Geometry of Linear 2-Normed Spaces [1 ed.] 9781621009276, 9781590330197

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Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved. Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved. Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

GEOMETRY OF LINEAR 2-NORMED SPACES

Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

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Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved. Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

GEOMETRY OF LINEAR 2-NORMED SPACES

Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

RAYMOND W. FREESE AND YEOL JE CHO

Nova Science Publishers, Inc. New York

Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

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The authors and publisher have taken care in preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained in this book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered herein. It is sold with the clear understanding that the publisher is not engaged in rendering legal or any other professional services. If legal or any other expert assistance is required, the services of a competent person should be sought. FROM A DECLARATION OF PARTICIPANTS JOINTLY ADOPTED BY A COMMITTEE OF THE AMERICAN BAR ASSOCIATION AND A COMMITTEE OF PUBLISHERS. Printed in the United States of America

Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

CONTENTS

PART I BASIC PROPERTIES OF LINEAR 2–NORMED SPACES

CHAPTER 1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 CHAPTER 2

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LINEAR 2–NORMED SPACES 2.1. 2.2. 2.3. 2.4. 2.5. 2.6. 2.7. 2.8. 2.9.

Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2–Norms and 2–Metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2–Norms and Bivectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Semi–2–Norms and Semi–2–Metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2–Metrics in the Restricted Sense . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Between Points and Midpoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Properties (U) and (L) in Linear 2–Normed Spaces . . . . . . . . . . . . 31 Properties (U) and (L) in 2–Metric Spaces . . . . . . . . . . . . . . . . . . . . . 33 Contributions to Non–Archimedean Functional Analysis . . . . . . . . . 39 CHAPTER 3 2–BANACH SPACES

3.1. Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.2. Bounded Linear 2–Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.3. More Extensions of Bounded Linear 2–Functionals . . . . . . . . . . . . . . 76

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CHAPTER 4 COMPLETION OF LINEAR 2–NORMED SPACES 4.1. Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 4.2. Completion of Linear 2–Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 85 4.3. Relations between Banach Spaces and 2–Banach Spaces . . . . . . . . 91 CHAPTER 5 2–INNER PRODUCT SPACES 5.1. 5.2. 5.3. 5.4.

2–Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Generating 2–Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .103 Generalizations of 2–Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 2–Inner Product Spaces and Gˆ ateaux Partial Derivatives . . . . . . . 118

PART II

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GEOMETRIC PROPERTIES OF LINEAR 2–NORMED SPACES

CHAPTER 6 STRICT CONVEXITY 6.1. 6.2. 6.3. 6.4. 6.5. 6.6.

Elementary Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 Strict Convexity by Duality Mappings . . . . . . . . . . . . . . . . . . . . . . . . . 126 Strict Convexity by p–Semi–Inner Products . . . . . . . . . . . . . . . . . . . . 133 Strict Convexity by Algebraic and 2–Norm Midpoints . . . . . . . . . . 140 Strict Convexity in Topological Vector Spaces . . . . . . . . . . . . . . . . . . 141 2–Norms Generated by Semi–Norms on the Space of Bivectors . 151

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CHAPTER 7 STRICT 2–CONVEXITY 7.1. Elementary Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 7.2. Strict 2–Convexity by Algebraic and 2–Norm Midpoints . . . . . . . . 169 7.3. Strict 2–Convexity by (α, β, γ)–2–Norm and (α, β, γ)–Algebraic Interior Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .173 7.4. Strict 2–Convexity by Duality Mappings . . . . . . . . . . . . . . . . . . . . . . . 182 7.5. Strict 2–Convexity by Extreme Points . . . . . . . . . . . . . . . . . . . . . . . . . 185 CHAPTER 8 UNIFORM CONVEXITY 8.1. 8.2. 8.3. 8.4. 8.5.

Elementary Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 Uniform Convexity in Quotient Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 199 Uniform Convexity in 2–Inner Product Spaces . . . . . . . . . . . . . . . . . 209 Uniform Convexity in the Space of Bivectors . . . . . . . . . . . . . . . . . . . 210 Uniform 2-Convexity in Linear 2-Normed Spaces . . . . . . . . . . . . . . 212

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CHAPTER 9 ISOMETRY CONDITIONS IN LINEAR 2-NORMED SPACES 9.1. 9.2. 9.3. 9.4.

Isometry Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 An Example of a Non–linear Isometry . . . . . . . . . . . . . . . . . . . . . . . . . 228 Weak Conditions of Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 Non–expansive Mappings in Linear 2–Normed Spaces . . . . . . . . . . 235 CHAPTER 10 ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS

10.1. Elementary Properties of Orthogonalities . . . . . . . . . . . . . . . . . . . . 241 10.2. Properties of Orthogonalities and The Characterization Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 10.3. The 2–Dimensional Case and Examples . . . . . . . . . . . . . . . . . . . . . . . 250

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10.4. Orthogonal Triples in Linear 2–Normed Spaces. . . . . . . . . . . . . . . .254 CHAPTER 11 QUADRATIC FORMS ON MODULES 11.1. 11.2. 11.3. 11.4.

Generalized A–Quadratic Forms of Type (P) . . . . . . . . . . . . . . . . . . 265 Generalized A–Quadratic Forms of Type (Q) . . . . . . . . . . . . . . . . . 270 Applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .275 Extensions to Linear 2–Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . 278

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

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SUBJECT INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

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PREFACE The purpose of this book is to give a comprehensive introduction to the study of 2–metric spaces and linear 2–normed spaces. Since 1963, when S. G¨ahler initially introduced the concepts of 2–metric spaces and linear 2–normed spaces, a number of authors have done considerable work on geometric structures of these spaces, for example, strict convexity, strict 2–convexity and uniform convexity, etc., and there are probably no books concerning the theory of 2–metric spaces and linear 2–normed spaces. This book is introdued as a survey of the latest results on the relations between linear 2–normed spaces and normed linear spaces, completion of linear 2–normed spaces, 2–inner product spaces, strict convexity, strict 2– convexity, uniform convexity, isometry conditions, orthogonal relations and quadratic forms, etc., which are mostly due to C.R. Diminnie, R.W. Freese, R.E. Ehret, S. G¨ ahler, M. Newton, Y.J. Cho, C.S. Lin and A. White, etc., and this compilation of results is intended as a reference work for scholars in the field and as an overview for those initiating research in this branch of mathematics. We hope this book will be devoted to the possible applications of metric geometry, functional analysis and topology, as a new tool. We are greatly indebted to Dr. C.R. Diminnie, Dr. K. Iseki, Dr. C.S. Lin and Dr. A. White for their kind comments and corrections for this book. We are also very grateful to Mrs. Y.S. Ahn, Miss S.J. Park, Mr. Y.H. Lee, Mr. K.M. Jang and, especially, Dr. J.H. Kim, Dr. S.S. Kim and Dr. S.M. Kang for their patience, excellent typing and adjustment of this book. Finally, we thank Dr. Frank Columbus, President and Editor-in-chief, Nova Science Publishers, Inc., New York, U.S.A., and his colleagues for their pleasant cooperation and encouragement for this book.

January, 2001 Raymond W. Freese Yeol Je Cho

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CHAPTER 1. INTRODUCTION In 1963, S. G¨ahler [95] published the first of several papers entitled “2– metric spaces and their topological structures” dealing with spaces on which is defined what he calls a 2–metric. The first extensive treatment of general n–metrics was made by K. Menger in 1928, and while S. G¨ ahler limits his considerations to n = 2, his study is more complete in view of the fact that he developes the topological properties of the spaces in question. The second article by S. G¨ahler [96] entitled “Linear 2–normed spaces” is limited to studying the special class of 2–metric spaces which are linear and have defined on them a 2–norm. He continues the development of the topological properties of such spaces and proves among other things that linear 2–normed spaces are normable and uniformizable provided the dimension of the space is greater than one. He also proves that if the space is a linear normed space, then it is possible to define a 2–norm on it. However, the converse is not true. Since 1963, E.Z. Andalafte, Y.J. Cho, C.R. Diminnie, R.E. Ehret, R.W. Freese, S. G¨ ahler, K. Is´eki, C.S. Lin, G. Murphy, M. Newton, A. White and many others have developed extensively the geometric structure of 2–metric spaces and linear 2–normed spaces. In Chapter II, we introduce the concepts of linear 2–normed spaces, 2– metric spaces, semi–2–normed spaces, semi–2–metric spaces, ultra–2–metric spaces, non–Archimedean 2–normed spaces and bivectors, the relations among these spaces, and their topological structures. Most of the results in this chapter are due to R.E. Ehret [72], R.W. Freese and S. G¨ ahler [93], S. G¨ ahler [96], S. G¨ ahler, A.H. Siddiqui and S.C. Gupta [108] and others. In Chapter III, we summarize many results in doctoral dissertations of A. White [242], [243], R.E. Ehret [72] and others. A. White, in his doctoral dissertation [243] entitled “2–Banach Spaces”, augments the concept of a linear 2–normed space by defining Cauchy sequences and convergent sequences for such spaces. These definitions lead directly to the theory of 2–Banach spaces in which he proves that every two dimensional linear 2–normed spaces is a 2–Banach space. He also considers the related idea of linear 2–functionals and their properties, for example, the Hahn–Banach theorem, etc. One consideration of R.E. Ehret’s doctoral dissertation [72] is an extended study of Cauchy sequences including a redefinition of a Cauchy sequence by means of neighborhoods. From this, equivalent sequences are derived, which in turn yield a complete linear space having at least a pseudo 2–norm defined Typeset by AMS-TEX Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

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R. W. FREESE AND Y. J. CHO

on it under the condition that the original 2–norm is uniformly continuous. S. G¨ ahler’s euclidean 2–norm may be defined on every finite dimensional linear space and such spaces are shown to be complete. Hence every finite dimensional space is a 2–Banach space. It is also proved that a space having Property (K), which is defined in Chapter IV, is a 2–Banach space under certain conditions. In [72], R.E. Ehret proved an important extension theorem for linear 2– functionals, which is an extension of A. White’s Hahn–Banach theorem in 2–Banach spaces. This extension involves a subadditive real–valued function p(x, y) which is homogenous in the sense that p(αx, βy) = αβp(x, y) when αβ > 0. A second extension theorem for linear 2–functionals is proved which involves extending a function defined on S × T to one defined on [S] × [T ], where S and T are subsets of a linear 2–normed space, [S] and [T ] are the subspaces generated by them. Most of the results in this chapter are due to R.E. Ehret [72], A. White [242], [243], S.N. Lal and M. Das [153] and others. In Chapter IV, we construct completions of linear 2–normed spaces and relations between Banach spaces and 2–Banach spaces. Most of the results in this chapter are due to R.E. Ehret [72] and others. In Chapter V, we introduce 2–inner product spaces (or 2–pre–Hilbert spaces) defined by C. Diminnie, S. G¨ ahler and A. White and some properties similar to the usual inner product spaces are given. These spaces satisfy a law quite similar to the Parallelogram law, namely,
a + b, c
2 +
a − b, c
2 = 2(
a, c
2 +
b, c
2 ). Linear 2–normed spaces which satisfy this Parallelogram law have defined on them a 2–inner product such that the
derived 2–norm,
a, c
= (a, a|c), is precisely the 2–norm given for the space. In the remainder of this chapter, we give some characterizations of 2–inner product spaces, generate 2–inner products on a linear 2–normed space and give generalizations of 2–inner products. Most of the results in this chapter are due to C.R. Diminnie, S. G¨ ahler and A. White [55], [57], [58], C.R. Diminnie and A. White [61], [62] and others. In Chapters VI and VII, we give many characterizations of strict convexity and strict 2–convexity in linear 2–normed spaces which have been proved by many mathematicians. C.R. Diminnie, S. G¨ ahler and A. White [56] introduced initially the concepts of strict convexity and strict 2–convexity for linear 2–normed spaces. They and many others have obtained significant results in strictly convex and strictly 2–convex linear 2–normed spaces. Especially, C.R. Diminnie and A. White [65] gave some geometric properties concerning strict 2–convexity

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INTRODUCTION

3

for linear 2–normed spaces and the relations between strict convexity and strict 2–convexity for linear 2–normed spaces. On the other hand, E.Z. Andalafte and R.W. Freese [6] and R.W. Freese and E.Z. Andalafte [87] gave some 2–metric properties of Euclidean geometry and 2–betweenness. R.W. Freese and S. G¨ahler [93] and R.W. Freese, Y. J. Cho and S.S. Kim [91], [144] derived some geometric characterizations of strictly 2–convex linear 2–normed spaces in terms of algebraic midpoints, algebraic between points, σ–between points and σ–midpoints, etc. The concept of 2–semi–inner products was introduced by A.H. Siddiqui and S.M. Riziv [209]. But Y. Ho and A. White [123] modified one part of their definition in 2–semi–inner products, gave an example of a 2–semi–inner product which is not a 2–inner product, and proved that if a 2–semi–inner product is homogenous and additive in the second argument and [a, a|c] = [c, c|a], then X is a 2–inner product space with (a, b|c) = [a, b|c], where (·, ·|·) is a 2–inner product. I. Franic [78] and S.A. Sahab and S.M. Khaleelulla [207] obtained some characterizations of strictly convex linear 2–normed spaces by means of 2–semi–inner products. M.E. Newton, in her doctoral dissertation [189], introduced the concepts of an (α, β, γ)–2–norm and (α, β, γ)–algebraic interior points and gave several characterizations of strict convexity and strict 2–convexity in linear 2–normed spaces, which are generalizations of the results of C.R. Diminnie and A. White [60]. Most of the results in this chapter are due to C.R. Diminnie, Y.J. Cho, R.W. Freese, S. G¨ ahler, C.S. Lin, M. Newton and others. In Chapter VIII, we give some characterizations of uniform convexity in linear 2–normed spaces, quotient spaces, 2–inner product spaces, BX – spaces, and relations among these characterizations. Most of the results in this chapter are due to M.E. Newton [189] and others. In Chapter IX, we introduce some results under weaker conditions than isometries in linear 2–normed spaces and examples of non–linear isometries, which were obtained by C.R. Diminnie and A. White [69] and Y.J. Cho and R.W. Freese [32]. Also, we discuss the structure of the set of fixed points of non–expansive mappings on linear 2–normed spaces introduced by C.R. Diminnie and A. White [63], [65] and K. Is´eki [126]. Most of the results in this chapter are due to J.A. Baker [9], C.R. Diminnie [50], C.R. Diminnie and A. White [63], K. Is´eki [126], Y.J. Cho and R.W. Freese [32] and others. In Chapter X, a new orthogonality relation for normed linear spaces is introduced using the concept of 2–norms given in C.R. Diminnie [51]. Several comparisons are drawn between this relation and earlier relations used by

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R. W. FREESE AND Y. J. CHO

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G. Birkhoff, etc. In addition, this new relation is utilized to obtain new characterizations of inner product spaces. Most of the results in this chapter are due to C.R. Diminnie [51] and others. Finally, in Chapter XI, we give a new generalization of a real 2–inner product space among linear 2–normed spaces by using the concepts of quadratic forms introduced by C.S. Lin [155]. Most of the results in this chapter are due to C.S. Lin [155] and others.

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CHAPTER 2. LINEAR 2–NORMED SPACES R.W. Freese and S. G¨ ahler [93], S. G¨ ahler [95], [96], S. G¨ ahler, A.H. Siddiqui and S.C. Gupta [108] introduced the concepts of linear 2–normed spaces, 2–metric spaces, semi–2–normed spaces, semi–2–metric spaces, ultra–2–normed spaces, non–Archimedean 2–normed spaces, their topological structures and relations between 2–norms and bivectors, etc. Particularly, in [90], R.W. Freese and Y.J. Cho showed that a 2–metric space in which unique “lines” exist and which satisfies an additional midpoint property, expressed entirely in terms of the 2–metric, is a linear 2–normed space.

2.1. Linear Spaces

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In this book, the letter K will stand either for the set R of all real numbers or for the set C of all complex numbers. The elements of K are called scalars. The purpose of this section is to introduce the concepts of linear spaces, normed linear spaces, metric spaces, inner product spaces and to present the basic properties of these spaces. Definition 2.1.1. A Linear space (or vector space) X over a field K is a set of elements together with a mapping (x, y) → x + y, from X × X into X, called addition, and a mapping (α, x) → αx, from K × X into X, called scalar multiplication, which satisfy the following conditions : (A1 ) (x + y) + z = x + (y + z), (A2 ) x + y = y + x, (A3 ) There is an element 0 in X such that x +0 = 0 +x = x for all x ∈ X, (A4 ) For each x ∈ X there is an element −x ∈ X such that x + (−x) = (−x) + x = 0, (S1 ) α(x + y) = αx + αy, (S2 ) (α + β)x = αx + βx, (S3 ) α(βx) = (αβ)x, (S4 ) 1 · x = x for all x, y, z ∈ X and α, β ∈ K. A linear space over R will be called a real linear space and a linear space over C a complex linear space. Properties (A1 )-(A4 ) imply that X is an abelian group under addition and Typeset by AMS-TEX Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

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R. W. FREESE AND Y. J. CHO

properties (S1 )-(S4 ) relate the operation of scalar multiplication to addition in X and to addition and multiplication in K. The elements of X are called points or vectors. Note that the symbol 0 is used to denote both the zero scalar and the zero vector and that a linear space over C can also be considered as a linear space over R since R is a subset of C. Example 2.1.1. The n–dimensional real euclidean space Vn (R) is the set of all n–tuples x = (α1 , α2 , · · · , αn ) of real numbers, where addition and scalar multiplication are defined by (α1 , · · · , αn ) + (β1 , · · · , βn ) = (α1 + β1 , · · · , αn + βn ),

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γ(α1 , · · · , αn ) = (γα1 , · · · , γαn ). That Vn (R) satisfies the conditions for a linear space over R is easily verified. Similarly, the set of all n–tuples of complex numbers with the above definitions of addition and scalar multiplication is also a linear space over C, which is denoted by Vn (C). In the case of n = 1, we understand that V1 (K) is the same as K and, hence, K is a linear space over K. Before giving more examples of linear spaces we obtain some elementary properties which can be derived directly from the definition of a linear space as follows: (1) 0x = 0 for all x ∈ X. (2) α0 = 0 for all α ∈ K. (3) The zero vector 0 is unique. (4) For each x ∈ X the element −x specified in (A4 ) is unique. (5) (−1)x = −x for all x ∈ X. (6) Given two vectors x, y ∈ X, there exists a unique vector z ∈ X such that x + z = y. Example 2.1.2. The set X of all functions from a non–empty set S into a field K with addition and scalar mutiplication defined by (f + g)(t) = f (t) + g(t), (αf )(t) = αf (t) for all f, g ∈ X, t ∈ S and α ∈ K is a linear space over K. If in Example 2.1.2, we let S = {1, · · · , n}, then we conclude that Vn (R) and Vn (C) are linear spaces over R and C, respectively. If S = N , the set

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LINEAR 2–NORMED SPACES

7

of all positive integers, then X is the set of all sequences of elements of K with addition and scalar multiplication defined by (αn ) + (βn ) = (αn + βn ), γ(αn ) = (γαn ). Thus these sequences with the given operation form a linear space over K which we denote by V∞ (K). If, in Example 2.1.2, S = {(i, j) : i = 1, · · · , m; j = 1, · · · , n}, then X is the set of all m × n matrices with entries in K. If we denote a matrix by  α1 1 α21  A = (αij ) =   .. .

αm 1

α12 α22 .. .

α1n α2n .. .

... ... .. .

αm 2

   

αm n

...

where αij ∈ K, then the operations in X can be written by  α1 + β 1 1 1 α21 + β12  A+B = (αij )+(βji ) =  ..  . Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

m αm 1 + β1

and

 γα1 1 γα21  γA = γ(αij ) =   .. .

γαm 1

α12 + β21 α22 + β22 .. .

m αm 2 + β2

γα12 γα22 .. .

γαm 2

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

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

γα1n γα2n .. .

α1n + βn1 α2n + βn2 .. .

   = (αij +βji ) 

m αm n + βn

   = (γαij ) 

γαm n

Thus, the set of all m × n matrices with these operations is a linear space over K which we denote by Knm . A subset E of a linear space X over a field K is called a linear subspace of X if αx + βy ∈ E whenever x, y ∈ E and α, β ∈ K, or equivalently, x + y ∈ E and αx ∈ E. The condition that αx + βy ∈ E for all x, y ∈ E and α, β ∈ K implies that the addition and scalar multiplication functions when restricted to E × E and K × E, respectively, map into E. It also implies that 0 ∈ E and −x ∈ E whenever x ∈ E. Since all the other conditions for a linear space will hold in E because they hold in X, we see that a subspace

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E is itself a linear space over K. Note that the singleton {0} is a subspace of any linear space and is called the trivial linear space. A subspace is also called a linear manifold. Consider now the space X of all real–valued functions defined on the closed interval [a, b] with the usual definition of addition and scalar multiplication. This is a linear space over R by Example 2.1.2. Let E be the subset of X consisting of all continuous functions in X. Since αf + βg is continuous if f and g are continuous, α and β are any real numbers, E is a subspace of X. This space will be denoted by C[a, b]. Given a linear space X and a subspace E of X, we can form another linear space as follows: In X, we define a relation a ≡ b if a − b ∈ E. It is easy to check that this is an equivalence relation. Also, this equivalence relation has the following properties: if a ≡ b and c ≡ d, then a + c ≡ b + d, and if a ≡ b, then αa ≡ αb for all α ∈ K. Let X/E denote the quotient set with respect to this equivalence relation and (a) denote an equivalence class with a representative a. Then X/E is a linear space when the operations of addition and scalar multiplication are defined by

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(a) + (b) = (a + b), α(a) = (αa). As usual in such a situation one must show that these definitions are independent of the representative used, for example, one must show that if (a) = (a ) and (b) = (b ), then (a) + (b) = (a + b) = (a + b ) = (a ) + (b ). The linear space X/E is called the quotient space of X modulo E. Definition 2.1.2. Given a linear space X, a norm on X is a real–valued function x → x satisfying the following conditions : (N1 ) x = 0 if and only if x = 0, (N2 ) αx = |α|x, (N3 ) x + y ≤ x + y for all x, y ∈ X and α ∈ K. A linear space X with a norm  ·  defined on it is called a normed linear space, which is denoted by (X,  · ). Example 2.1.3. In Example 2.1.1, Vn (R) is a linear space over R. The norm on Vn (R) is defined by

2  n  x2i x = x21 + x22 + · · · + x2n = 1

i=1

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for all x = (x1 , x2 , · · · , xn ) ∈ Vn (R). In fact, the conditions (N1 ), (N2 ) and (N3 ) can be readily verified. Therefore, (Vn (R),  · ) is a normed linear space. Example 2.1.4. C[a, b] is a linear space over R by Example 2.1.2. The norm of f ∈ C[a, b] is defined by f  = max |f (x)|. x∈[a,b]

In fact, the conditions (N1 ), (N2 ) and (N3 ) are easily verified. Therefore, (C[a, b],  · ) is a normed linear space. The norm of a linear space defines a metric in natural way as follows:

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Definition 2.1.3. A metric (or a distance function) on a set X is a real–valued function (x, y) → d(x, y) satisfying the following conditions: (D1 ) d(x, y) ≥ 0, (D2 ) d(x, y) = 0 if and only if x = y, (D3 ) d(x, y) = d(y, x), (D4 ) d(x, y) ≤ d(x, z) + d(y, z) for all x, y, z ∈ X. A set X with a metric d defined on it is called a metric space, which is denoted by (X, d). Example 2.1.5. Let X be a non–empty set and let a real–valued function d be defined by  0, if x = y, d(x, y) = 1, if x = y. Then d is a metric on X, which is called the discrete metric. Example 2.1.6. In Vn (R), let a real–valued function d be defined by d(x, y) = max{|xi − yi | : i = 1, 2, · · · , n} for all x = (x1 , x2 , · · · , xn ), y = (y1 , y2 , · · · , yn ) ∈ Vn (R). Then (Vn (R), d) is a metric space. Example 2.1.7. In C[a, b], let d(f, g) = max{|f (x) − g(x)| : x ∈ [a, b]}. Then (C[a, b], d) is a metric space.

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If (X,  · ) is a normed linear space, then d(x, y) = x − y defines a metric. Thus, we shall always consider a normed linear space as a metric space equipped with the metric d just defined. In a metric space (and thus in a normed linear space), we define convergence of sequences, Cauchy sequences and completeness as follows: Definition 2.1.4. A sequence {xn } in a metric space (X, d) is said to be convergent to a point x if for every ε > 0 there exists an integer N = N (ε) such that d(xn , x) < ε for all n ≥ N . Definition 2.1.5. A sequence {xn } is a metric space (X, d) is called a Cauchy sequence if for every ε > 0 there exists an integer N = N (ε) such that d(xn , xm ) < ε for all m, n ≥ N . We can easily prove that every convergent sequence is a Cauchy sequence. Definition 2.1.6. A metric space (X, d) is said to be complete if every Cauchy sequence in X converges to a point.

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It is a basic property of the real line R(and also of the complex plane C) that it is a complete metric space. But the set Q of rational numbers is not a complete metric space. Definition 2.1.7. A normed linear space (X,  · ) is called a Banach space if it is complete as a metric space. Example 2.1.8. In Example 2.1.3, (Vn (R), ·) is a normed linear space, which is also a Banach space. Example 2.1.9. In Exmple 2.1.4, (C[a, b],  · ) is a normed linear space, which is also a Banach space. Definition 2.1.8. Let (X1 , d1 ) and (X2 , d2 ) be two metric spaces. A function f from X1 into X2 is called an isometry if d2 (f (x), f (y)) = d1 (x, y) for all x, y ∈ X1 . Two metric spaces (X1 , d1 ) and (X2 , d2 ) are said to be isometric if there exists an isometry of X1 onto X2 . An application of isometry suggests a method for defining a metric for certain sets. Let (X2 , d2 ) be a metric space, X1 be a set and f be a one-to-

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one function from X1 into X2 . If we define d1 (x, y) = d2 (f (x), f (y)) for all x, y ∈ X1 , then (X1 , d1 ) is a metric space. Now, we define an inner product on a linear space and give some basic properties of inner products. Definition 2.1.9. An inner product on a linear space X is a scalar– valued function (x, y) → (x|y) from X × X into K satisfying the following conditions: (I1 ) (x|x) ≥ 0, (x|x) = 0 if and only if x = 0, (I2 ) (x|y) = (y|x), (I3 ) (x + y|z) = (x|z) + (y|z), (I4 ) (αx|y) = α(x|y) for all x, y, z ∈ X and α ∈ K, where (y|x) denotes the complex conjugate. A linear space X with an inner product (·|·) on it is called an inner product space (or pre–Hilbert space), which is denoted by (X, (·|·)). We have some basic properties of inner products:

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Theorem 2.1.1. If (X, (·|·)) is an inner product space, then x = (x|x) defines a norm on X. Theorem 2.1.2. If (X, (·|·)) is an inner product space, then |(x|y)| ≤ xy for all x, y ∈ X, where equality holds if and only if x and y are linearly dependent. The inequality in the Theorem 2.1.2 is well–known as the Schwarz inequality, the Cauchy–Schwarz inequality or the Cauchy–Bunyakovski–Schwarz inequality. Theorem 2.1.3. If (X, (·|·)) is an inner product space, then the inner product (·|·) is a continuous function on X × X. From Theorem 2.1.1, we know that an inner product space can be considered as a normed linear space with the norm defined by x = (x|x). This norm is called the norm associated with the given inner product (·|·). Definition 2.1.10. If an inner product space is complete in the associated norm, then it is called a Hilbert space.

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Example 2.1.10. In the linear space Vn (R), define an inner product by (x|y) = x1 y1 + x2 y2 + · · · + xn yn for all x = (x1 , x2 , · · · , xn ), y = (y1 , y2 , · · · , yn ) ∈ Vn (R). The conditions (I1 )-(I4 ) are trivial to check and the norm obtained from the inner product is the same as the norm in Example 2.1.3. Thus (Vn (R),  · ) is a real Hilbert space. Example 2.1.11. In the linear space C[a, b], define an inner product by

b (f |g) = a f (x)g(x)dx for all f, g ∈ C[a, b]. Then (C[a, b], (·|·)) is an inner product space but it is not complete and so not a Hilbert space. Theorem 2.1.4. If a normed linear space (X,  · ) is an inner product space, then its norm satisfies the Paralleogram Law: x + y2 + x − y2 = 2(x2 + y2 ) for all x, y ∈ X. It is also true that if the Paralleogram Law holds in a normed linear space, then it is an inner product space.

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Next we define various classes of subsets in linear spaces: Definition 2.1.11. A subset C of a linear space X is said to be convex if for x, y ∈ C and t ∈ [0, 1], tx + (1 − t)y ∈ C. Example 2.1.12. A linear subspace of a vector space is clearly a convex set. The intersection of an arbitrary family of convex sets is a convex set. Given an arbitrary set E in a linear space X, there exists a smallest convex set A containing E, that is, the intersection of all convex sets containing E. Thus we have following: Definition 2.1.12. If E is a subset of a linear space X, then the smallest convex set A containing E is called the convex hull of E, which is denoted by conv E. Theorem 2.1.5. If E is a subset of a linear space X, then convE = {z : z =

n i=1

ai xi , ai ≥ 0,

n i=1

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ai = 1, xi ∈ E}.

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Consider now other classes of sets in linear spaces: Definition 2.1.13. Let X be a linear space over K and E be a subset of X. (1) E is said to be symmetric if E = −E = {−x : x ∈ E}. (2) E is said to be absorbing (or absorbent) if for each x ∈ V there exists tx > 0 such that for all |t| ≤ tx , tx ∈ E. (3) E is said to be balanced if tE = {tx : x ∈ E, t ∈ K, |t| ≤ 1} ⊂ E. (4) E is said to be affine if tE + (1 − t)E = {tx + (1 − t)y : x, y ∈ E, t ∈ [0, 1]} ⊂ E. (5) E is called a line through x and y if E = {tx+(1−t)y : x, y ∈ E, t ∈ R}. (6) E is called the segment joining x and y if E = {tx + (1 − t)y : x, y ∈ E, t ∈ [0, 1]}. We now give examples of sets in Definition 2.1.4: Example 2.1.13. Let X = R2 = {(x1 , x2 ) : x1 , x2 ∈ R} and E = {(x1 , x2 ) : x21 +x22 = 1}. It is easy to see that X is symmetric and absorbing. Obviously, E is not convex.

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Example 2.1.14. Let X be a compact Hausdorff space and C(X) be the set of all real valued continuous function on X. Then clearly C(X) is convex and balanced. Let X be a linear space over K. An important class of functions on X is considered in the following definition: Definition 2.1.14. The real–valued function p : X → R is called a semi–norm if the following conditions hold: (1) p(x + y) ≤ p(x) + p(y), (2) p(αx) = |α|p(x) for all x, y ∈ X and all α ∈ K. The semi–norm is said to be a norm if p(x) = 0 if and only if x = 0. If X is a linear space over K and p is a semi–norm on X, then the set Bp (0, 1) = {x : p(x) < 1} is clearly convex, symmetric, balanced and absorbing. The following theorem gives the relation between convex sets which are balanced and absorbing and the semi–norms. Theorem 2.1.6. Let X be a linear space over K and E be a subset in X with the following properties :

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(a) E is convex, (b) E is balanced and absorbing. Then the function pE defined on X by pE (x) = inf{t > 0 : x ∈ tE} has the following properties : (1) pE (x) ≥ 0, (2) pE (x + y) ≤ pE (x) + pE (y), (3) pE (αx) = |α|pE (E), (4) {x : pE (x) < 1} ⊂ E ⊂ {x : pE (x) ≤ 1}. To define the notion of locally convex spaces, first we introduce the notion of topological spaces : Definition 2.1.15. If X is an arbitrary nonempty set, then a topology on X is a collection τ of subsets of X satisfying the following conditions: (1) X ∈ τ, φ ∈ τ, (2) The union of an arbitrary family of elements of τ is in τ , (3) If 01 , · · · , 0m ∈ τ , then 01 ∩ 02 ∩ · · · ∩ 0m ∈ τ . The pair (X, τ ) is called a topological space. For short, X itself is called a topological space.

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Definition 2.1.16. If x ∈ X, then a neighborhood of x is any subset in X containing a set 0 ∈ τ such that x ∈ 0. Definition 2.1.17. If X1 and X2 are two topological spaces and f : X1 → X2 is a mapping defined on X1 with values in X2 , then we say that f is continuous at the point x1 of X1 if for any neighborhood N2 of f (x1 ) there exists a neighborhood N1 of x1 such that f (x) ∈ N2 if x ∈ N1 . The function f is said to be continuous on X1 if it is continuous at every point of X1 . Definition 2.1.18. A linear space X over K is said to be a topological linear space if on X there exists a topology τ such that X × X and K × X with the product topology have the property that + and · are continuous functions. In this case, τ is called a linear topology on X. Definition 2.1.19. A linear topology on a linear topological space X is said to be a locally convex topology if every neighborhood of 0 includes a convex neighborhood of 0. Theorem 2.1.7. If X is a locally convex space over K, then the topology

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of X is determined by a family (pi )i∈I of semi–norms. 2.2. 2–Norms and 2–Metrics In [95] and [96], S. G¨ ahler introduced the concepts of 2–metric spaces and linear 2–normed spaces and their topological structures. By a 2–metric space (X, σ) (or a space X with a 2–metric σ) we understand a set X in which to each triple of elements a, b, c ∈ X there corresponds a real number σ(a, b, c) having the following conditions: (2M1 ) For two different elements a and b in X there is an element c in X such that σ(a, b, c) = 0, (2M2 ) σ(a, b, c) = 0 when two of the three elements are equal, (2M3 ) σ(a, b, c) = σ(a, c, b) = σ(b, c, a), (2M4 ) σ(a, b, c) ≤ σ(a, b, d) + σ(a, d, c) + σ(d, b, c) for any d in X. It is easily proven that the 2–metric σ is non–negative. Every euclidean space of finite dimension m ≥ 2 has the following 2–metric defined on it:

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   ai  1  bi σ(a, b, c) = 2 i 0, define the ε–neighborhood of two points a and b in X as the all points c in X such that σ(a, b, c) < ε. Let B be the set of all set Uε (a, b) of intersections i=1 Uεi (ai , bi ) of finitely many εi –neighborhoods of arbitrary points ai and bi in X. B forms a basis for the 2–metric topology of X while the system of all ε–neighborhoods of two points in X forms a subbasis for this topology. This topology is called the natural topology (or the topology generated by  the 2–metric) in X. The family of all sets WΣ (a) defined by WΣ (a) = i=1 Uεi (a, bi ) with arbitrary pairs Σ = {(b1 , ε1 ), · · · , (bn , εn )} forms a complete system of neighborhoods of the point a. A point a in a 2–metric space X is called a limit point of a set M in X if for an arbitrary Σ = {(b1 , 1 ), · · · , (bn , n )} there is a point aΣ in M , distinct from a, such that aΣ ∈ WΣ (a).

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A 2–metric space (X, σ) is said to be have Property (K) if the following holds: If for a sequence of points a, a1 , a2 , · · · of X there exist two points b and c of X with σ(a, b, c) = 0, lim σ(a, b, ai) = 0,

i→∞

lim σ(a, c, ai) = 0,

i→∞

then limi→∞ σ(a, a , ai ) = 0 for every point a of X. A 2–metric space (X, σ) has Property (S) if for any points a and b in X and every ε > 0 there exists neighborhoods Ua of a and Ub of b such that if points a ∈ Ua and b ∈ Ub , then σ(a, a , b ) < ε. Theorem 2.2.1. A 2-metric space (X, σ) has Property (K) if and only if, for each point triple a, b, c in X with σ(a, b, c) = 0, the collection of sets WΣi (a) with Σi = {(b, 1i ), (c, 1i )}, i = 1, 2, · · · , forms a complete system of neighborhoods of the point a. Theorem 2.2.2. A 2–metric space (X, σ) has Property (K) if and only if, for arbitrary points a, b, c in X with σ(a, b, c) = 0, a is a limit point of a set M in X in case there is a sequence of points x1 , x2 , · · · in M, distinct from a, such that lim σ(a, b, xi) = 0,

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i→∞

lim σ(a, c, xi) = 0.

i→∞

Theorem 2.2.3. A 2–metric space (X, σ) has Property (S) if and only if the triple measure σ(a, b, c) is a continuous area function of three points a, b, c in X. Let X be a real linear space of dimension greater than 1 and let ·, · be a real valued function on X × X satisfying the following conditions: (2N1 ) x, y = 0 if and only if x and y are linearly dependent, (2N2 ) x, y = y, x, (2N3 ) αx, y = |α|x, y, (2N4 ) x, y + z ≤ x, y + x, z for all x, y, z ∈ X and α ∈ R. ·, · is called a 2–norm on X and (X, ·, ·) is called a linear 2–normed space. It is easily proven that the 2–norm ·, · is non–negative. Theorem 2.2.4. In the definition of a linear 2–normed space (X, ·, ·), the condition (2N4 ) may be replaced by the following condition: x + z, y + z ≤ x, y + y, z + z, x.

