic valuation, p being prime and prove that any valuation of Q (the field of rational numbers) is either the trivial valuation, a p-adic valuation or a power of the usual absolute value, where the power is positive and less than or equal to 1. We discuss equivalent valuations too. Keywords Archimedean axiom · Ultrametric valuation · Ultrametric field · p-adic valuation · p-adic field · p-adic numbers · Equivalent valuations

The purpose of this book is to introduce a “NEW ANALYSIS” to students of Mathematics at the undergraduate and post graduate levels, which in turn introduces a geometry very different from the familiar Euclidean geometry and Riemannian geometry. Strange things happen: for instance, ‘every triangle is isosceles’ and ‘every point of a sphere is a centre of the sphere’!. ‘Analysis’ is that branch of Mathematics where we use the idea of limits extensively. A study of Analysis starts with limits, continuity, differentiability etc. and almost all mathematical models are governed by differential equations over the field R of real numbers. R has a geometry which is Euclidean. Imagine a pygmy tortoise trying to travel along a very long path; assume that its destination is at a very long distance from its starting point. If at every step, the pygmy tortoise covers a small distance , can it ever reach its destination, assuming that the tortoise has infinite life? Our common sense says “Yes”. It is one of the important axioms in Euclidean geometry that “Any large segment on a straight line can be covered by successive addition of small segments along the same line”. It is equivalent to the statement: “given any number M > 0, there exists an integer N such that N > M”. This is familiarly known as the “Archimedean axiom” of the real number field R. What would happen if we do not have this axiom? Are there fields which are non-archimedean? In the sequel, we will show that such fields exist and the metric on such fields introduces a geometry very different from the familiar Euclidean geometry and Riemannian geometry. In such a geometry, every triangle is isoceles; given two spheres, either © Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_1

1

2

1 Introduction and Preliminaries

they are disjoint or identical and consequently every point of a sphere is a centre of the sphere! Let us now revisit the field Q of rational numbers. The usual absolute value | · | on Q is defined as  x, if x ≥ 0; |x| = −x, if x < 0. We recall the process by which we get real numbers from rational numbers. Any real number x has a decimal expansion and we can write x = ±10β (x0 + x1 10−1 + x2 10−2 + · · · ), β ∈ Z,

(∗ )

x j = 0, 1, . . . , 9, j = 0, 1, 2, . . . , Z being the ring of integers. We note that the decimal expansion of a real number x enables us to look at x as the limit of a sequence of rational numbers. The metric space (Q, | · |), where | · | is the usual metric on Q, is not complete in the sense that there are Cauchy sequences in Q which do not converge in Q. We now consider the set ζ of all Cauchy sequences in Q which do not have a limit in Q. Declare that any two sequences {xn } and {yn } are equivalent if {xn − yn } converges to 0 in (Q, | · |). After such an identification, the set of all equivalence classes in ζ is precisely the set R of all real numbers. Note that Q is dense in R. The metric d in R is defined as follows: d(x, y) = lim |xn − yn |, n→∞

where {xn }, {yn } are the Cauchy sequences of rational numbers representing the real numbers x and y respectively. This procedure of embedding Q, as a dense subspace, in a complete metric space R is called the completion of Q. We can imitate this process and embed any given metric space, as a dense subspace, in a complete metric space and speak of the “completion of a metric space”. Given a field K , by a valuation of K , we mean a mapping | · | : K → R such that |x| ≥ 0; |x| = 0 if and only if x = 0;

(1.1)

|x y| = |x||y|;

(1.2)

|x + y| ≤ |x| + |y|,

(1.3)

and x, y ∈ K . The field K , with the valuation | · |, is called a valued field. Example 1.1 The usual absolute value function is a valuation of Q. Example 1.2 Given any field K , define a valuation | · | of K as follows:

1 Introduction and Preliminaries

3

 |x| =

0, if x = 0; 1, if x = 0.

In Example 1.2 above, | · | is called the trivial valuation of K . Any valuation of K which is not the trivial valuation is called a non-trivial valuation. If K is a valued field with valuation | · |, define d(x, y) = |x − y|,

(1.4)

x, y ∈ K . It is easy to check that d is a metric on K so that K is a metric space with respect to the ‘metric d induced by the valuation’ defined by (1.4). Consequently, we can define topological concepts like open set, closed set, convergence, etc. in valued fields. We now deal with Q in a different direction. Definition 1.1 Let c be any fixed real number such that 0 < c < 1 and p be a fixed prime number. We define a valuation, denoted by | · | p , of Q as follows: Define |0| p = 0; if x ∈ Q, x = 0, we write x in the form x = pα

a  b

,

where α, a ∈ Z, b ∈ Z+ = {1, 2, . . .}, p does not divide a, b and (a, b) = 1. Noting that the above form of x is unique, define |x| p = cα . It is now worthwhile to check that | · | p is a valuation of Q. From the definition, it is clear that |x| p ≥ 0 and |x| p = 0 if and only if x = 0. For y ∈ Q, we shall write y = pβ



a b

 ,

where β, a ∈ Z, b ∈ Z+ , p does not divide a , b and (a , b ) = 1 so that |y| p = cβ . We now have,   aa , x y = p α+β bb where α + β, aa ∈ Z, bb ∈ Z+ , p does not divide aa , bb and (aa , bb ) = 1, this being so since p is a prime number (this is the reason why p was chosen to be a prime number!). Thus |x y| p = cα+β = cα . cβ = |x| p |y| p . We now claim that |x + y| p ≤ max(|x| p , |y| p ), which is stronger than (1.3). To prove (1.5), we show that

(1.5)

4

1 Introduction and Preliminaries

|x| p ≤ 1 ⇒ |1 + x| p ≤ 1.

(1.6)

Leaving out the trivial case, let x = 0. It follows from the definition that |x| p ≤ 1 ⇒ α ≥ 0 and x can be written in the form x=

c , d

where c , d are integers which are relatively prime and p does not divide d . Now, 1+x =1+

c + d c = . d d

Since 1 + x has a denominator which is prime to p, we have |1 + x| p ≤ 1, proving our claim. Now, if y = 0, (1.5) is trivially true.   Let y = 0. Without loss of x  generality, we shall suppose that |x| p ≤ |y| p so that   ≤ 1. Using (1.6), we have, y p     x 1 +  ≤ 1 and so  y p        x  x   1 + |x + y| p =  y 1 + = |y| p y p y p ≤ |y| p = max(|x| p , |y| p ), establishing (1.5). If a valuation of K satisfies (1.5) too, it is called a ‘non-archimedean valuation’ of K and K , with such a valuation, is called a ‘non-archimedean valued field’ or just a ‘non-archimedean field’. The metric induced by a non-archimedean valuation satisfies the much stronger inequality d(x, y) ≤ max(d(x, z), d(z, y)),

(1.7)

which is known as the ‘ultrametric inequality’. Any metric which satisfies (1.7) is known as an ‘ultrametric’. Study of analysis in non-archimedean fields is known as ‘non-archimedean analysis’ or ‘ p-adic analysis’ or ‘ultrametric analysis’. A nonarchimedean valuation is called an ‘ultrametric valuation’ and a non-archimedean field is called an ‘ultrametric field’. We have thus proved that | · | p is a non-archimedean valuation of the rational number field Q. The completion of Q with respect to the p-adic valuation | · | p is called the p-adic field, denoted by Q p . Elements of Q p are called p-adic numbers.

1 Introduction and Preliminaries

5

We are very familiar with analysis in R (i.e., real analysis) or in C (i.e., complex analysis). We now prove some elementary results in ultrametric analysis. These results point out significant departures from real or complex analysis. In the sequel we shall suppose that K is a non-archimedean field with valuation | · |. Theorem 1.1 If K is a non-archimedean field and if |a| = |b|, a, b ∈ K , then |a + b| = max(|a|, |b|). Proof For definiteness, let us suppose that |a| > |b|. Now, |a| = |(a + b) − b| ≤ max(|a + b|, |b|) = |a + b|, for, otherwise, |a| ≤ |b|, a contradiction of our assumption. Consequently, |a| ≤ |a + b| ≤ max(|a|, |b|) = |a|, and so |a + b| = max(|a|, |b|), 

completing the proof.

Corollary 1.1 Any triangle is isosceles in the following sense: given a triad of points in an ultrametric field, at least two pairs of points have the same distance. Proof Let d be the ultrametric induced by the non-archimedean valuation | · | of K . Consider a triangle with vertices x, y and z. Cases 1 If the triangle is equilateral, i.e., d(x, y) = d(y, z) = d(z, x), the triangle is already isosceles and there is nothing to prove! Cases 2 Suppose the triangle is not equilateral (say) d(x, y) = d(y, z), i.e., |x − y| = |y − z|. Then d(x, z) = |x − z| = |(x − y) + (y − z)| = max(|x − y|, |y − z|) using Theorem 1.1 = max(d(x, y), d(y, z)), 

proving our claim. Theorem 1.2 Every point of the open sphere S (x) = {y ∈ K /|y − x| < },  > 0 is a centre, i.e., if y ∈ S (x), S (y) = S (x).

6

1 Introduction and Preliminaries

Proof Let y ∈ S (x) and z ∈ S (y). Then |y − x| <  and |z − y| <  so that |z − x| = |(z − y) + (y − x)| ≤ max(|z − y|, |y − x|) < . Thus z ∈ S (x). Consequently S (y) ⊆ S (x). The reverse inclusion is similarly  proved so that S (y) = S (x), completing the proof. Corollary 1.2 Any two spheres are either disjoint or one is contained in the other. More specifically, given two spheres S1 (x), S2 (y) with 1 ≤ 2 , either S1 (x) ∩ S2 (y) = ∅ or S1 (x) ⊆ S2 (y). Theorem 1.3 Any sequence {xn } in K is a Cauchy sequence if and only if |xn+1 − xn | → 0, n → ∞.

(1.8)

Proof If {xn } is a Cauchy sequence in K , it is clear that (1.8) holds. Conversely, let (1.8) hold. So given  > 0, there exists a positive integer N such that |xn+1 − xn | < , n ≥ N .

(1.9)

Now, for definiteness, let m > n. Then |xm − xn | = |(xm − xm−1 ) + (xm−1 − xm−2 ) + · · · + (xn+1 − xn )| ≤ max(|xm − xm−1 |, |xm−1 − xm−2 |, . . . , |xn+1 − xn |) < , m > n ≥ N , in view of (1.9). This proves that {xn } is a Cauchy sequence. In classical analysis i.e., real or complex analysis, we know that if

∞ 

 xn con-

n=1

∞  1 verges, then xn → 0, n → ∞. However, the converse is not true. For instance, n n=1

diverges, though n1 → 0, n → ∞. We will now prove that the converse is also true in the case of complete, non-archimedean fields. Theorem 1.4 Let K be a complete, non-archimedean field. Then n = 1, 2, . . . , converges if and only if an → 0, n → ∞.

∞  n=1

an , an ∈ K ,

1 Introduction and Preliminaries

Proof Let so

∞ 

7

an converge and sn =

n=1

n 

ak , n = 1, 2, . . . . Then {sn } converges and

k=1

|an | = |sn − sn−1 | → 0, n → ∞. Thus an → 0, n → ∞. Conversely, let an → 0, n → ∞. Now, |sn − sn−1 | = |an | → 0, n → ∞. In view of Theorem 1.3, {sn } is a Cauchy sequence. Since K is complete with respect ∞  to the valuation, lim sn exists in K and so an converges.  n→∞

n=1

Going deeper and deeper into ultrametric analysis, one can point out more and more results, which indicate significant departure from classical analysis—in fact ultrametric analysis is a “STRANGE AND RUGGED PATH”!. One is tempted to ask: “Are there natural non-archimedean phenomena?”. We have observed earlier, as a corollary of Theorem 1.2, that any two spheres in a non-archimedean field K are either disjoint or one is contained in the other. This situation is amply illustrated by two drops of mercury on a smooth surface, which is a typical non-archimedean phenomenon. When we face with such situations, ultrametric analysis would be appropriate. Bachman’s book [1] is an excellent intoduction to non-archimedean analysis. If K is a field with an ultrametric valuation, then, for any integer n, |n| ≤ max(|1|, |1|, . . . , |1|)

 n times

= |1| = 1 (note that |1| = |1.1| = |1||1| implying that |1| = 1 using the fact that |1| = 0). Thus {n}∞ n=1 is a bounded sequence (note that this is the reason why | · | is nonarchimedean and the field K is a non-archimedean field!). Converse is also true. This fact is established by the following result. Theorem 1.5 Let K be a field with a valuation | · | and let there exist d > 0 such that |m| ≤ d for all integers m of K , where the set of integers refers to the isomorphic image of Z in K , if K has characteristic 0 or the isomorphic image of the residue classes modulo p, if K has characteristic p. Then the valuation | · | is non-archimedean. Proof If a, b ∈ K and τ is a positive integer, |a + b|τ = |(a + b)τ |         τ τ τ −1 τ τ −2 2 τ  a b+ a b + ··· + b  = a + 1 2

8

1 Introduction and Preliminaries

   τ  τ −1  |a| |b| + · · · + |b|τ ≤ |a|τ +  1  ≤ (τ + 1)d max(|a|, |b|)τ . Taking τ th roots on both sides and allowing τ → ∞, we have, |a + b| ≤ max(|a|, |b|), proving that the valuation | · | is non-archimedean.



We already obtained several valuations of Q, viz., the p-adic valuations, the usual absolute value and the trivial valuation. We will prove that these are the only valuations of Q in a certain sense. So we wish to find all the valuations of Q. To achieve this, we need the following lemmas. Lemma 1.1 If 0 < r ≤ s and ai , i = 1, 2, . . . , n are non-negative real numbers, then 1 n 1

n s r   aks ≤ akr . i=1

Proof Let d =

n 

k=1

akr . Then

k=1

n 

1 s

aks

k=1

=

1

dr

n 1  as s k s

k=1

=

 n  s 1  ar r s k

 ≤

since

akr d

s r

≤

d

k=1 n  ar

k

k=1



dr

d

1 s

,

ar akr s , this being so since k ≤ 1 and ≥ 1. However, d d r n  ar

k

k=1

d

=

n 1 r 1 ak = d = 1, d d k=1

1 Introduction and Preliminaries

so that

9

n 

1 s

aks

k=1

≤ 1,

1

dr

i.e.,

n 

1 s

aks

1

≤ dr

k=1

=

n 

1 r

akr

,

k=1



completing the proof.

Lemma 1.2 If 0 < α ≤ 1 and bi , i = 1, 2, . . . , n are non-negative real numbers, then (b1 + b2 + · · · + bn )α ≤ b1α + b2α + · · · + bnα . Proof In Lemma 1.1, choose ai = bi , i = 1, 2, . . . , n and s = 1, r = α. Then, we have 1

b1 + b2 + · · · + bn ≤ (b1α + b2α + · · · + bnα ) α , i.e., (b1 + b2 + · · · + bn )α ≤ b1α + b2α + · · · + bnα .



Going back to our problem, let m, n be integers > 1, m ≥ n. Writing m using the base n, we have, m = a0 + a1 n + a2 n 2 + · · · + ak n k , where 0 ≤ ai ≤ n − 1, i = 0, 1, 2, . . . , k, n k ≤ m. So, k≤

log m . log n

Let | · | be a valuation of Q. Now, note that |ai | < n, i = 0, 1, 2, . . . , k so that |m| ≤ |a0 | + |a1 ||n| + |a2 ||n|2 + · · · + |ak ||n|k < n(1 + |n| + |n|2 + · · · + |n|k ) ≤ n(k + 1) max(1, |n|)k

(1.10)

10

1 Introduction and Preliminaries

 ≤n

 log m log m + 1 max(1, |n|) log n , using (1.10). log n

(1.11)

Since (1.11) holds for any integers m, n > 1, replace m by m τ , τ ∈ Z+ in (1.11). Thus   log m τ log m τ τ + 1 max(1, |n|) log n , |m | ≤ n log n 

τ

i.e., |m| ≤ n

 τ log m τ log m + 1 max(1, |n|) log n . log n

Taking τ th root on both sides,    1 log m τ τ log m |m| ≤ n +1 max(1, |n|) log n . log n 1

Allowing τ → ∞ and using the fact that lim τ τ = 1, we have, τ →∞

log m

|m| ≤ max(1, |n|) log n .

(1.12)

Cases 1 There exists n > 1 such that |n| ≤ 1. Now (1.12) implies that |m| ≤ 1 for all m > 1. Using | − 1| = 1, we see that |m| ≤ 1 for all m ∈ Z. In view of Theorem 1.5, it follows that the valuation in this case is non-archimedean. If |m| = 1 for all m ∈ Z, m = 0, it follows that | · | is the trivial valuation. On the other hand, let us suppose that |m| < 1 for some m ∈ Z, m > 1. Let p be the least positive integer such that | p| < 1. Then p must be a prime. For, otherwise, p = ab, where a, b are integers > 1, a, b < p. Then, | p| = |ab| = |a||b| = 1.1 = 1, which is a contradiction. Let m = qp + r, 0 ≤ r < p. We claim that r = 0. For, otherwise, i.e., if r = 0, since r < p, |r | = 1. But |qp| = | p + p + · · · + p | ≤ | p| < 1,

 q times

so that |m| = max(|qp|, |r |) = 1, in view of Theorem 1.1

1 Introduction and Preliminaries

11

which is a contradiction. Thus r = 0 and so p/m. Conversely, if p/m, then |m| < 1. Consequently, |m| < 1 if and only if p/m. If x ∈ Q, x = 0, then x can be written as a x = pα , b where a, b, α ∈ Z, b = 0, p  \a, p  \b. In view of the above discussion, |x| = | p|α

|a| = | p|α , since |a| = |b| = 1 |b| = cα ,

where c = | p| is some real number such that 0 < c < 1. Thus in this case, | · | is the p-adic valuation. Cases 2 For any integer n > 1, |n| > 1. From (1.12) we have, 1

1

|m| log m ≤ |n| log n . Since this is true for all integers m, n > 1, interchanging m and n, we obtain 1

1

1

1

|m| log m ≥ |n| log n . Thus

|m| log m = |n| log n , for all integers m, n > 1. In other words, 1

|m| log m = c, where c is independent of m. Noting that c > 1, we can write c = eα , for some α > 0. 1

i.e., |m| log m = eα α

i.e., |m| = eα log m = elog m = m α . In particular,

so that

|2| = 2α 2α = |2| = |1 + 1| ≤ |1| + |1| = 1 + 1 = 2.

(1.13)

12

1 Introduction and Preliminaries

Consequently α ≤ 1. For any integer m > 1, using (1.13), |m| = m α = |m|α∞ ,

(1.14)

where | · |∞ denotes the usual absolute value function on R. Note that (1.14) holds trivially for m = 0. Since | − 1| = 1, | − 1|∞ = 1, (1.14) is true for all m ∈ Z. So for any x ∈ Q, |x| = |x|α∞ , where 0 < α ≤ 1. We shall show that all functions of the form | · |α∞ , 0 < α ≤ 1 are valuations of Q. Clearly (1.1) and (1.2) are satisfied. Now, |x + y|α∞ ≤ (|x|∞ + |y|∞ )α ≤ |x|α∞ + |y|α∞ , since 0 < α ≤ 1, using Lemma 1.2. Thus, we have proved the following interesting result. Theorem 1.6 Any valuation of Q is either the trivial valuation, a p-adic valuation or a power of the usual absolute value, i.e., | · |α∞ , where 0 < α ≤ 1. We shall now prove that all valuations of the form |·|α∞ , 0 < α ≤ 1, are equivalent in a sense which will be made precise in the sequel. Definition 1.2 Let | · |1 , | · |2 be two non-trivial valuations of a field K . | · |1 is said to be equivalent to | · |2 , written as | · |1 ∼ | · |2 , if |a|1 < 1 implies |a|2 < 1, a ∈ K . Note that if every null sequence with respect to | · |1 is also a null sequence with respect to | · |2 then | · |1 ∼ | · |2 . Let |a|1 < 1. Then {a n } is a null sequence with respect to | · |1 . By hypothesis, {a n } is a null sequence with respect to | · |2 too and so we should have |a|2 < 1. Thus | · |1 ∼ | · |2 . We  also  note that if | · |1 ∼ | · |2 and |a|1 > 1, then |a|2 > 1. For, if |a|1 > 1, then  a1 1 < 1. Since | · |1 ∼ | · |2 ,  a1 2 < 1 and so |a|2 > 1. Theorem 1.7 If | · |1 ∼ | · |2 , then |a|1 = 1 implies |a|2 = 1. Proof Since | · |1 is non-trivial, there exists b ∈ K , b = 0 such that |b|1 < 1. Now, |a n b|1 = |a|n1 |b1 | = |b|1 < 1. Since | · |1 ∼ | · |2 , |a n b|2 < 1, i.e., |a|n2 |b|2 < 1,

1 Introduction and Preliminaries

13

 i.e., |a|2
1, then |a|2 > 1, using the comment preceding Theorem 1.7. In both the cases, we have a contradiction. So |a|1 < 1. Thus ∼ is symmetric and consequently ∼ is an equivalence relation.  γ

Theorem 1.8 If | · |1 ∼ | · |2 , then | · |2 = | · |1 , where γ is a positive real number. Proof Let b ∈ K be fixed such that |b|1 > 1. Let a ∈ K , a = 0. Then log |a|1 . |a|1 = |b|α1 , where α = log |b|1 Let n, m ∈ Z such that

n m

> α. Then n

|a|1 < |b|1m ,  m a  i.e.,  n  < 1. b 1 Since by hypothesis, | · |1 ∼ | · |2 ,  m a    < 1,  bn  2 from which we get

n

|a|2 < |b|2m . Similarly, if

n < α, we get, m

n

|a|2 > |b|2m . Consequently, we have,

|a|2 = |b|α2 .

Thus α=

log |a|2 log |a|1 = log |b|1 log |b|2

14

1 Introduction and Preliminaries

and so log |b|2 · log |a|1 log |b|1 = γ log |a|1 ,

log |a|2 =

where γ =

log |b|2 . Then log |b|1 γ

|a|2 = |a|1 (where γ =

log |b|2 is a positive real number). log |b|1

The above result is trivially true for a = 0. The proof of the theorem is now complete.  Theorem 1.8 yields an alternate characterization of equivalent valuations. Theorem 1.9 The non-trivial valuations | · |1 , | · |2 are equivalent if and only if any null sequence with respect to | · |1 is also a null sequence with respect to | · |2 . Proof If | · |1 ∼ | · |2 , in view of Theorem 1.8, any null sequence with respect to | · |1 is also a null sequence with respect to | · |2 . By the comment immediately following Definition 1.2, the converse is also true.  Using Theorem 1.8, we can “reformulate” Theorem 1.6 as follows. Theorem 1.10 Any valuation of Q is either the trivial valuation, a p-adic valuation or the usual absolute value (upto equivalent valuations). In the context of Theorem 1.8, we note that the choice of c, 0 < c < 1, used to define the p-adic valuation | · | p , has no impact on the study of the valuation | · | p in the sense that for different values of c, 0 < c < 1, the corresponding valuations | · | p are equivalent. It is worthwhile for the reader to try the following exercises. Exercise 1.1 Prove that any valuation of a field of characteristic not equal to 0 is non-archimedean. Exercise 1.2 Prove that the only valuation of a finite field is the trivial valuation. Exercise 1.3 Prove that Q is not complete with respect to the p-adic valuation | · | p .

Reference 1. Bachman, G.: Introduction to p-adic Numbers and Valuation Theory. Academic Press, New York (1964)

Chapter 2

Some Arithmetic and Analysis in Q p ; Derivatives in Ultrametric Analysis

Abstract In this chapter, we discuss some arithmetic and analysis in the p-adic field. We also introduce the concepts of differentiability and derivatives in ultrametric analysis and briefly indicate how ultrametric calculus is different from our usual calculus. Keywords Valuation ring · Residue class field · Canonical expansion · Differentiability · Derivatives We need the following results in the sequel. Theorem 2.1 Let | · | be an ultrametric valuation of K . Then the set V ⊆ K of all elements a such that |a| ≤ 1 is a ring with identity. The set P ⊆ V of all elements a such that |a| < 1 is the unique maximal ideal of V and P is also a prime ideal. Proof If a, b ∈ V , then |ab| = |a||b| ≤ 1 and so ab ∈ V . Again, |a − b| ≤ max(|a|, |b|) ≤ 1, so that a − b ∈ V . Thus V is a ring. Also 1 ∈ V . Now, if a, b ∈ P, |a + b| ≤ max(|a|, |b|) < 1, and so a + b ∈ P. If a ∈ P, b ∈ V , then |ba| = |b||a| < 1, so ba ∈ P. Thus P is an ideal of V . Further, if a ∈ V and if a ∈ / P, then |a| = 1. Now, 1 = |1| = |a.a −1 | = |a||a −1 |, so that |a −1 | = 1. It now follows that P is the unique maximal ideal of V . Since 1 ∈ V , P is also a prime ideal, completing the proof. 

© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_2

15

2 Some Arithmetic and Analysis in Q p …

16

The ring V is called the “valuation ring” associated with the ultrametric valuation | · |. The field V /P is called the associated “residue class field”.

Theorem 2.2 If | · | is an ultrametric valuation of K and Kˆ is the completion of K , then |K | = | Kˆ |, where |K | is the image of K in R under the valuation | · | and | Kˆ | is the image of Kˆ in R under the extended valuation (for the notion of extended valuation, see [1]), which we denote by | · | again. Proof Let α ∈ Kˆ . If α = 0, |α| = 0. So let α = 0. Since K is dense in Kˆ , there exists a sequence {an } in K such that lim an = α. Now, since | · | is an ultrametric n→∞ valuation, |an | = |α + (an − α)| = max(|α|, |an − α|) = |α|, for sufficiently large n, since |α| = 0, |an − α| can be made arbitrarily small for sufficiently large n. Thus | Kˆ | ⊆ |K |. The reverse inclusion is trivial. This proves the theorem.  Theorem 2.3 Any α ∈ Q p can be written as α=

∞ 

aj pj,

(2.1)

j=n

where a j ∈ Z, j = n, n + 1, . . . and n is such that |α| p = | p|np . Proof Let α ∈ Q p , α = 0. In view of Theorem 2.2, |Q p | p = |Q| p = {| p|np , n = 0, ±1, ±2, . . . } so that |α| p = | p|np for some n ∈ Z.

(2.2)

α pn

so that |β| p = 1. Let V, P respectively denote the valuation ring of | · | p on Q and the unique maximal ideal of V ; let Vˆ , Pˆ respectively denote the valuation ring of | · | p on Q p and the unique maximal ideal of Vˆ . Now, β ∈ Vˆ and so β = lim ck , ck ∈ Q, k = 0, 1, 2, . . . . There exists a positive integer N (depending Let β =

k→∞

on n) such that |β − ck | p < 1, k ≥ N . In particular, |β − c N | p < 1. Consequently, |c N | p = |β + (c N − β)| p = max(|β| p , |c N − β| p ) = 1, since | · | p is an ultrametric valuation. Since c N ∈ Q and |c N | p = 1, c N ∈ V . ˆ Thus Let us write c N = bn . |β − bn | p = |β − c N | p < 1 and so β − bn ∈ P. en ˆ ˆ β + P = bn + P. Since |bn | p = 1, bn ∈ Q, bn = dn , en , dn ∈ Z, en , dn are prime to p. Thus there exist integers x, y such that xdn + yp = 1,

2 Some Arithmetic and Analysis in Q p …

17

i.e., xdn ≡ 1(mod p). Then, en − en x dn   1 −x = en dn en (1 − dn x) = dn ≡ 0(mod p)

bn − en x =

ˆ Let an = en x. so that |bn − en x| p < 1. Thus bn − en x ∈ P and so bn − en x ∈ P. ˆ Already Then an ∈ Z and |bn − an | p < 1, i.e., bn − an ∈ Pˆ and so bn + Pˆ = an + P. ˆ ˆ ˆ ˆ ˆ β + P = bn + P so that β + P = an + P. Thus β − an ∈ P and so |β − an | p < 1. We note that |an | p = |(an − β) + β| p = max(|an − β| p , |β| p ) = 1. We now have α = β p n = an p n + (β − an ) p n = an p n + γ1 , where γ1 = (β − an ) p n . |γ1 | p = |(β − an ) p n | p = |β − an | p | p|np < | p|np . So |γ1 | p = | p|mp , where m > n,

(2.3)

which is similar to (2.2). Treating γ1 like α and continuing the process, after k steps, we get α = an p n + an+1 p n+1 + · · · + an+k−1 p n+k−1 + γk , where ai ∈ Z, i = n, n + 1, . . . , n + k − 1, |ai | p = 1 or 0 and |γk | p < | p|n+k → 0, k → ∞, p since | p| p < 1. This completes the proof of the theorem.