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Theorem 2.2.5. For every linear 2–normed space (X, ·, ·) the function defined on X × X × X by σ(x, y, z) = x − z, y − z is a 2–metric. Thus, every linear 2–normed space (X, ·, ·) will be considered to be a 2–metric space with this 2–metric, and the 2–metric σ defined thus is continuous in all three variables if the following holds : For every ε > 0 there exists a neighborhood U of 0 such that for all points a∗ and b∗ in U , a∗ , b∗  < ε. This property is equivalent to Property (S) and is simpler to apply. A set M in a linear 2–normed space (X, ·, ·) is said to be bounded relative to the 2–norm ·, · if σ(M ) < ∞, where σ(M ) = sup{σ(a, b, c) : a, b, c ∈ M }. Theorem 2.2.6. For any points a, b ∈ X and any γ ∈ K, a, b = a, b + γa. n n Theorem 2.2.7. For a = i=1 αi ei and b = i=1 βi ei , we have a, b = |α1 β2 − β1 α2 |e1 , e2  in case n = 2, and

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n n     (α2 βi − β2 αi )ei  |α1 β2 − β1 α2 |a, b =  (α1 βi − β1 αi )ei , i=1

i=1

in case n > 2. Theorem 2.2.8. Every linear 2–normed space (X, ·, ·) of dimension different from one is a locally convex topological vector space. As a consequence of Theorem 2.2.8, every linear 2–normed space of dimension diferent from one is uniformizable and thereby completely regular. A second consequence of Theorem 2.2.8 is that in a linear space X of finite dimension greater than one over which a norm (an ordinary norm) and a 2– norm have been defined and the corresponding metric (an ordinary metric) and a 2–metric associated with them, equal topologies are generated by the two metrics. Hence the topologies generated by the usual euclidean metric and the euclidean 2–metric are equal. 2.3. 2–Norms and Bivectors

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S. G¨ ahler [96] defined the space of bivectors over a linear space and gave the relations between the space of bivectors and linear 2–normed spaces as follows:  If X is a linear space mof dimension greater than 1, let BX be the set of all formal expressions i=1 xi × yi where xi , yi , i = 1, 2, . . . , m, are vectors  defined by in X. Let ∼ be the equivalence relation on BX m




xi × y i ∼

i=1

m

xi × yi

i=1

if and only if for arbitrary linear functions f and g on X,  m  m
  f (xi ) g(xi )   f (x ) i   =  f (yi ) g(yi )   f (y  ) i

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i=1

i=1

 g(xi )  . g(yi ) 

 / ∼. The elements of BX are called Let BX be the quotient space BX  belonging to a bivector are bivectors over X and the elements of BX called of this bivector. Thebivector with the representarepresentatives m m tive i=1 xi × yi will also be denoted by b( i=1 xi × yi ). If a bivector has 1 a representative of the form i=1 xi × yi = x1 × y1 , then it is said to be simple. Only in the case where X has dimension less than or equal to 3 does every bivector over X turn out to be simple. Such a bivector b(0 × 0) is called a null–bivector (or a zero bivector). The space BX is a linear space with n m     m+n   b xi × y i + b xi+n × yi+n = b xi × y i i=1

i=1

and αb

m  i=1

i=1

m    xi × y i = b xi × αyi , i=1

where α is a real number. Theorem 2.3.1. If  ·  is a norm on BX , then x, y = b(x × y) defines a 2–norm on X. Theorem 2.3.2. If all bivectors over X are simple, that is, if X has dimension less than or equal to 3, then for every 2–norm ·, · on X, there is a norm  ·  on BX with b(x × y) = x, y for all x, y in X.

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An example, by H. Busemann and E. Strauss [23], shows that it is not necessarily possible to define a norm · on BX such that b(a×b) = a, b if the dimension of X is greater than three. 2.4. Semi–2–Norms and Semi–2–Metrics In [93], R.W. Freese and S. G¨ ahler introduced the concept of a semi– normed linear space and its properties. A semi–norm on a linear space X is a mapping  ·  : X → R+ , the set of non–negative real numbers, with the following conditions: (SN1 ) x = 0 if and only if x = 0, (SN2 ) αx = |α|x for all x ∈ X, α ∈ R. (X,  · ) is called a semi–normed linear space.

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Let X be a linear space of dimension greater than 1 and ·, · be a mapping from X × X into R+ with the following conditions: (S2N1 ) a, b = 0 if and only if a and b are linearly dependent, (S2N2 ) a, b = b, a, (S2N3 ) αa, b = |α|a, b for every α ∈ R, (S2N4 ) a, b = a, b − a for all a, b ∈ X. ·, · is called a semi–2–morm on X and (X, ·, ·) is called a semi–2–normed space. Theorem 2.2.7 holds also in a semi–2–normed space as follows: Theorem 2.4.1. For arbitrary points a, b of a semi–2–normed space (X, ·, ·) and arbitrary points x = αa + βb, y = γa + δb of X where α, β, γ, δ are real numbers, x, y = |αδ − βγ|a, b. Proof. Because of properties (S2N2 ), (S2N3 ) and (S2N4 ) in the case β = 0, the assertion is immediate. If β = 0, it follows that   δ   x, y = αa + βb, γa + δb − (αa + βb) β   βγ − αδ   = αa + βb, a β  αδ − βγ    = a, αa + βb β = |αδ − βγ|a, b, which proves the theorem.

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For every semi–2–normed space (X, ·, ·) and arbitrary points b, b ∈ X for which b, b  = 0, the function defined by a = a, b + a, b  is a semi–norm on X. For every linear space X of dimension greater than 1 and arbitrary point b of X distinct from 0, denote by V (b) the linear subspace generated by b and let Xb be the quotient space X/V (b). For each a ∈ X let (a)b denote the equivalence class with respect to V (b) containing a. Xb is a linear space under addition (a)b +(a )b = (a+a )b and scalar multiplication (αa)b = α(a)b Theorem 2.4.2. If ·, · is a semi–2–norm on X, then for arbitrary a, a ∈ (a)b ∈ Xb we have a, b = a , b, and the function (a)b b = a, b, a ∈ (a)b ∈ Xb , defines a semi–norm  · b on Xb . Proof. Let ·, · be a semi–2–norm on X. For arbitrary a, a ∈ (a)b ∈ Xb there exists a real number α with a = a +αb, and hence from Theorem 2.4.1 we have a, b = a , b. Hence (a)b b = a, b, a ∈ (a)b ∈ Xb , determines a function  · b on Xb with range in R+ . For arbitrary a ∈ X, (a)b b = 0 if and only if a, b = 0, i.e., a ∈ V (b), or (a)b = (0)b . Therefore, for any real number α, we have α(a)b b = (αa)b b = αa, b = |α|a, b = |α|(a)b b ,

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which proves the theorem. Corollary 2.4.3. Let ·, · be a function on X × X with a, b = b, a for all a, b ∈ X and with 0, 0 = 0. For each b ∈ X, b = 0, and for each a, a ∈ (a)b ∈ Xb , suppose a, b = a , b and that (a)b b = a, b, a ∈ (a)b ∈ Xb , determines a semi–norm ·b on Xb . Then ·, · is a semi–2–norm on X. Theorem 2.4.4. For each semi–norm  ·  on BX , a, b = b(a × b) defines a semi–2–norm on X. If X has dimension less than or equal 3, then for every semi–2–norm ·, · on X, b(a × b) = a, b defines a semi–norm on BX . Proof. It is easily seen that for any semi–norm  ·  on BX , a, b = b(a×b) defines a semi–2–norm on X. Now, suppose that X has dimension less than or equal 3 and let ·, · be a semi–2–norm on X. By above remarks, every bivector on X is simple and hence can be represented in the form b(a × b). Using Theorem 2.3.2 and Theorem 2.4.1, we conclude that, by b(a × b) = a, b, a function  ·  is defined on BX . It is easy to see that this function is a semi–norm.

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A semi–2–metric space is a set X with a mapping σ : X × X × X → R+ satisfying the following conditions : (S2M1 ) For every two distinct points a and b of X there exists at least one point c in X such that σ(a, b, c) = 0, (S2M2 ) σ(a, b, c) = 0 if at least two of the three points a, b, c are equal, (S2M3 ) σ(a, b, c) = σ(a, c, b) = σ(b, c, a). σ is called a semi–2–metric and a semi–2–metric space is denoted by (X, σ). Theorem 2.4.5. In every semi–2–normed space (X, ·, ·), σ(a, b, c) = a − c, b − c defines a semi–2–metric. Proof. For every two distinct points a and b of X there exists a point c ∈ X such that a−c and b−c are linearly indepentent and hence σ(a, b, c) = a − c, b − c = 0. Furthermore, σ(a, b, c) = a − c, b − c = 0, if at least two of the three points a, b, c are equal. For arbitrary points a, b, c ∈ X, we have a − c, b − c = a − c − (b − c), c − b = a − b, c − b, and a − c, b − c = b − c − (a − c), c − a = b − a, c − a,

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from which it follow that σ(a, b, c) = σ(a, c, b) = σ(b, c, a), which completes the proof. In the following theorem, every semi–2–normed space X will be considered as a semi–2–metric space with the (natural) semi–2–metric given by σ(a, b, c) = a − c, b − c. Theorem 2.4.6. The natural semi–2–metric σ of a semi–2–normed space (X, ·, ·) possesses the following properties: (1) σ is translation invariant. (2) σ is invariant under reflection in an arbitrary point b ∈ X, i.e., under the mapping f : X → X with f (a) = 2b − a. (3) For arbitrary points a, b, c ∈ X and x = b + α(a − b), where α ∈ [0, 1], then we have σ(a, x, c) = (1 − α)σ(a, b, c), σ(x, b, c) = ασ(a, b, c)

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and σ(a, b, c) = σ(a, x, c) + σ(x, b, c). (4) For arbitrary points a, b, c ∈ X and x = αa + βb + γc, where α, β, γ ∈ [0, 1] and α + β + γ = 1, then we have σ(a, b, x) = γσ(a, b, c), σ(a, x, c) = βσ(a, b, c), σ(x, b, c) = ασ(a, b, c) and σ(a, b, c) = σ(a, b, x) + σ(a, x, c) + σ(x, b, c). Proof. Assertions (1) and (2) are obvious. Let a, b, c, x be points as in assertion (3). Then we have σ(a, x, c) = x − a, c − a = b − a − α(b − a), c − a = (1 − α)b − a, c − a = (1 − α)σ(a, b, c) and Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

σ(x, b, c) = x − b, c − b = αa − b, c − b = ασ(a, b, c), which completes the proof of assertion (3). Assertion (4) is proved similarly. Theorem 2.4.7. In a semi–2–normed space (X, ·, ·), the following statements are equivalent: (1) For all a, b, c ∈ X, a + b, c ≤ a, c + b, c. (2) For all a, b, c ∈ X, a + c, b + c ≤ a, b + b, c + c, a. (3) For all a, b, c, d ∈ X, σ(a, b, c) ≤ σ(a, b, d) + σ(a, d, c) + σ(d, b, c). Proof. The equivalence of the properties (1) and (2) was shown in Theorem 2.2.4. Given that the property (2) holds then we have σ(a, b, c) = a − c, b − c ≤ a − d, b − d + b − d, c − d + c − d, a − d = σ(a, b, d) + σ(a, d, c) + σ(d, b, c), which is the property (3).

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Conversely, assuming the property (3), then we have a + c, b + c = σ(a, b, −c) ≤ σ(a, b, 0) + σ(a, 0, −c) + σ(0, b, −c) = a, b + b, c + c, a, which is the property (2). Hence the the three statements are equivalent. A semi–2–normed space possessing the property (1) (and consequently the equivalent the properties (2) and (3)) of Theorem 2.4.7 is called a linear 2–normed space as in the section 2.2, its semi–2–norm being called a 2– norm. From Theorem 2.4.1 it follows that a semi–2–norm defined on a linear space X of dimension 2 is always a 2–norm. We now describe an example that shows this is no longer the case if the dimension of X is greater than or equal to 3. Let X = R3 . On X we define a semi– 2–norm ·, · under which, for arbitrary linearly independent vectors a = (α1 , α2 , α3 ) and b = (β1 , β2 , β3 ) ∈ X,  

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a, b =

i 0. Let x = a + ζ(b − a). Using the property (3) of Theorem 2.4.6, it follows that σ(a, x, −c) = ζσ(a, b, −c), σ(x, b, −c) = (1 − ζ)σ(a, b, −c),

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σ(a, x, 0) = ζσ(a, b, 0) = ζγσ(a, b, −c) and σ(x, b, 0) = (1 − ζ)σ(a, b, 0) = (1 − ζ)γσ(a, b, −c). Thus we obtain σ(a, −c, x) = σ(a, −c, 0) + σ(a, 0, x) and σ(b, −c, x) = σ(b, −c, 0) + σ(b, 0, x). Under the assumption of the property (2), there exists a real number φ ∈ [0, 1] with 0 = −φc + (1 − φ)x. We may suppose φ = 0 for in the contrary case x = 0 and hence a, b = 0. Thus we have c=

 1 − φ 1−φ x= (1 − ζ)a + ζb . φ φ

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Since φ < 1 and ζ < 1, the property (3) follows. (3) ⇒ (1): Now assume that the property (3) holds and let a, b, c be non– collinear points of X with d being a σ–between point of these points. Then we have σ(a, b, c) = σ(a, b, d) + σ(a, d, c) + σ(d, b, c), and letting a = a − d, b = b − d and c = −c + d, we have a + c , b + c  = a , b  + b , c  + c , a . If a , b b , c c , a  = 0, then from the assumption of the property (3), there exist real numbers α, β > 0 with c = αa + βb . Then −c + d = α(a − d) + β(b − d) and hence we have d=

αa + βb + c , α+β+1

thus verifying d to be an algebraic between point of a, b, c. Now consider the possibility that a , b b , c c , a  = 0. The three subcases a , b  = 0, b , c  = 0, c , a  = 0 are similar and thus we shall deal only with the case a , b  = 0. If b = 0, then we have d = b. Hence we may assume b = 0. Then there exists a real number β such that a = −βb . From a + c , b + c  = b , c  + c , a 

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it follows that β ≥ 0, and thus a − d = −β(b − d) yields d = (a + βb)/(1 + β). Thus we have shown that the property (1) follows from the property (3), which completes the proof of the theorem. Theorem 2.6.4. In a semi–2–normed space (X, ·, ·), the validity of any (and thus each) of the properties from Theorem 2.6.3 implies the validity of any (and thus each) of the properties given in Theorem 2.6.2. Proof. Assume that the properties from Theorem 2.6.3 hold in a space X and let a, b and c be arbitrary points of X with a + c, b + c = a, b = b, c = c, a = 1. 3 By property (3) of Theorem 2.6.3, there exist real numbers α, β > 0 with c = αa + βb. From a, b = b, c = c, a = 1, we have that α = β = 1, which proves the validity of property (2) (as well as property (1)) of Theorem 2.6.2.

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A semi–2–norm on a linear space X is said to be weakly continuous provided for every point a ∈ X, a, · is continuous on every line of X with respect to its natural topology. Every 2–norm is weakly continuous. Concerning stronger continuity properties of 2–norms, we refer to S. G¨ahler [96]. Corollary 2.6.5. Let (X, ·, ·) be a semi–2–normed space with weakly continuous semi–2–norm ·, ·. Assume that for a, b, c, x, y ∈ X with σ(a, b, c) = 0 if x is a σ–between point of a, b, c and y = va + (1 − v)x, v ∈ [0, 1], then y is a σ–between point of a, b, c. Then the properties from Theorem 2.6.2 and Theorem 2.6.3 are equivalent. Proof. Suppose that the properties given in Theorem 2.6.3 are not valid. Then there exist non–collinear points a, b, c of X and a σ–between point x of a, b, c which does not belong to the plane generated by a, b, c. Let σ(a, b, x) ≤ σ(a, x, c) ≤ σ(x, b, c). Under the assumption of the corollary, there exists a σ–between point y = vb + (1 − v)x, v ∈ [0, 1), of a, b, c, such that σ(y, b, c) = max{σ(a, b, y), σ(a, y, c)}. Moreover, there exists a σ–between point z = ζb + (1 − ζ)y or z = ζc + (1 − ζ)y, ζ ∈ [0, 1), of a, b, c, such that, using Theorem 2.4.1, σ(a, b, z) = σ(z, a, c) = σ(z, b, c).

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z is a σ–midpoint but not an algebraic midpoint of a, b, c. Hence the properties from Theorem 2.6.2 are not valid showing that the validity of any (and hence each) of the properties from Theorem 2.6.2 implies the validity of any (and hence each) of the properties given in Theorem 2.6.3. Because of Theorem 2.6.4, the corollary is proven. Corollary 2.6.6. In a linear 2–normed space the properties from Theorem 2.6.2 and Theorem 2.6.3 are equivalent. Proof. See Theorem 7.1.1. 2.7. Properties (U) and (L) in Linear 2–Normed Spaces

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G.P. Murphy [186], [187] showed the existence of subsets called “lines” in a class of spaces called planar. The purpose of this section is to observe that a 2–metric space in which such lines exist uniquely and which satisfy an additional midpoint property, stated entirely in terms of the 2–metric, is a linear 2–normed space. The results in the sections 2.7 and 2.8 were given by G.P. Murphy [186], R.W. Freese and Y.J. Cho [90]. A one–to–one function between subsets A, B of 2–metric spaces (X, σ) and (Y, σ ) that preserves the 2–metric is called a 2–congruence, denoted by A ≈2 B. For any two points x, y of a 2–metric space X, let m(x, y) denote the set of all points z such that σ(x, z, y) = 0 and σ(w, x, y) = 2σ(w, x, z) = 2σ(w, z, y) for all w ∈ X. We observe that trivially x ∈ m(x, x) for any x ∈ X. We note however that m(x, y) does not necessarily consist of a single element as is shown by the 2–metric space consisting of points a, b, c, d, e in which σ(a, b, c) = σ(a, d, c) = σ(a, b, d) = σ(b, c, d) = 0, σ(e, a, b) = σ(e, a, d) = σ(e, b, d) = σ(e, d, c) = 1 and σ(e, a, c) = 2. In this example, m(a, c) = {b, d}. Letting m∗ (x, y) denote any element of m(x, y), a 2–metric space is said to possess Property (U) provided m(x, y) is non–empty for all x, y ∈ X

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and for all a, b, c, d ∈ X, there exists an element m∗ (·, ·) of the respective elements such that m∗ (m∗ (a, b), m∗ (c, d)) = m∗ (m∗ (a, c), m∗ (b, d)). That is, for all a, b, c, d ∈ X and for all e ∈ m(a, b), f ∈ m(c, d), g ∈ m(e, f ), there exists h ∈ m(a, c), i ∈ m(b, d) with g ∈ m(h, i) Following G. Murphy, we define, for a, b ∈ X such that a = b, L(a, b) = {x ∈ X : σ(a, b, x) = 0}. Let X  be the euclidean 2–metric space, that is, the Cartesian plane equipped with the 2–metric A(·, ·, ·) defined by   A (ξ1 , ξ2 ), (η1 , η2 ), (ζ1 , ζ2 ) = |(ξ1 − ζ1 )(η2 − ζ2 ) − (ξ2 − ζ2 )(η1 − ζ1 )|

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and let L be the analog to L in X  . Under certain additional conditions, G.P. Murphy [186] was able to show the following : Property (L): for all q, a, b ∈ X such that σ(a, b, q) = 0, and all a , b , q  ∈ X  with A(a , b , q  ) = 0, {q} ∪ L(a, b)≈2 {q  } ∪ L (a , b ) where a , b , q  are points of the euclidean plane and the 2–metric in the plane is the euclidean area of the triple. It has been shown by S. G¨ ahler [96] that every linear 2–normed space is a 2–metric space. We now show that Properties (L) and (U) hold in every linear 2–normed space. Theorem 2.7.1. For all elements x, y in a linear 2–normed space (X, ·, ·), m(x, y) consists of a single point. Proof. Let x, y ∈ X and m1 and m2 ∈ m(x, y). Then either x = y in which case m(x, y) = {x} or x = y. In this latter case, σ(x, m1 , y) = σ(x, m2 , y) = 0 and hence y − x, m1 − x = y − x, m2 − x = 0. Thus y − x, m1 − x, and m2 − x are linearly dependent. From the fact that for all w ∈ X, 12 σ(w, x, y) = σ(w, x, m1 ) = σ(w, x, m2 ) = σ(w, y, m1) = σ(w, y, m2), we have that m1 and m2 are each the algebraic midpoint of x and y. Hence m1 = m2 . Theorem 2.7.2. A linear 2–normed space (X, ·, ·) possesses Property (L). Proof. In light of Theorem 2.7.1, it suffices to show that for any two points x, y ∈ X, the algebraic line determined by x and y satisfies the conditions required of L(x, y) in Property (L). That is, we shall show that given x, y, q ∈ X for which σ(x, y, q) = 0, there exists a triple x , y  , q  in the

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euclidean plane for which there exists a 2–congruence between {q} ∪ L(x, y) and {q  } ∪ L(x , y  ). We begin by observing that given σ(x, y, q), we can choose three points of the euclidean plane, denoted by x , y  , q  , such that the area A(x , y  , q  ) = σ(x, y, q). Using the fact that the euclidean plane is a linear space, we observe that each element of the line containing x , y  can be written uniquely as γ  x + (1 − γ  )y which is also the case with points of the algebraic line L(x, y) determined by x and y. This induces a natural correspondence between the two sets which we shall show is a 2–congruence. First for every γ in the unit interval, letting z = γx + (1 − γ)y, it is clear from property (2N3 ) of the 2–norn ·, · that σ(q, x, z) = γσ(q, x, y) = γA(q  , x , y  ) = A(q  , x , z  ) and σ(q, z, y) = (1 − γ)σ(q, x, y) = (1 − γ)A(q  , x , y  ) = A(q  , z  , y  ). A similar argument extends the 2–congruence to those points for which γ is not in the unit interval. Thus we have

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{q} ∪ L(x, y)≈2 {q  } ∪ L(x , y  ). Thoerem 2.7.3. A linear 2–normed space (X, ·, ·) possesses Property (U). Proof. In light of Theorem 2.7.1, it will be sufficent to show that algebraic midpoints satisfy the condition required of the points m∗ (x, y). Given a, b, c, d ∈ X, then we have  1 11 (a + b) + (c + d) 2 2 2 1 = [a + b + c + d] 4  1 11 (a + c) + (b + d) = 2 2 2 = m∗ (m∗ (a, c), m∗ (b, d))

m∗ (m∗ (a, b), m∗(c, d)) =

and the conclusion follows. 2.8. Properties (U) and (L) in 2–Metric Spaces

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In the remainder of this section “a space M” will refer to a 2–metric space that possesses Properties (L) and (U). The principal result shows that a space M is a linear 2–normed space and thus in light of Theorem 2.7.2 and Theorem 2.7.3, the Properties (L) and (U) characterize linear 2–normed spaces in the class of 2–metric spaces. In order to prove the main theorem, we shall first derive some properties of the sets L(x, y). The first two theorems follow immediately from the definition of those sets. Theorem 2.8.1. In a space M, if z ∈ L(x, y), x = y, z = x, then L(x, y) = L(x, z). Theorem 2.8.2. In a space M , if w, z ∈ L(x, y), x = y, z = w,then L(x, y) = L(w, z). Theorem 2.8.3. In a space M , if σ(a, b, p) = 0 = σ(a, b, q), then {p} ∪ L(a, b) ≈2 {q} ∪ L(a, b). Proof. This follows immediately from Property (L) and the fact that if / L and p” ∈ / L”, L and L” are lines in the euclidean plane with points p ∈   then there exists a 2–congruence in the plane such that {p }∪L ≈2 {p”}∪L”.

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Theorem 2.8.4. In a space M , every quadruple p, q, r, s is 2– congruently imbeddable in the euclidean 3–dimensional space. Proof. By Theorem 2.5.1, it suffices to show that there do not exist exactly 2 triples, say {p, q, r} and {p, q, s}, such that σ(p, q, r) = σ(p, q, s) = 0 nor that exactly one triple, say {p, q, r}, is such that σ(p, q, r) = 0 unless of the triple of real numbers σ(p, q, s), σ(p, r, s), σ(q, r, s), one is the sum of the other two. By Theorem 2.8.2, if two triples {p, q, r}, {p, q, s} of a quadruple are such that σ(p, q, r) = σ(p, q, s) = 0, then σ(p, r, s) = σ(q, r, s) = 0, and thus contrary to exactly 2 of the triples having that property. If exactly one triple {p, q, r} is such that σ(p, q, r) = 0, then σ(p, q, s) = 0 and by Theorem 2.8.3, {s} ∪ L(p, q) ≈2 {s } ∪ L(p , q  ) where the primed points are in the euclidean plane. Since σ(p, q, r) = 0, the point r  corresponding to r in the 2–congruence lies on the line L (p , q  ) and thus one of the areas A(p , q  , s ), A(p , r  , s ), A(q  , r  , s ) is the sum of

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the other two. Hence the same equality holds for the corresponding points p, q, r, s of M , and so the theorem holds. Let Θ ∈ M be a fixed element. We define scalar multiplcation in the space MΘ as follows : For x ∈ MΘ , 1x = x, and 2x is the element such that m(2x, Θ) = {x}. We know that this element exists by the 2–congruence {q}∪L(2x, Θ), for any q in M , with a subset of the euclidean plane consisting of a euclidean line and a point not contained in that line, and the fact in the euclidean plane such an element is unique. For all integers k > 1, define (k + 1)x as that element such that m([k + 1]x, [k − 1]x) = {kx}.

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Theorem 2.8.5. For all x ∈ MΘ , scalar multiplication kx is well–defined for all positive integers k. Proof. Let a, b1 , b2 , y with {y} = m(a, b1 ) = m(a, b2 ) be given. We shall show that then b1 = b2 . Assume in contrary that b1 = b2 . Then b1 = a = b2 . From {y} = m(a, b1 ) and b1 = a, we have a = y = b1 with σ(a, y, b1) = 0. Also {y} = m(a, b2 ) and a = b2 imply σ(a, y, b2) = 0. From Theorem 2.8.2, σ(a, y, b1) = 0 = σ(a, y, b2) implies σ(a, b1 , b2 ) = 0 = σ(y, b1, b2 ). Letting w ∈ M such that σ(w, a, y) = 0, then we have σ(a, b1, w) = σ(a, b2 , w) = 2σ(a, y, w) = 2σ(y, b1 , w) = 2σ(y, b2 , w) = 0. Since σ(y, b1, b2 ) = 0 and σ(y, b1, w) = σ(y, b2, w) = 0, then by Theorem 2.8.3 and Property (L) , it follows that σ(b1 , b2 , w) = σ(b1 , y, w) + σ(y, b2, w). However, since σ(a, b1 , b2 ) = 0 and σ(w, a, b1) = 0 and σ(w, a, b2) = 0, then by the same argument σ(b1 , b2 , w) = σ(b1 , a, w) + σ(a, b2, w), which is contrary to σ(w, a, bi) > σ(w, y, bi) for i ∈ {1, 2}. From the fact that m(·, ·) is a singleton set, Theorem 2.8.5 can be extended to all rationals of the form 12 k for k a positive integer by ( 12 k)x = y where

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{y} = m(Θ, kx). Thus, Theorem 2.8.5 can be extended to all positive dyadic rationals. Then we define λx for negative dyadic rational by the condition that m(λx, −λx) = Θ. We use the natural correspondence induced by the Property (L) to extend scalar multiplication to all reals. It is clear that this extended definition is well–defined and that the point λx in the above proof and subsequent discussion is independent of the element w used in the proof. Thus for all w ∈ M , σ(Θ, w, λx) = |λ|σ(Θ, w, x). We now define in MΘ the sum x + y of elements x and y of M to be the element such that m(x + y, Θ) = m(x, y).

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By arguments of the proof of Theorem 2.8.5, we see that x+y is well–defined. We have the following as immediate conseqences of the definitions and the Properties (L) and (U): Theorem 2.8.6. For elements x, y in MΘ and for real numbers λ, µ, (1) x + y = y + x, (2) λ(µx) = (λµ)x, (3) 1 · x = x, (4) (λ + µ)x = λx + µx, (5) 2(x + y) = 2x + 2y, (6) 2k (x + y) = 2k x + 2k y for all integers k. Proof. The proofs of parts (1)–(4) follow directly from the definitions and the Properties (L) and (U) along with (in the case of parts (2) and (4)) the uniqueness of lines in the space. Part (6) follows by induction from part (5). To observe the validity of part (5) we note that it would be sufficient to verify that m(Θ, 2(x + y)) = m(Θ, 2x + 2y). This follows from the definition of Property (U) after letting a = 2x, b = 2y and c = d = Θ, since then m(Θ, 2(x + y)) = {x + y} = m(Θ, 2x + 2y). Theorem 2.8.7. In MΘ if λ1 (x + y) = λ1 x + λ1 y and λ2 (x + y) = λ2 x + λ2 y, then we have λ1 + λ2 λ1 + λ2 λ1 + λ2 (x + y) = x+ y. 2 2 2

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Proof. Using Property (U) and letting a = λ1 x, b = λ2 x, c = λ1 y, d = λ2 y, then we have m(m(λ1 x, λ2 x), m(λ1 y, λ2 y)) = m(m(λ1 x, λ1 y), m(λ2 x, λ2 y)) and so we have λ + λ λ1 + λ2  1 2 x, y = m(m(Θ, λ1 x + λ1 y), m(λ2 x + λ2 y)). m 2 2 Thus we have

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λ + λ λ1 + λ2 λ1 + λ2  λ1 + λ2 1 2 x+ y = 2m x, y 2 2 2 2   = 2m m(Θ, λ1 (x + y), m(Θ, λ2 (x + y))   = 2m Θ, m(λ1 (x + y), λ2 (x + y)   λ +λ 1 2 (x + y) = 2m Θ, 2 λ1 + λ2 (x + y). = 2 Corollary 2.8.8. The set of all real numbers α for which α(x + y) = αx + αy is dense in the set of real numbers. Proof. This follows immediately from Theorem 2.8.7 and (6) of Theorem 2.8.6. Theorem 2.8.9. For all x, y in MΘ and for all real numbers λ, λ(x+y) = λx + λy. Proof. In light of Corollary 2.8.8, it suffices, given a sequence {λn } converging to λ, λ = 0 with λn (x + y) = λn x + λn y, to show that λ(x + y) = λx + λy, or equivalently, that m(λx, λy) = λm(x, y). We show first that σ(λx, λm(x, y), y) = 0. We have   σ(λx, λm(x, y), λy) ≤ σ m(λn x, λn y), λm(x, y), λy   + σ λx, m(λn x, λn y), λy   + σ λx, λm(x, y), m(λnx, λn y) .

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Of the three terms on the right, the first and third tend to zero as n increases since m(λn x, λn y) = λn m(x, y). Considering the second one, we have σ(λx, m(λn x, λn y), λy) ≤ σ(λn x, m(λn x, λn y), λy) + σ(λx, λn x, λy) + σ(λx, m(λn x, λn y), λnx). Of these terms, the second one tends to zero while the first and third are equal, respectively, to 12 σ(λy, λnx, λn y) and 12 σ(λx, λnx, λn y). But we have λ − λn σ(λx, Θ, λny) λ λ − λn λn σ(Θ, x, y) = λ λ

σ(λx, λn x, λn y) =

and

λ − λn σ(λy, Θ, λn x) λ λ − λn λn = σ(Θ, x, y) λ λ

σ(λy, λn x, λn y) =

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.

Therefore, (λx, λm(x, y), λy) is arbitrarily small and thus is zero. We note that there can only be one point z ∈ L(λx, λy) such that σ(Θ, z, λx) = 1 2 σ(Θ, λx, λy) and that m(λx, λy) is such a point. However, from the remarks following Theorem 2.8.5, we also have σ(Θ, λm(x, y), λx) = λσ(Θ, m(x, y), λx) = λ2 σ(Θ, m(x, y), x) = =

1 2 λ σ(Θ, y, x) 2

1 1 λσ(Θ, λy, x) = σ(Θ, λy, λx). 2 2

Thus the points m(λx, λy) and λm(x, y) must coincide and the theorem is proved. Theorem 2.8.10. MΘ is a linear space. Proof. In light of the previous theorems it suffices to show that the operation + is associative, i.e., for any x, y, z ∈ MΘ
, x+(y +z) = (x+y)+z. Thus in light of Theorem 2.8.2, we must show m(z, x + y) = m(x, y + z).

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By Property (U), given x, y, Θ, z, we have m(m(x, y), m(Θ, z)) = m(m(Θ, x), m(y, z)). Thus by (5) of Theorem 2.8.6, we have m(2m(x, y), 2m(Θ, z)) = m(2m(Θ, x), 2m(Θ, z)) and hence we have m(x + y, z) = m(x, y + z), which proves the theorem.

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Theorem 2.8.11. MΘ is a linear 2–normed space. Proof. Define a 2–norm ·, · on MΘ × MΘ by x, y = σ(Θ, x, y). Then (2N2 ) holds from the symmetry of the 2–metric function σ. Also if x, y are linearly dependent (i.e., if y = αx, x = Θ), then by definition of scalar mutiplication in MΘ we have σ(Θ, x, y) = 0. Further, if σ(Θ, x, y) = 0, then y ∈ L(Θ, x), which implies that y is is a scalar multiple of x and thus x, y are linearly dependent, which shows that property (2N1 ) holds. That property (2N3 ) holds has been implied earlier. To show property (2N4 ) let x, y, z be any elements of MΘ . By R.W. Freese and S. G¨ ahler [93], a space was defined to be a semi–2–normed space provived it satisfied properties (2N1 ), (2N2 ) and (2N3 ). Thus MΘ is a semi– 2–normed space and thus by Theorem 2.4.7, we have that x + y, z ≤ x, z + y, z. Therefore MΘ is a linear 2–normed space. Thus, from Theorems 2.7.2, 2.7.3 and 2.8.11, we have the following: Theorem 2.8.12. A 2–metric space is a linear 2–normed space if and only if it possesses Properties (L) and (U). 2.9. Contribution to Non–Archimedean Functional Analysis Recently, non–Archimedean funtional analysis and general non–Archimedean analysis have attracted the attention of many mathematicians, especially, A.W. Ingleton [125], A.F. Monna [182], A.C.M. van Rooij [203] and others. In [108], S. G¨ ahler, A.H. Siddiqui and S.C. Gupta introduced the non-archimedean aspect in the theory of 2–metric spaces and linear 2– normed spaces and obtained properties of ultra–2–metric spaces and non– Archimedean 2-normed spaces.

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An ultra–metric on a non–empty set X is a metric σ on X which satisfies the following condition: σ(a, b) ≤ max{σ(a, c), σ(c, b)} for every a, b, c ∈ X. Using the concept of an ultra–metric space as motivation, we give the definition of an ultra–2–metric space: A function σ : X × X × X → R is an ultra–2–metric if σ is a 2–metric and σ satisfies the following condition: (U2M4 )

σ(a, b, c) ≤ max{σ(a, b, d), σ(a, d, c), σ(b, d, c)}.

for a, b, c, d ∈ X. A space (X, σ) with an ultra–2–metric σ is called an ultra–2–metric space. By the definition of ultra–2–metric, we see that an ultra–2–metric σ on X is non–negative. Theorem 2.9.1. A 2–metric σ on X is an ultra–2–metric if and only if for any points a, b, c, d in X, the two largest of the numbers Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

σ(a, b, c), σ(a, b, d), σ(a, c, d), σ(b, c, d) are equal. Proof. Let σ be a 2–metric on X. If for any points a, b, c and d of X the two largest of the above mentioned numbers are equal, clearly σ satisfies the condition (U2M4 ), which implies that σ is an ultra–2–metric on X. Conversely, we now suppose that σ is an ultra–2–metric on X. For given points a, b, c, d ∈ X, let σ(a, c, b) ≤ σ(a, b, d) ≤ σ(b, c, d). Then we have σ(a, c, d) ≤ max{σ(a, c, b), σ(a, b, d), σ(b, c, d)} = σ(b, c, d). If σ(a, b, d) < σ(b, c, d), then we have σ(b, c, d) ≤ max{σ(b, c, a), σ(b, a, d), σ(a, c, d)} = max{σ(b, a, d), σ(a, c, d)} = σ(a, c, d)

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and therefore σ(b, c, d) = σ(a, c, d). Thus the two largest of the numbers σ(a, b, c), σ(a, b, d), σ(a, c, d) and σ(b, c, d) are equal and the theorem is proven. Let (X, σ) be an ultra–2–metric space and Uα (a, b) = {c ∈ X : σ(a, b, c) < α} for any real number α > 0 and a, b ∈ X. For any Σ = {(a1 , α1 ), (a2 , α2 ), · · · , (an , αn )}, ai ∈ X, αi ∈ R, αi > 0, i = 1, 2, . . . , n and a ∈ X, let WΣ (a) = ∩nn=1 {c ∈ X : σ(a, ai, c) ≤ αi } and WΣ (a) = ∩ni=1 Uα (a, ai). Theorem 2.9.2. Two points a and a ∈ X are equal if and only if for every α > 0 and every b ∈ X, Uα (a, b) ∪ Uα (a , b) ⊂ Uα (a, a ).