The integer coefficients of (2.1) are only unique modulo p. So if we agree to choose the ai ’s such that 0 ≤ ai ≤ p − 1, then (2.1) is called the “canonical representation or expansion” of α. We shall illustrate the above with an example. We shall find the canonical expansion of 38 in Q5 . We shall follow the notation used in Theorem 2.3. Since 3   = |5|0 = 1, we see that n = 0. A solution of 5 8 5

2 Some Arithmetic and Analysis in Q p …

18

8x ≡ 1(mod 5) is x = 2. Since 3 · 2 ≡ 1(mod 5), a0 = 1. Now,  γ1 =

 3 5 − 1 50 = − . 8 8

    We repeat the above procedure for γ1 . |γ1 |5 = − 58  = |5|15 . Now, − 58 · 5 Again, a solution of 8x ≡ 1(mod 5)

1 5

= − 18 .

is x = 2 · (−1) · 2 ≡ 3(mod 5) so that a1 = 3. Next,    25 1 5 γ2 = − − 3 5 = − 8 8 

and so |γ2 |5 = |5|35 , which implies that a2 = 0. Since γ532 = − 18 , proceeding as above, we see that a3 = 3. Continuing in this manner, we see that a4 = a6 = a8 = · · · = 0 and a5 = a7 = a9 = · · · = 3. We now follow the notation as under: The expansion of α ∈ Q p , say, α=

a−γ a−γ+1 + γ−1 + · · · + a0 + a1 p + a2 p 2 + · · · pγ p

is, for convenience, written as α = a−γ a−γ+1 . . . a0 , a1 a2 . . . ( p). We now write the canonical expansion of

3 8

in Q5 as

3 = 1, 30 30 30 . . . (5), 8 or, in a shorter form as

3 = 1, 30 . . . (5), 8

where the bar above denotes periodic repetition. If α ∈ Q p has an expansion of the form α = a0 , a1 a2 . . . ( p),

(2.4)

2 Some Arithmetic and Analysis in Q p …

19

then α is called a p-adic integer. Note that α is a p-adic integer if and only if |α| p = | p|np with n ≥ 0, i.e., if and only if α ∈ Vˆ , the valuation ring of | · | p on Q p . We shall now illustrate the arithmetic operations in Q p , using the notation introduced in (2.4). Addition (1) In Q7 , add the following 1 4 5 2, + 3 7, 4 1 3, (2) In Q5 ,

1 1 5 0

1 11 1 37612 213152 613303

 3 =  81  + 1 0, 0 0 0 0 0 0 . . . = 5 1 1, 3 0 3 0 3 0 . . . 1, 3 0 3 0 3 0 . . .

Subtraction (1) In Q7 , −

5 6, 3 5 2 4 1, 2 4 0 3 5 5, 1 1 2 1

(2) In Q5 , 7 5 3 56 2 2 1, 4 3 0 2 1 − 1 3 4, 2 3 1 4 2 2 1 4 1, 1 0 4 2 3 2 4 4 . . . Multiplication In Q7 , ×

1 2, 1, 1 2, 2,

3 2 3 4

14 03 14 6211 362451 1 4, 0 4 6 4 5 5 1

2 Some Arithmetic and Analysis in Q p …

20

Division In Q5 , divide 32,13 by 43,12 4 3, 1 2 ) 3 2, 1 3 ( 2, 0 2 4 4 2 0 3 2, 3 4 3 3 4 4 ... 3234 1 1 0 4 4 ... 102 4 1 1 3 4 2 4 4 ... 10 2 4 1 3 2 3 2 4 4 ... 3 2 3 4 3 3 4 4 ... In R, we write a given number as a decimal expansion. Its analogue in Q p is the canonical expansion. We recall that in R, a number is rational if and only if its decimal expansion is periodic. We have an analogue in Q p . We state the result (for proof, one can refer to [1]). Theorem 2.4 An element α ∈ Q p is rational if and only if its canonical expansion α=

∞ 

a j p j , 0 ≤ a j ≤ p − 1,

j=n

when n is such that |α| p = | p|np , is periodic. Example 2.1 Find the rational number represented by the canonical expansion 1, 30 in Q5 . 1, 30 = 1 + 3.51 + 0.52 + 3.53 + 0.54 + · · · = 1 + 3[51 + 53 + · · · ] 5 = 1 + 3· , since |52 |5 = |5|25 < 1 1 − 52 15 =1− 24 5 =1− 8 3 = . 8

2 Some Arithmetic and Analysis in Q p …

21

Exercise 2.1 Find the canonical expansion of (i)

1 5

in Q3 ;

(ii)

1 3

in Q2 ;

(iii) − 57 in Q5 .

Exercise 2.2 In Q5 , find (i)

1 2 3, 4 1 2 ; + 4 2 1, 0 3 2

(ii)

1 2 4, 1 3 1 ; − 3 2 1, 2 2 1

(iii)

(3 4, 1 2 1) ; × (0, 2 1 0 3)

(iv) (1 3 1, 2) ÷ (2, 4 2) Exercise 2.3 In Q3 , find the rational number whose canonical expansion is 2, 0121. As in the classical set up, in Q p too, we have the “exponential” and “logarithmic” series respectively defined by E(x) =

∞ ∞   xn xn and L(1 + x) = (−1)n−1 , n! n n=0

n=1

which converge for all x ∈ Q p with |x| p < 1. These series have properties which are very similar to their classical counterparts, say, for instance, E(x + y) = E(x)E(y); L((1 + x)(1 + y)) = L(1 + x) + L(1 + y); L(E(x)) = x; E(L(1 + x)) = 1 + x. In Q p , we have Binomial series too (for details, refer to [1]). Though not relevant to the present monograph, it is worth noting that the concept of derivative and its properties have been studied in ultrametric analysis (see [2]). With regard to derivatives, we need the following definition. Definition 2.1 If U is any subset of an ultrametric field K without isolated points and f : U → K , we say that f is differentiable at x ∈ U if lim

y→0

f (x + y) − f (x) exists. y

(2.5)

Whenever the limit (2.5) exists, it is called the derivative of f at x, denoted by f (x). It is immediate from the above definition that the characteristic function (K valued) χU of any clopen set (i.e., any set which is both open and closed) is differen = 0 everywhere. This shows that there are nonconstant tiable everywhere with χU functions whose derivatives are 0 everywhere contrary to the classical situation. There

2 Some Arithmetic and Analysis in Q p …

22

exist (1-1) functions too whose derivatives are 0 everywhere. If the characteristic of K is 2 (We recall that any valuation of such a field is ultrametric, in view of Exercise 1.1) and f : K → K is defined by f (x) = x 2 , x ∈ K , then f is (1-1) (since f (x) = f (y) ⇒ x 2 = y 2 ⇒ x = y, using the fact that the characteristic of K is 2) and x 2 − a2 = lim (x + a) = 2a = 0, f (a) = lim x→a x − a x→a at any a ∈ K . We can also give examples of functions which are continuous everywhere but not differentiable anywhere. For  instance, let U denote the closed unit disc   in Q p . Let f : U → Q p defined by f an p n = an p 2n , 0 ≤ an ≤ p − 1. n

n

Then f is continuous everywhere but not differentiable anywhere (see [3]). In classical analysis, functions which have antiderivatives do not have jump discontinuities and they are pointwise limits of continuous functions. However, both these conditions are not sufficient for the functions to have an antiderivative. Unlike the classical case in which sufficient conditions are not known, the situation in the ultrametric case is simpler: If U is a subset of K without isolated points, then f : U → K has an antiderivative if and only if f is the pointwise limit of continuous functions ([4], p. 283).

References 1. Bachman, G.: Introduction to p-adic Numbers and Valuation Theory. Academic Press, New York (1964) 2. Schikhof, W.H.: Ultrametric Calculus. Cambridge University Press, Cambridge (1984) 3. Narici, L., Beckenstein, E.: Strange terrain—non-archimedean spaces. Amer. Math. Monthly 88, 667–676 (1981) 4. Van Rooij, A.C.M.: Non-archimedean Functional Analysis. Marcel Dekker, New York (1978)

Chapter 3

Ultrametric Functional Analysis

Abstract In this chapter, we introduce ultrametric Banach spaces and mention that many results of the classical Banach space theory carry over in the ultrametric set up too. However, the Hahn–Banach theorem fails to hold. To salvage the Hahn– Banach theorem, the concept of a “spherically complete field” is introduced and Ingleton’s version of the Hahn–Banach theorem is proved. The classical “convexity” does not work in the ultrametric set up and it is replaced by the notion of “K convexity”, which is briefly discussed at the end of the chapter. Keywords Ultrametric Banach space · Spherically complete filed · Hahn–Banach theorem · K -convexity The definition of an ultrametric Banach space over an ultrametric field K is the same as for the classical case except that the norm satisfies the strong triangle inequality, i.e., ||x + y|| ≤ max(||x||, ||y||). We also require the condition that the valuation of K is non-trivial since with this assumption, a linear map between ultrametric Banach spaces is continuous if and only if it is bounded (note that this result may fail if the valuation of K is trivial). Even after excluding the trivial valuation of K , there may not exist unit vectors in X ; a vector x cannot be divided by ||x||, since ||x|| is a real number and not a scalar. Because of these deficiencies, even one-dimensional ultrametric Banach spaces need not be isometrically isomorphic, though they are linearly homeomorphic. Consider the following formulae for the norm of a bounded linear map A : X → Y where X, Y are ultrametric Banach spaces.  ||A(x)|| : 0 < ||x|| ≤ 1 ; ||A|| = sup ||x||

(3.1)

||A|| = sup {||A(x)|| : ||x|| = 1} ;

(3.2)

||A|| = sup {||A(x)|| : 0 ≤ ||x|| ≤ 1} .

(3.3)



© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_3

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24

3 Ultrametric Functional Analysis

We recall that in the classical case, all the above three formulae for norm hold. In the ultrametric case, if K is non-trivially valued, (3.1) holds but (3.2) and (3.3) may not hold ([1], p. 75). Many results about classical and ultrametric Banach spaces have exactly same formal statements but the proofs are entirely different. For instance, consider the classical theorem that locally compact Banach spaces must be finite-dimensional. The proof of this theorem involves use of unit vectors. In the ultrametric case, we cannot use this technique. We consider various cases and vectors whose norms are close to 1 and finally conclude that locally compact ultrametric Banach spaces must be finite-dimensional and the underlying field K must be locally compact ([1], p. 70). Some results of the classical Banach space theory which hold in the ultrametric setting too are the closed graph theorem, Banach-Steinhaus theorem and open mapping theorem. However, the classical Hahn–Banach theorem does not carry over to the ultrametric setting—this makes the situation more interesting. In the case of ultrametric Banach spaces, it may not be possible to extend a given continuous linear functional from a subspace to the entire space. The fault is not with the linear space X but with the underlying field K . With a view to retain the Hahn–Banach theorem in the ultrametric set up too, a new notion of “spherical completeness” is introduced. We recall that completeness means that every nested sequence of closed balls, whose diameters tend to 0, have non-empty intersection. Spherical completeness makes the same demand but drops the requirement that the diameters tend to 0, i.e., spherical completeness means that every nested sequence of closed balls have non-empty intersection. It is clear that spherical completeness is stronger than completeness. However, the converse is not true, a counterexample may not be easy! For a counterexample, see [1], Example 4, pp. 81–83. Ingleton proved the following result. Theorem 3.1 ([1], p.78) (The ultrametric Hahn–Banach theorem) Let X be a normed linear space and Y be an ultrametric normed linear space over K (K may even be trivially valued). A continuous linear mapping A defined on a subspace M of X into Y may be extended to a continuous linear mapping A∗ of X into Y with the same norm, as defined by (3.1), if and only if Y is spherically complete. We now prove the above theorem to exhibit the crucial role played by “Spherical Completeness” in proving the ultrametric version of the Hahn–Banach theorem. Proof of Theorem 3.1 Let A be a bounded linear transformation from M into Y , where M ⊆ X and let Y be spherically complete. We will prove that A can be extended to the whole space X in a norm-preserving fashion. Let now x0 ∈ M  (the complement of M). Consider the subspace [M, x0 ] of X spanned by M and x0 . Let Cx (Ax) = {y ∈ Y/||y − Ax|| ≤ x }, where x = ||A|| M ||x − x0 ||. For convenience, we shall use ||A|| for ||A|| M henceforth. We claim that the set of spheres {Cx (Ax)/x ∈ M} is a nest. Now,

3 Ultrametric Functional Analysis

25

||Ax1 − Ax2 || = ||A(x1 − x2 )|| ≤ ||A||||x1 − x2 || ≤ ||A|| max{||x1 − x0 ||, ||x2 − x0 ||}, i.e., ||Ax1 − Ax2 || ≤ max{x1 , x2 }. It now follows that Ax1 ∈ C2 (Ax2 ) or Ax2 ∈ C1 (Ax1 ). Consequently, the spheres {C  x (Ax)/x ∈ M} form a nest. Since Y is spherically complete, there exists z 0 ∈ ¯ x∈M C x (Ax). Define an extension A of A to [M, x 0 ] by ¯ + λx0 ) = Ax + λz 0 . A(x It is clear that A¯ is a linear extension of A. Now, ¯ + λx0 )|| = ||Ax + λz 0 || || A(x = |λ|||A(λ−1 x) + z 0 || = |λ|||A(−λ−1 x) − z 0 || ≤ |λ|−λ−1 x = |λ|||A|||| − λ−1 x − x0 || = ||A||||x + λx0 ||. ¯ = ||A||. At this stage, an application of Zorn’s lemma Thus A¯ is bounded with || A|| proves that A can be extended to the entire space X in a norm-preserving fashion. To prove the converse, we shall suppose that Y is not spherically complete and arrive at a contradiction, i.e., exhibit a bounded linear transformation defined on a subspace of a normed linear space X , which cannot be extended to X in a normpreserving fashion. Since we have assumed that Y is not spherically complete, there exists in Y a nonempty nest of spheres {Cα (yα )/α ∈ } with empty intersection. Let y ∈ Y . Then y ∈ / Cβ (yβ ) for some β ∈ . Define the function ϕ by ϕ(y) = ||y − yβ ||. We now claim that ϕ is well defined. First we note that for any z ∈ Cβ (yβ ), ||z − / Cβ (yβ ), ||y − yβ || > β so that ||y − yβ || > ||z − yβ ||. Thus yβ || ≤ β . Since y ∈ ||y − z|| = ||(y − yβ ) − (z − yβ )|| = ||y − yβ ||, using Theorem 1.1, so that ϕ(y) = ||y − z||. / C2 (y2 ). Since the spheres form a nest, we can Suppose y ∈ / C1 (y1 ) and y ∈ suppose that C1 (y1 ) ⊆ C2 (y2 ) and so 1 ≤ 2 . Now, ϕ(y) = ||y − y2 || > 2 and ||y1 − y2 || ≤ 2 so that

26

3 Ultrametric Functional Analysis

||y − y1 || = ||(y − y2 ) + (y2 − y1 )|| = ||y − y2 ||, in view of Theorem 1.1. Thus ϕ(y) = ||y − y2 || = ||y − y1 ||, proving that ϕ is well defined. Let now y ∈ Cα (yα ). We have already noted that there exists a sphere Cβ (yβ ) such that y ∈ / Cβ (yβ ). Since the spheres form a nest and y ∈ Cα (yα ), it follows that Cβ (yβ ) ⊆ Cα (yα ). Consequently, ϕ(y) = ||y − yβ || ≤ max{||y − yα ||, ||yα − yβ ||} ≤ α . Thus, for y ∈ Cα (yα ), ϕ(y) ≤ α . Let X be the direct sum of Y and F. For (y, λ) ∈ X , define  |λ|ϕ(λ−1 y), if λ = 0; ||(y, λ)|| = ||y||, if λ = 0. We prove that || · || defines a non-archimedean norm on X . Note that ϕ(y) > 0 implies that ||(y, λ)|| = 0 if and only if (y, λ) = (0, 0). It is also clear that ||μ(y, λ)|| = ||(μy, μλ)|| = |μ|||(y, λ)||. It remains to prove that || · || satisfies the ultrametric inequality. We have to consider several cases. We will consider one of the typical cases, leaving the remaining ones as exercise. Let (y1 , λ1 ), (y2 , λ2 ) be such that λ1 , λ2 , λ1 +λ2 = 0. We can find a sphere −1 −1 / C0 (y0 ). Now, C0 (y0 ) from the nest such that λ−1 1 y1 , λ2 y2 , (λ1 +λ2 ) (y1 + y2 ) ∈ ||(y1 , λ1 )|| = ||y1 − λ1 y0 ||, ||(y2 , λ2 )|| = ||y2 − λ2 y0 || and ||(y1 + y2 , λ1 + λ2 )|| = ||y1 + y2 − (λ1 + λ2 )y0 || ≤ max{||(y1 , λ1 )||, ||(y2 , λ2 )||}, completing the proof in this case. Note that X contains an isometrically isomorphic image of Y , viz., Yˆ = {(y, 0)/y ∈ Y }. Hereafter, we will replace Y by Yˆ . Consider the linear transformation 1, defined by Yˆ → Yˆ 1: (y, 0) → (y, 0).

3 Ultrametric Functional Analysis

27

ˆ = Suppose 1 has a norm-preserving extension 1ˆ to the entire space X . Note that ||1|| 1. Let z ∈ Y be such that ˆ −1) = (z, 0). 1(0, ˆ ˆ ˆ 1) = (y, 0) − (z, 0) = (y − z, 0). 1(y, 1) = 1(y, 0) + 1(0, Now, ˆ ˆ =1 ||y − z|| = ||1(y, 1)|| ≤ ||(y, 1)||, using ||1|| = ϕ(y). each of the spheres Cα (yα ) of the nest, we get If we consider the vector yα from ≤ ϕ(y ) ≤  . Thus z ∈ ||yα −z|| α α α∈ C α (yα ), a contradiction of our assumption   that α∈ Cα (yα ) = φ, completing the proof of the theorem. There are equivalent ways of describing spherical completeness ([1], Chap. 2). The idea that spherical completeness should be substituted in the ultrametric set up whenever completeness appears in the classical case is not, however, true. In fact, spherical completeness plays a very little role in the case of ultrametric Banach algebras and ultrametric measure theory. In fact, in some ways, it is disadvantageous: For instance, if K is spherically complete, then no infinite-dimensional ultrametric Banach space over K is reflexive ([2], Chap. 4). On the other hand, if K is not spherically complete, then c0 and ∞ are reflexive. Since ultrametric fields are totally disconnected, they are not totally ordered. The lack of ordering on K makes it difficult to find an analogue for classical “convexity”. Classical convexity is replaced, in the ultrametric setting, by a notion called “K convexity” (see [3]), which is defined as follows. Definition 3.1 A set S of vectors is said to be “absolutely K -convex” if ax + by ∈ S whenever |a|, |b| ≤ 1 and x, y ∈ S; translates w + S of such sets S are called “K convex”. A topological linear space X (which is defined as in the classical case) is said to be “locally K -convex” if its topology has a base of K -convex sets at 0. Using the above notion of K -convexity, a weaker form of compactness, called “ccompactness” could be defined. c-compactness demands that filterbases composed of K -convex sets possess adherence points rather than the usual requirement of compactness that all filterbases possess adherence points (see [4]). It is known that in the case of ultrametric fields, spherical completeness is equivalent to c-compactness. For a detailed study of K -convexity and c-compactness, one can refer to [3–5]. A seminorm p is ultrametric if it is a seminorm in the usual sense and if, in addition, satisfies p(x + y) ≤ max( p(x), p(y)). The connection between K -convex sets and ultrametric seminorms is very similar to the classical case. The notion of K -convexity leads to defining F-barrels etc. These notions are used in spaces of continuous functions and analogues of known classical theorems are obtained.

28

3 Ultrametric Functional Analysis

Besides its requirements about complex conjugates, the inner product ( , ) on a classical Hilbert space satisfies the condition (x, x) ≥ 0 for every x. Even assuming that an inner product on a linear space X over an ultrametric field K is scalar valued, the “≥ 0” cannot be carried over to the ultrametric set up. K is not totally ordered too. Because of these deficiencies, it seems that we do not have a meaningful analogue in the ultrametric setting of a classical Hilbert space. However, we can define the notion of “orthogonality” in ultrametric Banach spaces. This leads to the notion of an “orthogonal base”, which has many properties as its classical counterpart. For the theory of Banach algebras, spaces of continuous functions, Measure and integral, the reader can refer to [6] and the relevant references given in [6]. For some applications of ultrametric analysis to mathematical physics, one can refer to [7].

References 1. Narici, L., Beckenstein, E., Bachman, G.: Functional Analysis and Valuation Theory. Marcel Dekkar, New York (1971) 2. Van Rooij, A.C.M.: Non-archimedean Functional Analysis. Marcel Dekker, New York (1978) 3. Monna, A.F.: Ensembles convexes dans les espaces vectoriels sur un corps valué. Indag. Math. 61, 528–539 (1958) 4. Springer, T.A.: Une notion de compacité dans la theéorie des espaces vectoriels topologiques. Indag. Math. 27, 182–189 (1965) 5. van Tiel, J.: Espaces localement K -convexes (I–IV). Indag. Math. 27, 249–289 (1965) 6. Narici, L., Beckenstein, E.: Strange terrain—non-archimedean spaces. Amer. Math. Monthly 88, 667–676 (1981) 7. Vladimirov, V.S., Volovich, I.V., Zelenov, E.I.: p-adic analysis and mathematical physics. World Scientific, Singapore (1994)

Chapter 4

Ultrametric Summability Theory

Abstract Our survey of the literature on “ultrametric summability theory” starts with a paper of Andree and Petersen of 1956 (it was the earliest known paper on the topic) to the present. In the present chapter, Silverman–Toeplitz theorem is proved using the “sliding-hump method”. Schur’s theorem and Steinhaus theorem also find a mention. Core of a sequence and Knopp’s core theorem is discussed. It is proved that certain Steinhaus-type theorems fail to hold. Keywords Silverman–Toeplitz theorem · Schur’s theorem Steinhaus-type theorems · Knopp’s core theorem

· Steinhaus theorem ·

4.1 Classes of Matrix Transformations We now present for the first time a brief survey, though not exhaustive, of the work done so far on ultrametric summability theory. Divergent series have been the motivating factor for the introduction of summability theory both in classical as well as ultrametric analysis. Study of infinite matrix transformations in the classical case is quite an old one. Numerous authors have studied general matrix transformations or matrix transformations pertaining to some special classes for the past several decades. On the other hand, the study of matrix transformations in the ultrametric case is of a comparatively recent origin. In spite of the pioneering work of A.F. Monna relating to analysis and functional analysis over ultrametric fields dating back to 1940s, it was only in 1956 that Andree and Petersen [1] proved the analogue for the p-adic field Q p of the Silverman–Toeplitz theorem on the regularity of an infinite matrix transformation. Roberts [2] proved the Silverman–Toeplitz theorem for a general ultrametric field, while Monna ([3], p. 127) obtained the same theorem using the Banach–Steinhaus theorem. Later to Monna, we have only the papers by Rangachari and Srinivasan [4], Srinivasan [5] and Somasundaram [6, 7] till Natarajan took up a detailed study of matrix transformations, special methods of summability and other aspects of summability theory in the ultrametric set up.

© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_4

29

30

4 Ultrametric Summability Theory

In the sequel, K is a complete, non-trivially valued, ultrametric field unless otherwise stated. We shall suppose that the entries of sequences, series and infinite matrices are in K . Definition 4.1 If A = (ank ), ank ∈ K , n, k = 0, 1, 2, . . . , is an infinite matrix, by the A-transform Ax of a sequence x = {xk }, xk ∈ K , k = 0, 1, 2, . . . , we mean the sequence {(Ax)n }, where (Ax)n =

∞ 

ank xk , n = 0, 1, 2, . . . ,

k=0

it being assumed that the series on the right converge. If lim (Ax)n = , we say that n→∞ the sequence x = {xk } is A-summable or summable A to . Definition 4.2 Let X and Y be sequence spaces with elements whose entries are in K . The infinite matrix A = (ank ), ank ∈ K , n, k = 0, 1, 2, . . . is said to transform X to Y if whenever the sequence x = {xk } ∈ X , (Ax)n is defined, n = 0, 1, 2, . . . and the sequence {(Ax)n } ∈ Y . In this case, we write A ∈ (X, Y ). Definition 4.3 If A ∈ (c, c) (where c is the ultrametric Banach space consisting of all convergent sequences in K with respect to the norm defined by ||x|| = sup |xk |, k≥0

x = {xk } ∈ c), A is said to be convergence preserving or conservative. If, in addition, lim (Ax)n = lim xk , A is called a regular matrix or a Toeplitz matrix. If n→∞

k→∞

A is regular, we write A ∈ (c, c; P) (P standing for “preservation of limit”). Theorem 4.1 (Silverman–Toeplitz) ([3], p. 127) A ∈ (c, c), i.e. A is convergence preserving if and only if (4.1) sup |ank | < ∞; n,k

lim ank = δk exists, k = 0, 1, 2, . . . ;

n→∞

and lim

n→∞

∞ 

ank = δ exists.

(4.3)

k=0

In such a case lim (Ax)n = sδ + n→∞

(4.2)

∞  (xk − s)δk , lim xk = s. Further, A ∈ k=0

k→∞

(c, c; P), i.e. A is regular if and only if (4.1), (4.2) and (4.3) hold with the additional requirements δk = 0, k = 0, 1, 2, . . . and δ = 1. Remark 4.1 Monna [3] proved Theorem 4.1 using modern tools like the ultrametric version of Banach–Steinhas theorem. In this context, Natarajan [8] obtained Theorem 4.1, using the “sliding hump method”. This proof is an instance in which traditional tools like ‘signum function’ available in R or C are also avoided.

4.1 Classes of Matrix Transformations

31

We now present this proof. Proof of Theorem 4.1 (see [8]) Let (4.1), (4.2) and (4.3) hold. In view of (4.1), there exists H > 0 such that sup |ank | ≤ H. n,k

Using (4.2), |δk | ≤ H , k = 0, 1, 2, . . . . Let {xk } be such that lim xk = s. Since k→∞

xk − s → 0, k → ∞ and {δk } is bounded, δk (xk − s) → 0, k → ∞ and so ∞  δk (xk − s) converges. Again, since ank → 0, k → ∞ using (4.3) and {xk } is k=0

bounded,

∞ 

ank xk converges and so

k=0 ∞ 

(Ax)n =

ank xk , n = 0, 1, 2, . . .

k=0

is defined and (Ax)n =

∞ 

(ank − δk )(xk − s) +

k=0

∞ 

δk (xk − s) + s

k=0

∞ 

ank .

k=0

Since xk − s → 0, k → ∞, given  > 0, there exists a positive integer k0 such that |xk − s|
k0 . H

Now, k0 ∞ ∞    (ank − δk )(xk − s) = (ank − δk )(xk − s) + (ank − δk )(xk − s). k=0

k=0

k=k0 +1

By (4.2), ank − δk → 0, n → ∞, k = 0, 1, 2, . . . and so we can find a positive integer N such that |ank − δk |
N , k = 0, 1, 2, . . . , k0 , M

M = sup |xk − s|. Thus k≥0

k  0      . M = , n > N .  (ank − δk )(xk − s) <   M k=0

32

4 Ultrametric Summability Theory

Also     ∞     (ank − δk )(xk − s) ≤ sup |ank − δk | |xk − s|   k>k0 k=k0 +1  = , n = 0, 1, 2, . . . . N ,  

So

k=0

from which it follows that lim

n→∞

∞  (ank − δk )(xk − s) = 0. k=0

Consequently, lim (Ax)n =

n→∞

∞ 

δk (xk − s) + sδ,

k=0

using (4.3). Conversely, the necessity of conditions (4.2) and (4.3) are clear by considering the sequences {0, 0, . . . , 0, 1, 0, . . . }, 1 occurring in the kth place, k = 0, 1, 2, . . . and {1, 1, 1, . . . }. We will now prove the necessity of (4.1). Using (4.3), we first note that (4.4) sup |an+1,k − ank | < ∞, n = 0, 1, 2, . . . . k≥0

We now claim that sup |an+1,k − ank | < ∞.

(4.5)

n,k

If not, we can then choose a strictly increasing sequence {n(m)} of positive integers such that (4.6) sup |an(m)+1,k − an(m),k | > m 2 , m = 1, 2, . . . . k≥0

In particular, sup |an(1)+1,k − an(1),k | > 12 .