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Proof. If a = a , then it is clear that for every α > 0 and b ∈ X the desired relation is true. Conversely, assume that the above relation holds. For c ∈ Uα (a, b), we have σ(a , b, c) ≤ max{σ(a , b, a), σ(a, a, c), σ(a, b, c)} and so c ∈ Uα (a , b), which implies that Uα (a, b) ⊂ Uα (a , b). By interchanging a and a , we have Uα (a , b) ⊂ Uα (a, b) and so Uα (a, b) = Uα (a , b). Therefore, by S. G¨ahler, it follows that a = a . Theorem 2.9.3. Let a and b be any points in X. For a given Σ = {(a1 , α1 ), · · · , (an , αn )}, let Σ = Σ ∪ {(a, inf αi ), (b, inf αi )}. Then b ∈ WΣ
(a) if and only if WΣ
(a) = WΣ
(b). Proof. From WΣ
(a) = WΣ
(b), the relation b ∈ WΣ
(a) follows immediately. Conversely, let b ∈ WΣ
(a). For any c ∈ WΣ
(a), we have σ(b, ai, c) ≤ max{σ(b, ai, a), σ(b, a, c), σ(a, ai, c)} < αi , σ(b, a, c) < inf αi , σ(b, b, c) < inf αi , which imply that c ∈ WΣ
(b) and hence WΣ
(a) ⊂ WΣ
(b). Since b ∈ WΣ
(a), a ∈ WΣ
(b) and we can prove as above that WΣ
(b) ⊂ WΣ
(a). Therefore, we have WΣ
(a) = WΣ
(b).

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Theorem 2.9.4. The sets WΣ (a), WΣ (a) and WΣ (a) − WΣ (a) are both open and closed. Proof. 1. By the definition of the natural topology τ of (X, σ) (cf. Section 2.2 of Chapter II), every set WΣ (a), Σ = {(a1 , α1 ), . . . , (an , αn )}, is open. Let b be any point in WΣ (a) and Σ = Σ ∪ {(a, inf αi )}. There exists a point cΣ
∈ WΣ
(b) ∩ WΣ
(a). Therefore we get σ(a, ai, b) ≤ max{σ(a, ai, cΣ
), σ(a, cΣ
, b), σ(cΣ
, ai , b)} < αi and thus b ∈ WΣ (a), which shows that WΣ (a) is also closed. 2. Every set WΣ (a), Σ = {(a1 , α1 ), . . . , (an , αn )}, is closed by S. G¨ ahler. Let b be a point in WΣ (a) and Σ = Σ ∪ {(a, inf αi )}. For every c ∈ WΣ
(b), we have σ(a, ai, c) ≤ max{σ(a, ai, b), σ(a, b, c), σ(b, ai, c)} ≤ αi and thus c ∈ WΣ (a). Therefore, WΣ
(b) ⊂ WΣ (a). This proves that WΣ (a) is open. 3. By virtue of the first two statements, it is clear that every set WΣ (a) − WΣ (a) is both open and closed.

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Corollary 2.9.5. Every ultra–2–metric space which contains more than one point is totally disconnected. A topological space (X, τ ) is said to be ultra–metrizable,(resp., ultra–2– metrizable) if an ultra-metric (resp., an ultra–2–metric) compatible with τ exists. S. G¨ ahler [95] gave an example of a 2–metric space which is not metrizable. It can be seen without difficulty that the 2–metric given there is an ultra–2–metric. Thus we have an example of an ultra–2–metric space which is not metrizable and hence not ultra–metrizable. Theorem 2.9.6. Let (X, τ ) be a topological space with more than two points. If (X, τ ) is ultra–metrizable (that is, by A.C.M. van Rooij [203], homeomorphic to a subspace of a countable product of discrete spaces), then (X, τ ) is ultra–2–metrizable. There exists an ultra–2–metric σ on X, compatible with the given topology τ , such that (X, σ) has the Properties (K) and (S). ahler, Proof. Let σ ∗ be an ultra–metric on X compatible with τ . By S. G¨ σ(a, b, c) = inf{σ ∗ (a, b), σ ∗(b, c), σ ∗(c, a)}

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defines a 2–metric σ on X, compatible with τ , such that (X, σ) has Properties (K) and (S). We shall prove that σ is an ultra–2–metric. Let a, b, c, d be any points in X with σ ∗ (a, b) = inf{σ ∗ (x, y) : x, y ∈ {a, b, c, d}, x = y}. For such points, we have σ(a, b, c) = σ(a, b, d) = σ ∗ (a, b) ≤ inf{σ(a, c, d), σ(b, c, d)}.

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If σ ∗ (a, b) < σ ∗ (a, x) or σ ∗ (a, b) < σ ∗ (b, x) for x ∈ {c, d}, then by A.C.M. van Rooij [203], the relation σ ∗ (a, x) = σ ∗ (b, x) holds. Thus we have σ ∗ (a, c) = σ ∗ (b, c) and σ ∗ (a, d) = σ ∗ (b, d), which imply that σ(a, c, d) = σ(b, c, d). Therefore, by Theorem 2.9.1, σ is an ultra–2–metric. Next, we introduce the concept of a non–Archimedean 2–normed space: Let K be a non–trivial valued field, X be a linear space over K and  ·  be a non–negative function on X with the following conditions: (AN1 ) a = 0 if and only if a = 0, (AN2 ) αa = |α|a for all α ∈ K, (AN3 ) a + b ≤ max{a, b} for all a, b ∈ X.  ·  is called a non–Archimedean norm and (X,  · ) is called a non– Archimedean normed space. If the condition (AN1 ) on  ·  is dropped, then  ·  is called a non–Archimedean semi–norm. We call a linear 2–normed space (X, ·, ·) over a non–trivial valued field K non–Archimedean if the 2–metric σ associated with the 2–norm ·, · is an ultra–2–metric. Theorem 2.9.7. A linear 2–normed space (X, ·, ·) is non–Archimedean if and only if for any points a, b, c in X, a, b + c ≤ max{a, b, a, c}. Proof. 1. Assume a, b + c ≤ max{a, b, a, c} for any points a, b, c ∈ X. Then the 2–metric σ associated with the 2–norm ·, · satisfies the following:

σ(a, b, c) = (a − d) − (c − d), (b − d) − (c − d) ≤ max{a − d, b − d, b − d, c − d, c − d, a − d} = max{σ(a, b, d), σ(a, d, c), σ(d, b, c)}, that is, (X, ·, ·) is non–Archimedean.

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2. Suppose now that (X, ·, ·) is non–Archimedean. Let a, b, c be in X and α be an element in K. Then we have a, b + c = σ(−αc, b + c − a − αc, −(a + αc)) ≤ max{σ(−αc, b + c − a − αc, 0), σ(−αc, 0, −(a + αc), σ(0, b + c − a − αc, −a − αc)} = max{b + c, a + αc, |α|c, b − a, |α|a, c} and a + αc, αb + αc = σ(a, αb, −αc) ≤ max{σ(a, αb, 0), σ(a, 0, −αc), σ(0, αb, −αc)} = max{|α|a, b, |α|a, c, |α|b, c}. For α ∈ K, α = 0, we have a, b + c ≤ max{a, b, a, c, |α|b, c, |α|c, b − a, |α|c, a}

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Since the valuation of α is non–trivial, it is possible to have α → 0 and so we have a, b + c ≤ max{a, b, a, c}. Corollary 2.9.8. If (X, ·, ·) is non–Achimedean, then the valuation of K is also non–Archimedean. Proof. For any natural number n and the identity element e of K, since we have |ne|a, b = a, nb ≤ max{a, b, . . . , a, b} = a, b, we get |ne| ≤ 1. Theorem 2.9.9. A linear 2–normed space (X, ·, ·) is non–Archimedean if and only if for any points a, b, c in X with a, b < a, c, a, b+c = a, c. Proof. 1. Suppose that the condition given holds and let a, b, c be any points in X. If a, c < a, b + c, then a, c = a, b + c − b = a, b + c and a, b = a, b + c − c = a, b + c, which imply that a, b + c ≤ max{a, b, a, c}. Therefore, (X, ·, ·) is non–Archimedean by Theorem 2.9.7.

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2. If (X, ·, ·) is non–Archimedean, then for any points a, b, c in X with a, b < a, c, we have a, b + c ≤ max{a, b, a, c} = a, c = a, b + c − b ≤ max{a, b + c, a, b} = a, b + c and thus we have a, b + c = a, c.

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Let X be a linear space over a valued field K and T be a vector topology on X. A subset M of X is said to be convex (or K–convex) (cf. Definition 2.1.2.) if for any points a, b, c ∈ M and any α, β, γ ∈ K with α ≤ 1, β ≤ 1, γ ≤ 1 and α + β + γ = e (e is the indentity of K), αa + βb + γc ∈ M . The topological vector space (X, T ) is said to be locally convex (cf. Definition 2.1.10) if there exists a base of convex neighborhoods of 0. If (X, ·, ·) is a non–Archimedean linear 2–normed space, then for any a ∈ X, a function  · a defined by ba = a, b is a non–Archimedean semi–norm on X and the family of all semi–norms  · a , a ∈ X, generates the natural topology of (X, ·, ·). From this and A.F. Monna [183], [184], we have the following: Theorem 2.9.10. Every non–Archimedean linear 2–normed space is a locally convex topological vector space. Corollary 2.9.11. Every non–Archimedean linear 2–normed space is uniformizable and hence completely regular. Proof. See Theorem 2.9.10 and N. Bourbaki [19]. For every non–Archimedean linear 2–normed space (X, ·, ·) and every pair of points a, b ∈ X with a, b = 0, the function  · a,b defined by ahler ca,b = max{ca , cb } is a non–Archimedean norm on X. By S. G¨ [96], we can prove the following: Theorem 2.9.12. Let (X, ·, ·) be a non–Archimedean linear 2–normed space. Then (X, ·, ·) has Property (K) if and only if for every pair of points a, b ∈ X with a, b = 0 the non–Archimedean norm  · a,b generates the same topology as ·, ·.

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Recall that an ultra–metric space is spherically complete if the intersection of every shrinking sequence of balls is non–empty. Every spherically complete ultra–metric space is complete. Let (X,  · ∗ ) be a non–Archimedean normed space of dimension > 1 over K. If K is spherically complete, by A.W. Ingleton, an analogue of the Hahn–Banach theorem holds, that is, every continuous linear mapping f ∗ defined on a linear subspace X ∗ of X into K can be extended to a continuous linear mapping f defined on X with the same norm as f ∗ , where the norm f ∗ ∗ of f ∗ is defined by f ∗ ∗ = sup

a∈X ∗ a=0

|f ∗ (a)| . a∗

For every a ∈ X, a = 0, there exists a continuous linear mapping f : X → K with |f (a)| = f ∗ a∗ and 0 < f ∗ ≤ 1. In fact, let Xa be the linear subspace of X generated by a and µ be an element of K with 0 < |µ|/a∗ ≤ 1. A continuous linear mapping f ∗ from Xa into K can be introduced by f ∗ (a) = µ. It follows that

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|f ∗ (λa)| =

|µλ|a∗ |µ|λa∗ = a∗ a∗

and hence f ∗ = |µ|/a∗ ≤ 1 and f ∗ (λa) = f ∗ ∗ a∗ . Let f be an extension of f ∗ with f ∗ = f ∗ ∗ . Then we have |f (a)| = f ∗ a∗ and 0 ≤ f ∗ ≤ 1. Theorem 2.9.13. Let K be a spherically complete non–trivial non– Archimedean valued field and (X,  · ∗ ) be a non–Archimedean normed space over K of dimension > 1. Let F be the set of all continuous linear mappings of X into K with norm ≤ 1. Then there exists a non–Archimedean 2–norm ·, · on X, compatible with the topology generated by  · ∗ , such that (X, ·, ·) has Properties (K) and (S), where the 2–norm ·, · is defined by a, b = sup |f (a)g(b) − g(a)f (b)|. f,g∈F

Proof. For any points a, b ∈ X, 0 ≤ a, b ≤ 2a∗ b∗ and so a, b ∈ [0, ∞). It is clear that if a and b are linearly dependent, then a, b = 0. Now suppose that a, b = 0, that is, |f (a)g(b) − g(a)f (b)| = 0 for all f, g ∈ F .

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We shall show that a and b are linearly dependent. Let a be in X with a = 0. By the statement before Theorem 2.9.13, there is an f ∈ F such that |f (a)| = f ∗ a∗ = 0. Moreover, there exists a γ ∈ K with f (b + γa) = 0. Since we have |f (a)g(b + γa)| = |f (a)g(b + γa) − g(a)f (b + γa)| = |f (a)g(b) − g(a)f (b)| = 0, it follows that g(b + γa) = 0 for all g ∈ F , that is, b + γa = 0. This means that a and b are linearly dependent. Thus ·, · has the property (N1 ) of a non–Archimedean 2–norm. It is clear that ·, · also has the properties (N2 ) and (N3 ) of a non–Archimedean 2–norm. For f, g ∈ F , we have |f (a)g(b + c) − g(a)f (b + c)| = |(f (a)g(b) − g(a)f (b)) + (f (a)g(c) − g(a)f (c))|

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≤ max{|f (a)g(b) − g(a)f (b)|, |f (a)g(c) − g(a)f (c)|} and so we have a, b + c ≤ max{a, b, a, c}. Therefore, it follows that ·, · is non–Archimedean 2–norm on X. Let σ ∗ be the metric generated by  · ∗ and σ the 2–metric generated by ·, ·. We shall prove that for an arbitrary sequence of points a, a1 , a2 , . . . in X, lim σ ∗ (a, ai ) = 0,

(2.1)

i→∞

if there exist two points b and c in X with σ(a, b, c) = 0 such that (2.2)

lim σ(a, b, ai) = 0,

i→∞

lim σ(a, c, ai) = 0.

i→∞

σ(a, b, c) = 0 implies that b − a = 0, and so there exists an element f ∈ F such that |f (b − a)| = f ∗ b − a∗ = 0. For i = 1, 2, · · · , there is a γi ∈ K such that f (ai − a − γi (b − a)) = 0. Thus, since we have |f (b − a)g(ai − a) − g(b − a)f (ai − a)| = |f (b − a)g(ai − a − γi (b − a)) − g(b − a)f (ai − a − γi (b − a))| = |f (b − a)g(ai − a − γi (b − a))|

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48

R. W. FREESE AND Y. J. CHO

and f (b−a) = 0, it follows that |g(ai −a−γi (b−a))| converges uniformly to 0 for all g ∈ F . Similarly we can show that there exists a δi ∈ K, i = 1, 2, . . . , such that |g(ai − a − δi (c − a))| converges uniformly to 0 for all g ∈ F . Hence |g(γi(b − a) − δi (c − a))| tends to 0 uniformly for all g ∈ F . We define a linear mapping g ∗ from the linear subspace X ∗ of X generated by b − a and c − a into K by g ∗ (b − a) = e and g ∗ (c − a) = 0. g ∗ is continuous by A.C.M. van Rooij. Let µ be an element of K, different from 0, such that |µ| ≥ g ∗ ∗ . g + = g ∗ /µ is a continuous linear mapping of X ∗ into K with the norm of which is less than or equal to 1. Let g  be an extension of g + with g  ∗ = g + ∗ . We have

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|g  (γi (b − a) − δi (c − a))| =

|g ∗ (γi (b − a) − δi (c − a))| → 0. |µ|

Hence |γi g  (b − a)| → 0 and so γi → 0. It follows that ai → a, that is, (2.1) is proved. Using the inequality σ(a, b, c) ≤ 2σ ∗ (a, b)σ ∗(a, c), for every sequence of points a, a1 , a2 , . . . in X which satisfy (2.1) and for every a ∈ X, we have ahler, σ and σ ∗ thus generate the same limi→∞ σ(a, a , ai ) = 0. By S. G¨ topology and (X, ·, ·) has Property (K). We can see without difficulty that the analogue of Theorem 2.2.2 for non–Archimedean linear 2–normed spaces is true. By virtue of this analogue and the fact that a, b ≤ 2a∗ b∗ and the topologies generated by σ and σ ∗ coincide, (X, ·, ·) has Property (S).

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CHAPTER 3. 2–BANACH SPACES In [242] and [243], A. White presented some of the basic properties, examples and principles of 2–Banach spaces. Especially, he introduced the Hahn–Banach theorem in 2–Banach spaces which is similar to the Hahn– Banach theorem of functional analysis and its related properties. 3.1. Elementary Properties In this section, we introduce some definitions on convergence of a sequence and a Cauchy sequence in linear 2–normed spaces, basic properties and some examples of 2–Banach spaces in [242] and [243]. Definition 3.1.1. A sequence {xn } in a linear 2–normed space (X,
·, ·
) is called a Cauchy sequence if there exist two points y, z ∈ X such that y and z are linearly independent, lim
xn − xm , y
= 0,

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m,n→∞

lim
xn − xm , z
= 0.

m,n→∞

Theorem 3.1.1. Let (X,
·, ·
) be a linear 2–normed space. (1) If {xn } is a Cauchy sequence in X with respect to a and b, then {
xn , a
} and {
xn , b
} are real Cauchy sequences. (2) If {xn } and {yn } are Cauchy sequences in X with respect to a and b and {αn } is a real Cauchy sequence, then {xn + yn } and {αn xn } are Cauchy sequences in X. Proof. (1) By property (N4 ) of the 2–norm
·, ·
, since we have
xn , a
=
(xn − xm ) + xm , a
≤
xn − xm , a
+
xm , a
, it follows that
xn , a
−
xm , a
≤
xn − xm , a
. Similarly, we have
xm , a
−
xn , a
≤
xn − xm , a
. Thus, by the above inequalities, we have



xn , a
−
xm , a

≤
xn − xm , a
. Typeset by AMS-TEX

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50

R. W. FREESE AND Y. J. CHO

Therefore, we see that {
xn , a
} is a real Cauchy sequence since the limm,n→∞
xn − xm , a
= 0. Similarly, {
xn , b
} is also a real Cauchy sequence. (2) Since we have the following:
(xn + yn ) − (xm + ym ), a
=
(xn − xm ) + (yn − ym ), a
≤
xm − xn , a
+
yn − ym , a
→ 0 as m, n → ∞, it follows that
(xn + yn ) − (xm + ym ), b
→ 0 as m, n → ∞. Therefore {xn + yn } is a Cauchy sequence in X. Using the fact that {αn } and {
xn , a
} are real Cauchy sequences and hence bounded, we have, as m, n → ∞,
αn xn − αm xm , a
=
(αn xn − αn xm ) + (αn xm − αm xm ), a
≤
αn xn − αn xm , a
+
αn xm − αm xm , a
= |αn |
xn − xm , a
+ |αn − αm |
xm , a
≤ K1
xn − xm , a
+ K2 |αn − αm | → 0

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for some K1 , K2 ≥ 0, which implies that
αn xn − αm xm , a
→ 0 as m, n → ∞. Therefore, {αn xn } is a Cauchy sequence in X. Definition 3.1.2. A sequence {xn } in a linear 2–normed space (X,
·, ·
) is called a convergent sequence if there exists an x ∈ X such that limn→∞
xn − x, z
= 0 for all z ∈ X. If {xn } converges to x, write xn → x as n → ∞ and call x the limit of {xn }. Definition 3.1.3. A linear 2–normed space in which every Cauchy sequence is a convergent sequence is called a 2–Banach space. Theorem 3.1.2. In any linear 2–normed space (X,
·, ·
), we have the following : (1) If xn → x and yn → y as n → ∞, then xn + yn → x + y as n → ∞. (2) If xn → x and αn → α as n → ∞, then αn xn → αx as n → ∞. (3) If dim X ≥ 2, xn → x and xn → y as n → ∞, then x = y. Proof. (1) By property (N4 ) of the 2–norm
·, ·
, since we have
(xn + yn ) − (x + y), a
=
(xn − x) + (yn − y), a
≤
xn − x, a
+
yn − y, a
→0

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2–BANACH SPACES

51

as n → ∞, it follows that xn + yn → x + y as n → ∞. (2) Using the fact that a real convergent sequence is bounded, we have
αn xn − αx, a
≤
αn xn − αn x, a
+
αn x − αx, a
= |αn |
xn − x, a
+ |αn − α|
x, a
≤ K
xn − x, a
+ |αn − α|
x, a
for some K ≥ 0. Therefore, αn xn → αx as n → ∞ since we have lim
xn − x, a
= 0,

n→∞

lim |αn − α| = 0.

n→∞

(3) By property (N4 ) of the 2–norm
·, ·
, we have
x − y, a
≤
x − xn , a
+
xn − y, a
and so it follows that
x − y, a
= 0 for all a ∈ X since xn → x and xn → y as n → ∞. Hence, x − y and a are linearly dependent for all a ∈ X. Since the dim X ≥ 2, the only way that x − y can be linearly dependent with all vectors a ∈ X is for x − y = 0.

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Example 3.1.1. Let E 3 denote euclidean 3–dimensional linear space. Let x = ai + bj + ck and y = di + ej + f k. Define


i j k



x, y
= |x × y| = abs

a b c


d e f
= |(bf − ce)i + (cd − af )j + (ae − db)k| 1

= [(bf − ce)2 + (cd − af )2 + (ae − db)2 ] 2 . Then (E3 ,
·, ·
) is a 2–Banach space. From the vector analysis, it is clear that
·, ·
is a 2–norm. Let xn = an i + bn j + cn k be a Cauchy sequence in E3 . Therefore there exists y = di + ej + f k and z = pi + qj + rk, linearly independent vectors in E3 , such that limn,m→∞ |(xn − xm ) × y| = 0 and limn,m→∞ |(xn − xm ) × z| = 0. We shall show that {an }, {bn } and {cn } are real Cauchy sequences.



i j k

|(xn − xm ) × y| = abs

an − am bn − bm cn − cm



d e f  = [f (bn − bm ) − e(cn − cm )]2 + [d(cn − cm ) − f (an − am )]2 1 + [e(an − am ) − d(bn − bm )]2 2 .

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52

R. W. FREESE AND Y. J. CHO

Therefore, limn,m→∞ |(xn − xm ) × y| = 0 if and only if α:

lim [f (bn − bm ) − e(cn − cm )] = 0,

n,m→∞

lim [d(cn − cm ) − f (an − am )] = 0,

β:

n,m→∞

γ:

lim [e(an − am ) − d(bn − bm )] = 0.

n,m→∞

Similarly, since we have  |(xn − xm ) × z| = [r(bn − bm ) − q(cn − cm )]2 + [p(cn − cm ) − r(an − am )]2 + [q(an − am ) − p(bn − bm )]2 limn,m→∞ |(xn − xm ) × z| = 0 if and only if δ: :

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ζ:

lim [r(bn − bm ) − q(cn − cm )] = 0,

n,m→∞

lim [p(cn − cm ) − r(an − am )] = 0,

n,m→∞

lim [q(an − am ) − p(bn − bm )] = 0.

n,m→∞

Using δ and α, we have lim [−rf (bn − bm ) + qf (cn − cm )] = 0

n,m→∞

and lim (rf (bn − bm ) − re(cn − cm )] = 0.

n,m→∞

By addition, we have lim (qf − re)(cn − cm ) = 0.

n,m→∞

Similarly, using β and , we have lim [rd(cn − cm ) − rf (an − am )] = 0

n,m→∞

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 12

,

2–BANACH SPACES

53

and lim [−pf (cn − cm ) + rf (an − am )] = 0.

n,m→∞

By addition, we have lim (rd − pf )(cn − cm ) = 0.

n,m→∞

Assume that {cn } is not a Cauchy sequence. Then qf = re and rd = pf , that is, r/f = q/e = p/d, which is impossible since y and z are linearly independent. Therefore {cn } is a Cauchy sequence. In a similar manner, it can be shown that {an } and {bn } are Cauchy sequences. Since R is complete, there exists real numbers a, b and c such that limn→∞ an = a, limn→∞ bn = b and limn→∞ cn = c. Let x = ai + bj + ck. We want to show that xn → x as n → ∞. Let w = si + tj + uk be an element of E3 . Then we have


i
j k

lim |(xn − x) × w| = lim abs

an − a bn − b cn − c

n→∞ n→∞
s t u
 = lim [u(bn − b) − t(cn − c)]2 Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

n→∞

+ [s(cn − c) − u(an − a)]2 1 + [t(an − a) − s(bn − b)]2 2 =0 since limn→∞ an = a, limn→∞ bn = b and limn→∞ cn = c. Therefore, (E3 ,
·, ·
) is a 2–Banach space. Example 3.1.2. Let Pn denote the set of real polynomials of degree ≤ n on the interval [0,1]. Define vector addition and scalar multiplication in the usual manner. Hence Pn is a linear space over R. Let {xi }2n i=0 be 2n + 1 arbitrary but distinct fixed points in [0,1]. Let f, g ∈ Pn . Define
f, g
= 0 as follows:  0 if f and g are linearly dependent,   2n 
f, g
=  |f (xi )g(xi )| if f and g are linearly independent.  i=0

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54

R. W. FREESE AND Y. J. CHO

Then (Pn ,
·, ·
) is a 2–Banach space. In fact, if f and g are linearly dependent, then
f, g
= 0. Assume that 2n 

|f (xi )g(xi )| = 0.

i=0

Since the degree of f ≤ n, the degree of g ≤ n and f (xi )g(xi ) = 0 at 2n + 1 distinct points, we have f = 0 or g = 0. Therefore if
f, g
= 0, then f and g are linearly dependent. Clearly,
f, g
=
g, f
,
f, βg
= |β|
f, g
and
f, g + h
≤
f, g
+
f, h
for all h in Pn . Therefore, (Pn ,
·, ·
) is a linear 2–normed space. Next, we shall prove that every Cauchy sequence in Pn converges to an element in Pn , that is, (Pn ,
·, ·
) is a 2–Banach space. Let {fk } be a Cauchy sequence in Pn . Then there exist linearly independent elements g and h in Pn such that lim
fk − fi , g
= 0,

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k,i→∞

lim
fk − fi , h
= 0.

k,i→∞

Each term in the finite sum of
fk −fj , g
is non–negative and since g(x) is of degree ≤ n, g(xi) = 0 for at most n of the points x0 , x1 , · · · , x2n . Therefore, {fk (xi )}∞ k=0 is a real Cauchy sequence for n + 1 values of xi , say, x0 , · · · , xn . i Let Akj = fk (xi ) − fj (xi ), i = 0, 1, · · · , n. Considering xni , xn−1 , · · · , xi , 1 as i i coefficients and hence the constants as unknowns, Akj is an equation with n + 1 unknowns. Then using (n + 1) × (n + 1) matrices, it is possible to solve for each of the unknowns since the coefficient matrix is the Vandermonde matrix which in this case has a non–zero determinant. Since lim Aikj = 0 for i = 0, 1, · · · , n, the determinant in the numerator is a real Cauchy sequence. Therefore, {fk (x)} converges to a polynomial of degree ≤ n. Therefore, (Pn ,
·, ·
) is a 2–Banach space. The following gives an example of a linear 2–normed space of dimension 3 which is not a 2–Banach space. It should be mentioned, however, that in the definition of a linear 2–normed space the linear space is defined over all the reals. In the following example, the linear space will be restricted to the rationals. Example 3.1.3. Let Q3 denote euclidean 3–dimensional linear space, where all coefficients are rationals, over the field of rationals. n Q3 is a linear space. Define
·, ·
in Q3 as in Example 3.1.1. Let xn = k=0 10−k(k+1)/2 i.

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2–BANACH SPACES

55

Then we have
xn − xm , i
= 0 and hence limn,m→∞
xn − xm , i
= 0. It follows that lim
xn − xm , j
=

n,m→∞

n m



−k(k+1)/2 lim
10 − 10−k(k+1)/2
= 0

n,m→∞

k=0

k=0

n since { k=0 10−k(k+1)/2 } is a real Cauchy sequence. Since i and j are linearly independent, {xn } is a Cauchy sequence in Q3 . Assume that there is an x = ai + bj + ck ∈ Q3 such that xn → x as n → ∞. Therefore we have limn→∞
xn − x, j
= 0, that is, lim

n→∞

n



10

−k(k+1)/2

−a

2

2

+c

 12

= 0.

k=0

n Clearly, c must be 0. Hence we have limn→∞ k=0 10−k(k+1)/2 = a. n { k=0 10−k(k+1)/2 } converges in the real number system to an irrational number. Therefore, the number a must be irrational. Since Q3 is a vector space over the field of rationals, this is impossible. Therefore Q3 is not a 2–Banach space.

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By using Theorem 2.2.7, we have the following: Theorem 3.1.3. Every linear 2–normed space of dimension 2 is a 2– Banach space when the underlying field is complete. Proof. Let (X,
·, ·
) be a linear 2–normed space with basis {e1 , e2 }. Let {xn } be a Cauchy sequence in X. Therefore there exist linearly independent vectors a and b in B such that lim
xn − xm , a
= 0,

n,m→∞

lim
xn − xm , b
= 0.

n,m→∞

Let xn = xn1 e1 + xn2 e2 , a = a1 e1 + a2 e2 and b = b1 e1 + b2 e2 . Now by Theorem 2.2.7, we have
xn − xm , a
=
(xn1 − xm1 )e1 + (xn2 − xm2 )e2 , a1 e1 + a2 e2
= |a2 (xn1 − xm1 ) − a1 (xn2 − xm2 )|
e1 , e2
. Similarly, we have
xn − xm , b
= |b2 (xn1 − xm1 ) − b1 (xn2 − xm2 )|
e1 , e2
.

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56

R. W. FREESE AND Y. J. CHO

Since e1 and e2 are linearly independent,
e1 , e2
= 0. Therefore, lim |a2 (xn1 − xm1 ) − a1 (xn2 − xm2 )| = 0

n,m→∞

and lim |b2 (xn1 − xm1 ) − b1 (xn2 − xm2 )| = 0.

n,m→∞

Hence, we have lim [a2 b2 (xn1 − xm1 ) − a1 b2 (xn2 − xm2 )] = 0

n,m→∞

and lim [−a2 b2 (xn1 − xm1 ) + a2 b1 (xn2 − xm2 )] = 0.

n,m→∞

Therefore, by addition, we have lim (a2 b1 − a1 b2 )(xn2 − xm2 ) = 0.

n,m→∞

a2 b1 − a1 b2 = 0 implies aa12 = bb12 , which is impossible, since a and b are linearly independent. Hence, limn,m→∞ |xn2 − xm2 | = 0, that is, {xn2 } is a Cauchy sequence. Since we have lim [a2 b1 (xn1 − xm1 ) − a1 b1 (xn2 − xm2 )] = 0

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n,m→∞

and lim [−a1 b2 (xn1 − xm1 ) + a1 b1 (xn2 − xm2 )] = 0,

n,m→∞

by addition, we have also lim (a2 b1 − a1 b2 )(xn1 − xm1 ) = 0.

n,m→∞

Since a2 b1 − a1 b2 = 0, limn,m→∞ |xn1 − xm1 | = 0, that is, {xn1 } is a Cauchy sequence. Since {xn1 } and {xn2 } are real Cauchy sequences, there are real numbers y1 and y2 such that limn→∞ xn1 = y1 and limn→∞ xn2 = y2 . Let x = y1 e1 + y2 e2 . Now we shall prove that xn → x as n → ∞. Let c = c1 e1 + c2 e2 be an element of X. Then we have lim
xn − x, c
= lim
(xn1 − y1 )e1 + (xn2 − y2 )e2 , c1 e1 + c2 e2


n→∞

n→∞

= lim |c2 (xn1 − y1 ) − c1 (xn2 − y2 )|
e1 , e2
n→∞

=0

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2–BANACH SPACES

57

since limn→∞ xn1 = y1 and limn→∞ xn2 = y2 . Therefore, xn → x as n, m → ∞, that is, (X,
·, ·
) is a 2–Banach space. Theorem 3.1.4. Let (X,
·, ·
) be a linear 2–normed space. If we have limn,m→∞
xn − xm , d
= 0, then {
xn − x, d
} is a convergent sequence for each x ∈ X. Proof. By property (N4 ) of 2–norm
·, ·
, since we have
xn − x, d
=
xn − xm + xm − x, d
≤
xn − xm , d
+
xm − x, d
, it follows that
xn − x, d
−
xm − x, d
≤
xn − xm , d
. In fact, we have



xn − x, d
−
xm − x, d

≤
xn − xm , d
.

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Hence, {
xn − x, d
} is a convergent sequence by hypothesis. Theorem 3.1.5. If limn→∞
xn − x, d
= 0, then limn→∞
xn , d
=
x, d
. Proof. As in the proof of Theorem 3.1.1, since we have

lim

xn , d
−
x, d

≤ lim
xn − x, d
= 0,

n→∞

n→∞

it follows that limn→∞
xn , d
=
x, d
. Note that limn→∞
xn , x
=
x, x
= 0, if the hypothesis holds for d = x. 3.2. Bounded Linear 2–Functionals The aim of this section is to present the Hahn–Banach theorem in 2– Banach spaces and some properties related to this theorem. Definition 3.2.1. A 2–functional is a real valued mapping with domain A × C, where A and C are linear manifolds of a linear 2–normed space. Definition 3.2.2. Let F be a 2–functional with domain A × C. F is called a linear 2–functional (or bilinear 2–functional) if

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58

R. W. FREESE AND Y. J. CHO

(1) F (a + c, b + d) = F (a, b) + F (a, d) + F (c, b) + F (c, d), (2) F (αa, βb) = αβF (a, b) for α, β in the underlying field. Definition 3.2.3. Let F be a 2–functional with domain D(F ). F is said to be bounded if there is a real constant K ≥ 0 such that |F (a, b)| ≤ K
a, b
for all (a, b) ∈ D(F ). If F is bounded, define the norm of F,
F
, by
F
= glb{K : |F (a, b)| ≤ K
a, b
for all (a, b) ∈ D(F )}. If F is not bounded, define
F
= +∞. Example 3.2.1. Let (X,
·, ·
) be a linear 2–normed space with basis {e1 , e2 }. Define F (a, b) = α1 β2 − α2 β1 where a = α1 e1 + α2 e2 and b = β1 e1 + β2 e2 . Let c = µ1 e1 + µ2 e2 and d = δ1 e1 + δ2 e2 . Then we have F (a + c, b + d) = (α1 + µ1 )(β2 + δ2 ) − (α2 + µ2 )(β1 + δ1 ) = α1 β2 + α1 δ2 + µ1 β2 + µ1 δ2 − α2 β1 − α2 δ1 − µ2 β1 − µ2 δ1

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= (α1 β2 − α2 β1 ) + (α1 δ2 − α2 δ1 ) + (µ1 β2 − µ2 β1 ) + (µ1 δ2 − µ2 δ1 ) = F (a, b) + F (a, d) + F (c, b) + F (c, d), F (αa, βb) = αα1 ββ2 − αα2 ββ1 = αβ(α1 β2 − α2 β1 ) = αβF (a, b) and |F (a, b)| = |α1 β2 − α2 β1 | =

1
a, b
.
e1 , e2


Therefore, F is a bounded linear 2–functional. Example 3.2.2. Let (E3 ,
·, ·
) be the 2–Banach space defined in Example 3.1.1. Define F (x, y) = (x|y) where (·|·) is the dot product of vector analysis. Then F is an unbounded linear 2–functional. Define G(x, y) = (|x|2 |y|2 − |(x|y)|2 )1/2 , where |a| denotes the length of a. Then G is a bounded 2–functional since |a|2 |b|2 − |(a|b)|2 = |a × b|2 .

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Lemma 3.2.1. If F is a bounded linear 2–functional and a and b are linearly dependent with (a, b) ∈ D(F ), then F (a, b) = 0. Proof. Since F is bounded, |F (x, y)| ≤
F

x, y
for all (x, y) ∈ D(F ). Since a and b are linearly dependent,
a, b
= 0. Therefore we have |F (a, b)| ≤
F
· 0 = 0. Theorem 3.2.2. Let F be a bounded linear 2–functional with domain D(F ). Then we have
F
= sup{|F (x, y)| :
x, y
= 1, (x, y) ∈ D(F )}

|F (x, y)| :
x, y
= 0, (x, y) ∈ D(F ) . = sup
x, y
Proof. Let A = sup{|F (x, y)| :
x, y
= 1, (x, y) ∈ D(F )}. Then we have |F (x, y)| ≤
F

x, y
for all (x, y) ∈ D(F ). Hence A ≤
F
. Assume
x, y
= 0. Since

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   x    
x, y
, y  = 1,




F




x , y

≤ A.
x, y


Therefore, |F (x, y)| ≤ A
x, y
for (x, y) ∈ D(F ) with
x, y
= 0. If
x, y
= 0, then x and y are linearly dependent. By Lemma 3.2.1, F (x, y) = 0. Thus, |F (x, y)| ≤ A
x, y
for all (x, y) ∈ D(F ) and hence we have
F
≤ A. Let C = sup

|F (x, y)|
x, y


:
x, y
= 0, (x, y) ∈ D(F ) .

By the definition of
F
, we have |F (x, y)|/
x, y
≤
F
for all (x, y) ∈ D(F ) with
x, y
= 0. So we have C ≤
F
. |F (x, y)| ≤ C
x, y
for all (x, y) ∈ D(F ) by Lemma 3.2.1 and the definition of C. Hence we have
F
≤ C. Definition 3.2.4. A 2–functional F is said to be continuous at (a, b) if given  > 0 there is a δ > 0 such that |F (a, b) − F (c, d)| <  whenever
a − c, b
< δ and
c, b − d
< δ or
a − c, d
< δ and
a, b − d
< δ. F is continuous if it is continuous at each point of its domain. Theorem 3.2.3. The 2–norm
·, ·
is a continuous 2–functional.