(4.7)

k≥0

Since an(1)+1,k − an(1),k → 0, k → ∞, given  > 0 (we can suppose that  < 1), we can choose a positive integer k(n(1)) such that

4.1 Classes of Matrix Transformations

sup

33

|an(1)+1,k − an(1),k | < .

(4.8)

k≥k(n(1))

From (4.7) and (4.8), we have sup

|an(1)+1,k − an(1),k | > 12 ,

0≤k 12 .

(4.9)

sup |an(2)+1,k − an(2),k | > 22 .

(4.10)

From (4.6), k≥0

Since, by (4.2), lim (an+1,k − ank ) = 0, k = 0, 1, 2, . . . ,

n→∞

we may suppose that sup

|an(2)+1,k − an(2),k | < .

(4.11)

0≤k k(n(1)) such that (4.12) sup |an(2)+1,k − an(2),k | < . k≥k(n(2))

Using (4.10), (4.11) and (4.12), it follows that sup

|an(2)+1,k − an(2),k | > 22 .

k(n(1))≤k 22 .

(4.13)

Proceeding in this fashion, we have strictly increasing sequences {n(m)}, {k(n(m))}, {k(m)} of positive integers with k(n(m − 1)) ≤ k(m) < k(n(m)) such that ⎧ ⎪ (i) sup |an(m)+1,k − an(m),k | < ; ⎪ ⎪ ⎪ 0≤k m, m = 1, 2, . . . . So {(Ax)n } is not a Cauchy sequence and does not therefore converge, which is a contradiction. This proves that (4.5) holds. Let sup |an+1,k − ank | < M1 , M1 > 0. n,k

Since a1k → 0, k → ∞, |a1k | < M2 , for some M2 > 0, k = 0, 1, 2, . . . . Let M = max(M1 , M2 ). Then |a2k | ≤ max{|a2k − a1k |, |a1k |} ≤ M, k = 0, 1, 2, . . . ; |a3k | ≤ max{|a3k − a2k |, |a2k |} ≤ M, k = 0, 1, 2, . . . ; Inductively, sup |ank | ≤ M, n,k

which shows that (4.1) holds. The latter part of the theorem relating to the regularity of the method is easily deduced thereafter. This completes the proof of the theorem. 

4.1 Classes of Matrix Transformations

35

Definition 4.4 A is called a Schur matrix if A ∈ (∞ , c) (where ∞ denotes the ultrametric Banach space of all bounded sequences in K with respect to the norm defined by ||x|| = sup |xk |, x = {xk } ∈ ∞ ). k≥0

Natarajan [9] proved the following result. Theorem 4.2 (Schur) A ∈ (∞ , c), i.e. A is a Schur matrix if and only if lim ank = 0, n = 0, 1, 2, . . . ;

(4.16)

lim sup |an+1,k − ank | = 0.

(4.17)

k→∞

and n→∞ k≥0

The following result is immediately deduced [9]. Theorem 4.3 (Steinhaus) A matrix cannot be both a Toeplitz and a Schur matrix, or, equivalently, given any regular matrix A, there exists a bounded divergent sequence which is not summable A. Symbolically, we can write this statement as (c, c; P) ∩ (∞ , c) = φ. Somasundaram [6] obtained the Steinhaus theorem for a restricted class of regular matrices and observed that the theorem was not true in general. His restriction is in fact, meaningless and his observation is incorrect. Neither of the examples in Sect. 2 of [6] provide a regular matrix which sums all bounded sequences.

4.2 Steinhaus-Type Theorems In the context of Steinhaus theorem, we introduce the following. Definition 4.5 Whatever be K , i.e. K = R or C or a complete, non-trivially valued, ultrametric field, r is the space of all sequences x = {xk } ∈ ∞ such that |xk+r − xk | → 0, k → ∞, r ≥ 1 being a fixed integer. It is easily proved that r is a closed subspace of ∞ with respect to the norm defined for elements in ∞ . When K = R or C, the following result, improving Steinhaus theorem, was proved in [10] (it is worthwhile to note that a constructive proof was given). Theorem 4.4 (c, c; P) ∩ (r −

r −1

i , c) = φ.

i=1

Proof Let A = (ank ) be a regular matrix. We can now choose two sequences of positive integers {n(m)} and {k(m)} such that if

36

4 Ultrametric Summability Theory

m = 2 p, n(m) > n(m − 1), k(m) > k(m − 1) + (2m − 5)r, then k(m−1)+(2m−5)r 

|an(m),k |
k(m − 1) + (m − 2)r, then k(m−1)+(m−2)r 

|an(m),k |


k=k(m−1)+(m−2)r +1 ∞ 

|an(m),k |
1. Since ϕ is ultrametric full, there exists S ∈ ϕ with sup |bi | = ∞. Consequently, we have i∈S

sup

i∈S∩N( j)

for, otherwise,

sup

i∈S∩N( j)

|bi | > 1 for infinitely many j’s,

|bi | ≤ 1, j = 1, 2, . . . and so sup |bi | ≤ 1, a contradiction.

Hence, for these infinitely many j’s,

i≥1

40

4 Ultrametric Summability Theory

sup |an( j),i | ≥ i∈S

sup

i∈S∩N( j)

|an( j),i |

=

|bi | j i∈S∩N( j) ρ

>

1 1 → ∞, j → ∞, since > 1, j ρ ρ

sup

contrading the fact that  sup sup |ank | < ∞ for every S ∈ ϕ. n≥1

k∈S

Sufficiency. Let {tk } be any sequence in K such that sup |tk | < ∞ for every S ∈ k∈S

ϕ. Define the matrix (ank ), where ank = tk , k = 1, 2, . . . ; n = 1, 2, . . . . Then  sup sup |ank | < ∞ for every S ∈ ϕ. By hypothesis, sup |ank | < ∞. It now n≥1

k∈S

n,k≥1

follows that sup |tk | < ∞ and so ϕ is ultrametric full, completing the proof of the k≥1



theorem.

Corollary 4.1 ϕ is a class of subsets of N satisfying (i) and (ii) of Definition 4.6. Then ϕ is ultrametric full if and only if for any infinite matrix (ank ) for which      ank  < ∞ for every S ∈ ϕ, then sup |ank | < ∞. sup   n≥1  n,k≥1 k∈S

Proof Let ϕ be ultrametric full. Let (ank ) be an infinite matrix for which  Necessity.      sup  ank  < ∞ for every S ∈ ϕ. Let S ∈ ϕ and k0 ∈ S. Since ϕ is hereditary,   n≥1 k∈S

S  = S\{k0 } ∈ ϕ. So

       ank − ank  < ∞, sup    n≥1  k∈S

k∈S

i.e. sup |ank0 | < ∞, n≥1

 for every k0 ∈ S and so sup sup |ank | < ∞ for every S ∈ ϕ. Since ϕ is ultrametric n≥1

k∈S

full, it follows from Theorem 4.5 that sup |ank | < ∞. n,k≥1

Sufficiency. Let (ank ) be an infinite matrix such that sup n≥1

every S ∈ ϕ. Then

 k∈S

 |ank |

< ∞ for

4.2 Steinhaus-Type Theorems

41

       sup  ank  ≤ sup sup |ank |  n≥1 k∈S n≥1  k∈S

0, S ∈ ϕ and two strictly increasing sequences {n(i)} and {k(i)} of positive integers such that

42

4 Ultrametric Summability Theory

sup |an(i)+1,k − an(i),k | > ; k∈S

sup

|an(i)+1,k − an(i),k |
k(i)

 ; 8

 . 8

In view of the above inequalities, there exists k(n(i)) ∈ S, k(i − 1) < k(n(i)) ≤ k(i) such that |an(i)+1,k(n(i)) − an(i),k(n(i)) | > . Define x = {xk }, where

 1, if k = k(n(i)); xk = 0, otherwise.

Now, (Ax)n(i)+1 − (Ax)n(i) =

∞  {an(i)+1,k − an(i),k }xk k=1

=

k(i−1) 

{an(i)+1,k − an(i),k }xk

k=1 k(i) 

+ +

{an(i)+1,k − an(i),k }xk

k=k(i−1)+1 ∞ 

{an(i)+1,k − an(i),k }xk

k=k(i)+1

=

k(i−1) 

{an(i)+1,k − an(i),k }xk

k=1

+ {an(i)+1,k(n(i)) − an(i),k(n(i)) } +

∞ 

{an(i)+1,k − an(i),k }xk

k=k(i)+1

so that  < |an(i)+1,k(n(i)) − an(i),k(n(i)) |    ≤ max |(Ax)n(i)+1 − (Ax)n(i) |, , 8 8

4.2 Steinhaus-Type Theorems

43

which implies that |(Ax)n(i)+1 − (Ax)n(i) | > , i = 1, 2, . . . . Thus x ∈ / C A . Note, however, that x ∈ χϕ . Consequently, χϕ  C A , a contradiction. So (ii) holds.  ank converges for every S ∈ ϕ. Sufficiency. Let (i), (ii) hold. In view of (i), k∈S

Now,              an+1,k − ank  =  (an+1,k − ank )      k∈S

k∈S

k∈S

≤ sup |an+1,k − ank | k∈S

→ 0, n → ∞, using (ii), which implies that lim



n→∞

ank exists for every S ∈ ϕ. Thus C A ⊇ χϕ , completing

k∈S



the proof of the theorem.

Corollary 4.2 (Hahn’s theorem for the ultrametric case) An infinite matrix A = (ank ) sums all bounded sequences if and only if it sums all sequences of 0’s and 1’s. Proof Leaving out the trivial part of the result, suppose A sums all sequences of 0’s and 1’s, i.e. C A ⊇ χϕ , where φ = 2N . Since N ∈ ϕ, lim ank = 0, n = 1, 2, . . . .

k→∞

Also, lim sup |an+1,k − ank | = 0,

n→∞ k∈N

i.e. lim sup |an+1,k − ank | = 0. n→∞ k≥1

In view of Theorem 2 of [9], it follows that A sums all bounded sequences, completing the proof.  ∞  ∞   In the context of r (i.e. closure of r in ∞ ), we introduce the notion r =1

of “generalized semiperiodic sequences”.

r =1

Definition 4.7 x = {xk } is called a “generalized semiperiodic sequence”, if for any  > 0, there exist n, k0 ∈ N such that |xk − xk+sn | < , k ≥ k0 , s = 0, 1, 2, . . . .

44

4 Ultrametric Summability Theory

Let Q denote the set of all generalized semiperiodic sequences. One can prove that Q is a closed linear subspace of ∞ . Further, whatever be K , Q⊆

∞ 

 r .

r =1

When K is a complete, non-trivially valued, ultrametric field, Q=

∞ 

 r .

r =1

Whatever be K , we shall define for α > 0,  α = x = {xk }, xk ∈ K , k = 0, 1, 2, . . . ,

∞ 

 α

|xk | < ∞ .

k=0

One of the interesting results of ultrametric analysis is the characterization of infinite matrices in (α , α ), α > 0 as given by the following theorem. Theorem 4.8 (see [17], Theorem 2.1) If K is a complete, non-trivially valued, ultrametric field and A = (ank ), ank ∈ K , n, k = 0, 1, 2, . . . , then A ∈ (α , α ), α > 0, if and only if ∞  |ank |α < ∞. (4.18) sup k≥0 n=0

Proof Since | · | is a non-archimedean valuation, we first note that   α |a| − |b|α  ≤ |a + b|α ≤ |a|α + |b|α , α > 0. Sufficiency. If x = {xk } ∈ α ,

∞ 

ank xk converges, n = 0, 1, 2, . . . , since xk → 0,

k=0

k → ∞ and sup |ank | < ∞ by (4.18). Also, n,k ∞ 

α

|(Ax)n | ≤

n=0

∞  ∞ 

∞  k=0

< ∞, so that {(Ax)n } ∈ α .

|ank |α |xk |α

n=0 k=0

≤

(4.19)

|xk |

 α

sup

∞ 

k≥0 n=0

 |ank |

α

4.2 Steinhaus-Type Theorems

45

Necessity. Suppose A ∈ (α , α ). We first note that sup |ank |α = Bn < ∞, k≥0

n = 0, 1, 2, . . . . For, if for some m, sup |amk |α = Bm = ∞, then, we can choose k≥0

a strictly increasing sequence {k(i)} of positive integers such that |am,k(i) |α > i 2 , i = 1, 2, . . . . Define the sequence {xk }, where  xk = Then {xk } ∈ α , for,

1 am,k ,

k = k(i)

0,

k = k(i)

∞ 

|xk |α =

k=0

∞ 

, i = 1, 2, . . . .

|xk(i) |α
3. Then choose a positive integer n(1) such that ∞ 

|an,k(1) |α < 1,

n=n(1)+1

so that

n(1) 

|an,k(1) |α > 2.

n=0

More generally, given the positive integers k( j), n( j), j ≤ m − 1, choose positive integers k(m) and n(m) such that k(m) > k(m − 1), n(m) > n(m − 1), n(m−1) 

∞ 

n=n(m−2)+1 k=k(m)

Bn k −2 < 1,

46

4 Ultrametric Summability Theory

μk(m) > 2



n(m−1) 

−α

Bn + ρ

m

m−1 

2+

2

n=0

and

 i

−2

μk(i)

i=1

∞ 

|an,k(m) |α
2

n(m−1) 

−α

Bn + ρ

m

2+

2

m−1 

i

−2



μk(i)

i=1

n(m−1)  n=0

|an,k(m) |α

n=n(m)+1



n=0

−

∞ 

|an,k(m) |α −

Bn −



= ρ−α m 2 2 +

n(m−1) 

n=0 m−1  −2

i

Bn 

μk(i) .

i=1

For every i = 1, 2, . . . , there exists a non-negative integer λ(i) such that 2

ρλ(i)+1 ≤ i − α < ρλ(i) . Define the sequence x = {xk } as follows:  xk =

{xk } ∈ α , for,

∞ 

|xk |α =

k=0

π λ(i)+1 , k = k(i) , i = 1, 2, . . . . 0, k = k(i)

∞ 

|xk(i) |α ≤

i=1

∞  1 < ∞. i2 i=1

However, using (4.19), n(m)  n=n(m−1)+1

|(Ax)n |α ≥ 1 − 2 − 3 ,

4.2 Steinhaus-Type Theorems

47

where n(m) 

1 =

|an,k(m) |α |xk(m) |α ,

n=n(m−1)+1 n(m) 

2 =

m−1 

|an,k(i) |α |xk(i) |α ,

n=n(m−1)+1 i=1 ∞ 

n(m) 

3 =

|an,k(i) |α |xk(i) |α .

n=n(m−1)+1 i=m+1

Now, n(m) 

1 =

|an,k(m) |α ρ(λ(m)+1)α

n=n(m−1)+1 n(m) 

≥ ρα

|an,k(m) |α m −2

n=n(m−1)+1

>2+

m−1 

i −2 μk(i) ;

i=1

n(m) 

2 =

m−1 

|an,k(i) |α ρ(λ(i)+1)α

n=n(m−1)+1 i=1

≤

m−1 

i −2

i=1

≤

m−1 

n(m) 

|an,k(i) |α

n=n(m−1)+1

i −2 μk(i) ;

i=1

n(m) 

3 =

∞ 

|an,k(i) |α ρ(λ(i)+1)α

n=n(m−1)+1 i=m+1 n(m) 

≤

∞ 

n=n(m−1)+1 i=k(m+1)

< 1.

Bn i −2

48

4 Ultrametric Summability Theory

Using the above, we have n(m) 

|(Ax)n |α > 1, m = 2, 3, . . . .

n=n(m−1)+1

This shows that {(Ax)n } ∈ / α , while {xk } ∈ α , a contradiction. Thus condition (4.18) is also necessary, completing the proof of the theorem.  Because of the fact that there is, as such, no classical analogue for the above result, Theorem 4.8 is interesting. When K = R or C, a complete characterization of the class (α , β ) of infinite matrices, α, β ≥ 2, does not seem to be available in the literature. Necessary and sufficient conditions for A ∈ (1 , 1 ) when K = R or C are due to Mears [18] (for alternate proofs, see, for instance, Fridy [19]). From the characterization mentioned in Theorem 4.8, it is deduced that the Cauchy product of two sequences in α , α > 0, is again in α , a result which fails to hold for α > 1, when K = R or C. ∞ ∞   We write A = (ank ) ∈ (α , α ; P) if A ∈ (α , α ) and (Ax)n = xk , n=0

k=0

x = {xk } ∈ α ; A ∈ (α , α ; P) if A ∈ (α , α ; P) and (4.16) holds. When K = R or C, Fridy [20] proved a Steinhaus-type result in the form (1 , 1 ; P) ∩ (α , 1 ) = φ, α > 1. This result, as such, fails to hold in the ultrametric set up (see [17], Remark 4.1). However, in the ultrametric set up, Natarajan [17] proved that (α , α ; P) ∩ (β , α ) = φ, β > α. In the above context, it is worth noting that following the proof of Theorem 4.1 of [17], we can show that given any matrix A ∈ (α , α ; P), there exists a sequence of 0’s and 1’s whose A-transform is not in α . This is analogous to Schur’s version of the well-known Steinhaus theorem for regular matrices (see [17]).

4.3 Core of a Sequence and Knopp’s Core Theorem The core of a complex sequence is defined as follows: Definition 4.8 If x = {xk } is a complex sequence, we denote by K n (x), n = 0, 1, 2, . . . , the smallest closed convex set containing xn , xn+1 , . . . .

4.3 Core of a Sequence and Knopp’s Core Theorem ∞

K (x) =

49

K n (x)

n=0

is defined as the core of x. It is known [21] that if x = {xk } is bounded,

K (x) =

C

z∈C

lim |z−xn | (z),

n→∞

where Cr (z) is the closed ball centred at z and radius r . Sherbakoff [21] generalized the notion of the core of a bounded complex sequence by introducing the idea of the generalized α-core K (α) (x) of a bounded complex sequence as K

(α)

(x) =

z∈C

Cα lim

|z−xn | (z),

n→∞

α ≥ 1.

(4.20)

When α = 1, K (α) (x) reduces to the usual core K (x). When K = C, Sherbakoff [21] showed that under the condition

∞ 

lim

n→∞

 |ank | = α, α ≥ 1,

(4.21)

k=0

K (A(x)) ⊆ K

(α)

(x).

Natarajan [22] improved the result of Sherbakoff by showing that his result works with the less stringent precise condition

lim

n→∞

∞ 

 |ank | ≤ α, α ≥ 1,

(4.22)

k=0

(4.22) being also necessary besides the regularity of A for K (A(x)) ⊆ K

(α)

(x)

for any bounded complex sequence x. This result for the case α = 1 yields a simple proof of Knopp’s core theorem (see, for instance, [23]). Natarajan’s theorem is as follows. Theorem 4.9 ([22], Theorem 2.1) When K = R or C, A = (ank ) is such that K (A(x)) ⊆ K

(α)

(x), α ≥ 1,

for any bounded sequence x if and only if A is regular and satisfies (4.22),

50

4 Ultrametric Summability Theory

i.e. lim

∞ 

n→∞

 |ank | ≤ α.

k=0

Proof Let x = {xn } be a bounded sequence. If y ∈ K (A(x)), for any z ∈ K , |y − z| ≤ lim |z − (Ax)n |. n→∞

If A is a regular matrix satisfying (4.22), then |y − z| ≤ lim |z − (Ax)n | n→∞ ∞      = lim  ank (z − xk ) n→∞   k=0

≤ α lim |z − xk |, k→∞

i.e. y ∈ C

α lim |z − xk |

(z) for any z ∈ K ,

k→∞

which implies that K (A(x)) ⊆ K (α) (x). Conversely, let K (A(x)) ⊆ K (α) (x). Then it is clear that A is regular by considering convergent sequences for which K

(α)

(x) =



 lim xn .

n→∞

It remains to prove (4.22). Let, if possible,

∞ 

lim

n→∞

Then lim

n→∞

∞ 

 |ank | > α.

k=0

 |ank | = α + h, for some h > 0.

k=0

Using the hypothesis and the fact that A is regular, we can now choose two strictly increasing sequences {n(i)} and {k(n(i))} of positive integers such that k(n(i−1))  k=0 k(n(i))  k=k(n(i−1))+1

|an(i),k |
α +

h 4

4.3 Core of a Sequence and Knopp’s Core Theorem

and

∞ 

51

|an(i),k |
α+

∞ is a bounded sequence. It has a convergent By the regularity of A, {(Ax)n(i) }i=1 subsequence whose limit cannot be in Cα (0) in view of the above inequality, i.e. |(Ax)n(i) | > α, i = 1, 2, . . . . Using (4.20), we have K (α) (x) ⊆ Cα (0) for the sequence x chosen above. This leads to a contradiction of the fact that K (A(x)) ⊆  K (α) (x), completing the proof.

In [22], Natarajan also proved the analogue of Theorem 4.9 when K is a complete, locally compact, non-trivially valued, ultrametric field. Definition 4.9 If x = {xn }, x ∈ K , n = 0, 1, 2, . . . , we denote by K n (x), n = 0, 1, 2, . . . the smallest closed K -convex set containing xn , xn+1 , . . . and call K (x) =

∞

K n (x)

n=0

the core of x. K

(α) (x)

is defined as in (4.20) in this case too.

Theorem 4.10 ([22], Theorem 3.1) K is a complete, locally compact, non-trivially valued, ultrametric field. A = (ank ), ank ∈ K , n, k = 0, 1, 2, . . . is such that K (A(x)) ⊆ K (α) (x) for any sequence (bounded or unbounded) if and only if A is regular and satisfies 

lim

n→∞

sup |ank | ≤ α, α > 0.

(4.23)

k≥0

Note that the above theorem, for α = 1, yields a simple proof of the analogue of Knopp’s core theorem in the ultrametric set up.

52

4 Ultrametric Summability Theory

4.4 A Characterization of Regular and Schur Matrices When K = R or C, Maddox [24] obtained a characterization of Schur matrices in terms of the existence of a bounded divergent sequence all of whose subsequences are summable by the matrix. This characterization included the earlier one of Buck’s [25], viz., a sequence {xk }, summable by a regular matrix A, is convergent if and only if A sums each one of its subsequences. Fridy [26] showed that we can replace “subsequence” in Buck’s result by “rearrangement” to obtain that a sequence {xk }, summable by a regular matrix A, is convergent if and only if A sums each one of its rearrangements. In this connection, it is to be reiterated that it is not as if every result in ultrametric analysis has a proof analogous to its classical counterpart or even a simpler proof. The absence of analogues for the signum function, upper limit and lower limit of real number sequences, etc. forces on us to search for alternate devices. These devices provide an entirely different proof of even an exact analogue of a classical theorem. For instance, the proof of the following theorem, amply, illustrates our claim. Theorem 4.11 Let K be a complete, non-trivially valued, ultrametric field. Then A = (ank ), ank ∈ K , n, k = 0, 1, 2, . . . is a Schur matrix if and only if there exists a bounded divergent sequence x = {xk }, xk ∈ K , k = 0, 1, 2, . . . , each one of whose subsequences is summable A. Proof If A is a Schur matrix, the assertion of the theorem is a consequence of the definition of such a matrix. Conversely, let x = {xk } be a divergent sequence, each one of whose subsequences is summable A. For each p = 0, 1, 2, . . . , we can choose (1) (2) (1) (2) two subsequences of x (say) {xk }, {xk } such that if yk = xk − xk , yk = 0, k = p while y p = 0. Such choice is possible since x diverges and therefore has two unequal entries after any stage for k. The sequence {yk } is summable A. So it follows that lim anp exists, p = 0, 1, 2, . . . . Next, we show that lim anp = 0, n→∞

p→∞

n = 0, 1, 2, . . . . Otherwise, there exists  > 0 and a non-negative integer m such that |am, p(i) | >  , i = 1, 2, . . . , where { p(i)} is an increasing sequence of positive integers. Since x diverges, it is not a null sequence and so there exists  > 0 and an increasing sequence {( j)} of positive integers such that |x( j) | >  , j = 1, 2, . . . . Now, |am, p(i) x( p(i)) | > 2 , i = 1, 2, . . . , where  = min( ,  ). This means that the A-transform of the subsequence {x( j) } does not exist, a contradiction. Thus lim anp = 0, n = 0, 1, 2, . . . . Since x diverges, p→∞

4.4 A Characterization of Regular and Schur Matrices

53

|xk+1 − xk | → 0, k → ∞ so that there exists  > 0 and an increasing sequence {k( j)} of positive integers such that |xk( j)+1 − xk( j) | >  , j = 1, 2, . . . .

(4.24)

We may assume that k( j + 1) − k( j) > 1, j = 1, 2, . . . . We claim that if A is not a Schur matrix, then we should have an  > 0 and two increasing sequences {n(i)} and { p(n(i))} of positive integers with ⎧ 2 ⎪ ⎪ (i) sup , |a − a | < ⎪ n(i)+1, p n(i), p ⎪ ⎪ 4M ⎨ 0≤ p≤ p(n(i−1)) (ii) |an(i)+1, p(n(i)) − an(i), p(n(i)) | > , ⎪ ⎪ ⎪ 2 ⎪ ⎪ , |an(i)+1, p − an(i), p | < sup ⎩(iii) 4M p≥ p(n(i+1))

(4.25)

where M = sup |xk |. Before proving the claim, we show that if A is not a Schur k≥0

matrix, then x is necessarily bounded under the hypothesis of the theorem. Suppose x is unbounded. We consider two cases: Case (i) If A is such that ank = 0 for some n and k = k(i), i = 1, 2, . . . , choose a subsequence {xα(k) } of x, which is unbounded, such that |ank xα(k) | > 1, k = k(i), i = 1, 2, . . . . Hence, {xα(k) } is not summable A, a contradiction. Case (ii) If now, ank = 0, k > k(n), n = 0, 1, 2, . . . , A not being a Schur matrix, we can find two strictly increasing sequences of positive integers {n( j)} and {k( j)} such that an( j),k( j) is the last non-zero term in the n( j)th row. x, being unbounded, we can choose a subsequence {xα(k) } of x such that |An( j) ({xα(k) })| > j, j = 1, 2, . . . . To do this, choose xα(i) , i = 1, 2, . . . , k(1) such that |xα(1) | >

1 |an(1),1 |

, if an(1),1 = 0,

while xα(1) is chosen as a suitable xk , otherwise. Having chosen xα(1) , choose xα(2) , α(2) > α(1) such that |xα(2) | >

1 |an(1),2 |



 1 + |an(1),1 xα(1) | , if an(1),2 = 0.

54

4 Ultrametric Summability Theory

Otherwise, choose xα(2) such that α(2) > α(1). Now, |an(1),1 xα(1) + an(1),2 xα(2) | ≥ |an(1),2 xα(2) | − |an(1),1 xα(1) |, if an(1),2 = 0; = |an(1),1 xα(1) |, if an(1),2 = 0. Thus |an(1),1 xα(1) + an(1),2 xα(2) | = 0, if an(1),1 = 0 = an(1),2 ; > 1, if one of an(1),1 , an(1),2 = 0. We choose xα(i) , i = 1, 2, . . . , k(1) as above. Then      k(1)   an(1),k xα(k)  > 1.  k=1  If now, k(1) 

an(2),k xα(k) = α,

k=1

choose similarly xα(k) , k(1) < k ≤ k(2) with α(k(1)) < α(k(1) + 1) < · · · < α(k(2)) and       k(2)    a x n(2),k α(k)  > 2 + |α|.  k=k(1)+1  Then,               k(2)   k(2)   k(1)       an(2),k xα(k)  ≥  an(2),k xα(k)  −  an(2),k xα(k)   k=1  k=k(1)+1   k=1  > 2 + |α| − |α| = 2. Inductively, we can therefore choose xα(k) , k = 1, 2, . . . with     k( j)   |An( j) ({xα(k) })| =  an( j),k xα(k)  k=1  > j, j = 1, 2, . . . .

4.4 A Characterization of Regular and Schur Matrices

55

It now follows that {xα(k) } is not summable A, a contradiction. Thus in both cases it turns out that x is bounded, if A were not to be a Schur matrix. Next, we observe that since A is not a Schur matrix (see [9]), there exist  > 0 and an increasing sequence {n(i)} of positive integers such that sup |an(i)+1, p − an(i), p | > , i = 1, 2, . . . . p≥0

Hence, there exists p(n(i)) such that |an(i)+1, p(n(i)) − an(i), p(n(i)) | > , i = 1, 2, . . . .

(4.26)

Suppose { p(n(i))} is bounded, then there are only a finite number of distinct entries in that sequence. Consequently, there exists p = p(n(m)) which occurs in the sequence { p(n(i))} infinite number of times. For this p, (4.26) will then contradict the existence of lim anp , p = 0, 1, 2, . . . established earlier. Having chosen {n(i)} and { p(n(i))} n→∞

to satisfy (4.26), it is clear that by choosing a subsequence of {n(i)}, if necessary, we can suppose that (4.25) holds. Consider now the sequence {y p } defined as follows:  yp =

xk( p)+1 , p = p(n(i)), p = p(n(i)), i = 1, 2, . . . , xk( p) ,

where the sequence {k( j)} is already chosen as in (4.24). Thus      ∞  {an(i)+1, p − an(i), p }(y p − xk( p) )     p=0 > 2 − =

2 2 M− M, using (4.25) 4M 4M

2 , i = 1, 2, . . . , 2

where we can suppose that  ≥  . This contradicts the fact that ⎧ ∞ ⎨ ⎩

p=0

(an+1, p − an, p )(y p − xk( p) )

⎫∞ ⎬ ⎭ n=0

converges. This proves that A is a Schur matrix. The proof of the theorem is now complete.  The analogue of Buck’s result in the ultrametric case follows as a corollary of Theorem 4.11, viz., a sequence x = {xk }, xk ∈ K , k = 0, 1, 2, . . . , summable by a regular matrix A, is convergent if and only if every one of its subsequences is summable A.

56

4 Ultrametric Summability Theory

The following analogue in the ultrametric case can also be established by means of a “sliding hump method” as described by Fridy [26]. Theorem 4.12 ([28], Theorem 3) A sequence x = {xk }, xk ∈ K , k = 0, 1, 2, . . . , summable by a regular matrix A, is convergent if and only if every one of its rearrangements is summable A. When K = R or C, in the context of rearrangements of a bounded sequence, Fridy [26] proved the following result. Theorem 4.13 A null sequence x = {xk } is in 1 if and only if there exists a matrix A = (ank ) ∈ (1 , 1 ; P) which transforms all rearrangements of {xk } into sequences in 1 . We can combine Theorem 4.13 and a result of Keagy [27] to state Theorem 4.14 When K = R or C, a null sequence x = {xk } is in 1 if and only if there exists a matrix A = (ank ) ∈ (1 , 1 ; P) which transforms every subsequence or rearrangement of x into a sequence in 1 . In the ultrametric setting, Theorem 4.14 fails (see [28]). However, when K is a complete, non-trivially valued, ultrametric field, the following theorem, due to Natarajan ([28], Theorem 6), is an attempt to salvage Fridy’s result in a general form. Theorem 4.15 A null sequence x = {xk } is in α if and only if there exists a matrix A = (ank ) ∈ (α , α ; P) such that A transforms every rearrangement of x into a sequence in α , where we recall that A ∈ (α , α ; P) if A ∈ (α , α ; P) and (4.16) holds. Proof We recall (see [17]) that A ∈ (α , α ; P) if and only if (4.18) holds and ∞ 

ank = 1, k = 0, 1, 2, . . . .

n=0

Leaving out the trivial part of the theorem, suppose x ∈ c0 −α and A ∈ (α , α ; P) transforms every rearrangement of x into a sequence in α . Choose k(1) = 1 and a positive integer n(1) such that ∞ 

|an,1 |α < 2−1 ,

n=n(1)+1

so that n(1)  n=0

|an,1 |α =

∞ 

|an,1 |α −

n=0

1 1 ≥1− = . 2 2

∞  n=n(1)+1

|an,1 |α

4.4 A Characterization of Regular and Schur Matrices

57

Having defined k( j), n( j), j ≤ m −1, choose a positive integer k(m) > k(m −1)+1 such that n(m−1)  1 |an,k(m) |α < 2−m , |xk(m) | < 2 mα n=0 and then choose a positive integer n(m) > n(m − 1) such that n(m)  n=n(m−1)+1 ∞ 

|an,k(m) |α ≥

1 , 2

|an,k(m) |α < 2−m .

n=n(m)+1

Let U consist of all k(m), m = 1, 2, . . . . Let u m = xk(m) and V be the set of all non-negative integers which are not in U . Let v = {xk }k∈V . Let y be a rearrangement of x, where yk = vm , k = k(m); = u k , otherwise. Defining n(0) = 0, we have ⎧ α  n(M) n(m) M      ⎪ ⎪   α ⎪ ⎪ |(Ay) | ≥ a y   n nk k ⎪ ⎪   ⎪ ⎪ n=0 m=1 k∈U n=n(m−1)+1 ⎪ α   ⎪ ⎪   ⎪ ⎪   ⎪ ⎪ − ank yk  , using (4.19) ⎪ ⎪   ⎪ ⎪ k∈V ⎪ α   ⎪ n(m) M ⎪      ⎪ ⎪   α ⎪ ⎪ ≥ ank yk  − |ank yk |  ⎪ ⎪   ⎪ ⎪ m=1 n=n(m−1)+1 ⎧ k∈U k∈V ⎪ ⎪ ⎪ ⎪ ⎪ n(m) M ⎪ ⎨    ⎪ ⎨ ≥ |an,k(i) vi |α |an,k(m) vm |α − ⎪ ⎩ i=1 m=1 n=n(m−1)+1 ⎪ ⎪ i =m ⎪ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪  ⎪ α ⎪ ⎪ − |ank yk | ⎪ ⎪ ⎪ ⎪ ⎭ ⎪ k∈V ⎪ ⎪ ⎪ ⎪ ⎪ n(m) ∞ M M ⎪    1 ⎪ α ⎪ ⎪ ≥ |vm | − |an,k(i) vi |α ⎪ ⎪ 2 ⎪ ⎪ m=1 m=1 n=n(m−1)+1 i=1 ⎪ ⎪ ⎪ n(M) ⎪   ⎪ ⎪ ⎪ − |ank yk |α . ⎪ ⎩ n=0 k∈V

(4.27)

58

4 Ultrametric Summability Theory

However, M 

∞ 

n(m) 

m=1 n=n(m−1)+1

α

|an,k(i) vi | < ||x||

α

∞ 

2−m+1 ,

(4.28)

m=1

i=1 i =m

where ||x|| = sup |xk |, for, k≥0 ∞ 

n(m) 



|an,k(i) |α

m=1 n=n(m−1)+1 im

Also,

|an,k(i) |α
0 such that

|ak | < M, |bk | < M, k = 0, 1, 2, . . . . Given  > 0, choose a positive integer N1 such that |ak | < M , for all k > N1 . Since lim bk−r = 0, r = 0, 1, 2, . . . , we can k→∞  . Thus choose a positive integer N2 > N1 such that for k > N2 , sup |bk−r | < M 0≤r ≤N1 for k > N2 ,   k     |ck | =  bk−r ar    r =0

60

4 Ultrametric Summability Theory

 N k 1      = bk−r ar + bk−r ar    r =0 r =N +1  ≤ max max |bk−r ||ar |, max |bk−r ||ar | 0≤r ≤N1 N1 +1≤r ≤k       M, M < max M M = . Consequently, lim ck = 0 and so k→∞

case

ck converges. It is easy to check that in this

k=0 ∞  k=0

completing the proof.

∞ 

ck =

∞  k=0

ak

 ∞ 

 bk ,

k=0



References 1. Andree, R.V., Petersen, G.M.: Matrix methods of summation, regular for p-adic valuations. Proc. Amer. Math. Soc. 7, 250–253 (1956) 2. Roberts, J.B.: Matrix summability in F-fields. Proc. Amer. Math. Soc. 8, 541–543 (1957) 3. Monna, A.F.: Sur le théorème de Banach-Steinhaus. Indag. Math. 25, 121–131 (1963) 4. Rangachari, M.S., Srinivasan, V.K.: Matrix transformations in non-archimedean fields. Indag. Math. 26, 422–429 (1964) 5. Srinivasan, V.K.: On certain summation processes in the p-adic field. Indag. Math. 27, 319–325 (1965) 6. Somasundaram, D.: Some properties of T -matrices over non-archimedean fields. Publ. Math. Debrecen 21, 171–177 (1974) 7. Somasundaram, D.: On a theorem of Brudno over non-archimedean fields. Bull. Austral. Math. Soc. 23, 191–194 (1981) 8. Natarajan, P.N.: Criterion for regular matrices in non-archimedean fields. J. Ramanujan Math. Soc. 6, 185–195 (1991) 9. Natarajan, P.N.: The Steinhaus theorem for Toeplitz matrices in non-archimedean fields. Comment. Math. Prace Mat. 20, 417–422 (1978) 10. Natarajan, P.N.: A Steinhaus type theorem. Proc. Amer. Math. Soc. 99, 559–562 (1987) 11. Natarajan, P.N.: Some Steinhaus type theorems over valued fields. Ann. Math. Blaise Pascal 3, 183–188 (1996) 12. Natarajan, P.N.: Some more Steinhaus type theorems over valued fields. Ann. Math. Blaise Pascal 6, 47–54 (1999) 13. Natarajan, P.N.: Some more Steinhaus type theorems over valued fields II. Comm. Math. Analysis 5, 1–7 (2008) 14. Natarajan, P.N.: On certain spaces containing the space of Cauchy sequences. J. Orissa Math. Soc. 9, 1–9 (1990) 15. Sember, J.J., Freedman, A.R.: On summing sequences of 0’s and 1’s. Rocky Mountain J. Math. 11, 419–425 (1981)

References

61

16. Natarajan, P.N.: On sequences of zeros and ones in non-archimedean analysis—A further study. Afr. Diaspora J. Math. 10, 49–54 (2010) 17. Natarajan, P.N.: Characterization of a class of infinite matrices with applications. Bull. Austral. Math. Soc. 34, 161–175 (1986) 18. Mears, F.M.: Absolute regularity and the Nörlund mean. Ann. Math. 38, 594–601 (1937) 19. Fridy, J.A.: A note on absolute summability. Proc. Amer. Math. Soc. 20, 285–286 (1969) 20. Fridy, J.A.: Properties of absolute summability matrices. Proc. Amer. Math. Soc. 24, 583–585 (1970) 21. Sherbakoff, A.A.: On cores of complex sequences and their regular transform (Russian). Mat. Zametki 22, 815–828 (1977) 22. Natarajan, P.N.: On the core of a sequence over valued fields. J. Indian. Math. Soc. 55, 189–198 (1990) 23. Cooke, R.G.: Infinite Matrices and Sequence Spaces. Macmillan, London (1950) 24. Maddox, I.J.: A Tauberian theorem for subsequences. Bull. London Math. Soc. 2, 63–65 (1970) 25. Buck, R.C.: A note on subsequences. Bull. Amer. Math. Soc. 49, 898–899 (1943) 26. Fridy, J.A.: Summability of rearrangements of sequences. Math. Z. 143, 187–192 (1975) 27. Keagy, T.A.: Matrix transformations and absolute summability. Pacific J. Math. 63, 411–415 (1976) 28. Natarajan, P.N.: Characterization of regular and Schur matrices over non-archimedean fields. Proc. Kon. Ned. Akad. Wetens Ser. A 90, 423–430 (1987) 29. Natarajan, P.N.: Multiplication of series with terms in a non-archimedean field. Simon Stevin 52, 157–160 (1978)

Chapter 5

The Nörlund and The Weighted Mean Methods

Abstract In this chapter, we introduce the Nörlund and the Weighted Mean methods in the ultrametric set-up and their properties are elaborately discussed. We also show that the Mazur–Orlicz theorem and Brudno’s theorem fail to hold in the ultrametric case. Keywords The Nörlund method · The weighted mean method theorem · Brudno’s theorem · Translative (matrix)

· Mazur–Orlicz

5.1 The Nörlund Method In an attempt to introduce special summability methods in ultrametric analysis, Srinivasan [1] defined the Nörlund method of summability, i.e. the (N , pn ) method in K as follows: The (N , pn ) method is defined by the infinite matrix (ank ) where pn−k , k ≤ n; Pn = 0, k > n,

ank =

where | p0 | > | p j |, j = 1, 2, . . . , and Pn = that p0 = 0. Srinivasan noted that if

n 

pk , n = 0, 1, 2, . . . . It is to be noted

k=0

lim pn = 0,

n→∞

(5.1)

then the (N , pn ) method is regular. An example of a regular (N , pn ) method, in the p-adic field Q p , for a prime p, is given by pn = p n , n = 0, 1, 2, . . . . We now observe that (5.1) is also necessary for the (N , pn ) method to be regular. Thus we have

© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_5

63

64

5 The Nörlund and The Weighted Mean Methods

Theorem 5.1 ([2], Theorem 1) The method (N , pn ) is regular if and only if (5.1) holds. Proof If (5.1) holds, the method is regular (see [1], p. 323). Conversely, if the method is regular, then lim an0 = 0. But an0 = Ppnn so that |an0 | = || ppn0 || (since |Pn | = | p0 | n→∞



using Theorem 1.1), which implies that (5.1) holds. We now prove a limitation theorem for the (N , pn ) method. Theorem 5.2 ([2], Lemma 3) If {sn } is (N , pn ) summable, then {sn } is bounded. Proof Let {tn } be the (N , pn ) transform of {sk }, i.e. tn =

p0 sn + p1 sn−1 + · · · + pn s0 , n = 0, 1, 2, . . . . Pn

Since {tn } converges, let M = sup |tn |. Now, n≥0

t0 =

p 0 s0 p 0 s0 = = s0 , P0 p0

and so |s0 | ≤ M. Let |sk | ≤ M, k = 0, 1, . . . , (n − 1). Since Pn tn − p1 sn−1 − · · · − pn s0 , p0 | p0 |M = M, |sn | ≤ | p0 | sn =



completing the proof.

Definition 5.1 We write (N , pn ) ⊆ (N , qn ) if whenever {sn } is summable (N , pn ) to s then it is also summable (N , qn ) to s. We say that (N , pn ) and (N , qn ) are equivalent if (N , pn ) ⊆ (N , qn ) and vice versa. The following two results are essentially different from their counterparts in the classical case (see [3], p. 67, Theorem 2.1). It is worth noting in this context that these results are also instances in which absolute convergence in classical analysis is replaced by ordinary convergence in ultrametric analysis. Theorem 5.3 ([2], Theorem 3) Let (N , pn ), (N , qn ) be regular methods. Then ∞  q(x) = k(x) = (N , pn ) ⊆ (N , qn ) if and only if lim kn = 0, where kn x n , n→∞ p(x) p(x) =

∞  n=0

pn x n , q(x) =

∞  n=0

n=0

qn x n , and {kn } is defined by kn p0 + kn−1 p1 + · · · +

k0 pn = qn , n = 0, 1, 2, . . . by recursion.

5.1 The Nörlund Method

65

Proof Let (N , pn ) ⊆ (N , qn ). Let {tn }, {τn } be, respectively, the (N , pn ), (N , qn ) transforms of the sequence {sn }. If |x| < 1, ∞ 

Q n τn x = n

n=0

=

∞ 

Qn

 n  qn−k

n=0 ∞ 

k=0

Qn

 sk x n

(q0 sn + q1 sn−1 + · · · + qn s0 )x n

n=0

= q(x)s(x). Similarly,

∞ 

Pn tn x n = p(x)s(x); if |x| < 1.

n=0

Now, k(x) p(x) = q(x), k(x) p(x)s(x) = q(x)s(x),  ∞ ∞   n Pn tn x Q n τn x n . = i.e. k(x) n=0

n=0

Thus Q n τn = Pn tn k0 + Pn−1 tn−1 k1 + · · · + P0 t0 kn , i.e. τn =

∞ 

an j t j ,

j=0

k

where an j =

n− j P j

Qn

0,

, j ≤ n; j > n.

By hypothesis, (an j ) is a regular matrix and so lim an0 = 0. Thus n→∞

0 = lim |an0 | = lim n→∞

n→∞

|kn || p0 | , |q0 |

which, in turn, implies that kn → 0, n → ∞. Conversely, let kn → 0, n → ∞. For j = 0, 1, 2, . . . , |kn− j ||P j | n→∞ |Q n |

lim |an j | = lim

n→∞

66

5 The Nörlund and The Weighted Mean Methods

|kn− j || p0 | n→∞ |q0 | = 0.

= lim

|an j | =

|kn− j || p0 | |q0 |

≤ M , M  =

L| p0 | |q0 | ,

L = sup |kn |, n, j = 0, 1, 2, . . . so that n≥0

sup |an j | < ∞. Finally, n, j ∞ 

an j =

j=0

n 

an j =

j=0

=

so that lim

n→∞

∞ 

k0 Pn + k1 Pn−1 + · · · + kn P0 Qn Qn = 1, n = 0, 1, 2, . . . , Qn

an j = 1. Thus (an j ) is regular and consequently (N , pn ) ⊆ (N , qn ),

j=0



completing the proof.

Theorem 5.4 ([2], Theorem 4) The regular methods (N , pn ), (N , qn ) are equivalent if and only if lim kn = 0 = lim h n , where {kn } is defined as in Theorem 5.3 and n→∞

{h n } is defined by

n→∞

∞

 p(x) = h(x) = hn x n . q(x) n=0

The following results too are different from their classical counterparts just like Theorems 5.3 and 5.4. Theorem 5.5 ([4], Theorem 1) If ak = o(1), k → ∞, i.e. lim ak = 0, {bk } k→∞

is summable by a regular (N , pn ) method, then {ck } is (N , pn ) summable, where k  ai bk−i , k = 0, 1, 2, . . . . ck = i=0

Theorem 5.6 ([4], Theorem 2) If by a regular (N , pn ) method, then

∞ 

ak converges to A,

k=0 ∞ 

∞ 

bk is summable to B

k=0

ck is (N , pn ) summable to AB, where ck is

k=0

defined as in Theorem 5.5. In the classical case, the following result, due to Mears, is known ([5], Theorem 1). Theorem 5.7 If the real series

∞  k=0

ak ,

∞  k=0

bk are such that the former is summable

by regular (N , pn ) method to A, the latter is summable by a regular (N , qn ) method

5.1 The Nörlund Method

67

to B, one of the summabilities being absolute, i.e.

∞ 

|tn+1 − tn | < ∞, where {tn }

n=0

is the (N , pn ) transform of {Ak } or the (N , qn ) transform of {Bk }, Ak =

k 

aj,

j=0

Bk =

k 

b j , k = 0, 1, 2, . . . , then

∞ 

j=0

where ck =

ck is summable by the (N , rn ) method to AB,

k=0 k 

a j bk− j , rk =

j=0

k 

p j qk− j , k = 0, 1, 2, . . . .

j=0

In the ultrametric case, we have the following analogue of Theorem 5.7, which indicates significant departure in the sense that there is no need to bring in absolute summability (see [5], Theorem 2). Theorem 5.8 If

∞ 

ak is summable by a regular (N , pn ) method to A,

k=0

summable by a regular (N , qn ) method to B, then

∞ 

∞ 

bk is

k=0

ck is summable by the regular

k=0

(N , rn ) method to AB, where ck =

k 

a j bk− j , rk =

j=0

k 

p j qk− j , k = 0, 1, 2, . . . .

j=0

In the context of Theorem 5.8, we note that the hypothesis that “(N , pn ), (N , qn ) ∞  are regular for ck to be summable (N , rn ) to AB” cannot be dropped (see [6], k=0

pp. 50–52). However, we have the following result in which (N , pn ), (N , qn ) need not be regular. Theorem 5.9 ([6], Theorem 1) Let (N , qn ) summable to 0. Then

∞ 

∞ 

ak be (N , pn ) summable to 0 and

k=0

∞ 

bk be

k=0

ck is (N , rn ) summable to 0.

k=0

There is a natural isomorphism ϕ from c0 onto the ultrametric algebra A of power series converging in the closed unit disc B(0, 1) = {x ∈ K /|x| ≤ 1}, defined by ϕ({an }) =

∞  n=0

an x n

68

5 The Nörlund and The Weighted Mean Methods

(For instance, see [7]). Both c0 and A are ultrametric Banach spaces with respect ∞  to the sup norm, i.e. ||{an }|| = sup |an |. If f = ϕ({an }), then f (1) = an . n≥0

n=0

Given sequences {an }, {bn } ∈ c0 , we know that their Cauchy product {cn } ∈ c0 , by Theorem 4.16. Putting f = ϕ({an }), g = ϕ({bn }), h = ϕ({cn }), we have h(x) = f (x)g(x). Let M = (m nk ) ∈ (c0 , c0 ), i.e. sup |m nk | < ∞ and lim m nk = 0, n,k

n→∞

k = 0, 1, 2, . . . . Let {an } ∈ c0 and {αn } = M({an }), i.e. {αn } is the M-transform of {an }. We then know that {αn } ∈ c0 . Thus M defines a linear operator M˜ of A, defined by ∞  ∞   n M˜ an x αn x n . = n=0

n=0

This applies to regular Nörlund matrices in particular. In view of the above remarks, Theorem 5.9 brings out the following connection between regular Nörlund matrices and analytic functions in the closed unit disc. Theorem 5.10 (see [6], Theorem 2) Let M, N be regular Nörlund matrices and f, g ∈ A. If M˜ f (1) = N˜ g(1) = 0, then we have M˜ ◦ N˜ ( f g)(1) = 0. Definition 5.2 The convergence field (or summability field) c A of any infinite matrix A = (ank ) is defined as c A = {x = {xk } : Ax = {(Ax)n } ∈ c}. In [8], we prove two interesting results regarding the convergence field of a regular Nörlund method. Theorem 5.11 Given a regular Nörlund method (N , pn ), the convergence field of (N , pn ) is c if and only if p(z) =

∞ 

pn z n = 0, on |z| ≤ 1.

n=0

Theorem 5.12 There is no p = { pn } such that the convergence field of (N , pn ) is ∞ . The following result is well known in classical summability theory (see [9], p. 231). Theorem 5.13 (Mazur–Orlicz) If a conservative matrix sums a bounded divergent sequence, it sums an unbounded one. The above Theorem 5.13 fails to hold in ultrametric analysis in view of Theorem 5.2.

5.1 The Nörlund Method

69

In [10], Natarajan and Sakthivel proved some more results regarding the convergence fields of regular Nörlund means in the ultrametric case. For instance, the following theorems were proved. Theorem 5.14 ([10], Lemma 3.5) Consider a function p(x) = for |x| < 1 and a function r (x) =

∞ 

∞ 

pn x n regular

n=0

rn x regular for |x| ≤ 1. Define q(x) by n

n=0

q(x) = r (x) p(x). If (N , p) is regular, then (N , q) is regular and 0((N , p)) = 0((N , q)), where (N , p) denotes the Nörlund method (N , pn ), (N , q) denotes the Nörlund method (N , qn ) and 0((N , p)) denotes the set of all sequences which are (N , p) summable to 0 with similar meaning for 0((N , q)). The following result brings out an additive relation between convergence fields of Nörlund means. Theorem 5.15 ([10], Theorem 4.3) If r (x) = 0 for |x| = 1 and p(x) = 0 for |x| < 1, then {sn } ∈ 0((N , q)) if and only if sn = u n + vn , where {u n } ∈ 0((N , r )) and {vn } ∈ 0((N , p)). Definition 5.3 A conservative matrix A = (ank ) for which lim

n→∞

∞ 

 ank

∞   k=0

 lim ank

n→∞

=

is said to be “conull”.

k=0

Wilansky ([11] p. 34) proved the following theorem in the classical case. Theorem 5.16 Every conservative, non-regular real Nörlund matrix is conull. However, in the ultrametric case, we have the following result, which is a violent departure from the classical case. Theorem 5.17 ([12], Theorem 2) Any conservative, non-regular Nörlund matrix is never conull. Corollary 5.1 Since any Schur matrix is conull (see [13], p. 160), it follows that any conservative, non-regular Nörlund matrix is never a Schur matrix.

70

5 The Nörlund and The Weighted Mean Methods

5.2 Some More Properties of the Nörlund Method We now record some interesting properties of regular Nörlund method (see [14]). The following result is easily proved. Theorem 5.18 A = (ank ) ∈ (c0 , c0 ) if and only if (4.1) and (4.2) hold with δk = 0, k = 0, 1, 2, . . . . Consequently, we have Theorem 5.19 The Nörlund method (N , p) ∈ (c0 , c0 ) if and only if p = { pn } ∈ c0 . Theorem 5.20 ([14], Theorem 2.3) The following statements are equivalent: (i) (ii) (iii) and (iv)

(N , p) is regular; (N , p) ∈ (c0 , c0 ); p ∈ c0 ; lim Pn = P = 0.

n→∞

Let Nc0 denote the set of all Nörlund methods that are (c0 , c0 ), i.e. Nc0 denotes the set of all regular Nörlund methods. Let c0 ((N , p)) denote the set of all sequences x = {xk } such that (N , p)(x) ∈ c0 . Definition 5.4 Given Nörlund methods (N , p), (N , q) ∈ Nc0 , we say that (N , g) = (N , p ∗ q) is the symmetric product of (N , p) and (N , q) provided (N , g) ∈ N , where N denotes the set of all Nörlund methods and g = {gn } = p ∗ q is the Cauchy n  (symmetric) product of the sequences p = { pn } and q = {qn }, i.e. gn = pk qn−k , k=0

n = 0, 1, 2, . . . .

Lemma 5.1 ([14], Lemma 2.1) Let p = { pn }, q = {qn } be two sequences in K and let r = p ∗ q. Suppose (N , p), (N , r ) ∈ N . Then c0 ((N , p)) ⊆ c0 ((N , r )) if and only if (N , q) ∈ Nc0 . Proof For any sequence x = {xk } in K , ((N , r )x)n =

n 1  qn−k Pk ((N , p)x)k Rn k=0

=

n  k=0

nk ((N , p)x)k ,

5.2 Some More Properties of the Nörlund Method



where nk =

qn−k Pk Rn ,

0,

71

k ≤ n; k > n.

It is easy to show that |Rn | = |r0 |, n = 0, 1, 2, . . . . Now, if k ≤ n,    qn−k Pk  |q0 || p0 | ≤ |nk =  ; Rn  |r0 | if k > n, nk = 0. Thus sup |nk | < ∞. In view of Theorem 5.18, n,k

c0 ((N , p)) ⊆ c0 ((N , r )) if and only if lim nk = 0, k = 0, 1, 2, . . . ,

n→∞

   qn−k Pk    = 0, k = 0, 1, 2, . . . , i.e. lim  n→∞ Rn  i.e. lim

n→∞

| p0 | |qn−k | = 0, k = 0, 1, 2, . . . , |r0 |

i.e. lim qn−k = 0, k = 0, 1, 2, . . . , n→∞

i.e. lim qn = 0. n→∞

In other words, (N , q) ∈ Nc0 , completing the proof. The following result is an immediate consequence of the above lemma. Theorem 5.21 ([14], Theorem 2.7) Suppose (N , p), (N , q) ∈ Nc0 . Then c0 ((N , p)) ⊆ c0 ((N , q)) if and only if b = {bn } ∈ c0 , which is defined by ∞

b(x) =

q(x)  = bn x n , p(x) n=0

p(x) =

∞  n=0

pn x n , q(x) =

∞  n=0

qn x n .



72

5 The Nörlund and The Weighted Mean Methods

Proof In the above Lemma 5.1, replace q by b so that r = p ∗ b = q. 

The result now follows. Corollary 5.2 Let (N , p), (N , q) ∈ Nc0 . Then

(i) c0 ((N , p)) = c0 ((N , q)) if and only if both a = {an } ∈ c0 and b = {bn } ∈ c0 , where {an } is defined by ∞

a(x) =

p(x)  an x n ; = q(x) n=0

/ c0 and b = {bn } ∈ c0 . (ii) c0 ((N , p))  c0 ((N , q)) if and only if both a = {an } ∈ Corollary 5.3 Let (N , p) ∈ Nc0 and h(x) = only if h ∈ c0 .

1 p(x) .

Then c0 ((N , p)) = c0 if and

Proof Let I be the identity matrix so that c0 (I ) = c0 . Then I (x) = i n = 0, n = 1, 2, . . . . So a(x) = conclusion now follows.

p(x) I (x)

= p(x) and b(x) =

I (x) p(x)

=

∞  n=0 1 p(x)

i n x n , i 0 = 1, = h(x). The 

Corollary 5.4 Let (N , p), (N , q) ∈ Nc0 and r = p ∗ q. Then c0 ((N , p)) ⊆ c0 ((N , r )). The following result is easily established. Theorem 5.22 ([14], Theorem 2.8) Let (N , p) ∈ Nc0 . Let q = { p0 , p1 , p2 , . . . }, with | p0 | < | p0 |. Then (N , q) ∈ Nc0 and further c0 ((N , p)) ∩ c0 ((N , q)) = c0 . Natarajan proved that Nc0 is an ordered abelian semigroup, the order relation is the set inclusion between summability fields of type c0 ((N , p)) and the binary operation is the Cauchy product or symmetric product of sequences. He also proved that there are infinite chains of Nörlund methods from Nc0 (for details, refer to [14], Sect. 3). This is similar to the Cesáro methods (C, k), k = 0, 1, 2, . . . in the classical case, which form an infinite chain.