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60

R. W. FREESE AND Y. J. CHO

Proof. By the property (N4 ) of the 2–norm
·, ·
, since we have
a, b
=
(a − c) + c, b
≤
a − c, b
+
c, b
=
a − c, b
+
c, (b − d) + d
≤
a − c, b
+
c, b − d
+
c, d
, it follows that
a, b
−
c, d
≤
a − c, b
+
c, b − d
. On the other hand, since we have
c, d
=
c, (d − b) + b
≤
c, d − b
+
c, b
=
c, d − b
+
(c − a) + a, b
≤
c, d − b
+
c − a, b
+
a, b
, it follows that

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c, d
−
a, b
≤
a − c, b
+
c, b − d
. Thus, we have



a, b
−
c, d

≤
c, b − d
+
a − c, b
. Therefore, by Definition 3.2.4, the 2–norm
·, ·
is a continuous 2–functional. Theorem 3.2.4. If a linear 2–functional F is continuous at (0, 0), then it is continuous at each point in its domain D(F ). Proof. Note that if F is linear, then F (0, 0) = 0. Since F is continuous at (0, 0), given  > 0, there is a δ > 0 such that |F (c, d)| < 12  whenever
c, d
< δ. Let (a, b) ∈ D(F ). Then we have |F (a, b) − F (x, y)| = |F (a, b) − F (x, b) + F (x, b) − F (x, y)| ≤ |F (a, b) − F (x, b)| + |F (x, b) − F (x, y)| = |F (a − x, b)| + |F (x, b − y)|   ≤ + = 2 2

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61

whenever
a − x, b
< δ and
x, b − y
< δ. Hence F is continuous at (a, b). Theorem 3.2.5. A linear 2–functional F is continuous if and only if it is bounded. Proof. Assume that F is continuous. There exists a δ > 0 such that |F (a, b)| < 1 whenever
a, b
< δ for (a, b) ∈ D(F ). For (c, d) ∈ D(F ) with c and d being linearly independent, consider ((c/
c, d
)(δ/2), d). Then we have    c δ  1 δ δ   
c, d
2 , d =
c, d
2
c, d
= 2 .   Hence we have F (c/
c, d
)(δ/2), d < 1, that is, |F (c, d)| < ( 2δ )
c, d
. There exists a δn > 0 such that |F (a, b)| < n1 whenever
a, b
< δn . If c and d are linearly dependent,
c, d
= 0 < δn and hence |F (c, d)| = 0. Therefore, F is bounded.

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Conversely, assume that F is bounded. There is a K ≥ 0 such that |F (x, y)| ≤ K
x, y
for all (x, y) ∈ D(F ). Given  > 0, let δ = /(K + 1). Then we have |F (x, y)| ≤ K
x, y
< K

 K +1

whenever
x, y
< δ. Hence F is continuous at (0, 0) and so F is continuous by Theorem 3.2.4. Definition 3.2.5. Let (X,
·, ·
) be a 2–Banach space and X ∗ the set of bounded linear 2–functionals with domain X × X. Let F, G ∈ X ∗ . Define (1) F = G if F (a, b) = G(a, b), (2) (F + G)(a, b) = F (a, b) + G(a, b), (3) (αF )(a, b) = αF (a, b) for all (a, b) ∈ X × X. By using Definitions 3.2.3 and 3.2.5, we have the following: Theorem 3.2.6. (B ∗ ,
·
) is a Banach space

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R. W. FREESE AND Y. J. CHO

Proof. For F, G ∈ X ∗ , we have (F + G)(a + c, b + d) = F (a + c, b + d) + G(a + c, b + d) = F (a, b) + F (a, d) + F (c, b) + F (c, d) + G(a, b) + G(a, d) + G(c, b) + G(c, d) = (F (a, b) + G(a, b)) + (F (a, d) + G(a, d)) + (F (c, b) + G(c, b)) + (F (c, d) + G(c, d)) = (F + G)(a, b) + (F + G)(a, d) + (F + G)(c, b) + (F + G)(c, d), (F + G)(αa, βb) = F (αa, βb) + G(αa, βb) = αβF (a, b) + αβG(a, b) = αβ(F (a, b) + G(a, b)) = αβ(F + G)(a, b) and |(F + G)(a, b)| = |F (a, b) + G(a, b)| Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

≤ |F (a, b)| + |G(a, b)| ≤
F

a, b
+
G

a, b
= (
F
+
G
)
a, b
. Therefore, F + G ∈ X ∗ and
F + G
≤
F
+
G
. Similarly, αF ∈ X ∗ . Hence, X ∗ is a linear space.
·
defines a norm on X ∗ since we have (1) If
F
= 0, then F = 0. If F = 0, then
F
= 0, (2)
αF
= |α|
F
, (3)
F + G
≤
F
+
G
by the above argument. Assume that {Fn } is a Cauchy sequence in X ∗ . Then we have lim
Fn − Fm
= 0.

m,n→∞

Thus, from |Fn (a, b)−Fm (a, b)| ≤
Fn −Fm

a, b
, it follows that {Fn (a, b)} is a real Cauchy sequence for all (a, b) ∈ B × B. Define F (a, b) =

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63

limn→∞ Fn (a, b). Then we have F (a + c, b + d) = lim Fn (a + c, b + d) n→∞

= lim [Fn (a, b) + Fn (a, d) + Fn (c, b) + Fn (c, d)] n→∞

= lim Fn (a, b) + lim Fn (a, d) n→∞

n→∞

+ lim Fn (c, b) + lim Fn (c, d) n→∞

n→∞

= F (a, b) + F (a, d) + F (c, b) + F (c, d), F (αa, βb) = lim Fn (αa, βb) = lim αβFn (a, b) n→∞

n→∞

= αβ lim Fn (a, b) = αβF (a, b). n→∞

Therefore, F is a linear 2–functional. Since we have



Fn
−
Fm

≤
Fn − Fm
, it follows that {
Fn
} is a real Cauchy sequence. Therefore, there is a real constant K such that
Fn
≤ K for all n. F ∈ X ∗ since we have |F (a, b)| = | lim Fn (a, b)| = lim |Fn (a, b)| ≤ lim
Fn

a, b
≤ K
a, b
. Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

n→∞

n→∞

n→∞

Suppose
a, b
= 0. For every  > 0, there is an integer N such that
Fm − Fn
<  for n, m > N and so we have |Fm (a, b) − Fn (a, b)
≤
Fm − Fn

a, b
≤
a, b
 for all m, n > N . Since F (a, b) = limn→∞ Fn (a, b), there is an M = M (a, b) > N such that |FM (a, b) − F (a, b)| < 
a, b
. Therefore we have |Fn (a, b) − F (a, b)| ≤ |Fn (a, b) − FM (a, b)| + |FM (a, b) − F (a, b)| ≤ 
a, b
+ 
a, b
= 2
a, b
 for n > N . If
a, b
= 0, then Fn (a, b) = 0 = F (a, b) and so we have |Fn (a, b) − F (a, b)| ≤ 2
a, b


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64

R. W. FREESE AND Y. J. CHO

Both cases yield that |Fn (a, b) − F (a, b)| ≤ 2
a, b
for all n > N , which means that
Fn − F
≤ 2 for n > N . Therefore (X ∗ ,
·
) is a Banach space. In the following theorem, the statement, “X ∗ is a 2–Banach space up to linear dependence”, means that X ∗ satisfies all the conditions for being a 2– Banach space except F and G may be linearly dependent and yet
F, G
= 0. Theorem 3.2.7. (X ∗ ,
·, ·
) is a 2–Banach space up to linear dependence where
F, G
=
F

G
. Proof. By Theorem 3.2.6, X ∗ is a linear space.
·, ·
defines a 2–norm on X ∗ up to linear dependence since we have the following : (1) If
F, G
= 0, then F = 0 or G = 0 and hence F and G are linearly dependent. However
F, αF
is not zero unless F = 0 or α = 0. (2)
F, G
=
F

G
=
G

F
=
G, F
. (3)
F, αG
=
F

αG
=
F
|α|
G
= |α|
F, G
. (4) From Theorem 3.2.5, it follows that
G + H
≤
G
+
H
. Hence we have
F, G + H
=
F

G + H
≤
F
(
G
+
H
)

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=
F

G
+
F

H
=
F, G
+
F, H
. Assume that {Fn } is a Cauchy sequence in (X ∗ ,
·, ·
). Therefore there exist linearly independent G and H in X ∗ such that lim
Fn − Fm , G
= 0,

m,n→∞

lim
Fn − Fm , H
= 0.

m,n→∞

Since G and H are linearly independent, limm,n→∞
Fn − Fm
= 0. By the same argument as in Theorem 3.2.6, {Fn } is a convergent sequence in (X ∗ ,
·
) and so in (X ∗ ,
·, ·
). Hence (X ∗ ,
·
) is 2–Banach space up to linear dependence. If (X ∗ ,
·, ·
) is redefined by
F, G
= 0 if F and G are linearly independent and
F, G
=
F

G
if F and G are linearly independent, then (X ∗ ,
·, ·
) satisfies the properties of a 2–Banach space except that
F, G + H
≤
F, G
+
F, H


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2–BANACH SPACES

65

is not always true, for consider the case where F and G are linearly dependent and F and H are linearly independent. It should also be noted that S. G¨ ahler [96] proved that if (X,
·
) is a normed linear space, then a 2–norm can be defined on X. However attempts to prove that (X,
·, ·
) is a 2–Banach space when (X,
·
) is a Banach space have failed. The following theorem is similar to the Hahn–Banach theorem of functional analysis: Theorem 3.2.8. Let (X,
·, ·
) be a 2–Banach space and let M and V (b) be linear manifolds in X. Let F be a bounded linear 2–functional with the domain M × V (b). Then there exists a bounded linear 2–functional H with the domain X × V (b) such that (1) H(a, αb) = F (a, αb) for all (a, αb) ∈ M × V (b). (2)
H
=
F
. Proof. Let g ∈ X − M . Let N = {a + βg : a ∈ M, β ∈ R}. Let a and  a be elements of M . Since we have F (a , b) − F (a , b) = F (a − a , b) ≤
F

a − a , b


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=
F

(a + g) − (a + g), b
≤
F

a + g, b
+
F

a + g, b
, it follows that (3.1)

−
F

a” + g, b
− F (a , b) ≤
F

a + g, b
− F (a , b).

Let a and a” vary over M . Hence we have S = sup {−
F

a + g, b
− F (a , b)} a

∈M

{
F

a + g, b
− F (a , b)} ≤ inf
a ∈M

= I. Let r be any number such that S ≤ r ≤ I. Set a = a” = a. Then (3.1) becomes −
F

a + g, b
− F (a, b) ≤ r ≤
F

a + g, b
− F (a, b),

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66

R. W. FREESE AND Y. J. CHO

that is, (3.2)

|F (a, b) + r| ≤
F

a + g, b
.

Define G on N × V (b) by G(a + βg, αb) = αF (a, b) + αβr. We shall show that G is a linear 2–functional. In fact, we have G(a1 + β1 g + a2 + β2 g, α1 b + α2 b)   = G (a1 + a2 ) + (β1 + β2 )g, (α1 + α2 )b = (α1 + α2 )F (a1 + a2 , b) + (α1 + α2 )(β1 + β2 )r = α1 F (a1 , b) + α1 F (a2 , b) + α2 F (a1 , b) + α2 F (a2 , b) + α1 β1 r + α1 β2 r + α2 β1 r + α2 β2 r = (α1 F (a1 , b) + α1 β1 r) + (α1 F (a2 , b) + α1 β2 r) + (α2 F (a1 , b) + α2 β1 r) + (α2 F (a2 , b) + α2 β2 r) = G(a1 + β1 g, α1 b) + G(a2 + β2 g, α1 b) + G(a1 + β1 g, α2b) + G(a2 + β2 g, α2 b) and

 G δ(a + βg), αb) = G(δa + δβg, αb)

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= αF (δa, b) + αδβr = δ(αF (a, b) + αβr) = δG(a + βg, αb). Hence G is a linear 2–functional which is an extension of F . Replace a by a/β in (3.2) where β = 0. Then we have, for all β, |F (a, b) + βr| ≤
F

a + βg, b
, |G(a + βg, αb)| = |αF (a, b) + αβr| = |α||F (a, b) + βr| ≤ |α|
F

a + βg, b
=
F

a + βg, αb
. Hence G is bounded and
G
≤
F
. Since G = F on M × V (B),
G
=
F
. Consider pairs {N, G} where N is a linear manifold and G is a bounded linear 2–functional with domain N × V (b). Write {N, G} < {N  , G } if and

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2–BANACH SPACES

67

only if N ⊂ N  and G is an extension of G with
G
=
G
. Let T be a collection of {N, G} such that {M, F } < {N, G}. T is partially ordered by 0. y∈M

Then there exists a bounded linear 2–functional F with domain X × V (b) such that (1) F (xo , b) = δ, (2) F (m, yo ) = 0 for all m ∈ M and yo ∈ V (b), (3)
F
= 1. Proof. Let N = [M, xo ], that is, N is the linear manifold of X generated by M and xo . Define f : N × V (b) → R by f (m + αxo , βb) = αβδ. Then we shall prove that f is a bounded linear 2–functional on N × V (b). In fact, let m, m ∈ M and α, α , β, β  , µ, γ ∈ R. Then we have

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f ((m + αxo ) + (m + α xo ), βb) = f ((m + m ) + (α + α )xo , βb) = (α + α )βδ = αβδ + α βδ = f (m + αxo , βb) + f (m + α xo , βb), f (m + αxo , βb + β  b) = f (m + αxo , (β + β  )b) = α(β + β  )δ = αβδ + αβ  δ = f (m + αxo , βb) + f (m + αxo , β  b), f (γ(m + αxo ), βb) = f (γm + γαxo , βb) = γαβδ = γf (m + αxo , βb) and f (m + αxo , µ(βb)) = αµβδ = µf (m + αxo , βb). Thus f is a linear 2–functional on N × V (b).

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2–BANACH SPACES

69

Next, to prove that f is bounded, without loss of generality, let α ∈ R be non–zero. Note that
m + αxo , βb
= |β|
m + αxo , b


m    , b = |β|α xo + α 

−m     = |α||β|xo − , b α ≥ |α||β|δ since (−m/α) ∈ M . Now, since we have |f (m + αxo , βb)| = |αβδ| = |α||β|δ ≤
m + αxo , βb
, f is bounded and
f
≤ 1. On the other hand, clearly, we have f (xo , b) = δ and f (m, yo ) = 0 for m ∈ M and yo ∈ V (b). By definition of infimum, given  > 0 there exists a point y ∈ M such that
xo − y, b
< δ + . It is clear that  x −y  xo − y   o ∈ N,  , b = 1.
xo − y, b

xo − y, b
Now, we have

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 xo − y δ δ f ,b = ≥ .
xo − y, b

xo − y, b
δ+ Since  is arbitrary, we have

f

 xo − y , b ≥ 1.
xo − y, b


Thus it follows that  x −y  

x −y   o o , b ≤
f
 , b =
f
. 1≤f
xo − y, b

xo − y, b
Therefore, we have
f
= 1. Applying Theorem 3.2.8, we have the desired result. By Theorem 3.2.11, we have the following: Corollary 3.2.12. Let (X,
·, ·
) be a linear 2–normed space over R and M and V (b) be linear manifolds of X. Let xo ∈ X − M and δ =

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70

R. W. FREESE AND Y. J. CHO

inf y∈M
b, xo − y
> 0. Then there exists a bounded linear 2–functional F with domain V (b) × X such that (1) F (b, xo ) = δ, (2) F (yo , m) = 0 for all m ∈ M and yo ∈ V (b), (3)
F
= 1. Corollary 3.2.13. Assume the hypotheses of Theorem 3.2.11. Then there exists a bounded linear 2–Functional F on X × V (b) such that (1) F (xo , b) = 1, (2) F (m, yo ) = 0 for all m ∈ M and yo ∈ V (b), (3)
F
= 1. One major consequence of Theorem 3.2.11 is that it asserts that for a linear 2–normed space (X,
·, ·
) the bounded linear 2–functional exists in profusion as in the next theorem:

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Theorem 3.2.14. Let (X,
·, ·
) be a linear 2–normed space and xo ∈ X be a non–zero element. Let b ∈ X be such that xo and b are linearly independent. Then there exists a bounded linear 2–functional F with domain X × V (b) such that (1) F (xo , b) =
xo , b
, (2)
F
= 1. Proof. In Theorem 3.2.11, take M = {0}. Then xo ∈ X − M and δ =
xo , b
. Since xo and b are linearly independent, δ > 0. Hence, by Theorem 3.2.11, the theorem follows. Corollary 3.2.15. Assume the hypotheses of Theorem 3.2.14. If |F (xo , b)| ≤ K for every 2–functional F with domain X × V (b),
F
= 1 and K is a constant independent of F , then
xo , b
≤ K. Proof. Assume
xo , b
> K. By Theorem 3.2.4, there exists a bounded linear 2–functional F on X × V (b) such that
F
= 1 and F (xo , b) =
xo , b
> K, which is a contradiction. Thus we have
xo , b
≤ K. Denote by (X × V (b))∗ the space of all bounded linear 2–functionals with domain X × V (b). Let f, g ∈ (X × V (b))∗ and define (1) (f + g)(x, βb) = f (x, βb) + g(x, βb), (2) (αf )(x, βb) = αf (x, βb), (3) f = g if and only if f (x, βb) = g(x, βb) for all x ∈ X and β ∈ R.

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71

For f ∈ (X × V (b))∗ , define
f
= sup

|f (x, b)|
x, b


: (x, b) ∈ X × V (b),
x, b
= 0 .

An argument similar to Theorem 3.2.6 shows that ((X × V (b))∗ ,
·
) is a Banach space. Theorem 3.2.15. Let (X,
·, ·
) be a linear 2–normed space. Let xo ∈ X be such that xo and b are linearly independent. Define a mapping [xo , b] on (X × V (b))∗ by [xo , b](f ) = f (xo , b). Then we have (1) [xo , b] is a bounded linear functional on (X × V (b))∗ , (2)
[xo , b]
=
xo , b
. Proof. Let f, g ∈ (X × V (b))∗ . Then we have [xo , b](f + g) = (f + g)(xo , b) = f (xo , b) + g(xo , b) and

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[xo , b](αf ) = (αf )(xo , b) = αf (xo , b). Also, |[xo , b](f )| = |f (xo , b)| ≤
f

xo , b
means that
[xo , b]
≤
xo , b
. Thus [xo , b] is a bounded linear functional on (X × V (b))∗ . Next, we shall show that (2) holds. By Theorem 3.2.14, there exists a bounded linear 2– functional g such that
g
= 1 and g(xo , b) =
xo , b
. Thus |[xo , b](g)| = |g(xo , b)| =
xo , b
≥
xo , b

g
means that
[xo , b]
≥
xo , b
. Therefore we have
[xo , b]
=
xo , b
. Theorem 3.2.16. Let (X,
·, ·
) be an n–dimensional linear 2–normed space with basis {ei }ni=1 . If F ∈ X ∗ , then we have F

n

 i=1

αi ei ,

n  i=1



βi ei =

n 

(αi βj − αj βi )F (ei , ej ).

i,j=1 i 0, we have 

x 

x ri ≤ p + xo , ei − f , ei α α or

x + αx t e 

x t e  o i i i i ri ≤ p −f , , α ti α ti

or αti ri + f (x, ti ei ) ≤ p(x + αxo , ti ei ). For αti < 0, we have

−x x  − xo , ei ) − f ( , ei ≤ ri −p α α

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or

1 1 p(x + αxo , ti ei ) − f (x, tiei ) ≤ ri αti αti

or p(x + αxo , ti ei ) ≥ αti ri + f (x, ti ei ). The case αti = 0 is like the case αβ = 0 in the Theorem 3.3.2. Since p is linear in its second coordinate, we have n n n



    ti ei = f x, ti ei + α ti r i F (y, z) = F x + αxo ,

=

n 

i=1

f (x, ti ei ) + α

i=1

n  i=1

ti r i ≤

i=1 n 

i=1

p(x + αxo , ti ei )

i=1

n 

 ti ei = p(y, z). = p x + αxo , i=1

Applying Zorn’s lemma to the set of all such extensions of f , the maximal element of this set is the function G required.

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R. W. FREESE AND Y. J. CHO

The Hann–Banach theorem ensures the extension of non–trivial continuous linear functionals in a general locally convex linear topological space. The next theorem allows for an extension of a 2–functional F defined on S × T , where S and T are the subsets of a linear 2–normed space (X,
·, ·
), to [S] × [T ], where [S] and [T ] are the linear manifolds generated by S and T over R. This extension is subject to a boundedness condition. Theorem 3.3.3. Let (X,
·, ·
) be a linear 2–normed space, S and T be subsets of X. Let F be a real valued 2–functional defined on S × T . A necessary and sufficent condition for F to admit of a bounded linear extension G from [S] × [T ] to R is that there exists a constant K such that n m  

   

αi βj F (xi , yj )
≤ K  α i xi , βj yj 


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i,j

i=1

j=1

for each finite subset {x1 , . . . , xn } in S and {y1 , . . . , ym } in T and every choice of scalars. Proof. Assume that the extension G of F exists and is bounded. Let x be an arbitrary element in [S] and y be an arbitrary element in [T ]. Let n m α x and y = x = i=1 i i j=1 βj yj , where xi is in S and yj is in T for i = 1, . . . , n and j = 1, . . . , m, respectively. Then we have n m n m



     G(x, y) = G α i xi , βj yj = α i G xi , βj yj i=1

=

n  i=1

=



αi

m



j=1



i=1

βj G(xi , yj ) =

j=1



j=1

αi βj G(xi , yj )

i,j

αi βj F (xi , yj ).

i,j

Since G is bounded, we have also n m 

  

αi βj F (xi , yj )
= |G(x, y)| ≤
G

x, y
=
G
 α i xi , βj yj .
i,j

i=1

j=1

Therefore, there exists a constant K as required where K =
G
. To prove the sufficiency, assume there exists a constant K such that n m

  

  α β F (x , y ) α x , β y ≤ K
 i j i j
i i j j . i,j

i=1

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j=1

2–BANACH SPACES

81

n m n m If either i=1 αi xi = 0 or j=1 βj yj = 0, then
i=1 αi xi , j=1 βj yj
= 0, which in turn implies that i,j αi , βj F (xi , yj ) = 0. This remark will be used to show that the extension G is well–defined. Let nx and y be arbitrary elements in [S] and [T ], respectively, with x = i=1 αi xi and m y = j=1 βj yj , where xi in S and yj in T . Define a function G on [S] × [T ] by  G(x, y) = αi βj F (xi , yj ). i,j

The following shows that G is a well–defined bounded linear 2– n and n  functional. Assume that x = i=1 nαi xi = i=1 αi xi , where some of αi  or αi may be 0. Then we have i=1 (αi − αi )xi = 0. By the previous remark, this implies that  (αi − αi )βj F (xi , yj ) = 0, i,j

which in turn yields that   αi βj F (xi , yj ) = αi βj F (xi , yj ). i,j

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m

m

i,j

Similarly, if y = j=1 βj yj = j=1 βj yj , where again some βj or βj may be 0, then the result is as follows:   αi βj F (xi , yj ) = αi βj F (xi , yj ). i,j

i,j

Thus G is independent of the representation of x and y in [S] and [T ]. To show that G is a linear 2–functional, let x and y be defined as above, m n while z = i=1 ri xi and w = j=1 δj yj . Then we have x+z =

n 

(αi + ri )xi ,

i=1

G(x + z, y + w) =



y+w =



(αi + ri )(βj + δj )F (xi , yj ) αi βj F (xi , yj ) +



i,j

+

(βj + δj )yj ,

j=1

i,j

=

m 



αi δj F (xi , yj )

i,j

ri βj F (xi , yj ) +

i,j



ri δj F (xi , yj )

i,j

= G(x, y) + G(x, w) + G(z, y) + G(z, w),

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82

R. W. FREESE AND Y. J. CHO

G(ψx, ξy) =



(ψαi )(ξβj )F (xi , yj ) = ψξ

i,j



αi βj F (xi , yj ) = ψξG(x, y).

i,j

Thus G is a linear 2–functional defined on [S] × [T ]. On the other hand, we have



αi βj F (xi , yj )
|G(x, y)| =
i,j

n m      α i xi , βj yj  ≤ K i=1

j=1

= K
x, y
, which implies that
G
≤ K and hence G is bounded. Recall that a pseudo 2–norm is defined as a real valued function having all the properties of a 2–norm except the condition that
a, b
= 0 implies that a and b are linearly dependent. A bounded linear 2–functional, an element in X ∗ , can be used to define such a function on a Banach space.

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The following theorem proves this assertion: Theorem 3.3.4. Let (X,
·
) be a Banach space and let X ∗ be the space of bounded linear 2–functionals defined on X. Then, for any F in X ∗ , |F (a, b)| defines a pesudo 2–norm on X. Proof. (1) If a and b are linearly dependent, then |F (a, b)| = 0. (2) For all a, b in X, 0 = F (a, b) + F (b, a). Thus F (a, b) = −F (b, a) and hence |F (a, b)| = |F (b, a)|. (3) |F (a, βb)| = |βF (a, b)| = |β||F (a, b)|. (4) |F (a, b + c)| = |F (a, b) + F (a, c)| ≤ |F (a, b)| + |F (a, c)|. Thus |F (a, b)| is a pseudo 2–norm on X. A. White [242], [243] has proved that if the space X is a euclidean 2– dimensional linear space, then the given function actually is a 2–norm for the space. For space of any higher dimension, it is possible for |F (a, b)| to equal 0 while a and b are linearly independent.

Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

CHAPTER 4. COMPLETION OF LINEAR 2–NORMED SPACES The concept of a complete metric space is well–known. A compact metric space is always complete. A complete subset of a metric space is always closed. If a metric space (X, σ) is not complete, then we can construct
called the completion of X, such that X is a complete metric space X,
which is dense in X.
Most often X will be isometric with a subset Xo of X
In [72], identified with Xo , and hence X is considered to be a subset of X. R.E. Ehret constructed the completion of linear 2–normed spaces by using the already well–know methods. In this chapter, unless otherwise stated, the space X will be a space of dimension greater than one having the following: (1) a 2–metric defined on it by σ(a, b, c) =
b − a, c − a
, (2) the corresponding neighborhood system for any point a defined by n the family of all sets WΣ (a) = i=1 U
i (a, bi) with arbitrary pairs Σ = {(b1 , 1 ), . . . , (bn , n )}.

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4.1. Elementary Properties In Chapter III, we have defined a Cauchy sequence as follows: A sequence {xn } in a linear 2–normed space (X,
·, ·
) is called a Cauchy sequence if there exist y and z in X such that y and z are linearly independent, lim
xn − xm , z
= 0. lim
xn − xm , y
= 0, n,m→∞

n,m→∞

We call a sequence satisfying these conditions a weak Cauchy sequence and reserve the term Cauchy sequence for the slightly stronger notion: Definition 4.1.1. A sequence {xn } is called a Cauchy sequence in a linear 2–normed space (X,
·, ·
) if to every neighborhood U of the origin, there is an integer N (U ) such that n, m ≥ N (U ) implies that xn − xm ∈ U. This definition is justified because every linear 2–normed space of dimension greater than one is a topological vector space and such spaces have this stronger definition for Cauchy sequences. Theorem 4.1.1. A sequence {xn } is a Cauchy sequence in a linear 2– normed space (X,
·, ·
) if and only if limn,m→∞
xn − xm , z
= 0 for every Typeset by AMS-TEX

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84

R. W. FREESE AND Y. J. CHO

z in X. Proof. Assume that {xn } is a Cauchy sequence in X. Let z be an arbitrary point in X. U
(0, z) is a neighborhood of 0. By definition of Cauchy sequence, there exists an integer N such that xn − xm ∈ U
(0, z) for n, m ≥ N , which implies that σ(xn − xm , 0, z) <  or


xn − xm , z
< 

for n, m ≥ N . Hence we have limn,m→∞
xn − xm , z
= 0. Since z is arbitrary, the conclusion holds for all z in X. Conversely, assume limn,m→∞
xn − xm , z
= 0 for all z in X. Let WΣ (0) be an arbitrary neighborhood of 0 with Σ = {(b1 , α1 ), . . . , (br , αr )}. By hypothesis, we have lim
xn − xm , bj
= 0 n,m→∞

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for j = 1, · · · , r. Thus for each αj there exists an integer Nj such that
xn − xm , bj
< αj for n, m ≥ Nj , j = 1, . . . , r. Let N = max{N1 , . . . , Nr }. Then σ(xn − xm , bj , 0) < αj for n, m ≥ N implies that xn − xm ∈ WΣ (0) for n, m ≥ N and thus it follows that {xn } is a Cauchy sequence in X. Definition 4.1.2. Two Cauchy sequences {xn } and {yn } in a linear 2– normed space (X,
·, ·
) are are said to be equivalent, denoted by {xn } ∼ {yn }, if for every neighborhood U of 0 there is an integer N (U ) such that n ≥ N (U ) implies that xn − yn ∈ U. Theorem 4.1.2. The relation ∼ on the set of Cauchy sequences in X is an equivalence relation on X. Proof. (1) xn − xn = 0 ∈ U for every neighborhood U of 0 and for all n shows that {xn } ∼ {xn }. (2) Let U be an arbitrary neighborhood of 0 and {xn } ∼ {yn }. There exists a balanced neighborhood V of 0, in which for each x in V, αx ∈ V for |α| ≤ 1, such that V ⊂ U. For this V there exists an integer N (V ) such that xn − yn ∈ V for n ≥ N (V ). Therefore, we have yn − xn = −(xn − yn ) ∈ V ⊂ U for n ≥ N (V ) and so {yn } ∼ {xn }. (3) Let {xn } ∼ {yn } and {yn } ∼ {zn }. Let U be an arbitrary neighborhood of 0. There exists a neighborhood V of 0 such that V + V ⊂ U , and for

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COMPLETION OF LINEAR 2–NORMED SPACES

85

this V there exists an integer N such that xn − yn and yn − zn are elements in V for n ≥ N . Hence xn − zn = (xn − yn ) + (yn − zn ) is an elements of U for n ≥ N . Therefore, {xn } ∼ {zn } and thus it is an equivalence relation on X. The proof of the following theorem closely resembles that of Theorem 4.1.1 and so we omit the proof: Theorem 4.1.3. {xn } is equivalent to {an } in a linear 2–normed space (X,
·, ·
) if and only if limn→∞
xn − an , z
= 0 for every z in X. By using Theorem 4.1.3, we have the following: Theorem 4.1.4. If {an } and {bn } are equivalent to {xn } and {yn } in a linear 2–normed space (X,
·, ·
), respectively, then {an + bn } is equivalent to {xn + yn } and {αan } is equivalent to {αxn }. Proof. Since {an } ∼ {xn } and {bn } ∼ {yn }, by Theorem 4.1.3, we have
(xn + yn ) − (an + bn ), z
=
(xn − an ) + (yn − bn ), z
≤
xn − an , z
+
yn − bn , z
→ 0,

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αxn − αan , z
= |α|
xn − an , z
→ 0 as n → ∞. Therefore, {an + bn } ∼ {xn + yn } and {αan } ∼ {αxn }. 4.2. Completion of Linear 2–Normed Spaces
the set of all equivalence classes of Cauchy sequences in X. Denote by X
Define an addition and scalar Let x ˆ, yˆ, zˆ, etc., denote the elements of X.
as follows: multiplication on X (i) x ˆ + yˆ = the set of sequences equivalent to {xn + yn }, where {xn } is in x ˆ and {yn } is in yˆ, and ˆ. (ii) αˆ x = the set of sequences equivalent to {αxn }, where {xn } is in x By Theorem 4.1.4, these two operations are well–defined since they are independent of the choice of elements from xˆ and yˆ. With these two opera
is a linear space. tions, X Theorem 4.2.1. If a sequence {xn } is a Cauchy sequence in a linear 2–normed space (X,
·, ·
), then limn→∞
xn , z
exists for every z in X.

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86

R. W. FREESE AND Y. J. CHO

Proof. Assume that {xn } is a Cauchy sequence in a linear 2–normed space (X,
·, ·
) and let z be any element in X. Several applications of property (N4 ) of the 2–norm
·, ·
yield that   
xn , z
−
xm , z
 ≤
xn − xm , z
. By Theorem 4.1.1, the limit of the right side of this expression equals 0 as m, n → ∞. Therefore {
xn , z
} is a Cauchy sequence of real numbers and limn→∞
xn , z
exists. Theorem 4.2.2. If two Cauchy sequences {xn } and {yn } in a linear 2–normed space (X,
·, ·
) are equivalent, then limn→∞
xn , z
= limn→∞
yn , z
for every z in X. Proof. By property (N4 ) of the 2–norm
·, ·
, since we have
xn , z
=
xn − yn + yn , z
≤
xn − yn , z
+
yn , z
, it follows that
xn , z
−
yn , z
≤
xn − yn , z
. Similarly, we have

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yn , z
−
xn , z
≤
yn − xn , z
. By using Theorem 4.1.3, we have   lim 
xn , z
−
yn , z
 ≤ lim
xn − yn , z
= 0. n→∞

n→∞

Thus the conclusion follows. Theorem 4.2.3. If {xn } and {yn } are Cauchy sequences in a linear 2–normed space (X,
·, ·
), then {xn − yn } is a Cauchy sequence in X. Proof. Let U be an arbitrary neighborhood of 0. There exists a neighborhood V of 0 such that V + V ⊂ U. For this V there exists an integer N such that xn − xm ∈ V, yn − ym ∈ V for n, m ≥ N . Thus we have (xn − yn ) − (xm − ym ) = (xn − xm ) − (yn − ym ) ∈ V + V ⊂ U

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COMPLETION OF LINEAR 2–NORMED SPACES

87

for n, m ≥ N . By definition of a Cauchy sequence, {xn − yn } is a Cauchy sequence in X. One of the major problems in working toward the completion of a linear 2-normed space has been the question of the existence of limn→∞
xn , yn
for any two Cauchy sequences in X. The problem is solved for spaces which have uniformly continuous 2–norms defined on them. As will be shown subsequently, such spaces can be completed at least to a pseudo 2–normed space. Definition 4.2.1. A 2–norm
·, ·
defined on a linear space X is said to be uniformly continuous in both  variables if for  any  > 0 there exists a neighborhood U
of 0 such that 
a, b
−
a , b
 <  whenever a − a and b − b are in U
, which is independent of the choice of a, a , b, or b .