5.3 The Weighted Mean Method

73

5.3 The Weighted Mean Method In developing summability methods in ultrametric fields, Srinivasan [1] defined the analogue of the classical weighted means ( N¯ , pn ) under the assumption that the sequence { pn } of weights satisfies the conditions: | p0 | < | p1 | < | p2 | < · · · < | pn | < · · · ;

(5.2)

lim | pn | = ∞.

(5.3)

and n→∞

However, it turned out that these weighted means were equivalent to usual convergence. Natarajan [15] remedied the situation by assuming that the sequence { pn } of weights satisfies the conditions: pn = 0, n = 0, 1, 2, . . . ;

(5.4)

| pi | ≤ |P j |, i = 0, 1, 2, . . . , j; j = 0, 1, 2, . . . ,

(5.5)

and

where P j =

j 

pk , j = 0, 1, 2, . . . .

k=0

Note that (5.5) is equivalent to max | pi | ≤ |P j |, j = 0, 1, 2, . . . .

0≤i≤ j

Since | · | is an ultrametric valuation, |P j | ≤ max | pi |, 0≤i≤ j

so that (5.5) is equivalent to |P j | = max | pi |. 0≤i≤ j

(5.6)

Definition 5.5 [15] The weighted mean method in K , denoted by ( N¯ , pn ), is defined by the infinite matrix (ank ), where  ank =

pk Pn ,

0,

k ≤ n; k > n,

pn = 0, n = 0, 1, 2, . . . and | pi | ≤ |P j |, i = 0, 1, 2, . . . , j; j = 0, 1, 2, . . . , n  Pn = pk , n = 0, 1, 2, . . . . k=0

74

5 The Nörlund and The Weighted Mean Methods

    Remark 5.1 (see [15], p. 193, Remark 1) If  PPn+1  > 1, n = 0, 1, 2, . . . and n lim |Pn | = ∞, i.e. {|Pn |} strictly increases to infinity, then the method ( N¯ , pn ) n→∞

is trivial. For, | pn | = |Pn − Pn−1 | = |Pn |, since |Pn | > |Pn−1 | and | · | is an ultrametric valuation. So (5.2) and (5.3) are satisfied. Consequently, ( N¯ , pn ) is trivial, i.e. ( N¯ , pn ) is equivalent to convergence.

Remark 5.2 (see [15], p. 193, Remark 2) We note that (5.5) is equivalent to |Pn+1 | ≥ |Pn |, n = 0, 1, 2, . . . .

(5.7)

Proof Let (5.5) hold. Now, |Pn+1 | =

max | pi |

= max max | pi |, | pn+1 | 0≤i≤n+1

0≤i≤n

= max[|Pn |, | pn+1 |] ≥ |Pn |, n = 0, 1, 2, . . . , so that (5.7) holds. Conversely, let (5.7) hold. For a fixed integer j ≥ 0, let 0 ≤ i ≤ j. Then | pi | = |Pi − Pi−1 | ≤ max[|Pi |, |Pi−1 |] = |Pi | ≤ |P j |, 

so that (5.5) holds. It is now easy to prove the following result. Theorem 5.23 (see [15], p. 194, Theorem 1) ( N¯ , pn ) is regular if and only if lim |Pn | = ∞.

n→∞

(5.8)

Example 5.1 (see [15], p. 194, Remark 4) There are non-trivial ( N¯ , pn ) methods. Let α ∈ K such since K is non-trivially valued.  that 0 < |α| < 1, this being possible  1 1 1 1 3 2 4 Let { pn } = α, α2 , α , α4 , . . . and {sn } = α , α , α3 , α , . . . . It is clear that {sn } does not converge. However, {sn } is ( N¯ , pn ) summable to zero. Theorem 5.24 (Limitation theorem) (see [15], p. 195, Theorem 2) If {sn } is ( N¯ , pn ) summable to s, then

 Pn , n → ∞, (5.9) |sn − s| = o pn

5.3 The Weighted Mean Method

75

p 0 s0 + p 1 s1 + · · · + p n sn , n = 0, 1, 2, . . . . Then, by hypothesis, Pn lim tn = s. Now, n→∞ Pn tn − Pn−1 tn−1 = pn sn , Proof Let tn =

so that

pn Pn tn − Pn−1 tn−1 pn (sn − s) = −s Pn Pn pn 1 [Pn tn − Pn−1 tn−1 − pn s] = Pn 1 = [Pn tn − Pn−1 tn−1 − (Pn − Pn−1 )s] Pn 1 = [Pn (tn − s) − Pn−1 (tn−1 − s)] Pn Pn−1 (tn−1 − s). = (tn − s) − Pn Consequently,         pn  (sn − s) ≤ max(|tn − s|, |tn−1 − s|), since  Pn−1  ≤ 1   P  P n n → 0, n → ∞.

Thus |sn − s| = o

Pn pn

 , n → ∞, 

completing the proof of the theorem.

In [15], some interesting inclusion theorems involving weighted means were proved. We shall record them here without proofs. Theorem 5.25 ([15], p. 195, Theorem 3) [Comparison theorem for two weighted means] If ( N¯ , pn ), ( N¯ , qn ) are two regular weighted mean methods and if       Pn     ≤ H  Q n  , n = 0, 1, 2, . . . , q  p  n n where H > 0 is a constant, Pn = ( N¯ , pn ) ⊆ ( N¯ , qn ).

n  k=0

pk , Q n =

n  k=0

(5.10)

qk , n = 0, 1, 2, . . . , then

76

5 The Nörlund and The Weighted Mean Methods

Theorem 5.26 ([15], p. 196, Theorem 4) [Comparison theorem for a regular ( N¯ , pn ) method and a regular matrix A] Let ( N¯ , pn ) be a regular weighted mean method and A = (ank ) be a regular matrix. If lim

k→∞

and

ank Pk = 0, n = 0, 1, 2, . . . ; pk

    ank  an,k+1 sup  − Pk  < ∞, pk pk+1 n,k

(5.11)

(5.12)

then ( N¯ , pn ) ⊆ A. Theorem 5.27 ([15], p. 197, Theorem 5) Let ( N¯ , pn ) be a regular weighted mean method and A = (ank ) be a regular triangular matrix. Then ( N¯ , pn ) ⊆ A if and only if (5.12) holds. In the context of Theorem 5.17, we have the following interesting result about weighted mean methods: Theorem 5.28 ([12], p. 431, Theorem 3) Every conservative, non-regular weighted mean matrix is a Schur matrix and hence conull. In view of Steinhaus theorem (Theorem 4.3), we can reformulate the above theorem as follows: Theorem 5.29 ([12], p. 432, Theorem 4) A conservative weighted mean matrix is non-regular if and only if it is a Schur matrix. Theorem 5.30 ([12], p. 432, Theorem 5) Let ( N¯ , pn ), ( N¯ , qn ) be two weighted mean methods. If both of them sum the same bounded sequences and if one of them is regular, then the other is regular too. Definition 5.6 Infinite matrices A = (ank ), B = (bnk ) are said to be “consistent” if lim (Ax)n = lim (Bx)n whenever x ∈ c A ∩ c B . B is said to be stronger than A if n→∞ n→∞ cA ⊆ cB . Note that A ⊆ B if and only if B is stronger than A and consistent with A. As in the classical case ([11], p. 12), we can prove the following result in the ultrametric case too. Theorem 5.31 ([16], Theorem 5) Let A, B be triangular matrices. Then B is stronger than A, i.e. c A ⊆ c B if and only if B A−1 is conservative. Also A ⊆ B if and only B A−1 is regular. Brudno’s result in the classical case (see [17], p. 130) is:

5.3 The Weighted Mean Method

77

Theorem 5.32 If every bounded sequence which is summable by a regular matrix A is also summable by a regular matrix B, then A and B are consistent for these sequences. Brudno’s theorem fails to hold in the ultrametric case as the following counterexample using weighted means shows. In the p-adic field Q p , let ⎛

⎞ 1 0 0 0 0 ··· ⎜ 21 21 0 0 0 · · ·⎟ ⎜ ⎟ ⎜ ⎟ Y = ⎜ 0 21 21 0 0 · · ·⎟ , ⎜ ⎟ ⎝ 0 0 1 1 0 · · ·⎠ 2 2 ··· ··· ··· ··· ··· ··· ⎛

1

⎜ 1 ⎜ 1+ p ⎜ N¯ = ⎜ 1 ⎜ ⎝ 1+ p+ p12 ···

0

0 0

p 1+ p p 1+ p+

···

1 p2 1 p2

1+ p+

···

1 p2

⎞ 0 0 ··· 0 0 · · ·⎟ ⎟ ⎟ ⎟. 0 0 · · ·⎟ ⎠ ··· ··· ···

Simple computation shows that ⎛

Y −1

1 ⎜−1 ⎜ ⎜1 =⎜ ⎜−1 ⎜ ⎝1 ···

0 2 −2 2 −2 ···

0 0 2 −2 2 ···

0 0 0 2 −2 ···

0 0 0 0 2 ···

⎞ 0 ··· 0 · · ·⎟ ⎟ 0 · · ·⎟ ⎟ 0 · · ·⎟ ⎟ 0 · · ·⎠ ···

and N Y −1 = (ank ) ⎛

1

1− p ⎜ ⎜ 1+ p ⎜ ⎜ 1− p+ p12 ⎜ 1 =⎜ ⎜ 1+ p+ p2 ⎜ 1− p+ 1 − p3 ⎜ p2 ⎜ ⎝ 1+ p+ p12 + p3 ···

0

0 0

2p  1+ p  2 p− 12

2

p



1+ p+

1 p2



1 + p3 p2 1+ p+ 12 + p 3 p

2 p−

···

1 p2

1+ p+ 12 p   2 12 − p 3 p

1+ p+

1 + p3 p2

···

0 0 0 2 p3 1+ p+ 12 + p 3 p

···

⎞ 0 ··· 0 · · ·⎟ ⎟ ⎟ ⎟ 0 · · ·⎟ ⎟. ⎟ ⎟ ⎟ 0 · · ·⎟ ⎠ ··· ···

78

5 The Nörlund and The Weighted Mean Methods

It is clear that |ank | ≤ 1, n, k = 0, 1, 2, . . . and so that lim

n→∞

∞ 

∞ 

ank = 1, n = 0, 1, 2, . . .

k=0

ank = 1. Easy computation shows that lim ank exists but = 0, n→∞

k=0

k = 0, 1, 2, . . . . Using Theorem 4.1 N¯ Y −1 is conservative but not regular. In view of Theorem 5.31, cY ⊆ c N¯ but Y  N¯ . Hence every bounded sequence which is Y summable is also N¯ summable but Y and N¯ are not consistent for these sequences, proving that Brudno’s theorem fails to hold in the ultrametric set-up. In this context, Somasundaram [18] also claims that Brudno’s theorem fails to hold in the ultrametric set-up. However, his counterexample is incorrect and his proof is vitiated by an error involving the conclusion that if |yn | ≤ |yn | + λ, n = 0, 1, 2, . . . and {yn } converges, so does {yn }. Definition 5.7 Given a sequence {xk }, define the sequence {x¯k } by x¯0 = 0; x¯k = xk−1 , k ≥ 1. A = (ank ) is said to be left translative if the A summability of {xk } to s implies the A summability of {x¯k } to s. A is said to be right translative if the A summability of {x¯k } to s implies the A summability of {xk } to s. If A is both left and right translative, A is said to be translative. In [19], Natarajan obtained necessary and sufficient conditions for a regular ( N¯ , pn ) method to be left translative and right translative, which we shall record here without proofs. Theorem 5.33 ([19], Theorem 3) ( N¯ , pn ) is left translative if and only if 

   Pk pk+1 p k+2  < ∞. − sup sup  pk pk+1  n 0≤k≤n−2 Pn 

(5.13)

Theorem 5.34 ([19], Theorem 4) ( N¯ , pn ) is right translative if and only if 

  Pk+1 pk pk+1   < ∞. − sup sup  Pn pk+1 pk+2  n 0≤k≤n−1 

(5.14)

Analogous to the classical situation (see [20], p. 17), we call the following theorem the “high indices theorem” for weighted means. Theorem 5.35 ([21], Theorem 2) The ( N¯ , pn ) method is equivalent to convergence if and only if    Pn  (5.15) sup   < ∞. pn n

5.3 The Weighted Mean Method

79

We now prove a result which gives an equivalent formulation of summability by weighted mean methods. Incidentally, we note that this result includes the ultrametic analogue of a theorem proved by Môricz and Rhodes (see [22], Theorem MR, p. 188). Theorem 5.36 ([23], Theorem 3) Let (N , pn ), (N , qn ) be two regular weighted ∞  mean methods. Let bn converge to , where n=0

bn = qn

∞  xk , n = 0, 1, 2, . . . . Qk k=n

Then

∞ 

xn is (N , pn ) summable to  if and only if

n=0

   pk Q k+1    < ∞. sup   n,k Pn qk+1 Proof Let Bn =

n 

bk → , n → ∞. Now,

k=0 ∞ ∞   xk xk bn bn+1 xn − = − = , qn qn+1 Qk Qk Qn k=n

so that xn = Q n

k=n+1

bn bn+1 − qn qn+1

 , n = 0, 1, 2, . . . .

Consequently, sm =

m 

xk =

k=0

=

m  k=0 m 

Qk



 bk bk − Q k−1 qk qk m+1

Qk

k=0

= Q0

bk bk+1 − qk qk+1

k=1

b0 + q0

= b0 +

m  k=1

m  k=1

qk

(Q k − Q k−1 )

bm+1 bk − Qm qk qm+1

bk bm+1 − Qm qk qm+1

80

5 The Nörlund and The Weighted Mean Methods

= b0 +

m 

bk − Q m

k=1

=

m 

bk − Q m

k=0

= Bm − Q m

By hypothesis,

bm+1 qm+1

bm+1 qm+1

bm+1 . qm+1

(5.16)

∞  xk converges so that Qk k=0

∞

 xk bn = → 0, n → ∞. qn Qk k=n

Now,

sm Bm bm+1 = − , using (5.16). Qm Qm qm+1

Since {Bn } converges, it is bounded so that |Bn | ≤ M, n = 0, 1, 2, . . . for some M > 0. As (N , qn ) is regular, |Q n | → ∞, n → ∞ in view of Theorem 5.23. Thus    Bm  M    Q  ≤ |Q | → 0, m → ∞. m m Consequently,

sm → 0, m → ∞. Qm

For n = 0, 1, 2, . . . , bn = qn

∞ m   xk sk − sk−1 = qn lim m→∞ Qk Qk k=n

k=n

(where s−1 = 0)  m  m−1  sk  sk = qn lim − m→∞ Qk Q k=n k=n−1 k+1 m−1  m−1  sk  sk sm sn−1 = qn lim + − − m→∞ Qk Qm Q k+1 Qn k=n k=n m−1    1 1 sm sn−1 = qn lim sk + − − m→∞ Qk Q k+1 Qm Qn k=n

5.3 The Weighted Mean Method

81 ∞

= −qn

 sn−1 + qn Qn k=n

1 1 − Qk Q k+1

 sk ,

sm =0 Qm ∞  sn−1 1 1 + qn u k sk , where u k = − . = −qn Qn Qk Q k+1

since lim

m→∞

(5.17)

k=n

Now, Bn =

n−1 

bk + bn

k=0 n−1 

bk qk + bn qk k=0 ∞  n−1   xu = qk + bn Qu =

k=0

= q0

u=k

∞  u=0

∞ ∞ ∞    xu xu xu xu + q1 + q2 + · · · + qn−1 + bn Qu Qu Qu Qu u=1

u=2

∞ 

= (q0 + q1 + · · · + qn−1 )

u=n−1

u=n−1

xu + q0 Qu

n−2  u=0

 xu  xu xu + q1 + q2 Qu Qu Qu n−2

n−2

u=1

u=2

xn−2 + · · · + qn−2 + bn Q n−2 ∞  xu xn−2 xn−3 x0 = Q n−1 + bn + Q n−2 + Q n−3 + · · · + Q0 Qu Q n−2 Q n−3 Q0 u=n−1

= Q n−1

∞ n−2   xu + bn + xk Qu

u=n−1

k=0

= sn−2 + bn + Q n−1

∞  u=n−1

= sn−2 + qn

∞  u=n

xu Qu

∞  xu xu + Q n−1 Qu Qu u=n−1

∞ ∞   xu xu = sn−2 + (Q n − Q n−1 ) + Q n−1 Qu Qu u=n u=n−1

∞  xu xn−1 = sn−2 + Q n + Q n−1 Q Q u n−1 u=n

82

5 The Nörlund and The Weighted Mean Methods

= sn−1 + Q n = sn−1 + Q n

∞  xu Qu u=n

bn qn 

 ∞ sn−1  + u k sk , using (5.17) − Qn

= sn−1 + Q n

k=n

= Qn

∞ 

u k sk ,

k=n

so that

∞

 Bn = u k sk . Qn k=n

Consequently, u n sn =

Bn Bn+1 − , n = 0, 1, 2, . . . . Qn Q n+1

Let {Tn } be the (N , pn ) transform of {sk } so that Tn =

n 1  p k sk Pn k=0

  n 1 Bk 1  Bk+1 , using (5.18) = pk − Pn uk Qk Q k+1 k=0    n  1 p0 B0  pk pk−1 Bk pn Bn+1 = + − − Pn u 0 Q 0 uk u k−1 Q k u n Q n+1 k=1

=

∞ 

ank Bk ,

k=0

where ank =

⎧1 p 0 , ⎪ ⎪ ⎪ Pn u0 Q 0 ⎪ ⎨ 1 pk − Pn

pk−1 u k−1

uk pn 1 ⎪ , − ⎪ ⎪ ⎪ Pn u n Q n+1

⎩



k = 0; 1 Qk ,

0,

It is clear that lim ank = 0, k = 0, 1, 2, . . . . Also, n→∞

1 ≤ k ≤ n; k = n + 1; k ≥ n + 2.

(5.18)

5.3 The Weighted Mean Method ∞ 

ank =

k=0

n+1  k=0

1 = Pn =

1 Pn

=

1 Pn

83

ank 

  n   pk pk−1 1 pn p0 + − − u0 Q0 uk u k−1 Q k u n Q n+1 k=1  

p0 1 1 p1 p0 p2 p1 + − + − u0 Q0 u1 u0 Q1 u2 u1 Q2 

1 pn pn−1 pn +··· + − − un u n−1 Q n u n Q n+1 



 p0 1 p1 1 p2 1 1 1 1 + + − − − u0 Q0 Q1 u1 Q1 Q2 u2 Q2 Q3



1 pn 1 +··· + − un Qn Q n+1

p0 p1 pn u0 + u1 + · · · + un u0 u1 un

1 Pn 1 Pn = Pn = 1, n = 0, 1, 2, . . . , =

so that lim

∞ 

n→∞

ank = 1. By hypothesis, Bk → , k → ∞. Using Theorem 4.1,

k=0

Tn → , n → ∞, i.e.

∞ 

xn is (N , pn ) summable to  if and only if

n=0

   

   pk 1  pn  1  p0  1 1 pk−1   , < ∞, max sup −   , 1≤k≤n |Q k |  u k u k−1  |Q n+1 |  u n  n |Pn | |Q 0 | u 0 i.e. if and only if sup n

1 |Pn |

   

 pk Q k+1 pk−1 Q k−1   pn Q n  , max  −  q  < ∞, 1≤k≤n q q



k+1

k

n+1

which is equivalent to 1 sup n |Pn |



 

 pk Q k+1 pk−1 Q k−1   max −  0, n = 0, 1, 2, . . . . Let

(N , pn ) summable to . Then

∞ 

n=0

bn converges to  if and only if

n=0



   Pk q q k+1 k+2  < ∞, − sup |Q n | sup   Q p Q p Q n k+1 k k k+1 k+2 k≥n 

where bn = qn

∞  xk , n = 0, 1, 2, . . . . Qk k=n

Proof Let tn =

p0 s0 + p1 s1 +···+ pn sn , Pn

n = 0, 1, 2, . . . , where sn =

n 

xk , n =

k=0

0, 1, 2, . . . . Then s0 = t0 and sn = p1n (Pn tn − Pn−1 tn−1 ), n = 0, 1, 2, . . . . Let ∞  xn be (N , pn ) summable to  so that lim tn = . Now, n→∞

n=0

sn 1 = (Pn tn − Pn−1 tn−1 ) Qn pn Q n  1  Pn (tn − ) − Pn−1 (tn−1 − ) + (Pn − Pn−1 ) = pn Q n  1  = Pn (tn − ) − Pn−1 (tn−1 − ) + pn pn Q n Pn Pn−1  = (tn − ) − (tn−1 − ) + pn Q n pn Q n Qn so that   sn  Q

n



  ≤ max M (|tn − |, |tn−1 − |) , || ,  |Q | n

since |Pn−1 | ≤ |Pn | → 0, n → ∞, since lim tn =  and lim |Q n | = ∞, n→∞

n→∞

(N , pn ) being regular, using Theorem 5.23.

5.3 The Weighted Mean Method

85

As already worked out in the proof of Theorem 5.36, ∞

bn = −

 qn sn−1 + qn ck sk , Qn k=n

where ck =

1 1 − , k = 0, 1, 2, . . . . Qk Q k+1

Now, 

 ∞ sn−1  + ck sk (see proof of Theorem 5.36) − Qn

Bn = sn−1 + Q n

k=n

= sn−1 − sn−1 + Q n

∞ 

ck sk

k=n

= Qn

∞ 

ck sk

k=n

= Q n lim

m→∞

= Q n lim

m→∞

m  k=n m  k=n

 = Q n lim

m→∞

ck sk ck

1 {Pk tk − Pk−1 tk−1 } pk

cm Pm tm cn Pn−1 tn−1  − + Pk tk pm pn m−1 k=n

ck ck+1 − pk pk+1

 .

(5.20)

Let  A1 = {xk } :  A2 = {xk } :



∞  k=0 ∞ 

xk is (N , pn ) summable ;  bk converges

.

k=0

We note that A1 , A2 are ultrametric BK spaces with respect to the norms defined by

x A1 = sup |tn |, x = {xk } ∈ A1 ; n≥0

86

5 The Nörlund and The Weighted Mean Methods

and

x A2 = sup |Bn |, x = {xk } ∈ A2 , n≥0

respectively. In view of Banach–Steinhaus theorem (see [25]),

x A2 ≤ L x A1 , for some L > 0.

(5.21)

For every fixed k = 0, 1, 2, . . . , define the sequence x = {xn }, where ⎧ ⎪ if n = k; ⎨1, xn = −1, if n = k + 1; ⎪ ⎩ 0, otherwise. For this sequence x,

x A1

   pk  =   and x A2 = |Q k ck |. P k

| pk | , so that Using (5.21), we have, for k = 0, 1, 2, . . . , |Q k ck | ≤ L |P k|

   ck Pk  L   lim |Q k | = ∞,  p  ≤ |Q | → 0, k → ∞, since k→∞ k k in view of Theorem 5.23. Consequently, lim

k→∞

ck Pk = 0. pk

(5.22)

Using (5.22) in (5.20), we have, ∞

 cn Pn−1 tn−1 Bn = − Qn + Qn Pk tk pn

k=n

=

∞ 

ck ck+1 − pk pk+1

ank tk ,

k=0

where the infinite matrix (ank ) is defined by

ank

⎧ ⎪ ⎪ ⎨0, Q c P n n n−1 = − pn , ⎪ ⎪ ⎩ Q n Pk ck − pk

ck+1 pk+1



0 ≤ k < n − 1; k = n − 1; , k ≥ n.



5.3 The Weighted Mean Method

87

We first note that lim ank = 0, k = 0, 1, 2, . . . and n→∞

so that lim

n→∞

∞ 

∞ 

ank = 1, n = 0, 1, 2, . . .

k=0

ank = 1. Thus, appealing to Theorem 4.1,

∞ 

bn converges to  if

n=0

k=0

and only if      cn Pn−1    c c k k+1   , sup  Pk < ∞. sup |Q n | max  −   pn pk pk+1  n≥0 k≥n 

(5.23)

However,      Q n Pn−1 cn   Q n Pn cn   ≤  , since |Pn−1 | ≤ |Pn |    p  pn n   ∞   ck ck+1   − = |Q n |  Pn  , using (5.22)  pk pk+1  k=n ∞   c ck+1  k  ≤ |Q n |  Pk −  , since |Pn | ≤ |Pk |, k ≥ n  pk pk+1  k=n    ck ck+1  . (5.24) ≤ |Q n | sup  Pk − pk pk+1  k≥n Using (5.24), it is now clear that (5.23) is equivalent to    ck ck+1   sup |Q n | sup  Pk < ∞. − pk pk+1  n≥0 k≥n 

Now,



 ck 1 1 1 ck+1 1 1 1 − − = − − pk pk+1 pk Q k Q k+1 pk+1 Q k+1 Q k+2



 1 Q k+2 − Q k+1 1 Q k+1 − Q k − = pk Q k Q k+1 pk+1 Q k+1 Q k+2 qk+1 qk+2 = − . pk Q k Q k+1 pk+1 Q k+1 Q k+2 Thus

∞ 

bn converges to  if and only if

n=0

    qk+1 qk+2  < ∞.  − sup |Q n | sup  Pk pk Q k Q k+1 pk+1 Q k+1 Q k+2  n≥0 k≥n 

88

5 The Nörlund and The Weighted Mean Methods



   Pk qk+1 qk+2  < ∞,  i.e. sup |Q n | sup  − pk Q k pk+1 Q k+2  n≥0 k≥n Q k+1 

completing the proof of the theorem.



References 1. Srinivasan, V.K.: On certain summation processes in the p-adic field. Indag. Math. 27, 319–325 (1965) 2. Natarajan, P.N.: On Nörlund method of summability in non-archimedean fields. J. Anal. 2, 97–102 (1994) 3. Hardy, G.H.: Divergent Series. Oxford (1949) 4. Natarajan, P.N.: Some theorems on Cauchy multiplication for Nörlund means in nonarchimedean fields. J. Orissa Math. Soc. 12–15, 33–37 (1993–1996) 5. Natarajan, P.N.: A multiplication theorem for Nörlund means in non-archimedean fields. J. Anal. 5, 59–63 (1997) 6. Natarajan, P.N.: Multiplication theorems for Nörlund means in non-archimedean fields II. J. Anal. 6, 49–53 (1998) 7. Escassut, A.: Analytic Elements in p-adic Analysis. World Scientific (1995) 8. Natarajan, P.N., Srinivasan, V.: On the convergence field of Nörlund means in non-archimedean fields. J. Anal. 13, 25–30 (2005) 9. Wilansky, A.: Functional Analysis. Blaisdell (1964) 10. Natarajan, P.N., Sakthivel, S.: Convergence fields of Nörlund means in non-archimedean fields. J. Approx. Theor. Appl. 2, 133–147 (2006) 11. Wilansky, A.: Summability Through Functional Analysis. North Holland, Amsterdam (1984) 12. Natarajan, P.N.: On conservative non-regular Nörlund and weighted means in non-archimedean fields. Ramanujan J. 4, 429–433 (2000) 13. Natarajan, P.N.: A note on Schur matrices in non-archimedean fields. J. Ramanujan Math. Soc. 10, 157–162 (1995) 14. Natarajan, P.N.: Some properties of regular Nörlund methods in non-archimedean fields. Indian J. Math. 53, 287–299 (2011) 15. Natarajan, P.N.: Weighted means in non-archimedean fields. Ann. Math. Blaise Pascal 2, 191– 200 (1995) 16. Natarajan, P.N.: Failure of two classical summability theorems in non-archimedean analysis. J. Anal. 7, 1–5 (1999) 17. Cooke, R.G.: Infinite Matrices and Sequence Spaces. Macmillan (1950) 18. Somasundaram, D.: On a theorem of Brudno over non-archimedean fields. Bull. Austral. Math. Soc. 23, 191–194 (1981) 19. Natarajan, P.N.: Translativity of weighted means in non-archimedean fields. Indian J. Math. 47, 123–130 (2005) 20. Peyerimhoff, A.: Lectures on summability. Lecture Notes in Mathematics, vol. 107. Springer (1969) 21. Natarajan, P.N.: More about ( N¯ , pn ) methods in non-archimedean fields. Indian J. Math. 46, 87–100 (2004) 22. Móricz, F., Rhoades, B.E.: An equivalent reformulation of summability by weighted mean methods, revisited. Linear Algebra Appl. 349, 187–192 (2002) 23. Natarajan, P.N.: A theorem on weighted means in non-archimedean fields, p-adic Numbers. Ultrametric Anal. Appl. 2, 363–367 (2010) 24. Natarajan, P.N.: Another theorem on weighted means in non-archimedean fields, p-adic Numbers. Ultrametric Anal. Appl. 3, 81–85 (2011) 25. Monna, A.F.: Sur le théorème de Banach-Steinhaus. Indag. Math. 25, 121–131 (1963)

Chapter 6

The Euler and The Taylor Methods

Abstract In this chapter, we introduce the Euler and the Taylor methods and present a detailed study of their properties. Keywords The Euler method · The Taylor method

We now define the Euler and Taylor methods of summability in ultrametric analysis and record some of their properties (see [1]). Definition 6.1 Let r ∈ K be such that |1 − r | < 1. The Euler method of order r or (r ) the (E, r ) method is given by the matrix (enk ), n, k = 0, 1, 2, . . . which is defined as follows: If r = 1,

 (r ) enk

where cnk =

n! k!(n−k)! ,

=

cnk r k (1 − r )n−k , k ≤ n; 0, k > n,

k ≤ n;

If r = 1,

 (r ) enk

=

1, k = n; 0, k = n.