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Definition 4.2.2. A pseudo 2–norm is defined to be a real–valued function having all the properties of a 2–norm
·, ·
except the condition that
a, b
= 0 implies the linear dependence of a and b. Theorem 4.2.4. If X is a linear space having a uniformly continuous 2–norm
·, ·
defined on it, then for any two Cauchy sequences {xn } and {yn } in X, limn→∞
xn , yn
exists. Proof. Let {xn } and {yn } be arbitrary Cauchy sequences in X. Let  > 0 be given and U
be the corresponding neighborhood of 0 such that   
a, b
−
a , b
 <  whenever a − a and b − b are in U
. There exists an integer N () such that xn − xm ∈ U
,

yn − ym ∈ U


for n, m ≥ N (). Thus we have   
xn , yn
−
xm , ym
 <  for n, m ≥ N (), which means that {
xn , yn
} is a Cauchy sequence of real numbers and hence we see that the desired limit exists for any pair of Cauchy sequences in X. The proof of the following theorem is almost identical to that of Theorem 4.2.4 and so we omit the proof:

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88

R. W. FREESE AND Y. J. CHO

Theorem 4.2.5. If X is a linear space having a uniformly continuous 2–norm
·, ·
defined on it, then for pairs of equivalent Cauchy sequences, {xn } ∼ {an } and {yn } ∼ {bn }, lim
xn , yn
= lim
an , bn
.

n→∞

n→∞

Whenever X is a space having a uniformly continuous 2–norm defined on
The function it, it is possible to define a real-valued function on the space X. is defined as folows:
For any two elements x ˆ and yˆ in X,
ˆ x, yˆ
= lim
xn , yn
, n→∞

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ˆ and {yn } ∈ yˆ. The function is well–defined since, by where {xn } ∈ x Theorem 4.2.4, the limit exists and, by Theorem 4.2.5, it is independent of the choice of elements in x ˆ and yˆ. The most that can be said thus far regarding the function defined above is that it is a pseudo 2–norm for the space X. It remains an open question whether the function is a 2–norm. Theorem 4.2.6. If X is a linear space having a uniformly continuous ˆ 2–norm
·, ·
defined on it and {xn } and {yn } are Cauchy sequences in x and yˆ, respectively, then the function defined by
ˆ x, yˆ
= limn→∞
xn , yn

is a pseudo 2–norm on X. Proof. The function is well–defined and independent of the choice of elements in x ˆ and yˆ as has already been stated. (1) If x ˆ = αˆ y , then we have
ˆ x, yˆ
=
αˆ y , yˆ
= limn→∞
αyn , yn
= x, yˆ
= 0, then limn→∞ |α|
yn , yn
= 0. since
yn , yn
= 0 for all n. If
ˆ limn→∞
xn , yn
= 0, where {xn } is in x ˆ and {yn } is in yˆ. y, x ˆ
. (2)
ˆ x, yˆ
= limn→∞
xn , yn
= limn→∞
yn , xn
=
ˆ x, yˆ
. (3)
ˆ x, αˆ y
= limn→∞
xn , αyn
= limn→∞ |α|
xn , yn
= |α|
ˆ (4)
ˆ x, yˆ + zˆ
= limn→∞
xn , yn + zn
≤ limn→∞ (
xn , yn
+
xn , zn
) = x, yˆ
+
ˆ x, zˆ
. limn→∞
xn , yn
+ limn→∞
xn , zn
=
ˆ
o be the subset of X consisting of those equivalence classes which Let X contain a Cauchy sequence {xn } for which x1 = x2 = · · · = xn = · · · . At most one sequence of this kind can be in each equivalence class. If x ˆ and
yˆ are in Xo and if corresponding Cauchy sequence are {xn } and {yn } with xn = x and yn = y for every n, then we have
ˆ x, yˆ
= lim
xn , yn
= lim
x, y
=
x, y
. n→∞

n→∞

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COMPLETION OF LINEAR 2–NORMED SPACES

89


o and X
are isometric. This isometry will be used to show that X
o Thus X
is dense in X. Theorem 4.2.7. If X is a linear space having a uniformly continuous
o ) = X.
2–norm
·, ·
defined on it, then (X
−X
o and {xn } be a Cauchy Proof. Let x ˆ be an arbitrary element in X sequence in X such that {xn } ∈ x ˆ. For each n, define x ˆn to be the element
o which contains the repetitive sequence xn , xn , xn , · · · . For each zˆ in in X
and each {zn } ∈ zˆ, since the 2–norm
·, ·
is uniformly continuous and X {xn } is a Cauchy sequence in X, we have
(xn )m − xm , zm
=
xn − xm , zm
< 

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for n, m > N (), where (xn )m is the mth –component of the repetitive sexn − x ˆ, zˆ
<  for n ≥ N () quence in x ˆn . By definition, this implies that
ˆ
Thus X
o is dense in X.
xn − x ˆ, zˆ
= 0 for every zˆ in X. and hence limn→∞
ˆ


there It should be noted that for each neighborhood U( 0) of
0 in X, exists a sequence of neighborhoods, {Um (0)}, of 0 in X, independent of any Cauchy sequence, such that if xn − xm ∈ Um (0) for n, m ≥ N (Um ), then
(

The final step in the completion process will ˆ∈U 0) for n ≥ N (U). x ˆn − x
is complete. be taken in the proof of the fact that X Theorem 4.2.8. If X is a linear space having a uniformly continuous
is complete. 2–norm
·, ·
defined on it, then X
Let ˆb be any element in X
Proof. Let {ˆ yn } be a Cauchy sequence in X.
o such that
ˆ and for each n, choose zˆn in X yn − zˆn , ˆb
< n1 , which is possible
On the other hand, we have
o is dense in X. since X zn − yˆn , ˆb
+
ˆ yn − yˆm , ˆb
+
ˆ ym − zˆm , ˆb

ˆ zn − zˆm , ˆb
≤
ˆ 1 1 +
ˆ yn − yˆm , ˆb
. < + n m
and ˆb is arbitrary, Theorem 4.1.1 Since {ˆ yn } is a Cauchy sequence in X
o . Let {zn } be the sequence in provided that {ˆ zn } is a Cauchy sequence in X
and X
o . Then {zn } X corresponding to {ˆ zn } under the isometry between X
such is a Cauchy sequence in X and hence there exists an element yˆ in X

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that {zn } ∈ yˆ and
ˆ yn − yˆ, ˆb
≤
ˆ yn − zˆn , ˆb
+
ˆ zn − yˆ, ˆb
< 1/n +
ˆ zn − yˆ, ˆb
. ˆ As in the proof of Theorem 4.2.7,
ˆ zn − yˆ, b
<  for n ≥ N () provided
zn − zm , bm
<  for n, m ≥ N (). The latter inequality holds since {zn } is a Cauchy sequence in X and the 2–norm
·, ·
is uniformly continuous. yn − yˆ, ˆb
= 0 and since the arbitrary Cauchy sequence {ˆ yn } Hence limn→∞
ˆ


in X converges to a point yˆ in X, the space X is complete. An example of a uniformly continuous 2–norm
·, ·
is the function defined by

 n  1   ai
a, b
=  bi 2 i 0, U
(0) is a neighborhood of 0 relative to the norm. Since the topologies for the related metrics are equal, there is a neighborhood W of 0, relative to the 2–norm, such that W ⊂ U
(0). Since {xn } is a Cauchy sequence relative to the 2–norm, there exists an integer N such that xn − xm ∈ U
(0) for n, m ≥ N , which implies that
xn − xm
<  for n, m ≥ N . Thus {xn } is a Cauchy sequence in Rn relative to the norm. Since Rn is complete in this norm, there exists an x in Rn such that limn→∞
xn − x
= 0. Let W  be an arbitrary neighborhood of 0 relative to the 2–norm. Again, using the equivalent topologies, there is a δ > 0 such that Uδ (0) is a neighborhood of 0 relative to the norm and Uδ (0) ⊂ W  . For this, there exists an integer N such that
xn − x
< δ for n ≥ N and hence it follows that xn − x ∈ Uδ (0) ⊂ W  for n ≥ N. From Theorem 4.2.10, it then follows that {xn } converges to x in Rn and thus Rn is a 2–Banach space relative to the given 2–norm. 4.3. Relations between Banach Spaces and 2–Banach Spaces Consider the norm
·
defined on a linear 2–normed space (X,
·, ·
) by the function
a
=
a, y
+
a, z
for any fixed y, z ∈ X and
y, z
= 0. Theorem 4.3.1. The function
·
defined above is a norm on X.

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Proof. (1)
a
≥ 0. If
a
= 0, then
a, y
= 0 and
a, z
= 0. Under the assumption that a = 0, this implies that a = αy and a = βz with α = 0 and β = 0, which contradicts the linear independence of y and z. Thus
a
= 0 implies that a = 0. On the other hand, we have (2)


a + b
=
a + b, y
+
a + b, z
≤
a, y
+
b, y
+
a, z
+
b, z
=
a
+
b
,

(3)


αa
=
αa, y
+
αa, z


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= |α|(
a, y
+
a, z
) = |α|
a
. Theorem 4.3.2. If (X,
·, ·
) is a linear 2–normed space possessing Property (K) and having a norm defined on it by
a
=
a∗ , a
+
b∗ , a
for a∗ and b∗ in X such that
a∗ , b∗
= 0, then X is a 2–Banach space if and only if X is a Banach space relative to this norm. Proof. Assume that X is a Banach space relative to the norm defined above. Let {xn } be a Cauchy sequence in X relative to the 2–norm. By Theorem 4.1.1, limm,n→∞
xn − xm , z
= 0 for every z in X, in particular for z = a∗ and z = b∗ . Thus we have
xn − xm
=
a∗ , xn − xm
+
b∗ , xn − xm
→ 0 as n, m → ∞, and so {xn } is a Cauchy sequence relative to the norm. Since X is a Banach space relative to this norm, there exsists an x in X such that
xn − x
→ 0 as n → ∞. The fact that the 2–norm is non-negative implies that limn→∞
xn − x, a∗
= 0 and limn→∞
xn − x, b∗
= 0. Property (K) then provides that limn→∞
xn − x, z
= 0 for all z in X. Thus, {xn } converges to x in X relative to the 2–norm and X is a 2–Banach space. Conversely, assume that X is a 2–Banach space and {xn } is a Cauchy sequence relative to the norm given. Then we have
xn − xm , a∗
+
xn − xm , b∗
=
xn − xm
→ 0 as n, m → ∞. Hence we have
xn − xm , a∗
→ 0,


xn − xm , b∗
→ 0

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as n, m → ∞. Property (K) then provides that
xn − xm , z
→ 0 for every z in X and hence {xn } is a Cauchy sequence in X relative to the 2–norm. Since X is a 2–Banach space, there exists an x in X such that limm,n→∞
xn − xm , z
= 0 for all z in X. In particular, since we have lim
xn − x, a∗
= 0,

n→∞

lim
xn − x, b∗
= 0,

n→∞

it follows that lim
xn − x
= lim
xn − x, a∗
+ lim
xn − x, b∗
= 0.

n→∞

n→∞

n→∞

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Since the arbitrary Cauchy sequence {xn } converges to an element x in X, X is a Banach space relative to the given norm.

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CHAPTER 5. 2–INNER PRODUCT SPACES The concepts of 2–inner product and 2–inner product space (or 2–pre– Hilbert space) are 2–dimensional analogs of the concepts of inner product and inner product space (or pre–Hilbert space). In [55] and [58], C.R. Diminnie, S. G¨ ahler and A. White introduced the concept of 2–inner product spaces and gave some characterizations of 2–inner product spaces by using Gˆ ateaux derivatives of a 2–norm
·, ·
. In [57] and [62], they also generated 2–inner products on linear 2–normed spaces and generalized 2–inner products.

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5.1. 2–Inner Products Let X be a linear space of dimension greater than one and (·, ·|·) be a real valued function on X × X × X which satisfies the following conditions: (I1 ) (i) (a, a|b) ≥ 0, (ii) (a, a|b) = 0 if and only if a and b are linearly dependent, (I2 ) (a, a|b) = (b, b|a), (I3 ) (a, b|c) = (b, a|c), (I4 ) (αa, b|c) = α(a, b|c) for every real α, (I5 ) (a + a
, b|c) = (a, b|c) + (a
, b|c) for all a, a
, b, c ∈ X. (·, ·|·) is called a 2–inner product and (X, (·, ·|·)) is called a 2–inner product space (or a 2–pre–Hilbert space). Lemma 5.1.1. For every a, b, c ∈ X, |(a, b|c)| ≤





(a, a|c) (b, b|c).

Proof. For any real number β, we have (a + β(a, b|c)b, a + β(a, b|c)b|c) = (a, a|c) + 2β(a, b|c)2 + β 2 (a, b|c)2(b, b|c) ≥ 0. If (b, b|c) = 0, the above inequality follows immediately. If (b, b|c) = 0, then the inequality is obtained by substituting β = −1/(b, b|c). Typeset by AMS-TEX

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Recall that the inequality in Lemma 5.1.1 is the 2–dimensional analogue of the Cauchy–Buniakowski Inequality. Corollary 5.1.2. For every a, b ∈ X, (a, b|b) = 0. Lemma 5.1.3. For every a, b, c ∈ X and γ ∈ R, (a, b|γc) = γ 2 (a, b|c). Proof. For every a, b, c ∈ X and γ ∈ R, we have 1 [(a + b, a + b|γc) − (a − b, a − b|γc)] 4 1 = [(γc, γc|a + b) − (γc, γc|a − b)] 4 γ2 [(c, c|a + b) − (c, c|a − b)] = 4 = γ 2 (a, b|c).

(a, b|γc) =

It is easily shown by using simple properties of bivectors, as in Chapter II, that the following holds: Theorem 5.1.4. For every inner product (·|·) on BX , the function (·, ·|·) defined on X × X × X by

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(a, b|c) = (b(a × c)|b(b × c)) represents a 2–inner product on X. By Theorem 5.1.4, we have the following: Corollary 5.1.5. Let (X, (·|·)) be a Hilbert space. A 2–inner product is defined on X by    (a|b) (a|c)   = (a|b)
c
2 − (a|c)(b|c). (a, b|c) =  (b|c) (c|c)  Corollary 5.1.6. A 2–inner product over the pre–Hilbert space of sequences is given by  (αi γj − γi αj )(βi γj − γi βj ) (a, b|c) = i 1, F can not be identically 0 on X × X. Further, since |F (a, b)| ≤
F

a, b
, it follows that
F
= 0. Theorem 5.2.2. Let (X,
·, ·
) be a linear 2–normed space and let F be a positive, symmetric, non–degenerate and bounded 2–functional on X × X such that (5.2)

F (a + b, c) + F (a − b, c) = 2F (a, c) + 2F (b, c).

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Then [a, b|c] = 14 [F (a + b, c) − F (a − b, c)] defines a 2–inner product on X.

Proof. Since F is bounded, it follows immediately that F (0, c) = 0 for every c ∈ X. We have the following steps to prove this theorem: 1. By setting b = a in (5.2), we have F (2a, c) = 4F (a, c). Hence, [a, a|c] = F (a, c) = F (c, a) = [c, c|a]. 2. Also, by setting a = 0 in (5.2), we have F (−b, c) = F (b, c), which implies that [a, b|c] = [b, a|c]. 3. By (5.2), we have   b  b  F (a + a
+ b, c) = 2F a + , c + 2F a
+ , c − F (a − a
, c), 2 2   b  b  F (a + a
− b, c) = 2F a − , c + 2F a
− , c − F (a − a
, c). 2 2 By subtracting these equations, we obtain the equation (5.3)

 b   b   [a + a , b|c] = 2 a, c + 2 a
, c . 2 2


Since F (−b, c) = F (b, c), it follows that [0, b|c] = 0. Therefore, by setting a
= 0 in (5.3), we have [a, b|c] = 2[a, 2b |c] and hence [a + a
, b|c] = [a, b|c] + [a
, b|c].

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4. Define φ(α) = [αa, b|c]. By step 3, φ(α + γ) = φ(α) + φ(γ). Also, by the boundedness of F , we have |φ(α)| = |[αa, b|c]| ≤


F
(|α|
a, c
+
b, c
). 2

Since φ is additive and bounded on [0, 1], it follows that φ(α) = αφ(1) and hence we have [αa, b|c] = α[a, b|c]. 5. Since F (0, c) = 0 and F is positive, it follows easily that [a, a|c] ≥ 0 for every a, c ∈ X. 6. If [a, a|c] = 0, then by steps 2, 3, 4, and 5, we have γ 2 [b, b|c] + 2αγ[a, b|c] = [αa + γb, αa + γb|c] ≥ 0 for any b ∈ X and any real numbers α and γ. In particular, for α = −[b, b|c]– 1 and γ = [a, b|c], we have −[a, b|c]2 ([b, b|c] + 2) ≥ 0.

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Hence, we have [a, b|c] = 0 and F (a + b, c) = F (a − b, c) for every b ∈ X. Since F is non–degenerate, a and c are linearly dependent. Finally, if a and c are linearly dependent, then we have [a, a|c] =

1 F (2a, c) = F (a, c) ≤
F

a, c
= 0. 4

Therefore, [·, ·|·] is a 2–inner product on X. Corollary 5.2.3. If
·, ·
∗ is the 2–norm determined by [·, ·|·] in the preceding Theorem 5.2.2, then
a, b
∗ ≤
F

a, b
for every a, b ∈ X. Proof. If b = 0, then the result is obvious. If b = 0, then by the Lemma 5.2.1, p1 (a) =
a, b
∗ and p2 (a) =
F

a, b
are non–zero semi–norms on X. If p2 (a) ≤ 1, then we have p1 (a) = [a, a|b]1/2 = (F (a, b))1/2 ≤ (
F

a, b
)1/2 = (p2 (a))1/2 ≤ 1. Therefore, since p1 and p2 are semi–norms on X for which p1 (a) ≤ 1 whenever p2 (a) ≤ 1, it follows that p1 (a) ≤ p2 (a) for every a ∈ X.

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5.3. Generalizations of 2–Inner Products In [57], C. R. Diminnie, S. G¨ ahler and A. White showed that there are many occasions where the study of 2–inner product spaces and linear 2– normed spaces is facilitated by considering the bivectors over X. By Theorem 5.1.4, if (·|·) is an inner product on BX , then (a, b|c) = (b(a×c)|b(b×c)) defines a 2–inner product (·, ·|·) on X. If all bivectors over X are simple, i.e., dim X ≤ 3, then conversely every 2–inner product on X has a corresponding inner product on BX for which

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(a, b|c) = (b(a × b)|b(b × c)) for every a, b, c ∈ X. If
·
is a norm on BX , then
a, b
=
b(a × b)
defines a 2–norm
·, ·
on X. There is an example which shows that for every 2-norm
·, ·
on X, there need not exist a norm
·
on BX which satisfies
b(a×b)
=
a, b
for all a, b ∈ X. In case dim X ≤ 3, every 2–norm on X has a corresponding norm on BX for which
a, b
=
b(a × b)
for all a, b ∈ X. In this section, we consider a generalization of the concept of 2–inner product. If X and X
are linear spaces with dim X > 1, let ζ denote the collection of all mappings T : X × X × X → X
which are linear in the first two arguments and for which T (a, b, c) = 0 if a and c or b and c are linearly dependent. With the usual operations, ζ is a linear space. For the case X
= R, we will say that T ∈ ζ has Property (1) if T (a, a, b) ≥ 0 for every a, b ∈ X. For arbitrary X
, we will say that T has Property (2) (resp. Property (3)) if T (a, a, b) = T (b, b, a) (resp., T (a, b, c) = T (b, a, c)) for every a, b, c ∈ X. Theorem 5.3.1. If T ∈ ζ, then for arbitrary a, b, c ∈ X, we have (5.4)

T (a + b, a + b, c) + T (a − b, a − b, c) = 2[T (a, a, c) + T (b, b, c)].

If T has Property (3), then we have (5.5)

T (a, b, c) =

1 [T (a + b, a + b, c) − T (a − b, a − b, c)] . 4

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Further, if X
= R and T has Properties (1) and (3), |T (a, b, c)| ≤ T (a, a, c)1/2T (b, b, c)1/2

(5.6)

Proof. See Lemma 5.1.1 and Theorem 5.1.9. Relations (5.4), (5.5), and (5.6) are the analogues for the Parallelogram Law, the Polarization Identity, and the Cauchy–Bunjakowski Inequality for inner products. (5.4) and (5.5) are immediate. The proof of (5.6) proceeds according to the proof of the Cauchy–Bunjakowski Inequality. By (5.5), every T ∈ ζ having Properties (2) and (3) is quadratic in its third argument. Theorem 5.3.2. Let T ∈ ζ have Properties (2) and (3). For any e1 , e2 , e3 ∈ X, we have T

3 

3 

αi ei ,

i=1

(5.7)

i=1

βi ei ,

3 

γi ei



i=1

3 1  (αi γi − αj γj )(βi γi − βj γj )T (ei , ei , ej ) = 2 i,j=1 i=j

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+

3 

(αi γk − αk γi )(βj γk − βk γj )T (ei , ej , ek ).

i,j,k=1 i=j=k=i

Proof. 1. We will show first that for a, b, c ∈ X, T (a, b, a + c) = T (a, b, c) − T (c, b, a).

(5.8) By (5.5), we have

4T (a, b, a + c) = T (a + b, a + b, a + c) − T (a − b, a − b, a + c) = T (a + c, a + c, a + b) − T (a + c, a + c, a − b) = 2T (a, c, a + b) − 2T (a, c, a − b) + 4T (a, b, c). If we substitute −c for c and recall that T is quadratic in its third argument, we obtain 4T (a, b, a − c) = 2T (a, c, a − b) − 2T (a, c, a + b) + 4T (a, b, c).

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R. W. FREESE AND Y. J. CHO

Therefore, we have T (a, b, c) =

1 [T (a, b, a + c) + T (a, b, a − c)] 2

and, similarly, we have also T (c, b, a) =

1 [T (c, b, a + c) + T (c, b, a − c)]. 2

Finally, we obtain 1 T (2a, b, a + c) 2 1 = [T (a − c, b, a + c) + T (a − c, b, a − c)] 2 = T (a, b, c) − T (c, b, a)

T (a, b, a + c) =

and so (5.8) is proven. 2. Next, we will show that T (ei ,ej , γi ei + γj ej + γk ek ) (5.9)

= γk2 T (ei , ej , ek ) − γi γk T (ej , ek , ei ) − γj γk T (ek , ei , ej ) − γi γj T (ei , ei , ej )

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for distinct i, j, k ∈ {1, 2, 3}. (5.9) is clearly true if γi = γj = 0. If γi = 0 and γj = 0, then by (5.8), T (ei , ej ,γi ei + γj ej + γk ek ) 1 = T (γi ei , ej , γi ei + γk ek ) γi 1 = [T (γi ei , ej , γk ek ) − T (γk ek , ej , γi ei )] γi = γk2 T (ei , ej , ek ) − γi γk T (ej , ek , ei ). Similar arguments hold when γi = 0 and γj = 0. If γi = 0 and γj = 0, it follows by (5.8) that T (ei ,ej , γi ei + γj ej + γk ek ) 1 = [T (γi ei , ej , γj ej + γk ek ) − T (γj ej + γk ek , ej , γi ei )] γi = T (ei , ej , γj ej + γk ek ) − γi γj T (ei , ei , ej ) − γi γk T (ej , ek , ei ) 1 = [T (γj ej , ei , γk ek ) − T (γk ek , ei , γj ej )] γj − γi γj T (ei , ei , ej ) − γi γk T (ej , ek , ei )

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2–INNER PRODUCT SPACES

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= γk2 T (ei , ej , ek ) − γi γk T (ej , ek , ei ) − γj γk T (ek , ei , ej ) − γi γj T (ei , ei , ej ). Thus, the relation (5.9) is proven. 3. Finally, we move to the proof of (5.7). For arbitrary i ∈ {1, 2, 3} and 3 for c = k=1 γk ek , we have T (ei , ei , c) = T (c, c, ei ) =

3 

γj γk T (ej , ek , ei )

j,k=1 j=k 3 

=

γj2 T (ei , ei , ej ) +

i,j=1 j=i

If a =

3

i=1

αi ei and b =

T (a, b, c) =

3 

3 

γj γk T (ej , ek , ei ).

j,k=1 i=j=k=i

3

j=1

βj ej , then by (5.9) and the above relation,

αi βj T (ei , ej , c)

i,j=1 i=j

=

3 

αi βi T (ei , ei , c) +

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i=1

=

3 

αi βj T (ei , ej , c)

i,j=1 i=j

3 

(αi βi γj2 − αi βj γi γj )T (ei , ei , ej )

i,j=1 i=j

+

3 

αi βi γj γk T (ej , ek , ei ) + αi βj γk2 T (ei , ej , ek )

i,j,k=1 i=j=k=i

− αi βi γj γk T (ej , ek , ei ) − αi βj γj γk T (ek , ei , ej ) =



3 1  (αi γj − αj γi )(βi γj − βj γi )T (ei , ei , ej ) 2 i,j=1 i=j

+

3 

(αi γk − αk γi )(βj γk − βk γj )T (ei , ej , ek ).

i,j,k=1 i=j=k=i

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Corollary 5.3.3. Let dim X = 2 and {e1 , e2 } be a basis for X. If (·, ·|·) is a 2–inner–product on X,
·, ·
is the corresponding 2–norm, and T ∈ ζ has Properties (2) and (3), then for any a, b, c ∈ X, we have T (a, b, c) = (a, b|c)

T (e1, e1 , e2 ) .
e1 , e2
2

For the case X
= R, T is a 2–inner product if and only if T (e1 , e1 , e2 ) > 0. Proof. Let αi , βi , γi be the components of a, b, c with respect to ei (i = 1, 2). By Theorem 5.3.2, we have T (a, b, c) = (α1 γ2 − α2 γ1 )(β1 γ2 − β2 γ1 )T (e1 , e1 , e2 ) = (a, b|c)

(e1 , e1 |e2 )
e1 , e2
2

T (e1 , e1 , e2 ) .
e1 , e2
2

Thus this corollary is true. If S is a bilinear mapping from BX × BX into X
, then

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(5.10)

T (a, b, c) = S(b(a × c), b(b × c))

defines a function from X × X × X into X
which is linear in its first two arguments. Since b(a × b) is the zero bivector if a and b are linearly dependent, it follows that T (a, b, c) = 0 if a and c or b and c are linearly dependent. Hence, the mapping T defined by (5.10) is an element of ζ. Since b(a × b) = −b(b × a), T has Property (2). For Property (1) (resp., Property (3)) of T , positivity of S (with X
= R) (resp., symmetry of S (with X
arbitrary)), is sufficient and, as can be seen by S. G¨ ahler, in the case dim X ≤ 3, necessary. In Theorem 5.3.5, we will show that in the case dim X ≤ 3, every T ∈ ζ with Properties (2) and (3) can be expressed in the above way. For this we need the following: Theorem 5.3.4. Let T ∈ ζ have Properties (2) and (3). If a, b, c, a
, b
, c
∈ X satisfy b(a × c) = b(a
× c
) and b(b × c) = b(b
× c
), then T (a, b, c) = T (a
, b
, c
) (cf. Theorem 5.1.7). Proof. If either of the bivectors is 0, then T (a, b, c) = T (a
, b
, c
) = 0 since for each representative a∗ ×b∗ of the null–bivector, a∗ and b∗ are linearly

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111

dependent. If both bivectors are not null, then by S.G¨ ahler, there are real




numbers α, α , β, β , γ, γ , δ, δ such that a = αa + γc, b
= βb + δc, c
= α
a + γ
c = β
b + δ
c, and αγ
− α
γ = βδ
− β
δ = 1. When α
= 0 and β
= 0, we obtain T (a
, b
, c
) = T (αa + γc, βb + δc, α
a + γ
c)   c
c
= T −
,−
,α a +γ c α β 1 =

T (α
a + γ
c, α
a + γ
c, c) αβ 1 =
T (a, β
b + δ
c, c) = T (a, b, c). β If α
= 0 or β
= 0, then since α
a + γ
c = β
b + δ
c and a, c and b, c are linearly independent, it follows that α
= β
= 0 and γ
= δ
. Therefore, we have T (a
, b
, c
) = T (αa + γc, βb + δc, γ
, c)

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= (αγ
)(βδ
)T (a, b, c) = T (a, b, c). By S. G¨ ahler, it follows that if b1 and b2 are simple bivectors whose sum is simple, then there are a, b, c ∈ X for which b1 = b(a×c) and b2 = b(b×c). In this case, for a given T ∈ ζ with Properties (2) and (3), we have (5.11)

S(b1 , b2 ) = T (a, b, c)

is independent of the representatives a × c of b1 and b × c of b2 by Theorem 5.3.4. Theorem 5.3.5. If dim X ≤ 3, then for an arbitrary T ∈ ζ which has Properties (2) and (3), (5.11) defines a symmetric bilinear mapping of BX × BX into X
(cf. Theorem 5.1.8). Proof. As the properties S(b1 , b2 ) = S(b2 , b1 ) and S(αb1 , b2 ) = αS(b1 , b2 ) follow easily from the properties of T , we have only to show that S(b1 + b2 , b3 ) = S(b1 , b3 ) + S(b2 , b3 ). If b1 , b2 , or b3 = 0, the conclusion is obvious. If b1 , b2 and b3 = 0, let b1 = b(a1 × b1 ), b2 = b(a2 × b2 ), b3 = b(a3 × b3 ) and let E1 , E2 , and E3

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R. W. FREESE AND Y. J. CHO

denote the planes of X generated by a1 and b1 , a2 and b2 , and a3 and b3 , respectively. If E1 , E2 , and E3 have a non–zero point c in common, then by S. G¨ ahler [96], there are a
1 , a
2 , a
3 ∈ X such that b1 = b(a
1 × c), b2 = b(a
2 × c), b3 = b(a
3 × c). From this, the result follows immediately. For the case E1 ∩ E2 ∩ E3 = {0}, let e12 , e23 , e31 be non–zero points in E1 ∩ E2 , E2 ∩ E3 , E3 ∩ E1 , respectively. By S. G¨ ahler, there are real numbers α, β, and γ such that b1 = b(αe31 × e12 ),

b2 = b(βe12 × e23 ),

b3 = b(γe23 × e31 ).

Then, we have b1 + b2 = b(−e12 × αe31 ) + b(−e12 × (−βe23 )) = b(−e12 × (αe31 − βe23 )) and

γ  1 b(γe23 × αe31 ) = b e23 × (αe31 − βe23 ) . α α Therefore, by Theorem 5.3.2, we obtain   γ S(b1 + b2 , b3 ) = T − e12 , e23 , αe31 − βe23 α γ = − T (e23 , e12 , βe23 − αe31 ) α γ = − [α2 T (e23 , e12 , e31 ) + βαT (e31 , e12 , e23 )] α = T (−αe12 , γe23 , e31 ) + T (βe12 , −γe31 , e23 )

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b3 =

= S(b1 , b3 ) + S(b2 , b3 ). Let X be an arbitrary linear space with dim X > 1 and X
= R . Further, let S be an alternating bilinear functional on X × X. When a and b are linearly dependent, S(a, b) = 0 holds. Thus, the equation (5.12)

T (a, b, c) = S(a, c)S(b, c)

defines a T ∈ ζ which has Properties (1), (2), and (3). For such a T , equality holds in (5.4), i.e., (5.13)

T (a, b, c)2 = T (a, a, c)T (b, b, c)

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2–INNER PRODUCT SPACES

113

for arbitrary a, b, c ∈ X. By (5.5), it follows that in the case dim X = 2, (5.13) is satisfied for every T ∈ ζ with Properties (2) and (3). However, this is not true when dim X ≥ 3. Theorem 5.3.6. Let dim X ≤ 3 and X
= R. For every T ∈ ζ, which has Properties (1), (2), and (3) and for which (5.13) is satisfied for every a, b, c ∈ X, there exists an alternating bilinear functional S on X × X for which (5.12) holds for every a, b, c ∈ X. Proof. 1. If dim X = 2, let {e1 , e2 } be a basis for X and define S

2  i=1

αi ei ,

2 

 1 βi ei = (α1 β2 − β1 α2 )T (e1 , e1 , e3 ) 2 .

i=1

The assertion is proven by a direct application of Theorem 5.3.2. 2. Now, consider the case dim X = 3. (i) We will show first that under the given hypothesis, there exists a basis E = {e1 , e2 , e3 } for X and an alternating functional s on E × E such that

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T (ei , ej , ek ) = s(ei , ek )s(ej , ek ) for i, j, k ∈ {1, 2, 3}. Since the case T = 0 is trival, we will assume that T = 0. For an arbitrary basis {a1 , a2 , a3 } of X, Theorem 5.3.2 implies that at least one of the numbers T (ai , ai , aj ), i = j, or T (ai , aj , ak ), i = j = k = i, is non–zero. If T (ai , aj , aj ) = 0 for all distinct i, j and k, then one of the numbers T (ai , ai , aj ), i = j, is positive while by (5.13) the other two are 0. Therefore, there exists a basis {e1 , e2 , e3 } for X such that T (e1 , e1 , e2 ) = 1, T (e2 , e2 , e3 ) = T (e3 , e3 , e1 ) = 0, T (ei , ej , ek ) = 0 for distinct i, j, k. Under these circumstances, define s(e1 , e2 ) = 1 and s(e2 , e3 ) = s(e3 , e1 ) = 0. If there are distinct i, j, k such that T (ai , aj , ak ) = 0, then there is a basis {b1 , b2 , b3 } for X for which T (b1 , b2 , b3 ) = 1. There are two cases to consider: If T (b1 , b1 , b2 ) = 0, let e1 = b1 , e2 = b2 , and e3 = b3 and define s(e1 , e2 ) = 0, s(e2 , e3 ) = T (e2 , e2 , e3 )1/2 and s(e3 , e1 ) = −T (e1 , e1 , e3 )1/2 . By (5.13), T (e1 , e2 , e3 ) = T (e1 , e1 , e3 )1/2 T (e2 , e2 , e3 )1/2 = s(e1 , e3 )s(e2 , e3 ).

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R. W. FREESE AND Y. J. CHO

All other required properties are obvious. If T (b1 , b2 , b3 ) > 0, let e1 = b1 , e2 = T (b1 , b1 , b2 )−1/2 b2 , and e3 = T (b1 , b1 , b2 )1/4 b3 . Then, {e1 , e2 , e3 } is a basis such that T (e1 , e2 , e3 ) = T (e1 , e1 , e2 ) = 1. By (5.13), 1 = T (e1 , e2 , e3 )2 = T (e1 , e1 , e3 )T (e2 , e2 , e3 ) and hence T (e1 , e1 , e3 ) = T (e2 , e2 , e3 )−1 . By Theorem 5.3.2, we have T (e1 , e3 , e2 ) − T (e2 , e3 , e1 ) = T (e1 , e3 , e1 + e2 ). Squaring both sides and applying (5.13), we obtain T (e1 , e3 , e2 )T (e2 , e3 , e1 ) = −T (e1 , e1 , e2 )T (e1 , e1 , e3 ) = −1. Define s(e1 , e2 ) = 1, s(e2 , e3 ) = −T (e1 , e3 , e2 ) and s(e3 , e1 ) = −T (e2 , e3 , e1 ). Note that s(e1 , e3 ) = s(e2 , e3 )−1 . Hence, we have

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T (e1 , e1 , e2 ) = s(e1 , e2 )2 , T (e2 , e2 , e3 ) = T (e1 , e3 , e2 )2 T (e1 , e1 , e2 )−1 = s(e2 , e3 )2 , T (e3 , e3 , e1 ) = T (e2 , e2 , e3 )−1 = s(e2 , e3 )−2 = s(e1 , e3 )2 , T (e1 , e2 , e3 ) = s(e1 , e3 )s(e2 , e3 ), T (e2 , e3 , e1 ) = −s(e3 , e1 ) = s(e2 , e1 )s(e3 , e1 ) and T (e3 , e1 , e2 ) = −s(e2 , e3 ) = s(e3 , e2 ) = s(e1 , e2 )s(e3 , e2 ). (ii) Let E = {e1 , e2 , e3 } be the basis for X and let s be the functional on E × E given in part (i) of this proof. Define S on X × X so that S|E×E = s, i.e., 3 3 3  1    αi ei , βi ei = (αi βj − αj βi )s(ei , ej ). S 2 i=1 i=1 i,j=1 i=j

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2–INNER PRODUCT SPACES

115

By Theorem 5.3.2, we have T

3 

αi ei ,

3 

i=1

=

1 2

+

βi ei ,

i=1 3 

3 

γi ei



i=1

(αi γj − αj γi )(βi γj − βj γi )T (ei , ei , ej )

i,j=1 i=j 3 

(αi γk − αk γi )(βj γk − βk γj )T (ei , ej , ek )

i,j,k=1 i=j=k=i 3 1  = (αi γj − αj γi )(βi γj − βj γi )s(ei , ej )2 2 i,j=1 i=j

+

3 

(αi γk − αk γi )(βj γk − βk γj )s(ei , ek )s(ej , ek )

i,j,k=1 i=j=k=i

=S

3 

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i=1

αi ei ,

3 

3 3     γi ei S βi ei , γi ei

i=1

i=1

i=1

and thus this theorem is proven. In the following, let (X,
·, ·
) be a linear 2–normed space and (X
,
·
) a normed linear space. We will call a mapping T : X × X × X → X
bounded if there is a K ≥ 0 for which
T (a, b, c)
≤ K
a, c

b, c
for all a, b, c ∈ X. With each bounded mapping T , there is associated a real number (5.14)
T
= inf{K ≥ 0 :
T (a, b, c)
≤ K
a, c

b, c
for all a, b, c ∈ X}. For a bounded mapping T , if a and c or b and c are linearly dependent, then T (a, b, c) = 0. Thus, each bounded mapping T : X × X × X → X
which is bilinear in its first two arguments is in ζ.

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116

R. W. FREESE AND Y. J. CHO

Theorem 5.3.7. If dim X is finite, then every T ∈ ζ with Properties (2) and (3) is bounded. Proof. If T were not bounded, then there would exist sequences of points ai , bi , ci ∈ X such that
T (ai , bi , ci )
> i
ai , ci

bi , ci
for i = 1, 2, · · · . Let
·
e be the Euclidean norm on X with respect to an arbitrary basis of X. Since for arbitrary a
, b
∈ X and β ∈ R,
a
, b

=
a
+ βb
, b

, we can arrange that ai and bi are orthogonal to ci with respect to
·
e for i = 1, 2, · · · . Further, we may assume that
ai
e =
bi
e =
ci
e = 1 and that the sequences {ai }, {bi }, and {ci } converge to a, b, and c, respectively, relative to
·
e . Then, a and b are orthogonal to c relative to
·
e . This implies that
a, c

b, c
> 0. By S. G¨ ahler [96], it follows that the sequences {
ai , ci
} and {
bi , ci
} converge to
a, c
and
b, c
, respectively. Therefore, by an above inequality, the sequence {
T (ai , bi , ci )
} tends to ∞. However, since on the other other hand it is easily seen that T (ai , bi , ci ) converges to T (a, b, c), which is a contradiction. We will denote the set of all bounded mappings T ∈ ζ by ζ ∗ . Theorem 5.3.8. ζ ∗ is a linear space and the function
·
defined by (5.14) is a norm on ζ ∗ . Also,

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T
= sup{
T (a, b, c)
:
a, c
≤ 1 and
b, c
≤ 1} and for any a, b, c ∈ X,
T (a, b, c)
≤
T

a, c

b, c
. If (X
,
·
) is a Banach space, then so is (ζ ∗ ,
·
). The proof of Theorem 5.3.8 is analogous to the proof of the corresponding theorem about bounded linear mappings from a normed linear space into a normed linear space. Corollary 5.3.9. If (X,
·, ·
) is a 2–inner–product space (i.e., there is a 2–inner–product (·, ·|·) on X with
a, b
= (a, a|b)1/2 for all a, b ∈ X), and if T ∈ ζ ∗ has Property (3), then
T
= sup{
T (a, a, c)
:
a, c
≤ 1}. Proof. Let P = sup{
T (a, a, c)
:
a, c
≤ 1}. Since
T (a, a, c)
≤
T
whenever
a, c
≤ 1, it follows that P ≤
T
. If
a, c
≤ 1 and
b, c
≤ 1,

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2–INNER PRODUCT SPACES

117

then by (5.5) and Theorem 5.1.9, 1
T (a + b, a + b, c) − T (a − b, a − b, c)
4 1 ≤ [
T (a + b, a + b, c)
+
T (a − b, a − b, c)
] 4 P ≤ [
a + b, c
2 +
a − b, c
2 ] 4 P = [
a, c
2 +
b, c
2 ] ≤ P. 2


T (a, b, c)
=

Thus,
T
≤ P and consequently P =
T
. In conclusion, we will compute
T
for three interesting cases where T ∈ ζ : 1. Let (X,
·, ·
) be a 2-inner product space and (·, ·|·) the 2–inner product which generates
·, ·
. Let (X
,
·
) be R with the usual norm. Then, the mapping T (a, b, c) = (a, b|c) belongs to ζ ∗ and by (5.6) and (5.13) which is true in the case dim X = 2,
T
= 1. 2. Suppose that there is a norm
·
on BX for which
a, b
=
b(a × b)
for all a, b ∈ X. Let S be a bilinear mapping from BX × BX into X
and define T by (5.10). By the remarks preceding Theorem 5.3.4, T ∈ ζ. Let S be bounded in the sense that there is a K ≥ 0 such that Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

∗


S(b1 , b2 )
≤ K
b1

b2
for all b1 , b2 ∈ BX and let
S
= inf{K ≥ 0 :
S(b1 , b2 )
≤ K
b1

b2
for all b1 , b2 ∈ BX }. ahler, we get Then T ∈ ζ ∗ and
T
≤
S
. In the case dim X ≤ 3, by S. G¨
T
=
S
. 3. Let (X
,
·
) be R with the usual norm. Let S be a bilinear functional on X × X for which there exists a K ≥ 0 such that
S(a, b)
≤ K
a, b
for all a, b ∈ X. Since S is alternating, (5.12) defines a T ∈ ζ ∗ . If we define
S
= inf{K ≥ 0 :
S(a, b)
≤ K
a, b
for all a, b ∈ X}, then we have
T
=
S
2 .

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118

R. W. FREESE AND Y. J. CHO

5.4. 2-Inner Product Spaces and Gˆ ateaux Partial Derivatives In [60], C.R. Diminnie and A. White characterized 2–inner product spaces by means of partial derivatives of 2–functionals. If (X, (·, ·|·) is a 2–inner product space with 2–norm defined by
x, y
= (x, x|y)1/2, then (a, b|c) = lim

t→0+


a + tb, c
2 −
a, c
2 . 2t

R.A. Tapia [230] discussed a characterization of inner product spaces which involves the Gˆateaux derivative of a certain functional. Several of the results of that paper are useful in studying 2–inner product spaces as well. Let (X,
·, ·
) be a linear 2–normed space of dimension > 1. If F (x, y) is a real 2–functional on X, then the right partial derivative of F with respect to x (resp., y) at (x, y) in the direction of h ∈ X, F1+ (x, y)(h) (resp., F2+ (x, y)(h)), is defined by F1+ (x, y)(h) = lim

t→0+

1 {F (x + th, y) − F (x, y)} t

(resp., F2+ (x, y)(h) = lim Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

t→0+

1 {F (x, y + th) − F (x, y)}). t

Similar definitions are used for F1− and F2− . The partial derivative of F with respect to x in the direction of h ∈ X, F1 (x, y)(h), is defined by F1+ (x, y)(h) = F1 (x, y)(h) = F1− (x, y)(h) whenever the one–sided partials agree. F2 (x, y)(h) is defined similarly. The following are easily proved from the above definitions. Theorem 5.4.1. Let x, y, h ∈ X and F be a real 2–functional on X. (1) If F is linear in its first variable, then F1 (x, y)(h) = F (h, y). (2) If F is linear in its second variable, then F2 (x, y)(h) = F (x, h). (3) If F is bilinear, then F1 (x, y)(h) = F (h, y) and F2 (x, y)(h) = F (x, h). Theorem 5.4.2. If F is a symmetric 2–functional and F1 (x, y)(h) exist, then F2 (y, x)(h) exists also and F2 (y, x)(h) = F1 (x, y)(h). The remainder of the discussion will be devoted to the 2–functional

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2–INNER PRODUCT SPACES

F (x, y) =

(5.15)

119

1
x, y
2 . 2

If c = 0, F generates a functional Fc on Xc defined by (5.16)

Fc ((a)c ) = F (a, c) =

1 1
a, c
2 =
(a)c
2c . 2 2

ateaux derivatives of Fc , then If Fc1+ , Fc1− , and Fc1 denote the Gˆ it is easily seen that F1+ (x, c)(h) = Fc1+ ((x)c )((h)c ), F1− (x, c)(h) = Fc1− ((x)c )((h)c ), and F1 (x, c)(h) = Fc1 ((x)c )((h)c ) whenever these derivatives exist. For a, b, c ∈ X, define (5.17)

[a, b|c] = F1+ (a, c)(b).

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Then we have the following: Theorem 5.4.3. [·, ·|·], has the following properties: (1) [a, b|c] is defined for every a, b, c ∈ X. (2)
a, b
= [a, a|b]1/2. (3) |[a, b|c]| ≤
a, c

b, c
. (4) If X is a 2–inner–product space with 2–inner–product (·, ·|·), then [a, b|c] = (a, b|c). Proof. Properties (2) and (4) follow by direct computation. (1) If c = 0, then [a, b|c] = limt→0+ 1t 12
a + tb, 0
2 − 12
a, 0
2 = 0. If c = 0, then Fc1+ ((a)c )((b)c ) exists for every a, b ∈ X by R.A. Tapia [230]. Therefore, [a, b|c] = F1+ (a, c)(b) exists, too. Hence, [a, b|c] exists for every a, b, c ∈ X. (3) If c = 0, the result is obvious since [a, b|0] = 0. If c = 0, then by R. A. Tapia [230], we have |[a, b|c]| = |F1+ (a, c)(b)| = |Fc1+ ((a)c )((b)c )| ≤
(a)c
c
(b)c
c =
a, c

b, c
.

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120

R. W. FREESE AND Y. J. CHO

The last theorem is a direct result of R.A. Tapia [230] and Theorem 5.1.12. Theorem 5.4.4. The following are equivalent. (1) (X,
·, ·
) is a 2–inner product space. (2) [a, b|c] is linear in a. (3) [a, b|c] is symmetric in a and b.

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Note that, by Theorem 5.4.2, [·, ·|·] could also have been defined by [a, b|c] = F2+ (c, a)(b).

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CHAPTER 6. STRICT CONVEXITY In [56], C.R. Diminnie, S. G¨ ahler and A. White introduced the concept of strict convexity in normed linear spaces to the setting of linear 2–normed spaces and gave some characterizations of strict convexity in linear 2–normed spaces. Since then, they and many other authors also obtained many characterizations of strictly convex linear 2–normed spaces in [29], [59], [68], [69], [78], [121], [123], [126], [156], [157] and [189]. It is the purpose of this chapter to extend the concept of strict convexity in normed linear spaces to linear 2–normed spaces. First, we summarize some characterizations of strict convexity in linear 2–normed spaces by means of duality mappings, 2–semi–inner products, algebraic and 2–norm midpoints, and 2–norms generated by semi-norms on the space of bivectors. Secondly, we introduce several characterizations of strict convexity in topological vector spaces by semi-norms defined by 2–norms in [64] and [66]. 6.1. Elementary Characterizations

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First, we give the definition of strict convexity in normed linear spaces and its geometric properties. Definition 6.1.1. A normed linear space (X,
·
) is said to be strictly convex if
x + y
=
x
+
y
and
x
=
y
= 1 imply that y = x. Geometrically, strict convexity means that no line segment can be contained in the unit sphere, or equivalently, if x and y belong to the unit sphere, their midpoint, 12 (x + y), does not. It is well–known that the spaces lp , Lp (1 < p < +∞) are strictly convex but the spaces co and l are not strictly convex. The following equivalent characterizations of strict convexity in normed linear spaces are well-known: (1)
x + y
=
x
+
y
, x, y = 0, implies that y = αx for some α > 0. (2)
x
=
y
= 1, x = y, implies that
12 (x + y)
< 1. (3) Every element of the unit sphere is an extreme point of the unit ball. (4) Every non–zero continuous linear functional attains a maximum on at most one point of the unit sphere. (5) For any normalized linearly independent set {x1 , x2 } in X, the set of Typeset by AMS-TEX Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

122

R. W. FREESE AND Y. J. CHO

points in V (x1 , x2 ) equidistant from x1 and x2 is a subset of {a1 x1 + a2 x2 : a1 a2 ≥ 0}, where V (x1 , x2 ) denotes the subspace of X generated by x1 and x2 . (6) Let [·, ·] be a consistent semi–inner product on X. Whenever [x, y] =
x

y
, x = 0, then y = αx for some α ≥ 0. (7) Let [·, ·] be a consistent semi–inner product on X. Whenever
y+x
≤
y
and [z, y] = 0, then z = 0. These characterizations are very useful to obtain some properties of Banach spaces. Now, we give the definition of strict convexity in linear 2– normed spaces and extend the above characterizations of strict convexity in normed linear spaces to linear 2–normed spaces.

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Definition 6.1.2. A linear 2–normed space (X,
·, ·
) is said to be strictly convex if
a+b, c
=
a, c
+
b, c
,
a, c
=
b, c
= 1 and c ∈ / V (a, b) imply that a = b. Using the concept of bivectors, as in Chapter II, relations between normed linear spaces and linear 2–normed spaces are demonstrated as follows: If
·
is a norm on BX , then
a, b
=
b(a × b)
defines a 2–norm on X, and there is an example which shows that for every 2–norm
·, ·
on X, there need not exist a norm
·
on BX which satisfies
b(a×b)
=
a, b
for all a, b ∈ X. If all bivectors over X are simple, i.e., if X has dimension less than or equal to 3, then for every 2–norm
·, ·
on X there is a
·
on BX with
b(a × b)
=
a, b
for all a, b ∈ X. Hence we have a characterization of strict convexity in linear 2–normed space from such properties: Theorem 6.1.1. Let X be a linear space of dimension greater than 1,
·
be a norm on BX and
·, ·
be a 2–norm on X with
b(a × b)
=
a, b
for all a, b ∈ X. If (BX ,
·
) is strictly convex, then (X,
·, ·
) is strictly convex. If the dimension of X is less than or equal to 3 and (X,
·, ·
) is strictly convex, then (BX ,
·
) is strictly convex. / V (a, b), assume Proof. 1. Suppose (BX ,
·
) is strictly convex. For c ∈
a + b, c
=
a, c
+
b, c
and
a, c
=
b, c
= 1. Then, we have
b(a × c) + b(b × c)
=
b(a × c)
+
b(b × c)
and
b(a × c)
=
b(b × c)
= 1.

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Since (BX ,
·
) is strictly convex, it follows that b(a × c) = b(b × c), which implies that a − b ∈ V (c). Therefore, a = b, since c ∈ / V (a, b). 2. Suppose that X has dimension less than or equal to 3 and (X,
·, ·
) is strictly convex. Let
b1 + b2
=
b1
+
b2
and
b1
=
b2
= 1. Then there exist vectors a, b, c ∈ X such that b1 = b(a × c) and b2 = b(b × c). Thus,
a + b, c
=
a, c
+
b, c
and
a, c
=
b, c
= 1. If c ∈ V (a, b), these equations imply that c = α(a − b) for some real α and hence b1 = b2 . If c∈ / V (a, b), then the strict convexity of (X,
·, ·
) implies that a = b, i.e., b1 = b2 .

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If X is a 2–dimensional linear space, then BX is a 1–dimensional linear space and every 1–dimensional normed space is trivially strictly convex. Therefore, by Theorem 6.1.1, every 2–dimensional linear 2–normed space is strictly convex. Also, it follows from Theorem 6.1.1 that there are linear 2–normed spaces which are not strictly convex. Theorem 6.1.2. The following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) For every non–zero c ∈ X, (Xc ,
·
c ) is strictly convex. (3)
a + b, c
=
a, c
+
b, c
and c ∈ / V (a, b) imply that b = αa for some α > 0. (4)
a − d, c
= α
a − b, c
,
b − d, c
= (1 − α)
a − b, c
, α ∈ (0, 1), and c∈ / V (a − d, b − d) imply that d = (1 − α)a + αb. Proof. (1) ⇒ (2) : Let (X,
·, ·
) be strictly convex and let c be a fixed non–zero element of X. If
(a)c + (b)c
c =
(a)c
c +
(b)c
c and
(a)c
c =
(b)c
c = 1, then
a+b, c
=
a, c
+
b, c
and
a, c
=
b, c
= 1. For the case c ∈ / V (a, b), the strict convexity of (X,
·, ·
) implies that a = b and thus, (a)c = (b)c . When c ∈ V (a, b), i.e., c = αa + βb for some real numbers α and β, then (0)c = (c)c = α(a)c + β(b)c or α(a)c = −β(b)c . From
(a)c
c =
(b)c
c = 1, it follows that α = ±β. If α = β, then c = α(a + b), which contradicts
a + b, c
=
(a)c
c +
(b)c
c = 2. Therefore, α = −β and hence (a)c = (b)c . Therefore, (Xc ,
·
c ) is strictly convex. (2) ⇒ (3) : Assume that condition (2) holds. For c ∈ / V (a, b), let
a + b, c
=
a, c
+
b, c
, i.e.,
(a)c + (b)c
c =
(a)c
c +
(b)c
c By the strict convexity of (Xc ,
·
c ), it follows that (b)c = α(a)c , for some α > 0. Finally, c ∈ / V (a, b) implies that b = αa and condition (3) is satisfied.

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R. W. FREESE AND Y. J. CHO

(3) ⇒ (4) : Assume that condition (3) holds. From
a − d, c
= α
a − b, c
,
b − d, c
= (1 − α)
a − b, c
, c ∈ / V (a − d, b − d) and α ∈ (0, 1), it follows that d − b = β(a − d) for some β > 0. Further, we have (1 − α)
a − b, c
=
b − d, c
= β
a − d, c
= αβ
a − b, c
, which implies that d = (1 − α)a + αb. Therefore, condition (4) is established. (4) ⇒ (1) : Finally, assume that condition (4) holds and let
a + b, c
=
a, c
+
b, c
,
a, c
=
b, c
= 1, where c ∈ / V (a, b). By condition (4), 1 0 = 2 (a − b), i.e., a = b. Therefore, (X,
·, ·
) is strictly convex. Corollary 6.1.3. Every linear 2–normed space of dimension 2 and every 2–inner product space are strictly convex. Proof. See Theorem 6.1.1.

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The following theorems are due to C.S. Lin [156], [157] and K. Iseki [126], respectively. Theorem 6.1.4. The following seven statements are equivalent: (1) (X,
·, ·
) is strictly convex. / V (x, y), then x = y. (2) If 12
x + y, z
=
x, z
=
y, z
and z ∈ (3) If
x + αy, z
= 2
x, z
for some α > 0 and z ∈ / V (x, y), then x = αy, and α = 1 if
x, z
=
y, z
. (4) If
x + y, z
=
x, z
+
y, z
and z ∈ / V (x, y), then x = βy for some β > 0 (cf. Theorem 6.1.2). (5) If
x − w, z
=
x − y, z
+
y − w, z
and z ∈ / V (x − y, y − w), then y = (1 − γ)x + γw for some γ, 0 < γ < 1. (6) If
x + y, z
=
x − y, z
=
x, z
and z ∈ / V (x, y), then y = 0. 1 (7) If 2
x + y, z
=
x, z
=
y, z
= 0 and x = y, then z = δ(x − y) for some δ = 0. Proof. Consider the following three statements: (3
) If
x + αy, z
= 2
x, z
and z ∈ / V (x, y), where α =
x, z
/
y, z
, then x = αy. / V (x, y), then
y, z
x =
x, z
y. (4
) If
x + y, z
=
x, z
+
y, z
and z ∈ (7
) If 12
x + y, z
=
x, z
=
y, z
= 0 and x = y, then
x, y
= 0, and z = ±
x, z
(x − y)/
x, y
. We shall proceed with the proof as follows: (1) ⇒ (2) ⇒ (3
) ⇒ (4
) ⇒ (1), (3
) ⇒ (3) ⇒ (2), (4
) ⇒ (4) ⇒ (2) ⇒ (7
) ⇒ (7) ⇒ (2), (4) ⇔ (5), and (2) ⇔ (6).

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125

(1) implies (2) : If 12
x + y, z
=
x, z
=
y, z
= γ, γ = 0, since z∈ / V (x, y), then we have



z
1



z

z

x + y,
=
x,
=
y,
= 1. 2 γ γ γ Therefore, it follows from (1) that x = y. (2) implies (3
) : Since
αy, z
=
x, z
= 12
x + αy, z
, x = αy by (2). (3
) implies (4
) : Suppose that
x, z
≤
y, z
and α =
x, z
/
y, z
. Then we have

y


x + y, z
=
x + αy + (
y, z
−
x, z
) , z

y, z
≤
x + αy, z
+
y, z
−
x, z
≤ 2
x, z
+
y, z
−
x, z


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=
x, z
+
y, z
. If the identity in (4
) is satisfied, then
x + αy, z
= 2
x, z
. Thus, the desired result follows by (3
). In case
x, z
≥ 2
y, z
, then we may rewrite the statement (3
) as follows: If
(x/α)+y, z
= 2
y, z
and z ∈ / V (x, y), where α is as in (3
), then x = αy.
This statement implies (4 ) analogously. Note that the statement is obtained merely by interchanging x and y in (3
). (4
) ⇒ (1) : If 12
x + y, z
=
x, z
=
y, z
= 1, then
x + y, z
=
x, z
+
y, z
and hence
y, z
x =
x, z
y by (4
). Therefore, we have x = y. The implications (3
) ⇒ (3) ⇒ (2), (4
) ⇒ (4) and (7
) ⇒ (7) are obvious. / V (x, y). Then x = βy (4) ⇒(2) : Let 12
x + y, z
=
x, z
=
y, z
, z ∈ for some β > 0 by (4). By substituting this to the relation
x, z
=
y, z
, we have β = 1. (2) ⇒ (7
) : If x = y, then z = βx + γy for some real numbers β and γ by (2). Thus,
x, z
=
x, βx + γy
= |γ|
x, y
and
y, z
=
y, βx + γy
= |β|
x, y
. Hence,
x, y
= 0 and |β| = |γ| =
x, z
/
x, y
. It follows by suitable combinations that z = ±
x, z
(x − y)/
x, y
.

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126

R. W. FREESE AND Y. J. CHO

(7) ⇒ (2) : If z ∈ / V (x, y), then z = δ(x − y) and so x = y by (7). (4) ⇒ (5) : If the condition in (5) hold, then x − y = β(y − w) for some β > 0 by (4). Hence, y = (1 − γ)x + γw, where γ = β/(β + 1) and 0 < γ < 1. (5) ⇒ (4) : If the conclusion of (5) may be rewritten as y − w = (1 − γ)(x − w). Now, in (5) replacing x − y by x and y − w by y, respectively, and hence, x − w by x + y, we have y = (1 − γ)(x + y), i.e., x = βy, where β = γ/(1 − γ) > 0. (2)⇔(6) : In (2) replacing x by x + y and y by x − y, respectively, we have (6), and vice versa. Theorem 6.1.5. The following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2)
x, z
=
y, z
= 1, x = y and z ∈ / V (x, y) imply
12 (x + y), z
< 1. Proof. (1) ⇒ (2) : By the condition (N4 ), we have
x + y, z
≤
x, z
+
y, z
= 2. If
x, z
=
y, z
=
12 (x + y), z
= 1, then

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x, z
+
y, z
=
x + y, z
. From z ∈ / V (x, y) and (1), we have y = αx for some α > 0. 1 =
x, z
=
αx, z
= α
x, z
implies that α = 1. Therefore, (1) ⇒ (2) follows. (2) ⇒ (1) : Let
x + y, z
=
x, z
+
y, z
,
x, z
≤
y, z
and x, y = 0. If z ∈ / V (x, y), we have






x

y
y y y x




+ , z
≥
+ , z
−
− , z

x, z

y, z

x, z

x, z

x, z

y, z
 1 1 
x, z
+
y, z
−
y, z
− =
x, z

x, z

y, z
= 2.

Hence by (2), we have

y x =
x, z

y, z


Therefore, y = αx for some α > 0, which shows (1) ⇒ (2). 6.2. Strict Convexity by Duality Mappings

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STRICT CONVEXITY

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In [28], [29], [156] and [157], several characterizations of strict convexity in linear 2–normed spaces in terms of bounded linear 2–functionals and duality mapping types are introduced. Let Xz∗ be the space of all bounded linear 2–functionals on X × V (z) for every non–zero z in X. Define the duality mapping types (A
) and (B
), I ∗ and Jφ : X × V (z) → 2Xz as follows, respectively: (A
)

I(x, z) = {F ∈ Xz∗ : F (x, z) =
F

x, z
}

and (B
)

Jφ (x, z) = {F ∈ Xz∗ : F (x, z) =
F

x, z
,
F
= φ(
x, z
)}

for every x, z in X, where φ is a semi–positive function, that is, a mapping from R+ into R+ such that φ(λ) = 0 if and only if λ = 0.

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Note that, by Theorem 3.2.8, if x and z are linearly independent, then I(x, z) = Φ. It is easy to see that x and z are linearly dependent if and only if I(x, z) = Xz∗ . Notice that  if φ is a semi–positive function, then Jφ (x, z) ⊂ I(x, z) and I(x, z) = α>0 αJφ (x, z) for every x in X. Before some characterizations of strict convexity in linear 2–normed spaces in terms of duality mappings, we give the following definitions and theorems: We will say that a linear 2–normed space (X,
·, ·
) has Property (R) if for every x, y, z ∈ X with x = y,
x, z
=
y, z
= 1 and z ∈ / V (x, y), we have
αx + (1 − α)y, z
< 1 for some real number α. The following is a further extension of Theorem 6.1.5: Theorem 6.2.1. The following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) (X,
·, ·
) has Property (R). Proof. Clearly if (X,
·, ·
) is strictly convex, then by Theorem 6.1.5, Property (R) holds with α = 12 , so it is enough to prove the converse. For the converse, it is sufficient to prove that if (X,
·, ·
) is not strictly convex, then (X,
·, ·
) does not have Property (R). Suppose that x, y, z in X with / V (x, y). By Theorem x = y,
x, z
=
y, z
=
12 (x + y), z
= 1 and z ∈ ∗ 3.2.10, there exists an F ∈ Xz such that x + y 
x + y



F
= 1, F ,z =
, z
= 1. 2 2

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R. W. FREESE AND Y. J. CHO

So we have 1 1 F (x, z) + F (y, z) = 1, F (x, z) ≤ |F (x, z)| ≤
F

x, z
= 1 2 2 and F (y, z) ≤ |F (y, z)| ≤
F

y, z
= 1. From these properties, it follows that F (x, z) = F (y, z) = 1. Therefore, for any real number α, we have
αx + (1 − α)y, z
≥ |F (αx + (1 − α)y, z)| = |αF (x, z) + (1 − α)F (y, z)| = 1.

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and so Property (R) fails. Theorem 6.2.2. The following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) If z = 0, F ∈ Xz∗ ,
x, z
=
y, z
= 1 and F (x, z) = F (y, z) =
F
, then either x = y or
x, y
= 0 and z = ±(x − y)/
x, y
. Proof. (1) ⇒ (2) : Assume that (X,
·, ·
) is strictly convex. Let z = 0 and F ∈ Xz∗ . If F (x, z) = F (y, z) =
F
and
x, z
=
y, z
= 1, then we have 1 F (x + y, z) ≤
x + y, z
≤
x, z
+
y, z
= 2. 2=
F
Therefore,
x + y, z
= 2. If x = y, then z ∈ V (x, y) since otherwise the strict convexity of X would yield x = y. Hence, there are real numbers α and β for which z = αx + βy. Then 1 =
x, z
=
x, αx + βy
= |β|
x, y
. Similarly, |α|
x, y
= 1. Therefore,
x, y
= 0 and |α| = |β| = 1/
x, y
. From
x + y, z
= 2, it follows that z = ±(x − y)/
x, y
. (2) ⇒ (1) : Assume that condition (2) holds and let
x, z
=
y, z
= 1, x = y, and z ∈ / V (x, y). Then, we have
x + y, z
≤
x, z
+
y, z
= 2. If
x + y, z
= 2, then by Theorem 3.2.10, there is an F ∈ Xz∗ such that
F
= 1,

F

x + y 2


x + y


,z =
, z
= 1. 2

Note that F (x, z) ≤ |F (x, z)| ≤
F

x, z
= 1. If F (x, z) = 1, then since x = 12 (x + y), condition (2) with x = x and y = 12 (x + y) implies that

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STRICT CONVEXITY

129

z ∈ V (x, y), which is impossible. Thus, F (x, z) < 1. A similar argument shows that F (y, z) < 1 also. Therefore, we have 1=

1 1 1 F (x + y, z) = F (x, z) + F (y, z) < 1. 2 2 2

Hence,
x + y, z
< 2 and so, by Theorem 6.1.5, (X,
·, ·
) is strictly convex. From Theorem 6.2.2, we have the following: Theorem 6.2.3. The following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) I(x, z)∩I(y, z) = Φ and z ∈ V (x, y) for x, y, z in X imply that y = αx for some α > 0. Proof. (1) ⇒ (2) : Assume that condition (1) holds. Suppose I(x, z) ∩ I(y, z) = Φ and z ∈ V (x, y). Notice that z ∈ V (x) and z ∈ V (y). Let F ∈ I(x, z) ∩ I(y, z) and F = 0. Then F (x, z) =
F

x, z
and F (y, z) =
F

y, z
, from which

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F



  y  x ,z = F , z =
F
.
x, z

y, z


Put u = x/
x, z
and v = y/
y, z
. Then we have
u, z
=
v, z
= 1 and F (u, z) = F (v, z) =
F
. By (1) and Theorem 6.2.2, we have u = v or
u, v
= 0,

z = ±(u − v)/
u, v
.

If u = v, then y = (
y, z
/
x, z
)x. Hence y = αx for α =
y, z
/
x, z
> 0. If
u, v
= 0 and z = ±(u − v)/
u, v
, then z ∈ V (x, y), which contradicts z ∈ V (x, y). (2) ⇒ (1) : Assume that condition (2) holds. Let z = 0, F ∈ Xz∗ , F = 0,
x, z
=
y, z
= 1 and F (x, z) = F (y, z) =
F
. Then we have F (x, z) =
F

x, z
,

F (y, z) =
F

y, z


and so F ∈ I(x, z) ∩ I(y, z) = Φ. If z ∈ V (x, y), then, by (2), y = αx for some α > 0. By
x, z
=
y, z
, α = 1 and then y = x. If z ∈ V (x, y) then z = βx + γy for some β and γ. From

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130

R. W. FREESE AND Y. J. CHO

1 =
x, z
=
x, βx + γy
= |γ|
x, y
and 1 =
y, z
=
y, βx + γy
= |β|
y, z
, we have
x, y
= 0,

|β| = |γ| =

1 .
x, y


Since
F

x + y, z
≥ F (x + y, z) = F (x, z) + F (y, z) = 2
F
, it follows that
x + y, z
= 2. Hence z = ±(x − y)/
x, y
. Therefore, by Theorem 6.2.2, (X,
·, ·
) is strictly convex.

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Theorem 6.2.4. Let φ1 and φ2 be onto semi–positive functions. Then the following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) Jφ1 (x, z) ∩ Jφ2 (y, z) = Φ and z ∈ V (x, y) for x, y, z in X imply that y = αx for some α > 0. Proof. (1) ⇒ (2) : Assume that condition (1) holds. Suppose Jφ1 (x, z) ∩ Jφ2 (y, z) = Φ and z ∈ V (x, y). Since Jφ1 (x, z) ⊂ I(x, z) and Jφ2 (y, z) ⊂ I(y, z), we have I(x, z) ∩ I(y, z) = Φ. Therefore, by Theorem 6.2.3, y = αx for some α > 0. (2) ⇒ (1) : Assume that the condition (2) holds. Let F ∈ Xz∗ , F = 0, F (x, z) = F (y, z) =
F
and
x, z
=
y, z
= 1. Since φ1 and φ2 are onto semi–positive functions, for
F
> 0, there exist numbers a > 0 and b > 0 such that φ1 (a) =
F
and φ2 (b) =
F
. Then we have F (ax, z) =
F

ax, z
, F (by, z) =
F

by, z
, and φ1 (
ax, z
) =
F
, φ2 (
by, z
) =
F
. Hence we have F ∈ Jφ1 (ax, z) ∩ Jφ2 (by, z), that is, Jφ1 (ax, z) ∩ Jφ2 (by, z) = Φ. If z ∈ V (ax, by), then, by (2), y = αx for some α > 0. If z ∈ V (ax, by), then z = βx + γy for some β and γ. By the same method in the proof of Theorem 6.2.3, we have
x, y
= 0,

z = ±(x − y)/
x, y
.

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131

Therefore, by Theorem 6.2.2, (X,
·, ·
) is strictly convex. Recall that a pseudo–gauge function is a strictly increasing semi–positive function. Corollary 6.2.5. Let φ be an onto pseudo–gauge function. Then the following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) Jφ (x, z) ∩ Jφ (y, z) = Φ and z ∈ V (x, y) for x, y, z in X imply that y = x. If φ is the function φ(a) = a, we denote Jφ by J. In this case φ is an onto psedo-gauge function. Therefore we have the following:

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Corollary 6.2.6. The following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) J(x, z) ∩J(y, z) = Φ and z ∈ / V (x, y) for x, y, z in X imply that y = x. Theorem 6.2.7. Consider the following statements: (a) (X,
·, ·
) is strictly convex. (b) I(x, z) ⊂ I(y, z) and z ∈ V (x, y) imply that y = αx for some α > 0. (c) I(x, z) = I(y, z) implies that either y = αx for some α > 0 or z = βx + γy for some β and γ. Then we have the following statements: (1) (a) implies (b). (2) (b) implies (c). (3) If I(x, z) ∩ I(y, z) ⊂ I(x + y, z), then (c) implies (a). Proof. By using Theorem 6.2.3, it is easy to show that (1) and (2) are satisfied. For (3), assume that (c) and I(x, z) ∩ I(y, z) ⊂ I(x + y, z) hold. Let
x + y, z
=
x, z
+
y, z
and z ∈ V (x, y). Let F ∈ I(y, z), F = 0. Then, by assumption, we have F ∈ I(x + y, z), F (y, z) =
F

y, z
, F (x + y, z) =
F

x + y, z
, and so we have
F

x, z
+
F

y, z
=
F
(
x, z
+
y, z
) =
F

x + y, z
= F (x, z) + F (y, z) = F (x, z) +
F

y, z
.

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R. W. FREESE AND Y. J. CHO

Hence we have F (x, z) =
F

x, z
, that is, F ∈ I(x, z). This means I(y, z) ⊂ I(x, z). Similarly, I(x, z) ⊂ I(y, z). Therefore, by (c) y = αx for some α > 0 or z = βx + γy for some β and γ. If z = βx + γy for some β and γ, then this contradicts z ∈ V (x, y). Therefore, by Theorem 6.1.2, (X,
·, ·
) is strictly convex. ∗

Definition 6.2.1. A duality mapping type (A
) I : X × V (z) → 2Xz is said to be strictly monotone if for every (x, z), (y, z) in X × V (z), we have (F − G)(x − y, z) > 0, where x = y, z ∈ V (x, y), F ∈ I(x, z) and G ∈ I(y, z). For the proof of Theorem 6.2.9, we need the following Lemma 6.2.8: Lemma 6.2.8. If φ is a pseudo–gauge function, then for every (x, z), (y, z) in X × V (z), we have (F − G)(x − y, z) = (
F
−
G
)(
x, z
−
y, z
)

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+ (
F

y, z
− F (y, z)) + (
G

x, z
− G(x, z)), where x = y, z ∈ V (x, y), F ∈ Jφ (x, z) and G ∈ Jφ (y, z). Proof. Let F ∈ Jφ (x, z), G ∈ Jφ (y, z) and z ∈ V (x, y). Then for every (x, z), (y, z) in X × V (z), we have (F − G)(x − y, z) = F (x − y, z) − G(x − y, z) = F (x, z) − F (y, z) − G(x, z) + G(y, z) =
F

x, z
− F (y, z) − G(x, z) +
G

y, z
=
F

x, z
−
F

y, z
+
F

y, z
− F (y, z) +
G

x, z
− G(x, z) −
G

x, z
+
G

y, z
= (
F
−
G
)(
x, z
−
y, z
) + (
F

y, z
− F (y, z)) + (
G

x, z
− G(x, z)), which implies that (F − G)(x − y, z) = (
F
−
G
)(
x, z
−
y, z
) + (
F

y, z
− F (y, z)) + (
G

x, z
− G(x, z)).

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133

We note that, in Lemma 6.2.8, if (F − G)(x − y, z) = 0, then we have
x, z
=
y, z
,
F
=
G
and F (y, z) = G(x, z) =
F

y, z
=
G

x, z
. Theorem 6.2.9. Let φ be an onto pseudo–gauge function. Then the following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) A duality mapping type (B
), Jφ , is strictly monotone. Proof. (X,
·, ·
) is not strictly convex if and only if, by Corollary 6.2.5, there exist x, y ∈ X, x = y, and an F ∈ Xz∗ such that F ∈ Jφ (x, z) ∩ Jφ (y, z) if and only if we have x = y, F (x, z) =
F

x, z
, F (y, z) =
F

y, z
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F
= φ(
x, z
) = φ(
y, z
) if and only if, by Lemma 6.2.8, there exist x, y ∈ X, x = y, F ∈ Jφ (x, z) and G ∈ Jφ (y, z) such that (F − G)(x − y, z) = 0 if and only if a duality mapping type (B
), Jφ , is not strictly monotone. 6.3. Strict Convexity by p–Semi–Inner Products In [123], Y. Ho and A. White introduced some characterizations of strict convexity in linear 2–normed spaces in terms of the concepts of p–semi–inner product spaces as related to linear 2–normed spaces. The notion of a 2–semi–inner product was introduced by S. Rizvi and A. Siddiqui [209]. This was generalized to the concept of a p–semi–inner product space by I. Franic [78]. Due to a problem in the proof of their main theorem, A. White and Y. Ho [123] modified one part, (S3 ), of their definition

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R. W. FREESE AND Y. J. CHO

and then stated and proved their theorem with the modified definition. Then they obtain several new results regarding p–semi–inner products. Let [·, ·|·] be a real valued function defined on X × X × X such that (S1 ) (i) [a, a|c] ≥ 0, (ii) [a, a|c] = 0 if and only if a and c are linearly dependent, (S2 ) [αa, b|c] = α[a, b|c], where α is real, (S3 ) [a + a
, b|c] = [a, b|c] + [a
, b|c], (S4 ) [a, b|c] ≤ [a, a|c]1/p[b, b|c](p−1)/p for all p ∈ (1, +∞). [·, ·|·] is called a p–semi–inner product and (X, [·, ·|·]) is called a p–semi–inner product space (abbreviated by p–s.i.p. space). In case p = 2, p–semi–inner products induce 2–semi–inner products. Theorem 6.3.1. A p–s.i.p. space is a linear 2–normed space with the 2–norm
a, b
= [a, a|b]1/p for all p ∈ (1, +∞) provided [a, a|b] = [b, b|a]. Proof. Assume that X is a p–s.i.p. space and define
a, b
= [a, a|b]1/p. Because of (S1 ) and our assumption, we have the validity of (2N1 ) and (2N2 ). If a and b are linearly dependent or α = 0, we have (2N3 ). Thus, assume that a and b are linearly independent and α = 0. First, note [αa, αa|b] = |[αa, αa|b]| = |α||[a, αa|b]|. By (S4 ), we have

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|[a, αa|b]| ≤ [a, a|b]1/p[αa, αa|b](p−1)/p. Therefore, we have [αa, αa|b] ≤ |α|[a, a|b]1/p[αa, αa|b](p−1)/p. Dividing [αa, αa|b](p−1)/p on both sides of the above inequality, we have (6.1)

[αa, αa|b]1/p ≤ |α|[a, a|b]1/p.

On the other hand, we have [a, a|b]1/p =

1 α

αa,

1/p  1  1   αa|b ≤  [αa, αa|b]1/p α α

or (6.2)

|α|[a, a|b]1/p ≤ [αa, αa|b]1/p.

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Thus (2N3 ) follows from (6.1) and (6.2). Since (2N4 ) follows if a + b and c are linearly dependent, we assume that [a + b, a + b|c] = 0. Thus, we have [a + b, a + b|c] = [a, a + b|c] + [b, a + b|c] ≤ [a, a|b]1/p[a + b, a + b|c](p−1)/p + [b, b|c]1/p[a + b, a + b|c](p−1)/p = ([a, a|c]1/p + [b, b|c]1/p)[a + b, a + b|c](p−1)/p . Dividing [a + b, a + b|c](p−1)/p on both sides of the above inequality, we have [a + b, a + b|c]1/p ≤ [a, a|c]1/p + [b, b|c]1/p.

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Therefore, (2N4 ) follows from this. Theorem 6.3.2. Every linear 2–normed space is a p–s.i.p. space. Proof. Let x and y be two linearly independent vectors. Define f (αx, βy) = αβ
x, y
p . Then f is a bounded linear 2–functional defined on V (x) × V (y) with
f
=
x, y
p−1 . Thus, f can be extended to a linear 2–functional F with domain X × V (y) such that F (x, y) =
x, y
p and
F
=
x, y
p−1 . We shall denote Fx,y by F . For x, y, z ∈ X, define  if y and z are linearly independent, Fy,z (x, z) [x, y|z] = 0 if y and z are linearly dependent. All of the properties of a p–s.i.p. are easily shown except (S4 ). Here we have: |Fy,z (x, z)| ≤
Fy,z

x, z
=
y, z
p−1
x, z
, which means that |Fy,z (x, z)| ≤ |Fx,z (x, z)|1/p |Fy,z (y, z)|(p−1)/p or

[x, y|z] ≤ [x, x|z]1/p[y, y|z](p−1)/p.