(r )

(enk ) is called the (E, r ) matrix. Sometimes, for convenience, we also use the notation n ck for cnk . Theorem 6.1 ([1], Theorem 1.2) The (E, r ) method is regular. Proof Since |1 − r | < 1, |r | = |(r − 1) + 1| = 1, since the valuation is ultrametric. (r ) If enk = 0,

© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_6

89

90

6 The Euler and The Taylor Methods (r )

|enk | = |cnk r k (1 − r )n−k | ≤ |r |k |1 − r |n−k , since |cnk | ≤ 1 = |1 − r |n−k , since |r | = 1 < 1, and so

(r )

sup |enk | < 1. n,k

Now, for k = 0, 1, 2, . . . , (r )

lim |enk | = lim |cnk r k (1 − r )n−k |

n→∞

n→∞

≤ lim |1 − r |n−k n→∞

= 0, since |1 − r | < 1. Consequently,

lim e(r ) n→∞ nk

= 0, k = 0, 1, 2, . . . .

Also, using Binomial theorem (see [2], p. 51), we have, ∞ 

(r ) enk =

k=0

n 

cnk r k (1 − r )n−k

k=0

= (1 − r + r )n = 1, n = 0, 1, 2, . . . , since |n| ≤ 1, so that lim

n→∞

∞ 

(r )

enk = 1.

k=0



Thus the (E, r ) method is regular. (r )

(s)

Theorem 6.2 ([1], Theorem 1.3) (enk )(enk ) is the (E, r s) matrix. (r )

(s)

Proof Let (an j ) = (enk )(enk ). It is clear that an j = 0, j > n. If r or s = 1, the result follows since (E, 1) is usual convergence. Let now, r, s = 1 such that |1 − r |, |1 − s| < 1. Let j ≤ n. Now, n  (r ) (s) an j = enk ek j , k= j

6 The Euler and The Taylor Methods

91

(s)

noting that ek j = 0, k < j. If j = n, (r ) (s) ann = enn enn = r n s n = (r s)n ;

If j < n, an j = =

n  k= j n 

(r ) (s)

enk ek j

j

cnk r k (1 − r )n−k ck s j (1 − s)k− j

k= j

= s j (1 − r )n

n 

j

cnk ck r k

k= j j

= cn s j r j (1 − r )n− j

n 

(1 − s)k− j (1 − r )k (k− j)

c(n− j)r k− j

k= j n 

(1 − s)k− j (1 − r )k− j



r (1 − s) 1−r k= j   r (1 − s) n− j j j n− j 1+ = cn (r s) (1 − r ) 1−r j

= cn (r s) j (1 − r )n− j

(k− j)

k− j

c(n− j)

j

= cn (r s) j (1 − r s)n− j , completing the proof of the theorem.



Remark 6.1 According to Definition 6.1, r belongs to the disc D = {x ∈ K /|x − 1| < 1}. This disc is a multiplicative group, noting that |1−r |, |1−s| < 1 imply |1− r s| < 1. Now the above theorem says: The subjective mapping from D onto the set of Euler matrices, which associates (E, r ) to each r ∈ D, is a group homomorphism.  Corollary 6.1 The (E, r ) matrix is invertible and its inverse is the E, r1 matrix. Theorem 6.3 ([1], Theorem 1.5) If |r − s| < |r |, then (E, r ) ⊆ (E, s). Proof Let {z k } be (E, r ) summable to y, i.e. {tn } converges to y, where tn =

n  k=0

(r )

enk z k , n = 0, 1, 2, . . . .

92

6 The Euler and The Taylor Methods

Now, j 

(1) e jnr tn

=

j 

n=0

n=0

=

j 

(1) e jnr

n 

 (r ) enk zk

k=0

⎛ ⎞ j  1 ( ) ⎝ e r e(r ) ⎠ z k jn

k=0

nk

n=k

= z j, in view of corollary of Theorem 6.2. Again, σk =

k  j=0

=

k 

ek(s)j z k ⎛ ⎞ j  1 ( ) (s) e ⎝ e r tn ⎠ kj

jn

n=0

j=0

⎛ ⎞ k k   1 ( ) ⎝ = e(s) e r ⎠ tn kj

n=0

=

k 

jn

j=n (s)

eknr tn ,

n=0

   using Theorem 6.2. Since |r − s| < |r |,  rs − 1 < 1, the method E, rs is regular.  s Since {tn } converges to y and E, r is regular, {σk } converges to y, i.e. {z k } is (E, s) summable to y. Thus (E, r ) ⊆ (E, s), completing the proof of the theorem.  We now prove a pair of theorems on the Cauchy multiplication of Euler summable sequences and series [3]. Theorem 6.4 If xk = o(1), k → ∞ and {yk } is (E, r ) summable to σ , then {z k } is (E, r ) summable to   ∞  k−1 xk r , σ x0 + k=1

where z n =

n 

xk yn−k , n = 0, 1, 2, . . . .

k=0

Proof Let {σn } be the (E, r ) transform of {yk }. Then, σn =

n  k=0

n

ck r k (1 − r )n−k yk , n = 0, 1, 2, . . . .

(6.1)

6 The Euler and The Taylor Methods

93

By hypothesis, lim σn = σ . Let {τn } be the (E, r ) transform of {z k } so that n→∞

τn =

n 

n

ck r k (1 − r )n−k z k

k=0

= (1 − r )n z 0 + n c1r (1 − r )n−1 z 1 + n c2 r 2 (1 − r )n−2 z 2 + · · · + r n z n = (1 − r )n (x0 y0 ) + n c1r (1 − r )n−1 (x0 y1 + x1 y0 ) + n c2 r 2 (1 − r )n−2 (x0 y2 + x1 y1 + x2 y0 ) + · · · + r n (x0 yn + x1 yn−1 + · · · + xn y0 )) = x0 [(1 − r )n y0 + n c1 r (1 − r )n−1 y1 + n c2 r 2 (1 − r )n−2 y2 + · · · + r n yn ] + x1 [n c1r (1 − r )n−1 y0 + n c2 r 2 (1 − r )n−2 y1 + · · · + r n yn−1 ] + x2 [n c2 r 2 (1 − r )n−2 y0 + n c3r 3 (1 − r )n−3 y1 + · · · + r n yn−2 ] + · · · + xn r n y0  n   n    n n k n−k k n−k = x0 ck r (1 − r ) yk + x1 ck r (1 − r ) yk−1 k=0

+ x2

 n 

 n

k=1

ck r k (1 − r )n−k yk−2 + · · · + xn r n y0

k=2

= x0 σn + x1 + x2

 n 

 n 

 n

ck r (1 − r ) k

k=1 n

n−k

yk−1 

ck r (1 − r ) k

n−k

yk−2 + · · · + xn r n σ0 .

(6.2)

k=2

Now, n 

n

ck r (1 − r ) k

n−k

yk−1 =

k=1

n−1 

n

c j+1r j+1 (1 − r )n− j−1 y j

j=0

=

n−1  

n

c j+1r j+1 (1 − r )n− j−1

j=0

×

⎧ j ⎨ ⎩

k=0

⎫⎤  k   j−k ⎬ 1 1 j 1− ck σk ⎦ , ⎭ r r

using Corollary 6.1 and (6.1) ⎧ ⎡ n−1 n−1 ⎨  ⎣r (1 − r )n−k−1 σk = (−1) j−k ⎩ k=0

j=k

interchanging the order of summation.

⎫⎤ ⎬ n c j+1 j ck ⎦ , ⎭ (6.3)

94

6 The Euler and The Taylor Methods

Using the identity ⎛ n−1  n−1  ⎝ (−1) j−k k=0

we have,

⎞ n

c j+1 j ck ⎠ z k =

j=k

n−1 

zk ,

k=0

n−1  (−1) j−k

n

c j+1 j ck = 1, 0 ≤ k ≤ n − 1.

(6.4)

j=k

Thus, using (6.3), (6.4), we have, n 

n

ck r k (1 − r )n−k yk−1 =

k=1

n−1 

r (1 − r )n−k−1 σk .

(6.5)

k=0

Using (6.5) and similar results, (6.2) can now be written as τn = x0 σn + x1 + x2

n−1 

n−2 

 r (1 − r )

k=0

r (1 − r )

= x0 (σn − σ ) + x1 + x2



n−1 

r (1 − r )n−k−1 (σk − σ )

r (1 − r )n−k−1

r (1 − r ) 2

= x0 (σn − σ ) + x1 n−2 



+ · · · + xn r n σ0

k=0

k=0

+ x2

n−1 

σk

r 2 (1 − r )n−k−2 (σk − σ ) + · · · + xn r n (σ0 − σ )

+ σ x0 + x1 + x2

n−k−2

k=0

k=0

n−2 

σk



2

k=0

n−2 

n−k−1

 n−k−2

n−1 

+ · · · + xn r

n

r (1 − r )n−k−1 (σk − σ )

k=0

r 2 (1 − r )n−k−2 (σk − σ ) + · · · + xn r n (σ0 − σ )

k=0

+ σ x0 + x1r



1 − (1 − r )n 1 − (1 − r )



 + x2r

2

1 − (1 − r )n−1 1 − (1 − r )



 + · · · + xn r

n

6 The Euler and The Taylor Methods

= x0 (σn − σ ) + x1 + x2

n−2 



n−1 

95

r (1 − r )n−k−1 (σk − σ )

k=0

r 2 (1 − r )n−k−2 (σk − σ ) + · · · + xn r n (σ0 − σ )

k=0

+ σ x0 + x1 {1 − (1 − r )n } + x2 r {1 − (1 − r )n−1 } ! + · · · + xn r n−1 {1 − (1 − r )} n−1  = x0 (σn − σ ) + x1r (1 − r )n−k−1 (σk − σ ) k=0

n−2  2 + x2r (1 − r )n−k−2 (σk − σ ) + · · · + xn r n (σ0 − σ ) k=0

 + σ (x0 + x1 + x2 r + · · · + xn r n−1 )

! − {x1 (1 − r )n + x2 r (1 − r )n−1 + · · · + xn r n−1 (1 − r )} .

(6.6)

Now, |xn r n−1 | = |xn |, since |r | = 1 → 0, n → ∞ and |(1 − r )n | = |1 − r |n → 0, n → ∞, since |1 − r | < 1. Thus {xn r n−1 } and {(1 − r )n } are null sequences. Note that the sequence {x1 (1 − r )n + x2 r (1 − r )n−1 + · · · + xn r n−1 (1 − r )} is the Cauchy product of {xn r n−1 } and {(1 − r )n }. In view of Theorem 4.16 lim {x1 (1 − r )n + x2 r (1 − r )n−1 + · · · + xn r n−1 (1 − r )} = 0.

n→∞

Let αn = xn r n . Note that {αn } is a null sequence since |r | = 1 and xn → 0, n → ∞. Let n−1  (1 − r )n−k−1 (σk − σ ). βn = k=0

96

6 The Euler and The Taylor Methods

Now, {βn } is the Cauchy product of the null sequences {(1 − r )n } and {σn − σ }. In view of Theorem 4.16, βn → 0, n → ∞. We now note that n−1   (1 − r )n−k−1 (σk − σ ) lim x1 r n→∞

k=0

 n−2  2 n−k−2 n + x2r (1 − r ) (σk − σ ) + · · · + xn r (σ0 − σ ) = 0, k=0

since lim αn = 0 = lim βn , again appealing to Theorem 4.16. Thus, taking limit n→∞ n→∞ as n → ∞ in (6.6), we have,  lim τn = σ x0 +

n→∞

∞ 

 xk r

k−1

,

k=1

noting that the series on the right converges since |r | = 1 and xk → 0, k → ∞. In other words, {z k } is (E, r ) summable to  σ x0 +

∞ 

 xk r

,

k−1

k=1



completing the proof of the theorem. The following result can be proved in a similar fashion: Theorem 6.5 If

∞ 

xk converges and

k=0

is (E, r ) summable to

∞ 

yk is (E, r ) summable to σ , then

k=0

 σ x0 +

∞ 

∞ 

zk

k=0

 xk r k−1 ,

k=1

where z n =

n 

xk yn−k , n = 0, 1, 2, . . . .

k=0

The following result is easily proved: Theorem 6.6 (Limitation theorem) If

∞  k=0

bounded.

xk is (E, r ) summable, then {xk } is

6 The Euler and The Taylor Methods

97

Remark 6.2 We now given an example of a bounded, non-convergent sequence which is (E, r ) summable. Let {sn } = {1, 0, 1, 0, . . . } and r = 21 . Let {tn } be the (E, 21 )-transform of sequence {sn }. Now tn =

n  k=0

n

1 ck ( )n sk 2

n 1 n = n ck sk 2 k=0

1 = n [n c0 + n c2 + n c4 + · · · + the last term depending on whether n 2 is even or odd] 1 = n 2n−1 2 1 = , n = 0, 1, 2, . . . . 2 Thus the sequence {1, 0, 1, 0, . . . } is (E, 21 ) summable to 21 . Remark 6.3 It was pointed out earlier that the Mazur–Orlicz theorem (Theorem 5.13) fails to hold in the ultrametric case, a counterexample being any regular (N , pn ) method. Theorem 6.6 shows that any (E, r ) method is also a counterexample for the failure of the Mazur–Orlicz theorem in the ultrametric set up. We now have Theorem 6.7 Any two Euler methods are consistent. Proof Consider the Euler methods (E, r ) and (E, s). We then have |1 − r |, |1 − s| < 1. Let {σn (r )}, {τn (s)} be the (E, r ), (E, s) transforms of {xk }, respectively. Let lim σn (r ) = σ and lim τn (s) = τ . We claim that σ = τ . Now,

n→∞

n→∞

σn (r ) = (E, r )({xn }) and τn (s) = (E, s)({xn }). So σn (r ) = (E, r )(E, s)−1 ({τn (s)}) 1 = (E, r )(E, )({τn (s)}), using Corollary 6.1 s r = (E, )({τn (s)}), using Theorem 6.2. s

(6.7)

98

6 The Euler and The Taylor Methods

Note that   1 −

  r   s − r  = |s − r |, since |s| = 1, using |1 − s| < 1 = s s  = |(1 − r ) − (1 − s)| ≤ max(|(1 − r )|, |(1 − s)|) < 1,

so that (E, rs ) is regular, in view of Definition 6.1 and Theorem 6.1. Using (6.7), it follows that σ = τ , completing the proof.  Remark 6.4 In view of Theorem 6.7, we are able to define a parameterless Euler method E of summability as follows: A sequence {xn } is summable E to σ if there exists r ∈ K , |1 − r | < 1 such that {xn } is (E, r ) summable to σ . Definition 6.2 Let r ∈ K be such that |r | < 1. The Taylor method of order r or the (r ) (T, r ) method is given by the matrix (cnk ), n, k = 0, 1, 2, . . . which is defined as follows: If r = 0,

 (r ) cnk

=

0, k < n; cnk r k−n (1 − r )n+1 , k ≥ n.

If r = 0,

 (0) cnk

=

1, k = n; 0, k = n.

(r )

(cnk ) is called the (T, r ) matrix. Remark 6.5 We note that r = 1, since |r | < 1. Theorem 6.8 ([1], Theorem 1.6) Let π = sup{|x|/x ∈ K , |x| < 1}. Let r ∈ K be 1 such that |r | < π − π −1 . Then the (T, r ) method is regular. (r )

Proof When r = 0, the result clearly holds. Now, let r = 0. Since cnk = 0, n > k, it follows that (r ) = 0, k = 0, 1, 2, . . . . lim cnk n→∞

6 The Euler and The Taylor Methods

99

Now, ∞ 

(r )

cnk =

k=0

∞ 

(r )

cnk

k=n

=

∞ 

ckn r k−n (1 − r )n+1

k=n

= (1 − r )n+1

∞ 

ckn r k−n

k=n

= (1 − r )n+1 (1 − r )−(n+1) , using Binomial theorem 1

and |n + 1| ≤ 1 and |r | < π − π −1 (see [2], p. 51) = 1, n = 0, 1, 2, . . . , so that lim

n→∞

∞ 

(r )

cnk = 1.

k=0

(r ) = 0, When cnk (r )

|cnk | = |ckn r k−n (1 − r )n+1 | ≤ |r |k−n |1 − r |n+1 , since |ckn | ≤ 1 < |1 − r |n+1 , since |r | < 1 and k − n ≥ 0 = 1, since |1 − r | = max(1, |r |) = 1, so that

(r )

sup |cnk | < 1. n,k

Thus the (T, r ) method is regular. Remark 6.6 When K = Q p , it is worthwhile to mention that π = is just the radius of convergence of the exponential.

 1 p

1

and then π − π −1

Notation Let A denote the transpose of the matrix A. Theorem 6.9 ([1], Theorem 1.7) The product of the (T, r ) and (T, s) matrices is the matrix (1 − r )(1 − s)(E, (1 − r )(1 − s)) .

100

6 The Euler and The Taylor Methods (r )

(s)

Proof Let dnk = (cn j )(c jk ). Then dnk = 0, if k < n and if k ≥ n, then dnk =

k 

(r ) (s)

cn j c jk

j=n

=

k 

j

cnj r j−n (1 − r )n+1 ck s k− j (1 − s) j+1

j=n k 



r (1 − s) s j=n   r (1 − s) k−n = ckn {(1 − r )(1 − s)}n+1 s k−n 1 + s = ckn {(1 − r )(1 − s)}n+1 s k−n

( j−n)

 j−n

c(k−n)

= ckn {(1 − r )(1 − s)}n+1 (s + r − r s)k−n . Let t = (1 − r )(1 − s) and k ≥ n. Then (t)

ekn = ckn t n (1 − t)k−n = ckn {(1 − r )(1 − s)}n {1 − (1 − r )(1 − s)}k−n (n)

= ck {(1 − r )(1 − s)}n (s + r − r s)k−n and so

((1−r )(1−s))

dnk = (1 − r )(1 − s)ekn

,

which completes the proof.



Corollary 6.2 In general, multiplication of matrices is not commutative. However, in the case of (T, r ) matrices, it follows from Theorem 6.9 that multiplication is commutative. # " r Corollary 6.3 The (T, r ) matrix is invertible and its inverse is the T, − 1−r matrix. # " r matrices is the matrix Proof The product of the (T, r ), T, − 1−r  (1 − r ) 1 +

   r r E, (1 − r ) 1 + 1−r 1−r     1 1 E, (1 − r ) = (1 − r ) 1−r 1−r = (E, 1) .

Since the matrix (E, 1) is the matrix of the (E, 1) method, which is usual convergence, the result follows. 

6 The Euler and The Taylor Methods

101

Theorem 6.10 ([1], Theorem 1.10) If |s − r | < |1 − r |, then (T, r ) ⊆ (T, s). # " (r ) (s) r = (qnk ). Let {z k } be (T, r ) Proof Let (T, r ) = (cnk ), (T, s) = (cnk ), T, − 1−r summable to y, i.e. {σn } converges to y, where σn =

∞ 

(r )

cnk z k , n = 0, 1, 2, . . . .

k=n

Let τ j =

∞ 

q jk σk , j = 0, 1, 2, . . . . Now,

k= j

τj =

∞ 

q jk

∞ 

k= j

 (r ) ckn z n

n=k

⎛ ⎞ ∞ n   ⎝ = q jk c(r ) ⎠ z n kn

n= j

k= j

= z j, using Corollary 6.3 of Theorem 6.9 and the fact that convergence is equivalent to unconditional convergence (see [4], p. 133). Now, tn =

∞ 

(s)

cnk z k

k=n

=

∞  k=n

=

∞ 

(s) cnk τk

⎛ ⎞ ∞  c(s) ⎝ qk j σ j ⎠ nk

k=n

=

∞  ∞ 

j=k

ckn s k−n (1 − s)n+1 (1 − r )−(k+1) ckj (−1) j−k

k=n j=k

= (1 − s)n+1

∞  j=n

⎡ cnj (−r ) j−n (1 − r ) j+1 ⎣

j  k=n



r 1−r

σj

⎤ " s #k−n (k−n) ⎦ σj c( j−n) − r

(interchanging order of summation as before) ∞ "  s # j−n cnj (−r ) j−n (1 − r )−( j+1) 1 − σj = (1 − s)n+1 r j=n

 j−k

102

6 The Euler and The Taylor Methods

 =

1−s 1−r

n+1  ∞ j=n

 cnj

s −r 1−r

 j−n σj,

  # "  s−r  s−r transform of {σ j }. By hypothesis,  1−r which is the T, 1−r  < 1 so that the " # # " s−r s−r method T, 1−r is regular. Since {σ j } converges to y and T, 1−r is regular, {tn } converges to y, i.e. {z k } is (T, s) summable to y, i.e. (T, r ) ⊆ (T, s). The proof of the theorem is now complete. 

References 1. Natarajan, P.N.: Euler and Taylor methods of summability in complete ultrametric fields. J. Anal. 11, 33–41 (2003) 2. Bachman, G.: Introduction to p-adic Numbers and Valuation Theory. Academic Press, New York (1964) 3. Deepa, R., Natarajan, P.N., Srinivasan, V.: Cauchy multiplication of Euler summable series in ultrametric fields. Comment. Math. Prace Mat. 53, 73–79 (2013) 4. Van Rooij, A.C.M., Schikhof, W.H.: Non-archimedean analysis. Nieuw Arch. Wisk. 29, 121–160 (1971)

Chapter 7

Tauberian Theorems

Abstract In this chapter, we prove Tauberian theorems for the Nörlund, the Weighted Mean and the Euler methods. Keywords The Nörlund method · The Weighted mean method · The Euler method

We recall that when K is a complete, non-trivially valued, ultrametric field, limn→∞ ∞  an = 0 implies that the series an converges. Thus an = o(1), n → ∞ is a n=0

Tauberian condition for any method of summability in ultrametric analysis. Probably, this seems to be the reason for the dearth of meaningful Tauberian theorems in ultrametric analysis, while there is a rich theory of Tauberian theorems in the classical case. However, we now present a few Tauberian theorems in ultrametric analysis, due to Natarajan [1]. Theorem 7.1 ([1], Theorem 1) Let A be any regular matrix method and A-summable to s. Let limn→∞ an = . If A(n) diverges, then

∞ 

∞ 

an be

n=0

an converges to s.

n=0

In other words, limn→∞ an =  is a Tauberian condition provided A(n) diverges. Proof Let sn =

n 

ak , n = 0, 1, 2, . . . and xn = sn − n, n = 0, 1, 2, . . .. Now,

k=0

xn − xn−1 = sn − n − {sn−1 − (n − 1)} = (sn − sn−1 ) −  = an −  → 0, n → ∞. So {xn } converges to ∗ ∈ K (say), in view of Theorem 1.3. i.e., sn − n − ∗ → 0, n → ∞. © Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_7

103

104

7 Tauberian Theorems

Since A is regular,

An ({sk − k − ∗ }) → 0, n → ∞,

i.e., An ({sk }) − An ({k}) − ∗ An ({1}) → 0, n → ∞, i.e., s − An ({k}) − ∗ → 0, n → ∞, i.e., An ({k}) → s − ∗ , n → ∞. Since A({n}) diverges by hypothesis, this is possible only if  = 0. So converges. Since A is regular,

∞ 

∞ 

an

n=0

an converges to s, completing the proof.



n=0

Corollary 7.1 ([1], p. 299, Corollary) If

∞ 

an is (N , pn ) summable to s, (N , pn )

n=0 ∞ 

being regular and if limn→∞ an = , then

an converges to s.

n=0

Proof In view of the above Theorem 7.1, it suffices to prove that {n} is not summable by the regular (N , pn ) method. Let tn = Pn =

n 

p0 . n + p1 . (n − 1) + · · · + pn−1 . 1 + pn . 0 , Pn

pk , n = 0, 1, 2, . . .. Now,

k=0

   p0 . (n + 1) + p1 . n + · · · + pn . 1 p0 . n + p1 . (n − 1) + · · · + pn−1 . 1  |tn+1 − tn | =  −  Pn+1 Pn    Pn { p0 + p1 + · · · + pn } + pn+1 { p0 . n + p1 . (n − 1) + · · · + pn−1 . 1}  . =   P P n n+1

Since the valuation of K is ultrametric, | pi | < | p0 |, i = 1, 2, . . ., and |n| ≤ 1, n = 1, 2, . . ., |Pn | = |Pn+1 | = | p0 |, |Pn { p0 + p1 + · · · + pn }| = | p0 |2 , and | pn+1 { p0 . n + p1 . (n − 1) + · · · + pn−1 . 1}| < | p0 |2 so that |tn+1 − tn | =

| p 0 |2 = 1. | p 0 |2

7 Tauberian Theorems

105

Consequently, {tn } cannot be Cauchy and so diverges, i.e., {n} is not summable ∞  (N , pn ). In view of Theorem 7.1, an converges to s.  n=0

Remark 7.1 The hypothesis that A({n}) diverges in Theorem 7.1 cannot be dropped. Consider the matrix A = (ank ), where

ank

⎧ ⎪ k = n; ⎨n, = 1 − n, k = n + 1; ⎪ ⎩ 0, otherwise.

We can easily check that A is regular. Now, ∞ 

ank . k = n . n + (1 − n)(n + 1)

k=0

= n2 + 1 − n2 = 1, n = 0, 1, 2, . . . , so that {n} is transformed (by A) into the sequence {1, 1, 1, . . .}. Thus the series ∞ ∞   an = 1 + 1 + 1 + · · · is A-summable to 1. However, the series an = n=0

1 + 1 + 1 + · · · diverges.

n=0

The following result gives an abundance of regular translative methods of summability. Theorem 7.2 ([1], Theorem 2) Every regular (N , pn ) method is translative. Proof For convenience, we write A ≡ (N , pn ). Now, pn . 0 + pn−1 . s0 + pn−2 . s1 + · · · + p0 . sn−1 Pn Pn−1 pn−1 . s0 + pn−2 . s1 + · · · + p0 . sn−1 = . Pn Pn−1 Pn−1 = . αn−1 , Pn    pn  | pn | Pn − pn pn where (As)n = αn . PPn−1 = = 1 − → 1, n → ∞, since  Pn  = | p0 | → 0, P P n n n n → ∞, the (N , pn ) method being regular. So, if αn → , n → ∞, then βn → , n → ∞ too. Again, (A¯s )n ≡ βn =

106

7 Tauberian Theorems

pn . s1 + pn−1 . s2 + · · · + p0 . sn+1 Pn ( pn+1 . s0 + pn . s1 + · · · + p0 . sn+1 ) − pn+1 . s0 = Pn Pn+1 αn+1 − pn+1 s0 = Pn Pn+1 pn+1 = . αn+1 − . s0 . Pn Pn

(As ∗ )n ≡ γn =

Since PPn+1 → 1, n → ∞ and pPn+1 → 0, n → ∞, αn → , n → ∞ implies that n n  γn → , n → ∞, proving that the regular (N , pn ) method is translative. We now prove a Tauberian theorem for regular, translative matrix summability methods. ∞  Theorem 7.3 ([1], Theorem 3) If an is summable to s by a regular matrix method n=0

A which is translative and an+1 − an → , n → ∞, then

∞ 

an converges to s.

n=0

Proof Let A = (ank ) be regular and translative. We first prove that A({n}) diverges. If not, i.e., A({n}) converges. Since A is translative, ∞ 

ank (xk+1 − xk ) → 0, n → ∞,

k=0

whenever {xk } is A-summable. Since, by assumption, A({n}) converges, ∞ 

ank (k + 1 − k) → 0, n → ∞,

k=0

i.e.,

∞ 

ank → 0, n → ∞,

k=0

which contradicts the fact that ∞ 

ank → 1, n → ∞,

k=0

this being so since A is regular. Let, now, xn = an+1 − n, n = 0, 1, 2, . . . .