First we give an example of a p–s.i.p. space, p = 2, that is not a 2–inner product space. Define [·, ·|·] on E 2 by (6.3)

[a, b|c] = (a1 c2 − a2 c1 )(b1 c2 − b2 c1 )(c21 + c22 ).

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R. W. FREESE AND Y. J. CHO

Theorem 6.3.3. If a 2–s.i.p. is homogeneous and additive in the second argument and [a, a|b] = [b, b|a], then X is 2–inner product space with (a, b|c) = [a, b|c]. Proof. By Theorem 6.3.1, (X,
·, ·
) is a linear 2–normed space. By Theorems 5.1.9 and 5.1.10, it is established that
a + b, c
2 +
a − b, c
2 = 2
a, c
2 + 2
b, c
2 characterizes 2–inner product spaces. Using the fact that

[a, b|c] =

[a + b, a + b|c] − [a − b, a − b|c] 4

we are able to establish our theorem.

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It is interesting to note that a 2–s.i.p. space satisfies the definition of a 2–inner product space except for (I2 ) and (I3 ) and that if we define (a, b|c) = [a, b|c] from (6.3), then (a, b|c) satisfies the properties of a 2–inner product except for (I2 ). Using this concept we can establish a characterization theorem similar to a theorem by E. Torrance [233]. Theorem 6.3.4. Let (X, [·, ·|·]) be a p–s.i.p. space such that [a, a|b] = [b, b|a] with the 2–norm as defined in Theorem 6.3.1 and let p ∈ (1, +∞). Then the following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) If
y + z, x
≤
y, x
and [z, y|x] = 0, then z and x are linearly dependent. (3) If
y + x, x
=
y, x
and [z, y|x] = 0, then z and x are linearly dependent. (4) If T is a linear operator on X and if
w + T w, x
≤
w, x
for all w ∈ X and [T y, y|x] = 0, then T y and x are linearly dependent. Proof. (1) ⇒ (2): If
x, y
= 0, then y and x are linearly dependent and
y + z, x
= 0, which, in turn, implies that z and x are linearly dependent.

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Now, assume that
y, x
= 1 and 0 ≤ t ≤ 1. Thus, we have
y, x
p = |[y, y|x]| = |[y + tz, y|x]| ≤ [y + tz, y + tz|x]p [y, y|x](p−1)/p =
y + tz, x

y, x
p−1 =
t(y + z) + (1 − t)y, x

y, x
p−1 ≤ (t
y + z, x
+ (1 − t)
y, x
)
y, x
p−1 ≤ (t
y, x
+ (1 − t)
y, x
)
y, x
p−1 =
y, x
p Thus,
y + tz, x
=
y, x
for all t, 0 ≤ t ≤ 1. If x ∈ / V (y + z, y), since (X,
·, ·
) is strictly convex, if we let t = 1 and then t = 12 , it follows that z = 0 and thus x and z are linearly dependent. If x ∈ / V (y + z, x),
y + tz, x
= 1 for all t, 0 ≤ t ≤ 1, imples x = αz. (2) ⇒ (4): This is clear. (4) ⇒ (3): If condition (3) is not true, then there are y and z, which are linearly independent to x, such that
y + z, x
=
y, x
and [z, y|x] = 0. Hence we define a linear operator T as follows :

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Tw =

1 [w, y|x](y + z) − w.
y, x
p

Now, we have 1 |[w, y|x]|
y + z, x

y, x
p 1 = |[w, y|x]|
y, x

y, x
p 1
w, x

y, x
p ≤
y, x
p =
w, x



w + T w, x
=

and T y = z. Moreover, [T y, y|x] = [z, y|x] = 0. Thus the condition (4) does not hold. (3) ⇒ (1): If the condition (1) were not true, then there exist distinct points a, b and c ∈ / V (a, b) such that
a, c
=
b, c
= 1 and
a + b, c
=
a, c
+
b, c
. Set y = 12 (a + b) and and z = 12 (b − a). Thus we have
1
1


y + z, c
=
b, c
= 1,
y, c
=
(a + b), c
=
a + b, c
= 1. 2 2

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R. W. FREESE AND Y. J. CHO

Therefore,
y + z, c
=
y, c
and z and c are independent. While we have |[z, y|c] + 1| = |[z, y|c] + [y, y|c]| = |[z + y, y|c]| = |[b, y|c]| ≤
b, c

y, c
p−1 = 1 and

|[z, y|c] − 1| = |[z, y|c] − [y, y|c]| = |[−a, y|c]| ≤
a, c

y, c
p−1 = 1

It follows that [z, y|c] = 0. Hence the condition (3) does not hold. Theorem 6.3.5. Let (X,
·, ·
) be a linear 2–normed space and let [·, ·|·] be a 2–s.i.p. such that 2–norm
a, b
= [a, a|b]1/p for all p ∈ (1, +∞). Then the following statements are equivalent: (1) (X,
·, ·
) is strictly convex. (2) For a, b, c ∈ X, if [a, b|c] = [a, a|c]1/p[b, b|c](p−1)/p,
a, c
=
b, c
= 1 and c ∈ / V (a, b), then a = b. The proof of Theorem 6.3.5 is based on the following two lemmas:

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Lemma 6.3.6. Let a, b, c ∈ X with c ∈ / V (a, b). If [a, b|c] = [a, a|c]1/p[b, b|c](p−1)/p, then
a + b, c
=
a, c
+
b, c
. Proof. Assume that [a, b|c] =
a, c

b, c
p−1. Then we have
a + b, c

b, c
p−1 ≥ |[a + b, b|c]| = |[a, b|c] + [b, b|c]| =
a, c

b, c
p−1 +
b, c
p = (
a, c
+
b, c
)
b, c
p−1 ≥
a + b, c

b, c
p−1 . Hence,
a + b, c
=
a, c
+
b, c
. Lemma 6.3.7. Let a, b, c ∈ X with c ∈ / V (a, b). Then, for any real number t = 0, we have
a + b, c
=
a, c
+
b, c
if and only if
a + b
p−1  a+b 

|c =
a, c

, c
a, t t and


a + b
p−1  a+b 

|c =
b, c

, c
b, t t

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139

Proof. Assume first that
a + b, c
=
a, c
+
b, c
. Then we have
a + b
p  a + b a + b  

, c
= , c
t t t a a + b  b a + b    = , c + , c t t t t
a

a + b
p−1
b

a + b
p−1







≤
, c

, c
+
, c

, c
t t t t 
a

b

a + b
p−1





, c
=
, c
+
, c

t t t
a + b
p−1 1

, c
= (
a, c
+
b, c
)
t t
a + b
p

=
, c
. t Hence we have
a + b
p−1  a + b  
a + b
p−1  a + b 

 

b, , c
, , c
. a, c =
a, c

c =
b, c

t t t t Conversely, suppose that

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a + b
p−1  a + b 

, c
, a, c =
a, c

t t


a + b
p−1  a + b 

, c
. b, c =
b, c

t t

Then we have
a + b
p  a + b a + b  

, c
= , c
t t t a a + b  b a + b    = , c + , c t t t t 1  a + b    a + b   a, = c + b, c t t t


a + b
p−1  1
a + b
p−1


a, c

, c
+
b, c

, c
= t t t
a + b
p−1 1

, c
. = (
a, c
+
b, c
)
t t p−1 , we conclude that Dividing by
a+b t , c



a + b
1

, c
= (
a, c
+
b, c
)
t t

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R. W. FREESE AND Y. J. CHO

or
a + b, c
=
a, c
+
b, c
. Proof of Theorem 6.3.5. The implication “(1) ⇒ (2)” follows immediately from Lemma 6.3.6. For the converse, let
a + b, c
=
a, c
+
b, c
,
a, c
=
b, c
= 1 and c ∈ / V (a, b). Then by Lemma 6.3.7. we have
a + b
p−1  a + b  

, c
a, c =
a, c

2 2 and


a + b
p−1  a + b 

, c
. b, c =
b, c

2 2 Therefore, a = (a + b)/2 and b = (a + b)/2 and hence the desired conclusion follows.

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6.4. Strict Convexity by Algebraic and 2–Norm Midpoints In [59], C.R. Diminnie, S. G¨ ahler and A. White introduced the concepts of algenraic and 2–norm midpoints in linear 2–normed spaces and, using these concepts, obtained some characterizations of strict convexity in linear 2-normed spaces. In a linear 2–normed space (X,
·, ·
), a point b

∈ X is called the algebraic n 1 midpoint of the points a1 , a2 , · · · , an ∈ X if b = n i=1 ai . A point c ∈ X is called the 2–norm midpoint of a, b ∈ X if there is a point d ∈ / V (a − c, b − c) such that 1
a − c, d
=
c − b, d
=
a − b, d
. 2 Further, a point d ∈ X is called the 2–norm midpoint of 3 non–collinear points a, b, c ∈ X if
a − d, b − d
=
a − d, c − d
=
b − d, c − d
= or σ(a, b, d) = σ(a, c, d) = σ(b, c, d) =

1
a − c, b − c
, 3

1 σ(a, b, c). 3

Theorem 6.4.1. If a point of X is the algebraic midpoint of 2 points, or 3 non–collinear points, then it is also the 2–norm midpoint of these points.

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Proof. 1. If c = 12 (a + b), then for any d ∈ X,
a − c, d
= 12
a − b, d
and
c − b, d
= 12
a − b, d
. Therefore, c is also the 2–norm midpoint of the points a and b. 2. If d = 13 (a + b + c), then we have 1
a − b, 2b − a − c
3 1 1 =
a − b, b − c
=
a − c, b − c
3 3


a − d, b − d
=
a − b, b − d
=

The remaining conditions follow from the symmetry of a, b and c in the definition of 2–norm midpoint. Under the assumption that (X,
·, ·
) is strictly convex, the converse of Theorem 6.4.1 is true: Theorem 6.4.2. If (X,
·, ·
) is strictly convex, then every 2–norm midpoint of 2 points is also the algebraic midpoints of these points. Proof. Let (X,
·, ·
) be strictly convex and c be the 2–norm midpoint of the points a, b ∈ X. Then, there is a point d ∈ / V (a − c, b − c) such that

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a − c, d
=
c − b, d
=

1
a − b, d
. 2

Therefore, By Theorem 6.1.3, c = 12 (a + b). Theorem 6.4.3. A linear 2–normed space (X,
·, ·
) is strictly convex if and only if for any 2 points of X the algebraic and 2–norm midpoints are identical. Proof. If (X,
·, ·
) is strictly convex, then by Theorems 6.4.1 and 6.4.2, the algebraic and 2–norm midpoints of any 2 points are identical. Conversely, assume that in (X,
·, ·
) the algebraic and 2–norm midpoints of any 2 points / V (a, b), then 0 is coincide. If
a, c
=
b, c
= 12
a + b, c
= 1 and c ∈ the 2–norm midpoint and hence also the algebraic midpoint of a and −b. Therefore, 0 = 12 (a − b) or a = b and so (X,
·, ·
) is strictly convex. 6.5. Strict Convexity in Topological Vector Spaces In [64] and [66], C.R. Diminnie and A. White extended the concept of strict convexity in normed linear spaces to topological vector spaces and

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R. W. FREESE AND Y. J. CHO

gave some strict convexity conditions for semi-norms. In addition, many of the results which hold for strictly convex normed linear spaces and linear 2–normed spaces are proven in a more general setting. In a linear 2–normed space (X,
·, ·
), every non–zero element a ∈ X generates a semi–norm pa defined by pa (x) =
x, a
. Using this, the definition of strict convexity in linear 2–normed spaces may be extended to a more general situation. Again, let E be a vector space of dimension > 1 and let P = {pt : t ∈ A} be a collection of non–zero semi–norms on E. For p ∈ P , let Vp denote the null space of p, i.e., Vp = {x : p(x) = 0}. Definition 6.5.1. The pair (E, P ) is said to be strictly convex if for a, b ∈ E, and p ∈ P , the conditions 12 p(a + b) = p(a) = p(b) = 1 and Vp ∩ V (a, b) = {0} imply that a = b. Example 6.5.1. If P consists of a single norm
·
and if the pair (E,
·
) is a strictly convex normed linear space, then (E, P ) is strictly convex.

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Example 6.5.2. If (E,
·, ·
) is a strictly convex linear 2–normed space and if P = {pa : a ∈ E − {0}}, where pa (x) =
x, a
, then the pair (E, P ) is strictly convex. Example 6.5.3. If F is a collection of non–zero linear functionals on E, then for each f ∈ F , the function p(x) = |f (x)| is a semi–norm on E. If P is the collection of these semi–norms, then the pair (E, P ) is strictly convex. Proof. Let f ∈ F and a, b ∈ E satisfy the conditions 12 |f (a + b)| = |f (a)| = |f (b)| = 1 and Vf ∩ V (a, b) = {0}, where Vf = {x : |f (x)| = 0}. Since f is linear, it follows that 1 |f (a) + f (b)| = |f (a)| = |f (b)| = 1. 2 Hence, f (a) = f (b), which implies that f (a − b) = 0. Finally, since Vf ∩ V (a, b) = {0}, a = b. Example 6.5.4. If E is an arbitrary vector space of dimension > 1 and P contains a non–strictly convex norm p, then the pair (E, P ) is not strictly convex. Proof. Since p is not strictly convex, there are element a, b ∈ E such that 12 p(a + b) = p(a) = p(b) = 1, but a = b. Further, since p is a norm, Vp ∩ V (a, b) = {0}. Hence, the pair (E, P ) is not strictly convex.

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Our first result shows that these characterizations hold also for the pair (E, P ). For p ∈ P , let Ep denote the quotient space E/Vp and for a ∈ E, let (a)p represent the equivalence class of a with respect to Vp . Then Ep is a vector space under addition (a)p + (b)p = (a + b)p and scalar multiplication (αa)p = α(a)p . In addition, if we define
(a)p
p = p(a) for any a in the equivalence class (a)p , then (Ep ,
·
p ) is a normed linear space. Theorem 6.5.1. For the pair (E, P ), the following statements are equivalent: (1) the pair (E, P ) is strictly convex. (2) For every p ∈ P, (Ep ,
·
p ) is a strictly convex normed linear space. (3) If p(a + b) = p(a) + p(b) and Vp ∩ V (a, b) = {0}, then a = αb for some α > 0. (4) If α ∈ (0, 1), p(a − c) = αp(a − b), p(b − c) = (1 − α)p(a − b), and Vp ∩ V (a − c, b − c) = {0}, then c = (1 − α)a + αb. Proof. (1) ⇒ (2) : Let the pair (E, P ) be strictly convex. If a, b ∈ E and p ∈ P satisfy the condition 12
(a)p + (b)p
p =
(a)p
p =
(b)p
p = 1, then 1 p(a + b) = p(a) = p(b) = 1. 2 Since (E, P ) is strictly convex, either a = b or Vp ∩ V (a, b) = {0}. If a = b, then clearly (a)p = (b)p . If Vp ∩ V (a, b) = {0}, then there is a non–zero c ∈ Vp and there are real numbers α and β such that c = αa + βb. Since this implies that α(a)p = −β(b)p and since c = 0, it follows that α and β are non–zero. Further, 12
(a)p + (b)p
p =
(a)p
p +
(b)p
p = 1 implies that α = −β. Therefore in either case, (a)p = (b)p and the pair (Ep ,
·
p ) is strictly convex. (2) ⇒ (3) : Assume condition (2) and let p(a + b) = p(a) + p(b) and Vp ∩ V (a, b) = {0}. Since (Ep ,
·
p ) is strictly convex, there is an α > 0 such that (a)p = α(b)p . Finally, Vp ∩ V (a, b) = {0} implies that a = αb. (3) ⇒ (4) : Assume condition (3) and let the hypothesis of condition (4) hold. Then, since p(a−c+c−b) = p(a−c)+p(c−b) and Vp ∩V (a−c, b−c) = {0}, it follows by condition (3) that (a−c) = β(c−b) for some β > 0. Further, αp(a − b) = p(a − c) = βp(c − b) = β(1 − α)p(a − b). Since p(a − b) = 0, β = α/(1 − α) and therefore, c = (1 − α)a + αb. (4) ⇒ (1) : Assume that condition (4) holds and let a, b ∈ E and p ∈ P satisfy the conditions 12 p(a + b) = p(a) = p(b) = 1 and Vp ∩ V (a, b) = {0}.

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Then, by condition (4), 0 = 12 a − 12 b and hence a = b. Therefore, the pair (E, P ) is strictly convex.  Definition 6.5.2. (E, P ) is called a Hausdorff pair if p∈P Vp = {0}. For the remainder of this section, we will assume that (E, P ) is a Hausdorff pair. Let Tp denote the coarsest locally convex topology on E in which every p ∈ P is continuous. Tp is said to be generated by the collection P . In particular, Tp is a Hausdorff topology on E. Another set P
of non–zero semi–norms on E will be said to be equivalent to P if P and P
generate the same topology. Based on these considerations, we make the following definition: Definition 6.5.3. A locally convex topological vector space (E, T ) is said to be topologically strictly convex if there is a set P of non–zero semi–norms such that T = Tp and the pair (E, P ) is strictly convex.

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Example 6.5.5. Let (E, T ) be a locally convex topological vector space and let E
be the topological dual of E. If T is the weak topology σ(E, E
), then (E, T ) is topological strictly convex by Example 6.5.3. Theorem 6.5.4. If there is a continuous, one–to–one, linear mapping F from (E, T ) into a strictly convex normed linear space (N,
·
), then (E, T ) is topologically strictly convex. Further, if P = {pt : t ∈ A} is a family of non–zero semi–norms which generates T , then the set P
of semi–norms pt
(x) = pt (x) +
F (x)
is an equivalent family such that the pair (E, P
) is strictly convex. Proof. Let P = {pt : t ∈ A} be a family of semi–norms such that Tp = T . For each t ∈ A, define a new semi–norm pt
on E by pt
(x) = pt (x) +
F (x)
and let P
= {pt
: t ∈ A}. Since pt ≤ pt
, each pt is continuous in (E, Tp ). Also, since F is continuous, each pt
is continuous in T . Therefore, P
is equivalent to P and Tp = T . Next, suppose that 1
pt (a + b) = pt
(a) = pt
(b) = 1 2 and Vpt  ∩ V (a, b) = {0}.

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Then, it follows that pt (a + b) = pt (a) + pt (b),


F (a) + F (b)
=
F (a)
+
F (b)
.

Since (N,
·
) is strictly convex, there is a β > 0 such that F (a) = βF (b). Further, since F is linear and one–to–one, a = βb. Finally, the condition pt
(a) = pt
(b) = 1 implies that β = 1 and hence a = b. Example 6.5.6. Let E = C ∞ [0, 1], pn (x) = sup0≤t≤1 |x(n) t|, and P = {pn : n = 0, 1, · · · }. Since po is a non–strictly convex norm on E, Example 6.5.4 implies that (E, P ) is not strictly convex. Define
·
on E by
x
=

 0

1

|x(t)|2 dt

1/2

.

Then, (E,
·
) is strictly convex since (E,
·
) is an inner product space. The identity mapping from (E, Tp ) to (E,
·
) is clearly linear and one–to–one. Further, it is continuous since

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x
=

 0

1

|x(t)|2 dt

1/2

≤ sup |x(t)| = po (x). 0≤t≤1

Therefore, by Theorem 6.5.4, (E, Tp ) is topologically strictly convex and a suitable equivalent family of semi–norms is the set P
of semi–norms pn
(x) = pn (x) +
x
. Recall that a normed linear space (N,
·
) is said to be strictly normable if there is an equivalent norm
·
∗ on N such that (N,
·
∗ ) is strictly convex. This concept will now be used to characterize topologically strictly convex topological vector spaces. Theorem 6.5.5. If (Ep ,
·
p ) is strictly normable for every p ∈ P , then (E, Tp ) is topologically strictly convex Proof. For each p ∈ P , there is a strictly convex norm
·
∗p on Ep which is equivalent to
·
p . Therefore, there are numbers m, M > 0 such that m
(x)p
p ≤
(x)p
∗p ≤ M
(x)p
p for every x ∈ E. If p
is defined on E by p
(x) =
(x)p
∗p , then p
is a semi–norm on E which satisfies the condition mp(x) ≤ p
(x) ≤ M p(x) for

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R. W. FREESE AND Y. J. CHO

every x ∈ E. Further, note that
·
p =
·
∗p on Ep . If P
is the set of all p
’s, then clearly P
is equivalent to P . Finally, the pair (E, P
) is strictly convex since (Ep ,
·
p ) = (Ep ,
·
∗p ) is strictly convex for every p
∈ P
. Corollary 6.5.6. (E, Tp ) is topologically strictly convex if and only if there is an equivalent set P
of semi–norms such that (Ep ,
·
p ) is strictly normable for every p
∈ P
. For any p ∈ P , it is obvious that the canonical mapping Π(x) = (x)p is a continuous homomorphism from (E, Tp) onto (Ep ,
·
p ). Hence, if (E, Tp ) is separable, then (Ep ,
·
p ) is also separable for every p ∈ P . This leads directly to our next result: Corollary 6.5.7. If (E, Tp ) is separable, then (E, Tp ) is topologically strictly convex. Proof. By our remarks above, (Ep ,
·
p ) is separable for every p ∈ P . Then, by R. Holmes, each (Ep ,
·
p ) is strictly normable. Hence, by Theorem 6.5.5, (E, Tp) is topologically strictly convex.

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Corollary 6.5.8. If E is finite dimensional and T is any locally convex topology on E, then (E, T ) is topologically strictly convex. Proof. Since all locally convex topologies on E are equivalent, we may choose T to be the euclidean topology. Since (E, T ) is separable, it is topologically strictly convex. In the previous theorems, certain strict convexity conditions were introduced for a class of semi–norms on a vector space. Here, we restrict our attention to two strict convexity conditions on a single semi–norm. These conditions are characterized and several examples are presented for the more interesting concept. Let X be a vector space of dimension greater than 1 and let p be a non– zero semi–norm on X. Define B = {x : p(x) ≤ 1}, S = {x : p(x) = 1}, and V = {x : p(x) = 0}. Note that V is a subspace of X and V = {0} if and only if p is a norm. Definition 6.5.4. (1) A non–zero semi–norm p on X satisfies SC1 if S contains no line segments. (2) A non–zero semi–norm p on X satisfies SC2 if the conditions a, b, 12 (a+ b) ∈ S and V (a − b) ∩ V = {0} imply that a = b.

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Lemma 6.5.9. If z ∈ V , then for every x ∈ X, p(x + z) = p(x). Proof. We have p(x) = p(x + z − z) ≤ p(x + z) + p(z) = p(x + z) ≤ p(x) + p(z) = p(x). Therefore, we have p(x + z) = p(x). Theorem 6.5.10. If V = {0}, then p cannot satisfy SC1 . Proof. Let x ∈ S and z ∈ V − {0}. If y = x + z, y = x and Lemma 6.5.9 implies that y ∈ S also. Further, since αz ∈ V for every α, we have p(tx + (1 − t)y) = p(x + (1 − t)z) = 1 for every t ∈ [0, 1]. Therefore, the line segment from y to x is in S and p does not satisfy SC1 . Corollary 6.5.11. p satisfies SC1 if and only if p is a strictly convex norm.

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Corollary 6.5.12. If p satisfies SC1 , then p satisfies SC2 . Since Corollary 6.5.11 shows that SC1 reduces to strict convexity in a normed linear space, we will restrict the remainder of this section to SC2 . Among other things, we will demonstrate that the converse of Corollary 6.5.12 is false. Before proceeding further, we will relate SC2 to previous theorems on strict convexity. Theorem 6.5.13. Let p be a non–zero semi–norm on X and P = {p}. Then, p satifies SC2 if and only if (X, P ) is strictly convex. Proof. The result will follow if we show that, under the condition p(a) = p(b) = 12 p(a + b) = 1, V (a, b) ∩ V = {0} if and only if V (a − b) ∩ V = {0}. Since V (a − b) ⊂ V (a, b), the condition V (a, b) ∩ V = {0} always implies that V (a − b) ∩ V = {0}. To show the converse, assume that p(a) = p(b) = 12 p(a + b) = 1 and V (a − b) ∩ V = {0}. If x ∈ V (a, b) ∩ V , then there are real numbers α, β such that x = αa + βb. Further, since x ∈ V , |α| = p(αa) = p(x − βb) = p(−βb) = |β|.

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The conditions αa + βb ∈ V and p(a + b) = 2 imply that α = −β. Therefore, x ∈ V (a − b) ∩ V and hence x = 0. ˆ denote the equivalence class of Xp generated Let Xp = X/V and let a by a. If we define
·
p on Xp by
ˆ a
p = p(a) for any representative a of a ˆ, it is easily shown that
·
p is a well–defined norm on Xp . For the results that follow, we will also use certain geometric concepts in X. If f is a non–zero linear functional on X, then for every real t, Ht = {x : f (x) = t} is a hyperplane in X. Ht is called a support hyperplane for B if Ht ∩ B = Φ where B ⊆ {x : f (x) ≤ t} or B ⊆ {x : f (x) ≥ t}. Since B is absolutely convex, it can be shown that for any support hyperplane H for B, there is a linear functional f and a t > 0 such that H = {x : f (x) = t} and B ⊆ {x : f (x) ≤ t}. More information on support hyperplanes is given by G. K¨ othe [147], [148]. The following technical lemma will be used in the proof of the main result for SC2 :

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Lemma 6.5.14. If H is a support hyperplane for B, then H ∩ B ⊂ S. Proof. By the comments made above, there is a linear functional f and a t > 0 such that H = {x : f (x) = t} and B ⊆ {x : f (x) ≤ t}. Suppose there is an element x ∈ H ∩ B for which p(x) < 1. If p(x) = 0, then since f (x) = t > 0, it follows that 2x ∈ / {x : f (x) ≤ t} while 2x ∈ V ⊆ B. If 0 < p(x) < 1, then x/p(x) ∈ B but f (x/p(x)) = t/p(x) > t. Since either possibility violates the condition that B ⊆ {x : f (x) ≤ t}, we must have p(x) = 1 for all x ∈ H ∩ B. Theorem 6.5.15. The following statements are equivalent: (1) p satisfies SC2 . (2) (Xp ,
·
p ) is a strictly convex normed linear space. (3) There is a strictly convex normed linear space (Y,
·
) and a linear function F : X → Y such that p(x) =
F (x)
for all x ∈ X. (4) For any support hyperplane H for B and any xo ∈ H ∩ B, H ∩ B = xo + V . Proof. (1) ⇒ (2): By Theorem 6.5.1, it follows that the statement (1) implies the statement (2). (2) ⇒ (3): If we assume the statement (2), let (Y,
·
) be (Xp ,
·
p ) and define F be the canonical mapping F (x) = x ¯. The statement (3) then follows immediately. (3) ⇒ (4): Assuming the statement (3), let H be a support hyperplane

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for B and let xo ∈ H ∩ B. Again, there is a linear functional f and a t > 0 such that H = {x : f (x) = t} and B ⊆ {x : f (x) ≤ t}. If x ∈ H ∩ B, then since H ∩ B is convex, Lemma 6.5.14 implies that x, xo and 12 (x + xo ) ∈ S. Therefore, by the statement (3), we have
F (x)
=
F (xo )
=

1
F (x) + F (xo )
= 1. 2

Using the strict convexity of (Y,
·
), we obtain

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F (x) = F (xo ) or
F (x − xo )
= 0. Then, p(x − xo ) =
F (x − xo )
= 0, which implies that x − xo ∈ V . Hence, H ∩ B ⊆ xo + V . Conversely, if x ∈ xo + V , then Lemma 6.5.9 implies that p(x) = p(xo ) and hence x ∈ B. If f (x − xo ) = 0, then there is a real number β such that f [β(x − xo )] > t. Since β(x − xo ) ∈ V ⊆ B, this would violate the condition that B ⊆ {x : f (x) ≤ t}. Therefore, 0 = f (x − xo ) = f (x) − t and x ∈ H also. It follows that xo + V ⊆ H ∩ B and the statement (4) is proven. (4) ⇒ (1) : Finally, assume that the statement (4) holds and let a, b ∈ X satisfy a, b, 12 (a + b) ∈ S and V (a − b) ∩ V = {0}. By the Hahn–Banach Theorem, there is a linear functional f on X such that f 12 (a + b) = 1 and f (x) ≤ p(x) for x ∈ X. Then H = {x : f (x) = 1} is a support hyperplane for B and further, we have 2 = f (a + b) = f (a) + f (b) ≤ p(a) + p(b) = 2. Therefore, f (a) = p(a) = 1 and f (b) = p(b) = 1, which imply that a, b ∈ H ∩ B. By the statement (4), it follows that a − b ∈ V and hence a = b since V (a − b) ∩ V = {0}. Thus, p satisfies SC2 and the proof is completed. Example 6.5.7. p is called a Hilbertian semi–norm on X if p satisfies the “Parallelogram Law”:   (p(x + y))2 + (p(x − y))2 = 2 (p(x))2 + (p(y))2 . These spaces are studied by T. Precupanu [195], [196]. Further examples are nuclear spaces and the spaces described by C.R. Diminnie, S. G¨ ahler and A. White [55]. If p is a Hilbertian semi–norm on X, then (Xp ,
·
p ) is an

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inner product space. Since inner product spaces are always strictly convex, p satisfies SC2 by Theorem 6.5.15. The spaces studied by C.R. Diminnie, S. G¨ ahler and A. White [55] demonstrate that there are Hilbertian semi–norms which are not norms. Therefore, there are semi–norms which satisfy SC2 but not SC1 . Eaxmple 6.5.8. If the dimension of X is 2 and V is 1–dimensional, then (Xp ,
·
p ) is a 1–dimensional normed linear space which is easily shown to be strictly convex. Again, by Theorem 6.5.15, p satisfies SC2 but not SC1 . Example 6.5.9. If X = C (n) [a, b], f is a linear differential operator of order ≤ n and
·
is a strictly convex norm on C (n) [a, b], then p(x) =
f (x)
satisfies SC2 by Theorem 6.5.15. Note that V is the set of all solutions to the differntial equation f (x) = 0.

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The last result presented here characterizes the elements of “minimal norm” in a convex subset of X. It is an analogue of the uniqueness theorem for minimal norm elements in strictly convex normed linear spaces. Theorem 6.5.16. If p satisfies SC2 , M is a non–empty convex set, xo ∈ X, and yo ∈ M satisfies the condition p(xo − yo ) ≤ p(xo − y) for all y ∈ M , then {y ∈ M : p(xo − y) = p(xo − yo )} = (yo + V ) ∩ M . Proof. If y ∈ (yo + V ) ∩ M , then y ∈ M and yo − y ∈ V . By Lemma 6.5.9, p(xo − y) = p(xo − yo + yo − y) = p(xo − yo ). On the other hand, suppose that y ∈ M satisfies p(xo − y) = p(xo − yo ). Since M is convex, y1 = 12 y + 12 yo ∈ M and hence p(xo − y1 ) ≥ p(xo − yo ). Also, we have 1 p(xo − y + xo − yo ) 2 1 ≤ [p(xo − y) + p(xo − yo )] 2 = p(xo − yo ).

p(xo − y1 ) =

If xo − yo ∈ / V , then by Theorem 6.5.1, either V (y − yo ) ∩ V = {0} or xo − y = α(xo − yo ) for some α > 0. The condition V (y − yo ) ∩ V = {0} implies directly that y − yo ∈ V . If xo − y = α(xo − yo ) for some α > 0, / V and αp(xo − yo ) = p(xo − y) = p(xo − yo ), we must then since xo − yo ∈ have α = 1 and y = yo . Finally, if xo − yo ∈ V , then xo − y ∈ V also and hence y − yo = (xo − yo ) − (xo − y) ∈ V . Therefore, in all cases, if y ∈ M and p(xo − y) = p(xo − yo ), then y ∈ (yo + V ) ∩ M .

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6.6. 2-Norms Generated by Seminorms on the Space of Bivectors In [67], C.R. Diminnie and A. White introduced the 2–norms generated by semi–norms on the space of bivectors and their properties. As in Theorem 5.1.4, if (·|·) is an inner product on BX , then (a, b|c) = (b(a × c)|b(b × c)) defines a 2–inner product (·, ·|·) on X for all a, b, c ∈ X. In case dim X ≤ 3, every 2–inner product on X has a corresponding inner product on BX for which (a, b|c) = (b(a × c)|b(b × c))

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for all a, b, c ∈ X. Let N be a vector space. By TX,N we denote the set of all mappings T : X × X → N such that for every such T there exists a linear mapping ξ : BX → N with ξ(b(a × c)) = T (a, b) for every a, b ∈ X. The mapping ξ is uniquely determined. Theorem 6.6.1. T ∈ TX,N if and only if T is bilinear and T (a, b) = 0 for all linearly dependent points a, b ∈ X. Proof. 1. Let T ∈ TX,N . Since ξ is linear and the space BX has the properties b(a × b) = −b(b × a) and α1 b(a1 × b) + α2 b(a2 × b) = b((α1 a1 + α2 a2 ) × b), T is bilinear. Further, since b(a × b) = 0 when a and b are linearly dependent, it follows that T (a, b) = ξ(b(a × c)) = 0. Hence, the condition is satisfied. 2. Assume that the condition holds. For any a, b ∈ X, 0 = T (a+b, a+b) = T (a, b) + T (b, a)

nDefine ξ : BX → N by

nand thus, T (a, b) = −T (b, a). setting ξ(b) = i=1 T (ai , bi ) for every b = b( i=1 ai × bi ) in BX . By the characterization of bivectors, it follows that ξ(b) is independent of the chosen representative of b. The linearity of ξ is obvious. Therefore, T ∈ TX,N and the proof is complete. Using Theorem 6.6.1, we immediately have the following: Theorem 6.6.2. Let T ∈ TX,N be such that T (a, b) = 0 when a and b are linearly independent. For every norm
·
on N,
a, b
=
T (a, b)
defines a 2–norm on X. For every inner product (·, ·) on N, (a, b|c) = (T (a, c), T (b, c)) defines a 2–inner product on X. If N = BX , then by T (a, b) = b(a × b) we get a T ∈ TX,N for which T (a, b) = 0 when a and b are linearly independent. In this case, the associated ξ : BX → N = BX is the identity mapping. If, in Theorem 6.6.2,

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we choose N = BX and T as given above, we see that Theorem 6.6.1 is a generalization of a part of Theorem 5.1.4. If X
is a linear subspace of X, a bilinear 2–functional F on X
× X
is said to be bounded if there is a K > 0 such that |F (a, b)| ≤ K
a, b
for every a, b ∈ X
. If F is such a functional, define the norm of F ,
F
by
F
= inf{K : |F (a, b)| ≤ K
a, b
for every a, b ∈ X
}. For every a, b ∈ X
, it follows that |F (a, b)| ≤
F

a, b
. Note that in this situation, if a and b are linearly dependent, then F (a, b) = 0. Additional properties of bounded bilinear 2–functionals may be found in Chapter IV. Theorem 6.6.3. Let (X,
·, ·
) be a linear 2–normed space of dimension ≤ 3 and
x, y
> 0. Then there exists a bounded bilinear 2–functional F on X × X such that
F
= 1 and F (x, y) =
x, y
.

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Proof. Since the dimension of X is ≤ 3, there exists a norm on BX such that
a, b
=
b(a × b)
for every a, b ∈ X. Letting b
= b(x × y) and applying the Hahn–Banach Theorem to (BX ,
·
), we obtain a bounded linear functional f on BX such that
f
= 1 and f (b
) =
b

. If we define F (x, y) = f (b(x × y)), then F is a bilinear 2–functional on X × X with the desired properties. Corollary 6.6.4. Every linear 2–normed space of dimension 2 and every 2–inner product space is strictly convex. Theorem 6.6.5. Let (X,
·, ·
) be a linear 2–normed space and (N,
·
) be a strictly convex normed linear space. If there is a T ∈ TX,N such that
a, b
=
T (a, b)
for every a, b ∈ X, then (X,
·, ·
) is strictly convex. / V (a, b). Proof. Assume that
a, c
=
b, c
= 12
a + b, c
= 1 and c ∈ 1 Then, since T is bilinear,
T (a, c)
=
T (b, c)
= 2
T (a, c) + T (b, c)
= 1. By the strict convexity of (N,
·
). T (a, c) = T (b, c). Then, since c ∈ / V (a, b) and
a − b, c
=
T (a, c) − T (b, c)
= 0, it follows that a = b. If
·
is a norm on BX , then
a, b
=
b(a × b)
defines a 2–norm on X. More generally, if p is a semi–norm on BX with the property that p[b(a×b)] = 0 implies that a and b are linearly dependent,
a, b
= p[b(a×b)] also defines a 2–norm on X. The converse of this situation is false in that there is an example which demonstrates the existence of linear 2–normed space (X,
·, ·
) for which no such semi–norm p can exist on BX . This section

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is concerned with characterizing those linear 2–normed spaces (X,
·, ·
) for which
·, ·
is “generated” by a semi–norm on BX . Definition 6.6.1. For a vector space X of dimension greater than 1, let S denote the set of those 2–norms
·, ·
on X for which there is a semi–norm p on BX such that
a, b
= p[b(a × b)] for all a, b ∈ X. In this case,
·, ·
is said to be generated by p. This topic is discussed by H. Busemann and E.G. Straus [22]. Here, we will develop some additional criteria for
·, ·
to be element of S and examine some of the natural candidates for generators of
·, ·
. In addition, we will demonstrate the connection between the bounded bilinear 2–functionals on a linear 2–normed space and the bounded linear functionals on the space of bivectors. Define the semi–norm ρ on BX as follows: (6.4)

ρ(b) = inf

 n

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i=1

n  
ai , bi
: b = b ai × bi . i=1

It is obvious that ρ[b(a × b)] ≤
a, b
for all a, b ∈ X. By H. Busemann and E.G. Straus [22], this is the first result which shows when ρ generates
·, ·
.

n Theorem 6.6.6. ρ generates
·, ·
if and only if
a, b
≤ i=1
ai , bi


n whenever b(a × b) = b( i=1 ai × bi ). We will also consider an alternate formulation for ρ which is mentioned by H. Busemann and E. G. Straus [22], but in a more general setting. In the space BX , let (6.5)

U = {b(a × b) :
a, b
≤ 1}

and (6.6)

AC(U ) = the absolutely convex envelope of U.