7 Tauberian Theorems

107

Then xn+1 − xn = {an+1 − n} − {an − (n − 1)} = (an+1 − an ) −  → 0, n → ∞. So {xn } converges to ∗ (say), i.e., an+1 − n − ∗ → 0, n → ∞, i.e., sn+1 − sn − n − ∗ → 0, n → ∞. Since A is regular, An ({sk+1 − sk − k − ∗ }) → 0, n → ∞, i.e., An ({sk+1 }) − An ({sk }) − An ({k}) − ∗ An ({1}) → 0, n → ∞. Since A is translative, s − s − An ({k}) − ∗ → 0, n → ∞. ∗

If  = 0, then An ({k}) → −  , n → ∞, which contradicts the fact that A({n}) ∞  an diverges. So  = 0. Thus {an } converges. Now, in view of Theorem 7.1, n=0

converges to s, completing the proof of the theorem.



In view of Theorems 7.1 and 7.3, we have the following result. Theorem 7.4 ([1], Theorem 4) If

∞ 

an is summable by a regular and translative

n=0

matrix method A, then the Tauberian conditions (i) an → , n → ∞; and (ii) an+1 − an →  , n → ∞ are equivalent. In the case of regular (N , pn ) methods, we have the following. Theorem 7.5 ([1], Theorem 5) If

∞ 

an is summable by a regular (N , pn ) method,

n=0

then the following Tauberian conditions are equivalent:

108

7 Tauberian Theorems

(i) an → , n → ∞; (ii) an = an+1 − an →  , n → ∞; If, further, an = 0, n = 0, 1, 2, . . ., each of (iii)

an+1 an

→ 1, n → ∞;

and (iv)

an+2 +an an+1

→ 2, n → ∞

is a weaker Tauberian condition for the summability of

∞ 

an by a regular (N , pn )

n=0

method. Proof The first part follows from Theorem 7.4. We now prove that (iii) implies (iv) and (iv) implies (ii). If (iii) holds, then an+2 + an an+2 an = + an+1 an+1 an+1 → 2, n → ∞, so that (iv) holds. Let (iv) hold. Since

∞ 

an is (N , pn ) summable, {an } is bounded, in

n=0

view of Theorem 5.2 so that there exists M > 0 such that |an | ≤ M, n = 0, 1, 2, . . .. Now, 1 1 2 | an | = |an+2 − 2an+1 + an | M  M  an+2 − 2an+1 + an   ≤   an+1    an+2 + an  =  − 2 an+1 → 0, n → ∞ so that an+2 − 2an+1 + an → 0, n → ∞, i.e., (an+2 − an+1 ) − (an+1 − an ) → 0, n → ∞. Thus {an+1 − an } is Cauchy and so an+1 − an →  , n → ∞ for some  ∈ K so that (ii) holds.  Remark 7.2 It is clear that (ii) does not imply (iii) or (iv), as is seen by choosing an = p n in Q p , for a prime p. We now prove a Tauberian theorem for weighted mean methods (see [2, 3]).

7 Tauberian Theorems

Theorem 7.6 If

∞ 

109

ak is (N , pn ) summable to s, where (N , pn ) is regular and if

k=0

an = O(

pn ), n → ∞; Pn

(7.1)

and an → , n → ∞, ∞ 

then

ak converges to s.

k=0

Proof We can suppose that s = 0. We now claim that  = 0. If not, choose  > 0 such that  < ||. We can now choose a positive integer N such that    pn  |sn |   < , n ≥ N , Pn using (5.9) |an − | < , n ≥ N , using limn→∞ an = . Now, |an | = |(an − ) + | = max(|an − |, ||) = ||, n ≥ N . Also,

   pn  |an | ≤ M   , M > 0, P n

in view of (7.1). Thus, for n ≥ N ,    pn  || ≤ M   , Pn    pn  || i.e.,   ≥ . Pn M Using (7.2), for n ≥ N ,

   pn  ||  > |sn |   ≥ |sn | , Pn M i.e., |sn |
0, the set {(m, n) ∈ N2 : |xm,n − x| ≥ } is finite,

m+n→∞

N being the set of all positive integers. In such a case, x is unique and we say that x is the limit of {xm,n }.

Definition 8.2 Let {xm,n } be a double sequence. We say that s=

∞ 

xm,n

m,n=0

if s= where sm,n =

m,n 

lim

m+n→∞

sm,n ,

xk, , m, n = 0, 1, 2, . . . .

k=0,=0

© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1_8

115

116

8 Silverman-Toeplitz Theorem for Double Sequences and Double Series

In such a case, we say that the double series

∞ 

xm,n converges to s.

m,n=0

Remark 8.1 If

lim

m+n→∞

xm,n = x, then the double sequence {xm,n } is bounded.

It is now easy to prove the following results. Lemma 8.1

lim

m+n→∞

xm,n = x if and only if

(i) lim xm,n = x, m = 0, 1, 2, . . . ; n→∞ (ii) lim xm,n = x, n = 0, 1, 2, . . . ; m→∞

and (iii) for each  > 0, there exists N ∈ N such that |xm,n − x| < , m, n ≥ N which we write as lim xm,n = x. m,n→∞

Lemma 8.2

∞ 

xm,n converges if and only if

m,n=0

lim

m+n→∞

xm,n = 0.

(8.1)

Definition 8.3 Given, the 4-dimensional infinite matrix A = (am,n,k, ) and a double sequence {xk, }, we define (Ax)m,n =

∞ 

am,n,k, xk, , m, n = 0, 1, 2, . . . ,

k,=0

it being assumed that the double series on the right converge. The double sequence {(Ax)m,n } is called the A-transform of the double sequence {xk, }. As in the case of simple sequences, if lim (Ax)m,n = s, we say that {xk, } is summable A or m+n→∞

A-summable to s. If

lim (Ax)m,n = s whenever

m+n→∞

lim

k+→∞

xk, = t, we say that A

is convergence preserving. If, further, s = t, we say that A is regular.

8.2 Silverman-Toeplitz Theorem Natarajan and Srinivasan proved the following theorem (see [1]). Theorem 8.1 (Silverman-Toeplitz theorem) A = (am,n,k, ) is regular if and only if lim

m+n→∞

am,n,k, = 0, k,  = 0, 1, 2, . . . ;

(8.2)

8.2 Silverman-Toeplitz Theorem

117 ∞ 

lim

m+n→∞

am,n,k, = 1;

(8.3)

k,=0

lim

sup |am,n,k, | = 0,  = 0, 1, 2, . . . ;

(8.4)

lim

sup |am,n,k, | = 0, k = 0, 1, 2, . . . ;

(8.5)

m+n→∞ k≥0

m+n→∞ ≥0

and sup |am,n,k, | < ∞.

(8.6)

m,n,k,

Proof Proof of necessity part Define the sequence x = {xk, } as follows: For any fixed p, q, let  xk, =

1, when k = p,  = q; 0, otherwise.

Then (Ax)m,n = am,n, p,q . Since {xk, } has limit 0, it follows that (8.2) is necessary. Define the sequence x = {xk, }, where xk, = 1, k,  = 0, 1, 2, . . . . Now, ∞,∞  (Ax)m,n = am,n,k, , m, n = 0, 1, 2, . . . . k=0,=0

This shows that the series on the right converge and since {xk, } has limit 1, it follows that ∞,∞  am,n,k, = 1, lim m+n→∞

k=0,=0

so that (8.3) is necessary. We now show that (8.4) holds. Suppose not. Then, there exists 0 ∈ N such that lim sup |am,n,k,0 | = 0 does not hold. So, there exists an  > 0 such that

m+n→∞ k≥0

{(m, n) : sup |am,n,k,0 | > } is infinite.

(8.7)

k≥0

Let m 1 = n 1 = r1 = 1. Choose m 2 , n 2 ∈ N such that m 2 + n 2 > m 1 + n 1 and sup |am 2 ,n 2 ,k,0 |
, using (8.7). k≥0

Then choose r2 ∈ N such that r2 > r1 and sup |am 2 ,n 2 ,k,0 |
r2

 , using (8.3). 8

Inductively choose m p + n p > m p−1 + n p−1 such that sup 0≤k≤r p−1

|am p ,n p ,k,0 |
;

(8.8) (8.9)

k≥0

and then choose r p > r p−1 such that sup |am p ,n p ,k,0 |
r p

 . 8

(8.10)

In view of (8.8), (8.9), (8.10), we have, sup

r p−1  −

  3 − = . 8 8 4

So, there exists k p , r p−1 < k p ≤ r p , such that |am p ,n p ,k p ,0 | >

3 . 4

(8.11)

Define the sequence x = {xk, } as follows:  xk, = We note that

lim

k+→∞

0, if  = 0 ; 1, if  = 0 , k = k p , p = 1, 2, . . . .

xk, = 0. Now, in view of (8.8),

r p−1       am p ,n p ,k,0 xk,0  ≤ sup |am p ,n p ,k,0 | < ;    0≤k≤r p−1 8 k=0

(8.12)

8.2 Silverman-Toeplitz Theorem

119

Using (8.10), we have,        ∞   am p ,n p ,k,0 xk,0  ≤ sup |am p ,n p ,k,0 | < ;  8  k>r p k=r p +1

(8.13)

and using (8.11), we get,      rp   3  am p ,n p ,k,0 xk,0  = |am p ,n p ,k,0 | > .  4 k=r p−1 +1 

(8.14)

Thus  (Ax)m

∞       = a x  m p ,n p ,k,0 k,0  p ,n p   k=0      rp  r p−1       ≥  am p ,n p ,k,0 xk,0  −  am p ,n p ,k,0 xk,0    k=r p−1 +1  k=0       ∞   − am p ,n p ,k,0 xk,0  k=r p +1  ≥ |am p ,n p ,k p ,0 | −

sup 0≤k≤r p−1

|am p ,n p ,k,0 | − sup |am p ,n p ,k,0 | k>r p

  3 − − , using (8.8), (8.9) and (8.10) 4 8 8  = , p = 1, 2, . . . . 2 >

Consequently,

lim (Ax)m,n = 0 does not hold, which is a contradiction. Thus

m+n→∞

(8.4) is necessary. The necessity of (8.5) follows in a similar fashion. To establish (8.6), we shall suppose that (8.6) does not hold and arrive at a contradiction. Since K is non-trivially valued, there exists π ∈ K such that 0 < ρ = |π| < 1. Choose m 1 = n 1 = 1. Using (8.2), (8.3), choose m 2 + n 2 > m 1 + n 1 such that sup

0≤k+≤m 1 +n 1

|am 2 ,n 2 ,k, | < 2, using (8.2);

 6 2 sup |am 2 ,n 2 ,k, | > ; ρ k+≥0 and sup

k+>m 1 +n 1

|am 2 ,n 2 ,k, | < 22 ,

120

8 Silverman-Toeplitz Theorem for Double Sequences and Double Series

using (8.3), Lemmas 8.1 and 8.2. It now follows that sup

k+>m 2 +n 2

|am 2 ,n 2 ,k, | < 22 .

Choose m 3 + n 3 > m 2 + n 2 such that sup

0≤k+>m 2 +n 2

|am 3 ,n 3 ,k, | < 22 ;

sup |am 2 ,n 2 ,k, | >

k+≥0

 8 2 ; ρ

and sup

k+>m 3 +n 3

|am 3 ,n 3 ,k, | < 24 .

Inductively, choose m p + n p > m p−1 + n p−1 such that sup

0≤k+≤m p−1 +n p−1

|am p ,n p ,k, | < 2 p−1 ;

(8.15)

 2 p+2 2 ; ρ

(8.16)

|am p ,n p ,k, | < 22 p−2 .

(8.17)

sup |am p ,n p ,k, | >

k+≥0

and sup

k+>m p +n p

Using (8.15), (8.16), (8.17), we have, sup

m p−1 +n p−1 1 ρ ρ ρ ρ    p−1   p+3   p−1 2 2 2 = − −1 ρ ρ ρ   p−1   p+3   p−1   p−1    p−1 2 2 2 2 2 ≥ − − ≥1 , since ρ ρ ρ ρ ρ   p−1  4   p−1   p−1  2 2 2 2 = −2 ρ ρ ρ ρ >

8.2 Silverman-Toeplitz Theorem

121

  p−1  4   p−1     p−1  2 2 2 2 2 2 > − , since > 2 ρ ρ ρ ρ ρ ρ   2 p−1  3 2 2 = −1 ρ ρ  2 p−1 2 2 > [23 − 1], since > 2 ρ ρ  2 p−1 2 =7 ρ  2 p−1 2 >4 ρ 22 p+1 ρ2 p−1 22 p+1 1 > , since > 1. p ρ ρ =

(8.18)

Thus, there exists k p and  p , m p−1 + n p−1 < k p +  p ≤ m p + n p such that |am p ,n p ,k p , p | >

22 p+1 . ρp

(8.19)

Now, define the sequence x = {xk, } as follows:  xk, = We note that

lim

k+→∞

π p , if k = k p ,  =  p , p = 1, 2, . . . ; 0, otherwise.

xk, = 0. Now,

   ∞,∞     |(Ax)m p ,n p | =  am p ,n p ,k, xk,  k=0,=0        m p−1  +n p−1 m p +n p         ≥ am p ,n p ,k, xk,  −  am p ,n p ,k, xk,  k+=(m p−1 +n p−1 )+1   k+=0      ∞    −  am p ,n p ,k, xk,  k+=(m p +n p )+1  ≥ |am p ,n p ,k p , p ||xk p , p | − −

sup

m p +n p 13 · 22 p−3 , p = 1, 2, . . . , i.e. lim (Ax)m,n = 0 does not hold, which is a contradiction. Thus (8.6) is m+n→∞ necessary. Proof of the sufficiency part Let x = {xm,n } be such that

lim

m+n→∞

xm,n = s. Then ∞,∞ 

(Ax)m,n − s =

am,n,k, xk, − s.

k=0,=0

Using (8.3), we have,

∞,∞ 

am,n,k, − rm,n = 1,

k=0,=0

where lim

m+n→∞

Thus (Ax)m,n − s =

∞,∞ 

rm,n = 0.

(8.20)

am,n,k, (xk, − s) + rm,n s.

k=0,=0

Given  > 0, we can choose sufficiently large p, q such that sup

|xk, − s|
p+q

where H =

sup m,n,k,≥0

 , 5H

|am,n,k, | > 0. Let L = sup |xk, − s|. k,≥0

(8.21)

8.2 Silverman-Toeplitz Theorem

123

We now choose N ∈ N such that whenever m + n ≥ N , the following are satisfied.  , using (8.2); (8.22) |am,n,k, | < sup 5 pq L 0≤k+≤ p+q sup |am,n,k, | < k≥0

sup |am,n,k, | < ≥0

 ,  = 0, 1, 2, . . . , q, using (8.4); 5q L

(8.23)

 , k = 0, 1, 2, . . . , p, using (8.5); 5 pL

(8.24)

and |rm,n |
0, there exists a positive integer N such that the set {(m, n), (k, ) ∈ N2 : |xm,n − xk, | ≥ , m, n, k,  ≥ N } is finite. It is now easy to prove the following result. Theorem 8.2 The double sequence {xm,n } in K is Cauchy if and only if lim

|xm+1,n − xm,n | = 0;

lim

|xm,n+1 − xm,n | = 0.

m+n→∞

and m+n→∞

Definition 8.6 If every Cauchy double sequence of an ultrametric normed linear space X converges to an element of X , then X is said to be double sequence complete or ds-complete. ∞ For x = {xm,n } ∈ ∞ ds , define x = sup |x m,n |. One can easily prove that ds is m,n

an ultrametric normed linear space which is ds-complete. With the same definition of norm for elements of cds , cds is a closed subspace of ∞ ds . In this context, we recall that Natarajan (Theorems 4.2 and 4.3) proved Schur’s theorem and Steinhaus’ theorem for 2-dimensional matrices over complete, non-trivially valued, ultrametric fields. In the sequel, we shall suppose that K is a non-trivially valued ultrametric field which is ds-complete. Theorem 8.3 (Schur) The necessary and sufficient conditions for a 4-dimensional infinite matrix A = (am,n,k, ) to transform double sequences in ∞ ds into double are: sequences in cds , i.e. {(Ax)m,n } ∈ cds whenever x = {xk, } ∈ ∞ ds lim am,n,k, = 0, m, n = 0, 1, 2, . . . ;

(8.26)

lim

sup |am+1,n,k, − am,n,k, | = 0;

(8.27)

lim

sup |am,n+1,k, − am,n,k, | = 0.

(8.28)

k+→∞

m+n→∞ k,≥0

and m+n→∞ k,≥0

8.3 Schur’s and Steinhaus Theorems

125

Proof Sufficiency. Let (8.26), (8.27), (8.28) hold. Let x = {xk, } ∈ ∞ ds . We first note that (8.26), (8.27) and (8.28) together imply that (8.6) holds. In view of (8.6) and (8.26), ∞  am,n,k, xk, , m, n = 0, 1, 2, . . . (Ax)m,n = k,=0

is defined, the double series on the right being convergent. Now,    ∞     (am+1,n,k, − am,n,k, )xk,  |(Ax)m+1,n − (Ax)m,n | =  k,=0  ≤ M sup |am+1,n,k, − am,n,k, | k,≥0

→ 0, m + n → ∞, using (8.27), where |xk, | ≤ M, k,  = 0, 1, 2, . . . , M > 0. Similarly, it follows that |(Ax)m,n+1 − (Ax)m,n | → 0, m + n → ∞, using (8.28). Thus {(Ax)m,n } is a Cauchy double sequence in K . Since K is ds-complete, {(Ax)m,n } converges and so {(Ax)m,n } ∈ cds , completing the sufficiency part of the proof. Necessity. Let {(Ax)m,n } ∈ cds whenever x = {xk, } ∈ ∞ ds . Consider, the double sequence {xk, } where xk, = 1, k,  = 0, 1, 2, . . . . {xk, } ∈ ∞ ds so that, by hypothesis, ∞  am,n,k, , m, n = 0, 1, 2, . . . (Ax)m,n = k,=0

is defined. Since the series on the right converges, (8.26) holds. Suppose (8.27) does not hold. Then, there exists 0 ∈ N such that lim

sup |am+1,n,k,0 − am,n,k,0 | = 0

m+n→∞ k≥0

does not hold. So, there exists  > 0 such that the set 



(m, n) ∈ N : sup |am+1,n,k,0 − am,n,k,0 | >  2

k≥0

is infinite. Thus, we can choose pairs of integers m p , n p ∈ N such that m 1 + n 1 < m 2 + n 2 < · · · < m p + n p < · · · and sup |am p +1,n p ,k,0 − am p ,n p ,k,0 | > , p = 1, 2, . . . . k≥0

(8.29)

126

8 Silverman-Toeplitz Theorem for Double Sequences and Double Series

Using (8.26), lim |am 1 +1,n 1 ,k,0 − am 1 ,n 1 ,k,0 | = 0.

k→∞

Consequently, there exists r1 ∈ N such that sup |am 1 +1,n 1 ,k,0 − am 1 ,n 1 ,k,0 |
,

(8.31)

0≤k .

(8.32)

By hypothesis, (8.2) holds. So we can suppose that sup |am 2 +1,n 2 ,k,0 − am 2 ,n 2 ,k,0 |
r1 such that sup |am 2 +1,n 2 ,k,0 − am 2 ,n 2 ,k,0 |
.

r1 ≤k .

(8.36)

Inductively, we choose strictly increasing sequences {r p }, {k p } such that r p−1 ≤ k p < r p,

8.3 Schur’s and Steinhaus Theorems

sup 0≤k M1 , n > N1 . Since lim sm−k,n− = 0 for every fixed m+n→∞

k,  = 0, 1, 2, . . . , we can choose positive integers M2 > M1 , N2 > N1 such that for all m > M2 , n > N2 ,  sup |sm−k,n− | < ; M 0≤k≤M1 0≤≤N1

sup

|sm−k,n− |
N2 ,      m,n  |z m,n | =  sm−k,n− tk,  k,=0        =  sm−k,n− tk, + sm−k,n− tk,  0≤k≤M1 0≤k≤M1  0≤≤N1 >N1       + sm−k,n− tk, + sm−k,n− tk,   k>M1 k>M1  0≤≤N1 >N1

9.2 Cauchy Multiplication of Double Series

137

⎡ ⎢ ≤ max ⎣ sup |sm−k,n− tk, |, 0≤k≤M1 0≤≤N1

|sm−k,n− tk, |,

sup 0≤k≤M1 N1 +1≤≤n

sup M1 +1≤k≤m 0≤≤N1

sup M1 +1≤k≤m N1 +1≤≤n

≤ max

|sm−k,n− tk, |, ⎤ ⎥ |sm−k,n− tk, |⎦

      . M, . M, . M, .M M M M M

= . ∞ 

Thus

z m,n converges. Now, we will prove that

m,n=0 ∞ 

⎛ z m,n = ⎝

m,n=0

Now, ⎛ ⎝

∞  m,n=0

⎞⎛ sm,n ⎠ ⎝

∞ 

∞ 

⎞⎛ sm,n ⎠ ⎝

m,n=0

⎛

tm,n ⎠ =

lim

m+n→∞

⎝

lim

m+n→∞

⎝

=

lim

m+n→∞

lim

∞ 

⎝

m,n 

⎞ sk, ⎠

⎛ lim

m+n→∞

⎞⎛ sk, ⎠ ⎝ ⎛ ⎝

k,=0 m,n 

m+n→∞

=

m,n 

k,=0

⎛ =

m,n 

k,=0

⎛ =

tm,n ⎠ .

m,n=0

⎞

m,n=0

⎞

∞ 

m,n 

k,=0 k, 

⎝

m,n 

⎞ tk, ⎠

k,=0

⎞

tk, ⎠ ⎞⎞

sk−i,− j ti, j ⎠⎠

i, j=0

z k,

k,=0

z m,n ,

m,n=0

completing the proof of the theorem. We need the following definitions in the sequel (see [2]).



138

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

Definition 9.3 Any matrix A = (am,n,k, ) for which sup |am,n,k, | < ∞ is called m,n,k,

an N matrix.

Definition 9.4 Any matrix A = (am,n,k, ) is called an N (1) matrix if for m = 0, 1, 2, . . . , there exists a positive integer km such that am,n,k, = 0, k > km , n,  = 0, 1, 2, . . . . Definition 9.5 An infinite matrix A = (am,n,k, ) is called an N (2) matrix if for n = 0, 1, 2, . . . , there exists a positive integer n such that am,n,k, = 0,  > n , m, k = 0, 1, 2, . . . . Definition 9.6 Given two infinite matrices A = (am,n,k, ), B = (bm,n,k, ), m, n, k,  = 0, 1, 2, . . . , their convolution product is defined as the matrix C = (cm,n,k, ), where, for k,  = 0, 1, 2, . . . cm,n,k, =

k, 

am,n,k−i,− j bm,n,i, j ,

i, j=0

m, n = 0, 1, 2, . . . . In such a case, we write C = A ∗ B. Theorem 9.6 If A = (am,n,k, ), B = (bm,n,k, ) are both N (1) or both N (2) or both N , then, their convolution C is also N (1) or N (2) or N , respectively. Further, for m, n = 0, 1, 2, . . . ⎛ ⎞⎛ ⎞ ∞ ∞ ∞    cm,n,k, = ⎝ am,n,k, ⎠ ⎝ bm,n,k, ⎠ , k,=0

k,=0

k,=0

provided the series on the right converge. Proof Let A = (am,n,k, ), B = (bm,n,k, ) be N (1) . Since A is N (1) , for m = (1) (1) 0, 1, 2, . . . , there exists a positive integer km such that am,n,k, = 0 for k > km , (2) such that n,  = 0, 1, 2, . . . . Also, since B is N (1) , there exists a positive integer km (2) (1) (2) bm,n,k, = 0 for k > km , n,  = 0, 1, 2, . . . . Let km = max(km , km ). Then, am,n,k, = 0 = bm,n,k, for k > km , n,  = 0, 1, 2, . . . . Let now, k > 2km . Then for n,  = 0, 1, 2, . . . , cm,n,k, =

k,  i, j=0

am,n,k−i,− j bm,n,i, j = 0.

9.2 Cauchy Multiplication of Double Series

139

Thus C is an N (1) matrix. Similarly, we can prove that if A, B are both N (2) , then C is also N (2) . Now, let A, B be N matrices. Then sup |am,n,k, | < ∞; m,n,k,

and sup |bm,n,k, | < ∞. m,n,k,

Thus    k,     sup |cm,n,k, | = sup  am,n,k−i,− j bm,n,i, j  m,n,k, m,n,k, i, j=0  ⎤ ⎡ ≤ sup ⎣ max |am,n,k−i,− j bm,n,i, j |⎦ m,n,k,

0≤i≤k 0≤ j≤

< ∞. Consequently, C is an N matrix too. The remaining part of the theorem follows from the latter part of Theorem 9.5.  If A = (am,n,k, ), B = (bm,n,k, ) are both regular, it is worthwhile to check whether the convolution of A and B is also regular or otherwise. We leave the details to the reader.

9.3 The Weighted Mean Method for Double Sequences We now introduce weighted means for double sequences and extend theorems dealing with weighted means for simple sequences (for details, refer to [3]). Definition 9.7 The (N , pm,n ) method is defined by the infinite matrix (am,n,k, ), where p k, , if k ≤ m,  ≤ n; am,n,k, = Pm,n 0, if k > m or  > n, Pm,n =

m,n  i, j=0

pi, j , m, n = 0, 1, 2, . . . ,

140

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

with the sequence { pm,n } of weights satisfying the conditions pm,n = 0, m, n = 0, 1, 2, . . . ,

(9.3)

| pk, | ≤ |Pi, j |, k = 0, 1, 2, . . . , i; i = 0, 1, 2, . . . ;  = 0, 1, 2, . . . , j; j = 0, 1, 2, . . . .

(9.4)

for each fixed pair (i, j),

Remark 9.1 From (9.4), it is clear that for every fixed i = 0, 1, 2, . . . , | pi, | ≤ |Pi, j |,  = 0, 1, 2, . . . , j; j = 0, 1, 2, . . .

(9.5)

and for every fixed j = 0, 1, 2, . . . , | pk, j | ≤ |Pi, j |, k = 0, 1, 2, . . . , i; i = 0, 1, 2, . . . .

(9.6)

Note that (9.4) is equivalent to max | pk, | ≤ |Pi, j |, i, j = 0, 1, 2, . . . ;

(9.7)

max | pi, | ≤ |Pi, j |, j = 0, 1, 2, . . . ;

(9.8)

max | pk, j | ≤ |Pi, j |, i = 0, 1, 2, . . . .

(9.9)

0≤k≤i 0≤≤ j

(9.5) is equivalent to 0≤≤ j

(9.6) is equivalent to 0≤k≤i

Since the valuation is ultrametric, |Pi, j | ≤ max | pk, |, i, j = 0, 1, 2, . . . . 0≤k≤i 0≤≤ j

(9.10)

Combining (9.7) and (9.10), we have for every fixed pair (i, j), |Pi, j | = max | pk, |, i, j = 0, 1, 2, . . . . 0≤k≤i 0≤≤ j

(9.11)

9.3 The Weighted Mean Method for Double Sequences

141

Using (9.11), we have, Pm,n = 0, m, n = 0, 1, 2, . . . .

(9.12)

Remark 9.2 (9.11) implies |Pm+1,n+1 | ≥ |Pm,n |;

(9.13)

|Pm,n+1 | ≥ |Pm,n |;

(9.14)

|Pm+1,n | ≥ |Pm,n |.

(9.15)

and

Proof From (9.11), |Pm+1,n+1 | = max | pk, | 0≤k≤m+1 0≤≤n+1



= max



max | pk, |, | pm,n+1 |, | pm+1,n |, | pm+1,n+1 |

0≤k≤m 0≤≤n

  = max |Pm,n |, | pm,n+1 |, | pm+1,n |, | pm+1,n+1 | ≥ |Pm,n |. In a similar fashion, we can prove that (9.11) implies (9.14) and (9.15).



Theorem 9.7 (N , pm,n ) is regular if and only if lim

m+n→∞

|Pm,n | = ∞;

(9.16)

max | pk, |

lim

m+n→∞

and

0≤k≤m

|Pm,n |

= 0,  = 0, 1, 2, . . . ;

(9.17)

= 0, k = 0, 1, 2, . . . .

(9.18)

max | pk, |

lim

m+n→∞

0≤≤n

|Pm,n |

Proof Necessity part. Let (N , pm,n ) be regular. Using (8.2), |am,n,0,0 | = 0,    p0,0   = 0. i.e., lim  m+n→∞ Pm,n  lim

m+n→∞

142

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

Thus, lim

m+n→∞

|Pm,n | = ∞,

since p0,0 = 0, using (9.3). Also, using (8.4), for  = 0, 1, 2, . . . , lim

sup |am,n,k, | = 0,

m+n→∞ k≥0

| pk, | = 0,  = 0, 1, 2, . . . , m+n→∞ 0≤k≤m |Pm,n | max | pk, | 0≤k≤m i.e., lim = 0,  = 0, 1, 2, . . . . m+n→∞ |Pm,n |

i.e.,

lim

max

Thus (9.17) holds. In the same manner, we can prove (9.18). Sufficiency part. Let (9.16), (9.17), (9.18) hold. For every fixed k,  = 0, 1, 2, . . . , lim

m+n→∞

am,n,k, =

lim

m+n→∞

pk, = 0, Pm,n

in view of (9.16). Now, m,n 

lim

∞ 

m+n→∞

am,n,k, =

k,=0

=

lim

Pm,n

m+n→∞

lim

pk,

k,=0

m+n→∞

Pm,n = 1, using (9.12). Pm,n

Also, for every fixed  = 0, 1, 2, . . . , lim

sup |am,n,k, | =

m+n→∞ k≥0

lim

sup |am,n,k, |

m+n→∞ 0≤k≤m

max | pk, |

=

lim

0≤k≤m

|Pm,n | = 0, using (9.17). m+n→∞

In a similar fashion, we can prove that for k = 0, 1, 2, . . . ,

lim

sup |am,n,k, | = 0,

m+n→∞ ≥0

using (9.18). Finally, |am,n,k, | = 0, if k > m or  > n; However, if k ≤ m,  ≤ n, | pk, | |Pm,n | ≤ 1, in view of (9.4).

|am,n,k, | =

9.3 The Weighted Mean Method for Double Sequences

143

Thus, sup |am,n,k, | < ∞. Consequently, the method (N , pm,n ) is regular.



m,n,k,

Example 9.1 Let K = Q p , the p-adic field for a prime p. Then, 0 < c = | p| p < 1, | · | p denoting the p-adic valuation. Let  p m+n , if m + n is odd; pm,n = (9.19) 1 , if m + n is even, p m+n 

and sm,n =

1 , p m+n p m+n ,

if m + n is odd;

(9.20)

if m + n is even.

It is clear that {sm,n } does not converge. Let {tm,n } be the (N , pm,n ) transform of {sm,n }, i.e. ∞  am,n,k, sk, , m, n = 0, 1, 2, . . . , tm,n = k,=0

where  am,n,k, = m,n 

Pm,n =

pk, Pm,n ,

if k ≤ m,  ≤ n;

0,

if k > m or  > n,

pi, j , m, n = 0, 1, 2, . . . .

i, j=0

Now,   2k,2+1    pi, j si, j    |t2k,2+1 | p =    i, j=0 P2k,2+1 

p

|2k(2 + 1)| p = . |P2k,2+1 | p If k,  are both odd or both even, then |t2k,2+1 | p =

|2k(2 + 1)| p |P2k,2+1 | p

=   1 + 2 p +

|2k(2 + 1)| p 3 p2

1 + · · · + (k +  + 1) pk+1

+ (k +  + 1) p k++1 + · · · + 2 .

1 p 2k+2

  + 1 . p 2k+2+1 

p

144

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

   2k(2 + 1)    =  1   2 . 2k+2 p

≤ |p

2k+2

p

|p

= | p|2(k+) ; p If k is even and  is odd or k is odd and  is even, then |t2k,2+1 | p =

|2k(2 + 1)| p |P2k,2+1 | p

=   1 + 2 p +

|2k(2 + 1)| p 3 p2

+ · · · + (k +  + 1) p k+

1 + (k +  + 1) pk++1 + ··· + 2.    2k(2 + 1)    =  2   2k+2 p

≤

1 p 2k+2

  + 1 . p 2k+2+1 

p

p

| p|2(k+) ; p

Similarly, we can prove that ; |t2k+1,2 | p ≤ | p|2(k+) p |t2k,2 | p ≤ | p|2(k+) p and . |t2k+1,2+1 | p ≤ | p|2(k++1) p It now follows that

lim

m+n→∞

tm,n = 0. Thus, {sm,n }, through non-convergent, is sum-

mable (N , pm,n ) to 0. Now, we will prove that the (N , pm,n ) method corresponding to { pm,n } defined in (9.19) is regular. Case 1 When m + n is even; By definition and using (9.11), lim

m+n→∞

|Pm,n | p = =

lim

m+n→∞

max | pi, j | p

0≤i≤m 0≤ j≤n

   1    m+n→∞  p m+n  lim

1 = lim m+n→∞ | p|m+n p = ∞.

p

9.3 The Weighted Mean Method for Double Sequences

145

Also, for  = 0, 1, 2, . . . , max0≤i≤m | pi, | p max0≤i≤m | pi, | p = lim m+n→∞ m+n→∞ max 0≤i≤m | pi, j | p |Pm,n | p lim

0≤ j≤n

= =

lim

max0≤i≤m | pi, | p

1 | p|m+n p lim | p m+n | p max | pi, | p . m+n→∞ 0≤i≤m m+n→∞

Subcase (1) If both m,  are even or both are odd, then     max0≤i≤m | pi, | p m+n  1  lim = lim | p | p  m+  m+n→∞ m+n→∞ |Pm,n | p p p = =

lim

| p n− | p

lim

| p|n− p

m+n→∞ m+n→∞

= 0; Subcase (2) When m is even and  is odd or vice versa, then lim

m+n→∞

max0≤i≤m | pi, | p = lim | p|m+n max | pi, | p p m+n→∞ 0≤i≤m |Pm,n | p    1  m+n  = lim | p | p  m+−1  m+n→∞ p p = =

lim

| p n−+1 | p

lim

| p|n−+1 p

m+n→∞ m+n→∞

= 0. Case 2 When m +n is odd. This case can be discussed in a similar fashion. (9.18) can be proved similarly. Consequently, (N , pm,n ), where { pm,n } is defined as in (9.19), is regular. Theorem 9.8 (Limitation theorem) If {sm,n } is (N , pm,n ) summable to s, then  sm,n − s = o in the sense that

Pm,n pm,n

 , m + n → ∞,

pm,n (sm,n − s) → 0, m + n → ∞. Pm,n

(9.21)

146

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

Proof If {tm,n } is the (N , pm,n ) transform of {sm,n }, then   (P t − Pm,n−1 tm,n−1 − Pm−1,n tm−1,n + Pm−1,n−1 tm−1,n−1 )    m,n m,n  pm,n   − (Pm,n − Pm,n−1 − Pm−1,n + Pm−1,n−1 )s     P (sm,n − s) =  Pm,n m,n      (tm,n − s) − Pm,n−1 (tm,n−1 − s) − Pm−1,n (tm−1,n − s)    Pm,n Pm,n =  P   + m−1,n−1 (t − s) m−1,n−1 Pm,n     ⎤ ⎡  Pm,n−1   Pm−1,n  |tm,n − s|,  Pm,n  |tm,n−1 − s|,  Pm,n  |tm−1,n − s|,   ⎦ ≤ max ⎣  Pm−1,n−1   Pm,n  |tm−1,n−1 − s|

       

≤ max[ |tm,n − s|, |tm,n−1 − s|, |tm−1,n − s|, |tm−1,n−1 − s| ]

in view of Remark 9.2. Since

lim

m+n→∞

tm,n = s, it follows that

   pm,n   lim  (sm,n − s) = 0, m+n→∞ Pm,n  i.e. sm,n − s = o

Pm,n pm,n

 , m + n → ∞,

completing the proof of the theorem.



Definition 9.8 We say that (N , pm,n ) is included in (N , qm,n ) (or (N , qm,n ) includes (N , pm,n )), written as (N , pm,n ) ⊆ (N , qm,n ) if sm,n → s(N , pm,n ) implies sm,n → s(N , qm,n ). We now prove a pair of inclusion theorems involving weighted means for double sequences. Theorem 9.9 (Comparison theorem for two weighted means for double sequences) Let (N , pm,n ), (N , qm,n ) be two weighted mean methods such that qm,n = O( pm,n ), m + n → ∞

(9.22)

in the sense that there exists M > 0 such that    qm,n     p  ≤ M, m, n = 0, 1, 2, . . . m,n and Pm,n = o(Q m,n ), m + n → ∞ in the sense that

(9.23)

9.3 The Weighted Mean Method for Double Sequences

  Pm,n  Q

m,n

where Pm,n =

m,n 

   → 0, m + n → ∞,  m,n 

pk, , Q m,n =

k,=0

147

qk, , m, n = 0, 1, 2, . . . .

k,=0

Then (N , pm,n ) ⊆ (N , qm,n ). Proof Using (9.22), (9.23),

   qm,n Pm,n    = 0 and so there exists H > 0 m+n→∞  pm,n Q m,n  lim

such that    qm,n Pm,n     p Q  ≤ H, m, n = 0, 1, 2, . . . . m,n m,n

(9.24)

Using (9.22), there exists M > 0 such that    qm,n     p  ≤ M, m, n = 0, 1, 2, . . . . m,n

(9.25)

Let, for a double sequence {sm,n }, m,n 

tm,n =

i, j=0

u m,n =

i, j=0

pi, j si, j

Pm,n m,n  qi, j si, j Q m,n

,

, m, n = 0, 1, 2, . . . .

Then, p0,0 s0,0 = P0,0 t0,0 ; pm,n sm,n = Pm,n tm,n − Pm−1,n tm−1,n − Pm,n−1 tm,n−1 + Pm−1,n−1 tm−1,n−1 , so that Pm,n tm,n − Pm−1,n tm−1,n − Pm,n−1 tm,n−1 + Pm−1,n−1 tm−1,n−1 , pm,n Pm,0 tm,0 − Pm−1,0 tm−1,0 = , pm,0 P0,n t0,n − P0,n−1 t0,n−1 = , p0,n

sm,n = sm,0 s0,n

148

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

where we suppose that P−1,n = 0, Pm,−1 = 0, P−1,−1 = 0. Now, ⎛ u m,n = = =

1 Q m,n

⎝

qi, j si, j ⎠

i, j=0

1

m,n 

Q m,n

i, j=0

∞ 

⎞

m,n 



qi, j

Pi, j ti, j − Pi−1, j ti−1, j − Pi, j−1 ti, j−1 + Pi−1, j−1 ti−1, j−1 pi, j

cm,n,k, tk, ,

k,=0

where,

cm,n,k,

⎧ qk, q q ⎪ − pk,+1 − k+1, + ⎪ ⎪ k,+1  pk+1, ⎪  pk, ⎪ ⎪ qk, qk,+1 Pk, ⎪ ⎪ ⎨ pk, − pk,+1  Q m,n , qk, qk+1, Pk, = pk, − pk+1, Q m,n , ⎪ ⎪ ⎪ qk, Pk, ⎪ ⎪ ⎪ pk, Q k, , ⎪ ⎪ ⎩0,

qk+1,+1 pk+1,+1



Pk, Q m,n ,

if k < m,  < n; if k = m,  < n; if k < m,  = n; if k = m,  = n; if k > m or  > n.

Using (9.22), (9.23), we get lim

m+n→∞

cm,n,k, = 0, k,  = 0, 1, 2, . . . .

If sm,n = 1, m, n = 0, 1, 2, . . . , then m,n 

tm,n =

i, j=0

=

i, j=0

pi, j si, j

Pm,n m,n  pi, j

Pm,n Pm,n = Pm,n = 1, using (9.12).



9.3 The Weighted Mean Method for Double Sequences

149

Similarly, u m,n = 1, m, n = 0, 1, 2, . . . so that u m,n =

∞ 

cm,n,k, tk,

k,=0

which yields 1=

∞ 

cm,n,k, (1),

k,=0 ∞ 

i.e.,

cm,n,k, = 1, m, n = 0, 1, 2, . . . .

k,=0

Consequently,

⎛ lim

m+n→∞

⎝

∞ 

⎞ cm,n,k, ⎠ = 1.

k,=0

If k < m and  < n, then     qk, qk,+1 qk+1, qk+1,+1 Pk,  |cm,n,k, | =  − − + pk, pk,+1 pk+1, pk+1,+1 Q m,n               qk,   Pk,   qk,+1   Pk,   qk+1,   Pk,   qk+1,+1   Pk,   ,  ,  ,   ≤ max                pk, Q m,n pk,+1 Q m,n pk+1, Q m,n pk+1,+1 Q m,n               qk,   Pk,   qk,+1   Pk,+1   qk+1,   Pk+1,   qk+1,+1   Pk+1,+1   ,  ,  ,   ≤ max  pk,   Q k,   pk,+1   Q k,+1   pk+1,   Q k+1,   pk+1,+1   Q k+1,+1  ≤ H, by (9.24),

since k < m,  < n imply |Q k, |, |Q k,+1 |, |Q k+1, |, |Q k+1,+1 | ≤ |Q m,n | and so

1 1 1 1 1 ≤ , , , |Q m,n | |Q k, | |Q k,+1 | |Q k+1, | |Q k+1,+1 |

and |Pk, | ≤ |Pk+1, |, |Pk,+1 |, |Pk+1,+1 |. The cases when k = m,  < n and k < m,  = n can be similarly discussed. If k = m,  = n, then    qm,n Pm,n    |cm,n,m,n | =  pm,n Q m,n  ≤ H, by (9.24)

150

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

and |cm,n,k, | = 0 ≤ H if k > m or  > n. Thus sup |cm,n,k, | < ∞. m,n,k,

We shall now prove that for every fixed  = 0, 1, 2, . . . , sup |cm,n,k, | = 0.

lim

m+n→∞ k≥0

For every fixed  = 0, 1, 2, . . . , there are three cases, viz.,  < n,  = n and  > n. Case 1 When  < n, lim

sup |cm,n,k, | =

m+n→∞ k≥0

max |cm,n,k, |

lim

m+n→∞ 0≤k≤m



=



max

lim

m+n→∞

max

0≤k≤m−1

 |cm,n,k, |, |cm,n,m, |

    qk, qk,+1 qk+1, qk+1,+1 Pk,   , − − +  m+n→∞ 0≤k≤m−1 pk, pk,+1 pk+1, pk+1,+1 Q m,n      qm, qm,+1 Pm,   −  p pm,+1 Q m,n  m,    Pm,n   , using (9.25) ≤ lim M  m+n→∞ Q m,n 

=

lim



max

max

= 0, using (9.23).

Therefore, sup |cm,n,k, | = 0 if  < n.

lim

m+n→∞ k≥0

Case 2 The case  = n can be proved similarly. Case 3 If  > n, then cm,n,k, = 0, m, n = 0, 1, 2, . . . , by definition. Thus

sup |cm,n,k, | = 0.

lim

m+n→∞ k≥0

Consequently, lim

sup |cm,n,k, | = 0.

m+n→∞ k≥0

Similarly, we can prove that lim

sup |cm,n,k, | = 0, k = 0, 1, 2, . . . .

m+n→∞ ≥0

9.3 The Weighted Mean Method for Double Sequences

151

The method (cm,n,k, ) is thus regular and so (N , pm,n ) ⊆ (N , qm,n ), 

completing the proof of the theorem.

Theorem 9.10 (Comparison theorem for a (N , pm,n ) method and a regular matrix) Let (N , pm,n ) be a weighted mean method and A = (am,n,k, ) be a regular matrix. If am,n,k, Pk, = 0, m, n = 0, 1, 2, . . . k+→∞ pk, lim

(9.26)

and Pm,n = O( pm,n ), m + n → ∞,    Pm,n    ≤ M, M > 0, m, n = 0, 1, 2, . . . , i.e.,  pm,n 

(9.27)

then (N , pm,n ) ⊆ A. Proof Let {sm,n } be any double sequence, {tm,n }, {τm,n } be its (N , pm,n ), Atransforms, respectively. Then, m,n 

tm,n = τm,n =

pi, j si, j

i, j=0

Pm,n ∞ 

,

am,n,k, sk, , m, n = 0, 1, 2, . . . .

k,=0

Now, sm,n =

Pm,n tm,n − Pm−1,n tm−1,n − Pm,n−1 tm,n−1 + Pm−1,n−1 tm−1,n−1 , pm,n

m, n = 0, 1, 2, . . . , where t−1,n = tm,−1 = t−1,−1 = 0.

152

Let

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

lim

m+n→∞

tm,n = s. Then,

∞ 

τm,n =

am,n,k, sk,

k,=0 ∞ 

 am,n,k,  Pk, tk, − Pk−1, tk−1, − Pk,−1 tk,−1 + Pk−1,−1 tk−1,−1 pk, k,=0 ∞   am,n,k, am,n,k+1, am,n,k,+1 am,n,k+1,+1 Pk, tk, , = − − + pk, pk+1, pk,+1 pk+1,+1

=

k,=0

(9.28) noting that τm,n exists, m, n = 0, 1, 2, . . . , since the right-hand side of (9.28) exists using (9.26) and using the fact that {tk, } is convergent and so bounded and   Pk,  P

k,+1

   ≤ 1, 

  Pk,  P

k+1,

We now write τm,n =

   ≤ 1 and 

∞ 

  Pk,  P

k+1,+1

   ≤ 1. 

bm,n,k, tk, ,

k,=0

where  bm,n,k, =

am,n,k, am,n,k+1, am,n,k,+1 am,n,k+1,+1 − − + pk, pk+1, pk,+1 pk+1,+1

 Pk, .

By using (9.27), sup |am,n,k, | < ∞ and the fact that m,n,k,

  Pk,  P

k,+1

   ≤ 1, 

  Pk,  P

k+1,

   ≤ 1 and 

  Pk,  P

k+1,+1

   ≤ 1, 

we have, sup |bm,n,k, | < ∞. m,n,k,

Also, using (9.27) and the regularity of A, we get lim

m+n→∞

bm,n,k, = 0, k,  = 0, 1, 2, . . . .

Let sm,n = 1, m, n = 0, 1, 2, . . . . Then tm,n = 1, m, n = 0, 1, 2, . . . . It now follows that

9.3 The Weighted Mean Method for Double Sequences ∞ 

bm,n,k, =

k,=0

∞ 

153

am,n,k, , m, n = 0, 1, 2, . . . .

k,=0

Consequently, ⎛ lim

m+n→∞

⎝

∞ 

⎞

⎛

bm,n,k, ⎠ =

k,=0

lim

m+n→∞

⎝

∞ 

⎞ am,n,k, ⎠

k,=0

= 1, since A is regular. Now, for  = 0, 1, 2, . . . ,   am,n,k, am,n,k+1, am,n,k,+1 lim sup |bm,n,k, | = lim sup  − − m+n→∞ k≥0 m+n→∞ k≥0 pk, pk+1, pk,+1    am,n,k+1,+1 + Pk,  pk+1,+1       am,n,k,   am,n,k+1,  ≤ lim sup max  Pk,  ,  Pk,  , m+n→∞ k≥0 pk, pk+1,      am,n,k,+1   am,n,k+1,+1   Pk,  ,  Pk,  .  p pk+1,+1 k,+1 Now,      am,n,k,   Pk,      lim sup Pk,  = lim sup |am,n,k, |  m+n→∞ k≥0  pk, m+n→∞ k≥0 pk,  ≤M

lim

sup |am,n,k, |, using (9.27)

m+n→∞ k≥0

= 0, since A is regular and so lim

sup |am,n,k, | = 0.

m+n→∞ k≥0

Again,      am,n,k+1,   Pk+1,      lim sup Pk,  ≤ lim sup |am,n,k+1, |  m+n→∞ k≥0  pk+1, m+n→∞ k≥0 pk+1,  ≤M

lim

sup |am,n,k+1, |, using (9.27)

m+n→∞ k≥0

= 0, since A is regular.

154

9 The Nörlund Method and The Weighted Mean Method for Double Sequences

Similarly, we can prove that    am,n,k,+1   Pk,  = 0 lim sup  m+n→∞ k≥0 pk,+1    am,n,k+1,+1   lim sup  Pk,  = 0. m+n→∞ k≥0 pk+1,+1

and

Consequently, lim

sup |bm,n,k, | = 0,  = 0, 1, 2, . . . .

m+n→∞ k≥0

Similarly we can prove that lim

sup |bm,n,k, | = 0, k = 0, 1, 2, . . . .

m+n→∞ ≥0

The method (bm,n,k, ) is thus regular and so lim

m+n→∞

τm,n = s. In other words,

lim

m+n→∞

tm,n = s implies that

(N , pm,n ) ⊆ A, completing the proof of the theorem.



We conclude, this book by remarking that there are other aspects of ultrametric analysis concerning summability theory, viz., sequence spaces, matrix transformations between sequence spaces, etc. We have not included these aspects in the current edition. For such topics, interested readers can refer to [4–20].

References 1. Natarajan, P.N., Srinivasan, V.: Silverman-Toeplitz theorem for double sequences and series and its application to Nörlund means in non-archimedean fields. Ann. Math. Blaise Pascal 9, 85–100 (2002) 2. Natarajan, P.N., Sakthivel, S.:Multiplication of double series and convolution of double infinite matrices in non-archimedean fields. Indian J. Math. 50, 125–133 (2008) 3. Natarajan, P.N., Sakthivel, S.: Weighted means for double sequences in non-archimedean fields. Indian J. Math. 48, 201–220 (2006) 4. Rangachari, M.S., Srinivasan, V.K.: Matrix transformations in non-archimedean fields. Indag. Math. 26, 422–429 (1964) 5. Katok, S.: p-adic analysis compared with real, Student Mathematical Library, Amer. Math. Soc. 37 (2007) 6. Natarajan, P.N., Rangachari, M.S.: Matrix transformations between sequence spaces over nonarchimedean fields. Rev. Roum. Math. Pures Appl. 24, 615–618 (1979)

References

155

7. Natarajan, P.N.: On a scale of summation processes in the p-adic field. Bull. Soc. Math. Belgique 31, 67–73 (1979) 8. Natarajan, P.N.: Continuous duals of certain sequence spaces and the related matrix transformations over non-archimedean fields. Indian J. Pure Appl. Math. 21, 82–87 (1990) 9. Natarajan, P.N.: Characterization of some special classes of infinite matrices over nonarchimedean fields. Indian J. Math. 34, 45–51 (1992) 10. Natarajan, P.N.: Matrix transformations between certain sequences spaces over valued fields. Indian J. Math. 39, 177–182 (1997) 11. Natarajan, P.N.: On the Algebras (c, c) and (lα , lα ) in Non-archimedean Fields. Lecture Notes in Pure and Applied Mathematics, pp. 225–231. Marcel Dekker, New York (1999) 12. Natarajan, P.N.: Matrix transformations between certain sequence spaces over valued fields II. Indian J. Math. 43, 353–358 (2001) 13. Natarajan, P.N.: Some Properties of Certain Sequence Spaces Over Non-archimedean Fields, p-adic Functional Analysis. Lecture Notes in Pure and Applied Mathematics, pp. 227–232. Marcel Dekker, New York (2001) 14. Natarajan, P.N.: On the algebra (c0 , c0 ) of infinite matrices in non-archimedean fields. Indian J. Math. 45, 79–87 (2003) 15. Natarajan, P.N.: Some results on certain summability methods in non-archimedean fields. J. Comb. Inf. Syst. Sci. 33, 151–161 (2008) 16. Natarajan, P.N.: Some more results on the Nörlund and Y methods of summability in nonarchimedean fields. J. Comb. Inf. Syst. Sci. 35, 81–90 (2010) 17. Natarajan, P.N.: More properties of c0 ( p) over non-archimedean fields. J. Comb. Inf. Syst. Sci. 38, 121–127 (2013) 18. Bhaskaran, R., Natarajan, P.N.: The space c0 ( p) over valued fields. Rocky Mt. J. Math. 16, 129–136 (1986) 19. Raghunathan, T.T.: On the space of entire functions over certain non-archimedean fields. Boll. Un. Mat. Ital. 1, 517–526 (1968) 20. Raghunathan, T.T.: The space of entire functions over certain non-archimedean fields and its dual. Studia Math. 33, 251–256 (1969)

Index

A Absolute value function, 2 Absolutely K-convex, 27 Analytic function, 68 Archimedean axiom, 1

B Banach space, 23 Banach-Steinhaus theorem, 24 Binomial series, 21 Bounded linear map, 23 Brudno’s theorem, 77

C Cauchy double sequence, 124, 125 Cauchy product, 49 Cauchy sequence, 2 c-compactness, 27 c0 -equivalent, 61 Chain, 72 Characteristic function, 41 Characteristic of a field, 7 Circle of convergence, 99 Closed graph theorem, 24 Comparison theorem, 76, 146, 151 Completion of a metric space, 2 Consistent, 76–78, 97, 134 Conull matrix, 69 Convergence field, 68, 69 Convergence preserving method, 30 Convolution product, 138 Core of a sequence, 49 Covering, 39 c0 -stronger, 76

D Derivative, 21 Differentiable, 21 Disc, 91 Double sequence, 115, 116, 124 Double series, 116, 125 ds-complete, 124, 128 E Equivalence theorem, 130 Euler summability method, 89 Exponential series, 21 F Field complete, 6 non-archimedean, 4 spherically complete, 24 ultrametric, 4 valued, 2 Filterbase, 27 Full non-archimedean, 39 ultrametric, 39 G Generalized semiperiodic sequence, 44 H Hahn–Banach theorem, 24 Hahn’s theorem, 44 Hereditary, 40 High indices theorem, 78 Hilbert space, 28

© Springer India 2015 P.N. Natarajan, An Introduction to Ultrametric Summability Theory, Forum for Interdisciplinary Mathematics 2, DOI 10.1007/978-81-322-2559-1

157

158 I Inclusion theorem, 75, 146 Ingleton’s theorem, 24 Invertible, 91, 100 Isomorphic image, 7

K K-convex, 27 Knopp’s core theorem, 49, 50, 53

L Left translative matrix, 78 Limitation theorem, 74 Locally compact, 24, 52 Locally K-convex, 27 Logarithmic series, 21

M Maximal ideal, 15 Mazur–Orlicz theorem, 68 Metric, 3 Multiplicative group, 91

N Nested sequence, 24 N matrix, 138, 139 N (1) matrix, 138, 139 N (2) matrix, 138, 139 Nörlund method, 63 Norm of a bounded linear map, 23 O Open mapping theorem, 24 Ordered abelian semigroup, 72 Orthogonal base, 28 Orthogonality, 28 P P-adic field, 4 integer, 19 number, 4 valuation, 4 Prime ideal, 15 Product theorem, 99 R Radius of convergence, 99 Rearrangement, 58

Index Reflexive Banach space, 27 Residue class field, 16 Right translative matrix, 78

S Schur matrix, 35 Schur’s theorem, 35 Silverman-Toeplitz theorem, 30, 116 Spherically complete, 24 Steinhaus theorem, 35 Steinhaus-type theorem, 38 Subsequence, 53 Summability factor, 29 Summability field, 72 Summability matrix conservative, 30 convergence preserving, 30 equivalent, 64 Euler, 89 left translative, 78 Nörlund, 63 regular, 30 right translative, 78 Taylor, 89, 98 translative, 78 weighted mean method, 73 Symmetric product, 70

T Tauberian theorem, 103, 106, 108, 112 Taylor method, 89, 98 Toeplitz matrix, 30 Translative matrix, 78

U Ultrametric algebra, 27 Banach space, 30 BK space, 85 field, 4 full, 39–41 inequality, 4 seminorm, 27 valuation, 4

V Valuation equivalent, 12 extension, 16 non-archimedean, 4

Index non-trivial, 3 p-adic, 4 ring, 16 trivial, 3 ultrametric, 4 Valued field, 2

159 W Weighted mean method, 73

Z Zorn’s lemma, 25