Lemma 6.6.7. AC(U ) is absorbent in BX . Proof.

nSince the zero bivector is in U , let us consider a non–zero bivector b = b( i=1 ai × bi ). Further, we may assume that the pairs ai , bi are linearly independent for i = 1, 2, · · · , n. If we let ci = bi /
ai , bi
for i =

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R. W. FREESE AND Y. J. CHO

1, 2, ·

· · , n, then each b(ai × bi ) ∈ U and b = n µ = i=1
ai , bi
, we see that when |λ| ≥ µ,

n

i=1


ai , bi
b(ai × ci ). For

n  1
ai , bi
b(ai × ci ) ∈ λAC(U ), b=λ λ i=1

n since each b(ai × ci ) ∈ U and i=1 |
ai , bi
/λ| = µ/|λ| ≤ 1. Therefore, AC(U ) is absorbent. Since AC(U ) is also absolutely convex, its gauge σ(b) = inf{λ > 0 : b ∈ λAC(U )} is a semi–norm on BX . Theorem 6.6.8. ρ = σ. Proof. Since ρ(b) = 0 = σ(b) when b is the zero bivector, we need only consider non–zero bivectors. 1. Let b ∈ BX and let λ > 0 be chosen so that b ∈ λAC(U ). Then there , an , bn ∈ X and real numbers α1 , α2 , · · · , αn such that each are a1 , b1 , · · ·

n
ai , bi
≤ 1, i=1 |αi | ≤ 1, and

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b=λ

n 

αi b(ai × bi ) = b

i=1

n 

 (λαi ai ) × bi .

i=1

n

n Since i=1
λαi ai , bi
≤ λ i=1 |αi | ≤ λ, it follows that ρ(b) ≤ λ. Further, since this is true whenever λ > 0 and

n b ∈ λAC(U ), ρ(b) ≤ σ(b). 2.

On the other hand, if b = b( i=1 ai × bi ) is a non–zero bivector, then n the proof of Lemma 6.6.7, b ∈ µAC(U ) µ = i=1
ai , bi
> 0. As shown in

n n and hence, σ(b) ≤ µ. Since σ(b) ≤ i=1
ai , bi
whenever b = b( i=1 ai × bi ), σ(b) ≤ ρ(b) and the proof is complete. Definition 6.6.2. A bilinear 2–functional B on X is said to be bounded if there is a K > 0 such that |B(a, b)| ≤ K
a, b
for all a, b ∈ X. By A. White [242], it is shown that (6.8)


B
= inf{K > 0 : |B(a, b)| ≤ K
a, b
for all a, b ∈ X}

is a norm on the space of bounded bilinear functionals on X.

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Theorem 6.6.9. A linear 2–functional B is bounded on X if and only if B satisfies Theorem 6.6.1 and the mapping T from Theorem 6.6.1 is bounded on BX . In this case,
B
=
T
. Proof. 1. Assume that there is a K > 0 such that |B(a, b)| ≤ K
a, b
for all a, b ∈ X. Then B satisfies Theorem 6.6.1 and there is a unique linear mapping

nT : BX → R for which B(a, b) = T [b(a × b)] for all a, b ∈ X. If b = b( i=1 ai × bi ), then we have |T (b)| ≤

n 

|T [b(ai × bi )]| =

i=1

n 

|B(ai , bi )| ≤ K

i=1

n 


ai , bi
,

i=1

which implies that |T (b)| ≤ Kρ(b) and hence T is bounded. 2. If B satisfies Theorem 6.6.1 and the mapping T from Theorem 6.6.1 is bounded on BX , then there is a K > 0 such that |T (b)| ≤ Kρ(b) for all b ∈ BX . Recall that ρ[b(a × b)] ≤
a, b
for all a, b ∈ X. Therefore,   |B(a, b)| = T [b(a × b)] ≤ Kρ[b(a × b)] ≤ K
a, b
and hence B is bounded on X. 3. By parts 1, 2 and the equation (6.8), it follows easily that
B
=
T
.

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This leads us to the main result: Theorem 6.6.10. The following statements are equivalent: (1)
·, ·
∈ S. (2) There is a normed linear space (N,
·
) and a bilinear mapping B : X × X → N such that
a, b
=
B(a, b)
for all a, b ∈ X. (3) If b(a × b) ∈ AC(U ), then b(a × b) ∈ U. (4) ρ generates
·, ·
. (5) For each a, b ∈ X, there is a bounded bilinear 2–functional B on X such that
B
≤ 1 and B(a, b) =
a, b
. Proof. (1) ⇒ (2): If the statement (1) holds, then there is a semi–norm p on BX such that p[b(a×b)] =
a, b
for all a, b ∈ X. For V = {b : p(b) = 0}, let N = BX /V . Define
·
on N by
b
= p(b), where b denotes the equivalence class of N determined by b. It is easily shown that
·
is a well–defined norm on N . Define B : X × X → N by B(a, b) = b(a × b) for all a, b ∈ X. Then B is bilinear and for any a, b ∈ X,
a, b
= p[b(a × b)] =
b(a × b)
=
B(a, b)
.

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R. W. FREESE AND Y. J. CHO

(2) ⇒ (3): Assume that the statement (2) holds and let b(a×b) ∈ AC(U ). i = 1, 2, · · · , n, such that Then there exist

n ai , bi ∈ X and real numbers αi ,

n b(a × b) = i=1 αi b(ai × bi ),
ai , bi
≤ 1, and i=1 |αi | ≤ 1. Theorem 6.6.1 implies that for the bilinear mapping B of the statement (2), there is a unique linear T : BX → N for which B(a, b) = T [b(a × b)] for all a, b ∈ X. Then we have n



αi T [b(ai × bi )]

a, b
=
T [b(a × b)]
=
≤

n 

|αi |
ai , bi
≤

i=1 n 

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i=1

|αi | ≤ 1

i=1

and hence, b(a × b) ∈ U. (3) ⇒ (4): Next, assume that the statement (3) holds. By the remarks preceding Theorem 6.6.1, it suffices in the statement (4) to show that
a, b
≤ ρ[b(a × b)] for all a, b ∈ X. If there are a, b ∈ X for which ρ[b(a × b)] <
a, b
, then for ρ[b(a × b)] < α <
a, b
, ρ[b((a/α) × b)] < 1 implies that b((a/α) × b) ∈ AC(U ) by Theorem 6.6.8 and a well–known property of gauges. The statement (3) implies that b((a/α) × b) ∈ U and hence
a, b
≤ α, which is impossible. Therefore, ρ[b(a × b)] =
a, b
for all a, b ∈ X and ρ generates
·, ·
. (4) ⇒ (5): Assume that the statement (4) is true and let a, b ∈ X. By the Hahn–Banach Theorem, there is a linear functional T on BX such that |T (b)| ≤ ρ(b) for all b ∈ BX and T [b(a × b)] = ρ[b(a × b)]. Then, B(x, y) = T [b(x × y)] is a bilinear functional on X with all the required properties. (5) ⇒ (1): Finally, assume the statement (5) and let a, b ∈ X. Applying Theorem 6.6.9 to the bilinear functional B given by the statement (5), there |T (b)| ≤ ρ(b) for all b ∈ BX and is a linear functional T on BX such that

n T [b(a × b)] =
a, b
. Then, if b(a × b) = i=1 b(ai × bi ), the remark before Theorem 6.6.6 implies that
a, b
= T [b(a × b)] = ≤

n  i=1

n 

T [b(ai × bi )]

i=1

ρ[b(ai × bi )] ≤

n 


ai , bi
.

i=1

By Theorem 6.6.6, ρ generates
·, ·
and the proof is completed.

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Corollary 6.6.11. If the dimension of X is less than or equal to 3, then ρ is a norm and ρ generates
·, ·
. Proof. By S. G¨ ahler [96], it is shown that there is a norm on BX which generates
·, ·
. By Theorem 6.6.10, ρ also generates
·, ·
. Since all bivectors are simple when the dimension of X is less than or equal to 3, it follows that ρ =
·
. For the last result of this section, definitions and properties of strictly convex semi–norms and Hilbertian semi–norms are given in the section 6.5.

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Corollary 6.6.12. If
·, ·
∈ S and if
·, ·
is generated by a semi–norm p which is strictly convex (resp., Hilbertian), then (X,
·, ·
) is strictly convex (resp., a 2–inner product space). Proof. If p generates
·, ·
and p is also a strictly convex seminorm, then using the notation of part 1 of the proof of Theorem 6.6.10, Theorem 6.5.15 implies that (X,
·
) is a strictly convex normed linear space. Further, for the B given in part 1 of the proof of Theorem 6.4.10,
a, b
=
B(a, b)
for all a, b ∈ X. Then, by Theorem 6.6.5, (X,
·, ·
) is a strictly convex linear 2– normed space. The rest follows from the definition of Hilbertian semi–norm and Theorem 5.1.10.

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CHAPTER 7. STRICT 2–CONVEXITY C.R. Diminnie and A. White [65], C.R. Diminnie, S. G¨ ahler and A. White [59], R.W. Freese and A. White [93], R.W. Freese, Y.J. Cho and S.S. Kim [91], M.E. Newton [189] introduced the concepts of algebraic and 2-norm midpoints, (α, β, γ)–2–norm and (α, β, γ)–algebraic interior points and extreme points, respectively, and gave some characterizations of strict 2–convexity in linear 2–normed spaces by using these concepts. In this chapter, we summarize many chatacterizations of strict 2–convexity in linear 2–normed spaces.

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7.1. Elementary Characterizations We will say that a normed linear space (X,
·
) has Property (H) if
x
=
y
=
z
= 13
x + y + z
= 1 implies that x, y and z are collinear. It is easily seen that every strictly convex normed linear space has Property (H), but the converse is not necessarily true (cf. Example 7.1.1). Motivated by these properties, C.R. Diminnie, S. G¨ ahler and A. White [59] introduced initially the concept of strict 2–convexity for linear 2–normed spaces and investigated several relationships between strict convexity and strict 2–convexity for linear 2–normed spaces. For example, every strictly convex linear 2–normed space is strictly 2–convex, but the converse is not necessarily true (cf. Theorem 7.1.3.). In this section, we introduce some elementary characterizations of strict 2–convexity in linear 2-normed spaces given in C.R. Diminnie, S. G¨ ahler and A. White [59], C.S. Lin [156] and A. White [245]. Definition 7.1.1. A linear 2–normed space (X,
·, ·
) is said to be strictly 2–convex if
x, y
=
y, z
=
z, x
= 13
x + z, y + z
= 1 for x, y, z in X implies that z = x + y. Theorem 7.1.1. The following statements are equivalent: (1) (X,
·
) is strictly 2-convex, (2) If
a, b
=
a, c
=
b, c
= 1 and there exists a non–zero bounded bilinear 2–functional F on V (a, b, c)×V (a, b, c) satisfying F (a, b) = F (a, c) = F (c, b) =
F
, then c = a + b. (3) If
a + c, b + c
=
a, b
=
c, a
=
c, b
and
a, b

b, c

c, a
= 0, Typeset by AMS-TEX Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

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R. W. FREESE AND Y. J. CHO

then c = αa + βb for some α > 0 and β > 0. Proof. (1) ⇒ (2): Assume that (X,
·, ·
) is strictly 2–convex,
a, b
=
a, c
=
b, c
= 1, and there is a non–zero bounded bilinear 2–functional F on V (a, b, c) × V (a, b, c) such that F (a, b) = F (a, c) = F (c, b) =
F
. Then we have  1
F (a, b) + F (a, c) + F (c, b) 3=
F
1 F (a + c, b + c) ≤
a + c, b + c
=
F
≤
a, b
+
a, c
+
b, c
= 3. Therefore, 13
a + c, b + c
=
a, b
=
a, c
=
b, c
= 1. The strict 2– convexity of (X,
·, ·
) then implies that c = a + b. Hence, the condition (2) is satisfied. (2) ⇒ (3): Assume the condition (2). Suppose
a + c, b + c
=
a, b
=
a, c
=
b, c
with
a, b

a, c

b, c
= 0. Then by Theorem 6.6.3, there is a bounded bilinear 2–functional F on V (a, b, c) × V (a, b, c) such that
F
= 1 and F (a + c, b + c) =
a + c, b + c
. Since
F
= 1, we have
a, b
+
a, c
+
b, c
≥ F (a, b) + F (a, c) + F (c, b) Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

= F (a + c, b + c) =
a + c, b + c
=
a, b
+
a, c
+
b, c
. Therefore, F (a, b) =
a, b
, F (a, c) =
a, c
and F (b, c) =
b, c
. Let 1 δ = (
a, b

a, c

b, c
)− 2 , α =
b, c
δ, β =
a, c
δ and γ =
a, b
δ. Then,
αa, βb
=
αa, γc
=
βb, γc
= 1 and F (αa, βb) = F (αa, γc) = F (γc, αa) = 1 =
F
. By the condition (2), γc = βb + βb and hence, the condition (3) is satisfied. (3) ⇒ (1): Assume the condition (3) and suppose that
a, b
=
a, c
=
b, c
= 13
a + c, b + c
= 1. Then by the condition (3), there are positive numbers α and β such that c = αa+βb. Since
a, b
=
a, c
=
b, c
= 1, it follows that α = β = 1 and hence c = a + b. Therefore, (X,
·, ·
) is strictly 2–convex and thus the proof is complete. It is easily seen that every strictly convex normed space has Property (H) since, in a strictly convex normed space (X,
·
),
a
=
b
=
c
=

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1 3
a + b + c


= 1 implies that a = b = c. The converse is not necessarily true as in the following example.
Example 7.1.1. On R3 , define a norm
·
by
(x, y, z)
= |x| + y 2 + 1 z 2 2 . Under this norm, R3 is not strictly convex. Let a = (x1 , y1 , z1 ), b = (x2 , y2 , z2 ), and c = (x3 , y3 , z3 ) and assume that 1
a + b + c
=
a
=
b
=
c
= 1. 3 Then we have (7.1) (7.2) (7.3) (7.4)

1
|x1 + x2 + x3 | + (y1 + y2 + z2 )2 + (z1 + z2 + z3 )2 2 = 3, 1
|x1 | + y12 + z12 2 = 1, 1
|x2 | + y22 + z22 2 = 1,
1 |x3 | + y32 + z32 2 = 1.

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By the equations (7.1) through (7.4), it follows that (7.5)

|x1 + x2 + x3 | = |x1 | + |x2 | + |x3 |,

(7.6)

1
(y1 + y2 + y3 )2 + (z1 + z2 + z3 )2 2 1
1
1
= y12 + z12 2 + y22 + z22 2 + y32 + z32 2 .

We now consider three cases: 1 1

1. If x1 = x2 = 0, then y12 + z12 2 = y22 + z22 2 = 1. By (7.6), it follows that (y1 , z1 ) = (y2 , z2 ) and hence a = b. Similar arguments show that if x2 = x3 = 0 or x3 = x1 = 0, then at least two of the vectors a, b, c are equal. 2. If y12 + z12 = y22 + z22 = 0, then |x1 | = |x2 | = 1. By (7.5), x1 = x2 and hence a = b. Similarly, if y22 + z22 = y32 + z32 = 0 or y32 + z32 = y12 + z12 = 0, then at least two of the vectors a, b, c are equal. 3. Therefore, we may assume that x1 and y12 + z12 are non–zero. Then, by (7.5) and (7.6), there are numbers α, β, γ, δ ≥ 0 such that x2 = αx1 , x3 = βx1 , (y2 , z2 ) = γ(y1 , z1 ), and (y3 , z3 ) = δ(y1 , z1 ). Using the equations (7.2), (7.3) and (7.4), we obtain 1
(7.7) 1 = α|x1 | + γ y12 + z12 2 = (α − γ)|x1 | + γ,
1 1 = β|x1 | + δ y12 + z12 2 = (β − δ)|x1 | + δ. (7.8)

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R. W. FREESE AND Y. J. CHO

If α = γ, then the equation (7.7) implies that α = γ = 1 and hence a = b. Similarly, if β = δ, then a = c. If α = γ and β = δ, than by (7.7) and (7.8), γ = 1 and (7.9)

(1 − δ)/(1 − γ) = (β − δ)/(α − γ).

The equation (7.9) may be rewritten as  β −δ β −δ + γ, δ = 1− α−γ α−γ  β −δ β −δ + α. β = 1− α−γ α−γ

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Then, we have c = (x3 , y3 , z3 ) = (βx1 , δy1 , δz1 )  β −δ   β −δ  β−δ β−δ 1− x1 + α x1 , 1 − y1 + γ y1 , = α−γ α−γ α−γ α−γ  β −δ   β−δ 1− z1 + γ z1 α−γ α−γ β −δ  β −δ a+ b. = 1− α−γ α−γ Therefore, in all cases, a, b and c are collinear and this space has Property (H). Theorem 7.1.2. Let (X,
·, ·
) be a linear 2–normed space and (N,
·
) be a normed linear space with Property (H). If there is a mapping T ∈ TX,N (cf. see the section 6.6) such that
a, b
=
T (a, b)
for every a, b ∈ X, then (X,
·, ·
) is strictly 2–convex. Proof. If
a, b
=
a, c
=
b, c
= 13
a + c, b + c
= 1, then
T (a, b)
=
T (a, c)
=
T (c, b)
1 =
T (a, b) + T (a, c) + T (c, b)
= 1. 3 Since (N,
·
) has Property (C), the vectors T (a, b), T (a, c) and T (c, b) are collinear. If T (a, b) = T (a, c), then since
b, c
= 1 and
a, b − c
=

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STRICT 2–CONVEXITY

163

.


T (a, b − c)
= 0, a = α(b − c) for some real number α. Further, since
a, b
=
b, c
=
b, c
= 13
a+c, b+c
= 1, it follows that α = −1 and hence c = a + b. Therefore, we may assume that T (c, b) = βT (a, b) + (1 − β)T (a, c) for some real number β. Then, since T is alternating, T (c − βa, b) = (1 − β)T (a, c) = (1 − β)T (a, c − βa) = T (c − βa, (β − 1)a), which implies that
c − βa, b − (β − 1)a
=
T (c − βa, b − (β − 1)a)
= 0.  Since
a, b
=  0, there is a real number γ such that c − βa = γ b − (β −    1)a or c = γb + β + γ(1 − β) a. Then, T (a, c) = γT (a, b) and hence, 1 =
a, c
= |γ|
a, b
= |γ|. If γ = −1, then T (a, c) = −T (a, b) and
T (a, b) + T (a, c) + T (c, b)
=
T (c, b)
= 1, which is impossible. Thus, γ = 1 and c = a + b. Theorem 7.1.3. If (X,
·, ·
) is strictly convex, then (X,
·, ·
) is also strictly 2–convex.

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Proof. If
a, b
=
a, c
=
b, c
= 13
a+c, b+c
= 1, then
a, b+c
= 2. Since (X,
·, ·
) is strictly convex, and b = c, it follows that there are real numbers α, β such that a = αb+βc. By substituting these into the equations
a, b
=
a, c
=
b, c
= 13
a + c, b + c
= 1, we see that α = −1 and β = 1, i.e., c = a + b. Hence (X,
·, ·
) is strictly 2–convex. The converse of Theorem 7.1.3 is not necessarily true as the following shows: Example 7.1.2. Let X = N = R3 and let
·
be the norm on N given in Example 7.1.1. Define T : X × X → N by setting T (a, b) = a × b, the cross product of a and b. By Theorem 6.4.1. T ∈ TX,N . If we let
a.b
=
T (a, b)
, then by Theorem 6.4.2, Theorem 7.1.2 and Example 7.1.1, (X,
·, ·
) is a strictly 2–convex 2–normed space. However, if we use a = (1, 0, 0), b = (0, 1, 0) and c = (0, 0, 1), we see that (X,
·, ·
) is not strictly convex. Corollary 7.1.4. Every linear 2–normed space of dimension 2 and every 2–inner product space is strictly 2–convex. Proof. Corollary 6.1.3 to Theorem 6.1.2 and Theorem 7.1.3. A linear 2–normed space which is not strictly 2–convex is given in the following:

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Example 7.1.3. Let X, N , and T be the same as in Example 7.1.2. but define
·
on N by
(x1 , y1 , z1 )
= |x1 | + |y1 | + |z1 |. Then, for
a, b
=
T (a, b)
for every a, b ∈ X, (X,
·, ·
) is a linear 2–normed space. In this case, the vectors a = (1, 0, 0), b = (0, 1, 0) and c = (0, 0, 1) show that (X,
·, ·
) is not strictly 2–convex. Theorem 7.1.5. The following seven statements are equivalent: (1) (X,
·, ·
) is strictly 2–convex. (2) If 13
x + z, y + z
=
x, y
=
y, z
=
z, x
= 0, then z = x + y. (3) If
x+z, αy+z
= 3
x, z
= 0 for some α > 0, and
x, y
=
y, z
= 0, then z = x + αy, and α = 1 if
x, z
=
y, z
. (4) If
x + z, y + z
=
x, z
+ 2
y, z
(or = 5
x, z
− 2
y, z
),
y, z

x, z
= 0 and
x, y
=
y, z
, then z = x + αy for some α > 0. (5) If
βx + z, γy + z
= 3
βx, z
and
x, y

y, z

x, z
= 0 for some β > 0 and γ > 0, then z = βx + γy, and β = 1 if
y, z
=
x, y
, and γ = 1 if
x, z
=
x, y
. (6) If
x + z, y + z
=
x, y
+
y, z
+
z, x
and
x, y

y, z

z, x
= 0, then z = βx + γy for some β > 0 and γ > 0. (7) If
x+z, y+z
=
x, y
(8βγ+1)−(3(
y, z
+
z, x
)(or =
x, y
(2βγ− 1) + 3
x, z
−
y, z
) for some β > 0, γ > 0 and
x, y

y, z

z, x
= 0, then z = βx + γy, and β = 1 if
y, z
=
x, y
, and γ = 1 if
z, x
=
x, y
. Proof. Let us consider the following four statements: (3
) If
x + z, αy + z
= 3
x, z
= 0 and
x, y
=
y, z
= 0, where α =
x, z
/
y, z
, then z = x + αy. (4
) If
x + z, y + z
=
x, z
+ 2
y, z
or = 5
x, z
− 2
y, z
),
y, z

x, z
= 0 and
x, y
=
y, z
, then z = x + αy, where α is as in (3
). (5
) If
βx + z, γy + z
= 3
βx, z
and
x, y

y, z

z, x
= 0, where β =
y, z
/
x, y
and γ =
x, z
/
x, y
, then z = βx + γy. (7
) If
x + z, y + z
=
x, y
+
y, z
+
z, x
or =
x, y
(8βγ + 1) − 3(
y, z
+
x, z
), or =
x, y
(2βγ − 1) + 3
x, z
−
y, z
) and
x, y

y, z

z, x
= 0, then z = βx + γy, where β and γ are as in (5
). First of all, we want to prove that the six statements (1), (2), (3
), (4
), (5
) and (7
) are all equivalent to one another. (1) ⇒ (2): If 13
x + z, y + z
=
x, y
=
y, z
=
z, x
= δ 2 = 0, then 1 x + z , y + z = x , y = y , z = x , z = 1. 3 δ δ δ δ δ δ δ δ δ δ Hence, z/δ = (x/δ) + (y/δ) by (1), i.e., z = x + y.

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(2) ⇒ (3
): Since
x, αy
=
αy, z
=
x, z
= 13
x + z, αy + z
, z = x + αy by (2). (3
) ⇒ (4
): Consider the case
y, z
≥
x, z
first, and let α =
x, z
/
y, z
, so, 0 < α < 1. Then we have y
x + z, y + z
= x + z, αy + z + (
y, z
−
x, z
)
y, z
y ≤
x + z, αy + z
+ x + z, (
y, z
−
x, z
)
y, z
≤ (
x, αy
+
x, z
+
z, αy
) + (1 − α)
x + z, y
≤ 3
x, z
+ (1 − α)
x, y
+ (1 − α)
y, z


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=
x, z
+ 2
y, z
. If the identity in (4
) is satisfied, then
x + z, αy + z
= 3
x, z
= 0. Hence, z = x + αy by (3
). The same conclusion is valid for the case
y, z
≤
x, z
, since
x + z, y + z
≤ 5
x, z
− 2
y, z
by the above computation. (4
) ⇒ (1): If 13
x + z, y + z
=
x, y
+
y, z
=
z, x
= 1, then
x + z, y + z
=
x, z
+ 2
y, z
(or = 5
x, z
− 2
y, z
). Hence, α = 1 and z = x + y by (4
). (3
) ⇒ (5
): Assume (3
),
βx+z, γy+z
= 3
βx, z
= 0. Since
βx, z
=
y, z
= 0 and
βx, z
/
y, z
= γ, we have z = βx + γy by (3
). (5
) ⇒ (7
): Consider the case
x, y
≥
x, z
and
y, z
first, and let β =
y, z
/
x, y
and γ =
x, z
/
x, y
. Then we have
x + z, y + z
y x , γy + z + (
x, y
−
x, z
) = βx + z + (
x, z
−
y, z
)
x, y

x, y
y ≤
βx + z, γy + z
+ βx + z, (
x, y
−
x, z
)
x, y
x + γy + z, (
x, y
−
y, z
)
x, y
x y + (
x, y
−
y, z
) , (
x, y
−
x, z
)
x, y

x, y
3
x, z

y, z

x, y
−
x, z

x, y
−
y, z
≤ + 2
y, z
+ 2
x, z

x, y

x, y

x, y

x, z

y, z
+
x, y
−
y, z
−
x, z
−
x, y
=
x, y
+
y, z
+
x, z


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R. W. FREESE AND Y. J. CHO

by straightforward simplifications. The identity in (7
) implies that
βx + z, γy + z
=

3
x, z

y, z
= 3
βx, z
.
x, y


Hence, by (5
), we have z = βx + γy. By similar computations as in the above, we have the following two cases: (a). If
x, y
≤
x, z
and
y, z
, then
x + z, y + z
≤
x, y
(8βγ + 1) − 3(
y, z
+
x, z
). (b). If
x, z
≥
x, y
≥
y, z
, then
x + z, y + z
≤
x, y
(2βγ − 1) + 3
x, z
−
y, z
.

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It is easily seen that both cases come to the same conclusion. (7
) ⇒ (2): If 13
x + z, y + z
=
x, y
=
y, z
=
z, x
= 0, then
x + z, y + z
=
x, y
=
y, z
=
z, x
. Hence β = γ = 1 and z = x + y by the first case in (7
). The other two cases in (7
) imply (2) analogously. Next, it is easily seen that the following implications hold, and we shall omit the details:(3
) ⇒ (3) ⇒ (2), (4
) ⇒ (4) ⇒ (2), (5
) ⇒ (5) ⇒ (2), (7
) ⇒ (6) ⇒ (2), and (7
) ⇒ (7) ⇒ (2). The proof is now completed. Theorem 7.1.6. The following statements are equivalent: (1) (X,
·, ·
) is strictly 2–convex. (2) ( 13
x + z, y + z
)p < 13 (
x, y
p +
y, z
p +
z, x
p ) for p > 1 and z∈ / V (x, y) with
x, y
= 0. Proof. Assume that (X,
·, ·
) is strictly 2–convex, p > 1 and z ∈ / V (x, y) with
x, y
= 0. From the triangle inequality and the fact that the real variable function f (t) = tp is convex on (0, ∞), we have p  1 p 1
x + z, y + z
≤ (
x, y
+
y, z
+
z, x
) 3 3 1 ≤ (
x, y
p +
y, z
p +
z, x
p ). 3 If ( 1
x+z, y +z
)p = 13 (
x, y
p +
y, z
p +
z, x
p ), then ( 31
x+z, y +z
)p = 1 3 p 3 (
x, y
+
y, z
+
z, x
) , that is,
x+z, y+z
=
x, y
=
y, z
=
z, x
. Since
x, y

y, z

z, x
= 0, by (6) of Theorem 7.1.1, there exist positive real numbers α, β such that z = αx + βy. Thus, z ∈ V (x, y), which contradicts the assumption. Therefore, p 1 1
x + z, y + z
< (
x, y
p +
y, z
p +
z, x
p ) 3 3

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STRICT 2–CONVEXITY

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for p > 1 and z ∈ / V (x, y) with
x, y
= 0. Conversely, suppose that
x, y
=
y, z
=
z, x
= 13
x + z, y + z
= 1. Assume that z ∈ / V (x, y). Since p > 1 and
x, y
= 0, by hypothesis, 1 3


x + z, y + z


p


0. Since 1 =
a, c
=
b, c
, α = t and β = 1 − t. Hence c = a + b, which is a contradiction. Consequently,
a + tc, b + (1 − t)c
< 2. 2. Assume
a, b
=
a, c
=
b, c
= 13
a + c, b + c
= 1. By Theorem 7.1.8 (5),
a + tc, b + (1 − t)c
=
a, b
+ t
b, c
+ (1 − t)
a, c
= 2. Hence c = a + b. 7.2. Strict 2–Convexity by Algebraic and 2–Norm Midpoints In this section, we introduce some geometric structures of strictly 2– convex linear spaces in terms of algebraic and 2–norm midpoints defined in the section 6.4 of Chapter VI, which are due to C.R. Diminnie and A. White [64].

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If (X,
·, ·
) is strictly 2–convex, the converse of Theorem 6.4.1 is true. Theorem 7.2.1. If (X,
·, ·
) is strictly 2–convex, then every 2–norm midpoint of 3 non–collinear points is also the algebraic midpoint of these points. Proof. Assume (X,
·, ·) is strictly 2–convex and let d be the 2–norm midpoint of the non–collinear points a, b and c. Then, if we let t2 =
a − d, b − d
=
a − d, c − d
=
b − d, c − d
=

1
a − c, b − c
, 3

it follows that t = 0 since a, b and c are non–collinear. Therefore, we have a − d b − d a − d d − c b − d d − c , , , = = = 1 t t t t t t and

a − d d − c b − d d − c a − c d − c + , + , = = 3. t t t t t t

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Since (X,
·, ·
) is strictly 2–convex, (d − c)/t = (a − d)/t + (b − d)/t and hence d = 13 (a + b + c). Theorem 7.2.2. A linear 2–normed space (X,
·, ·
) is strictly 2–convex if and only if for any 3 non–collinear points of X the algebraic and 2–norm midpoints are identical. Proof. If (X,
·, ·
) is strictly 2–convex, then by Theorems 6.7.1 and 7.2.1 the algebraic and 2–norm midpoints of any 3 non–collinear points are identical. Conversely, suppose that in (X,
·, ·
) the algebraic and 2–norm midpoints of X coincide. Then, if
a, b
=
a, c
=
b, c
= 13
a + c, b + c
= 1, then a, b and c are non–collinear and 0 is the 2–norm midpoint of a, b, and −c. Since 0 is also the algebraic midpoint of a, b and −c, 0 = 13 (a + b − c) or c = a + b. Therefore, (X,
·, ·
) is strictly 2–convex. For non–collinear a, b, c ∈ X, let T (a, b, c) = {x : σ(a, b, c) = σ(a, b, x) + σ(a, x, c) + σ(x, b, c)}.

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T (a, b, c) will be called the triangle with vertices a, b, c. Further, we will designate the area of T (a, b, c) to be σ(a.b.c). A point d will be called a center of a, b, c if d is a 2–norm midpoint of a, b, c. By Theorem 7.1.1, T (a, b, c) has a unique center if and only if (X,
·, ·
) is strictly 2–convex. Theorem 7.2.3. If a, b, c are non–collinear, then T (a, b, c) is convex. Proof. Let x1 , x2 ∈ T (a, b, c) and let x = tx1 +(1−t)x2 for some t ∈ [0, 1]. Then we have σ(a, b, x) = σ(x, b, a) =
tx1 + (1 − t)x2 − a, b − a
≤ t
x1 − a, b − a
+ (1 − t)
x2 − a, b − a
= tσ(x1 , b, a) + (1 − t)σ(x2 , b, a) = tσ(a, b, x1) + (1 − t)σ(a, b, x2). Similarly, it follows that σ(a, x, c) ≤ tσ(a, x1 , c) + (1 − t)σ(a, x2, c) and σ(x, b, c) ≤ tσ(x1 , b, c) +(1 −t)σ(x2, b, c). Therefore, since x1 , x2 ∈ T (a, b, c), we have σ(a, b, c) ≤ σ(a, b, x) + σ(a, x, c) + σ(x, b, c) ≤ tσ(a, b, x1) + (1 − t)σ(a, b, x2) + tσ(a, x1 , c) + (1 − t)σ(a, x2 , c) + tσ(x1 , b, c) + (1 − t)σ(x2 , b, c) = σ(a, b, c).

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Hence, x ∈ T (a, b, c) and so T (a, b, c) is convex. Before proceeding to the main result of this section, we present the following technical lemma. Its proof follows from the usual techniques of the calculus and will be omitted here. Lemma 7.2.4. Subject to the conditions α + β + γ = 1 and α, β, γ ≥ 0, the maximum value of f (α, β, γ) = αβ + αγ + βγ is 13 . Further, this value is attained only when α = β = γ = 13 . If a, b, c are non–collinear, let C(a, b, c) denote the convex envelope of {a, b, c}, that is, C(a, b, c) is the smallest convex set containing {a, b, c}. In particular, we will use the result that

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C(a, b, c) = {αa + βb + γc : α, β, γ ≥ 0, α + β + γ = 1}. Theorem 7.2.5. The following statements are equivalent: (1) (X,
·, ·
) is strictly 2–convex. (2) If a, b, c are non–collinear, then T (a, b, c) = C(a, b, c). (3) If a, b ∈ X with
a, b
> 0, then there is a unique c ∈ X such that 0 is a center of T (a, b, c). (4) If
a, b
=
a, c
=
b, c
= 1, c = −(a+b), α, β, γ > 0 and α+β +γ = 1, then σ(αa, βb, γc) < 13 .

Proof. (1) ⇒ (2): Assume that (X,
·, ·
) is strictly 2–convex and let a, b, c be non–collinear points of X. Since a, b, c ∈ T (a, b, c), Theorem 7.2.3 implies that C(a, b, c) ⊂ T (a, b, c). Let x ∈ T (a, b, c), that is, let σ(a, b, c) = σ(a, b, x) + σ(a, x, c) + σ(x, b, c). If σ(a, b, x) = 0, then either x = a, x = b, or a − x = α(b − x) for some α = 0, 1. This implies that x = βa + (1 − β)b with β = 1/(1 − α). Then we have 0 < σ(a, b, c) = σ(a, x, c) + σ(x, b, c) =
a − c, (1 − β)(b − c)
+
β(a − c), b − c
= (|1 − β| + |β|)σ(a, b, c). Therefore, |1 − β| + |β| = 1 from which it follows that 0 < β < 1. Hence, if σ(a, b, x) = 0, then x ∈ C(a, b, c). Similar arguments hold if σ(a, x, c) = 0 or σ(x, b, c) = 0. If σ(a, b, x)σ(a, x, c)σ(x, b, c) = 0, then since σ(a, b, c) =

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R. W. FREESE AND Y. J. CHO

σ(a, b, x)+σ(a, x, c)+σ(x, b, c), Theorem 7.1.1 implies that there are γ, δ > 0 such that x − c = γ(a − x) + δ(b − x) or

     δ 1 γ a+ b+ c. x= 1+γ +δ 1+γ+δ 1+γ+δ 

Therefore, once again x ∈ C(a, b, c) and hence T (a, b, c) = C(a, b, c). (2) ⇒ (3): Assume the statement (2) and let a, b ∈ X with
a, b
> 0. If c = −(a + b), then it is easily seen that a, b, c are non–collinear and 0 is a center of T (a, b, c). If d is any point with the same property, then

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a, b
=
a, d
=
b, d
=

1
a − d, b − d
. 3

Since 0 ∈ T (a, b, d), the statement (2) implies that there are numbers α, β, γ ≥ 0 such that α + β + γ = 1 and 0 = αa + βb + γd. Further, since
a, b
> 0 implies that γ > 0, we may write d = α1 a + β1 b with α1 , β1 ≤ 0. Then, from the conditions
a, b
=
a, d
=
b, d
and
a, b
> 0, it follows that α1 = β1 = −1. Therefore, d = −(a + b) = c and so the statement (3) is established. (3) ⇒ (4): Assume the statement (3) and let
a, b
=
a, c
=
b, c
= 1, c = −(a + b), α, β, γ > 0, and α + β + γ = 1. Then, by Lemma 7.2.4, σ(αa, βb, γc) =
αa − γc, βb − γc
≤ αβ
a, b
+ αγ
a, c
+ βγ
b, c
1 = αβ + αγ + βγ ≤ . 3 If σ(αa, βb, γc) = 13 , then Lemma 7.2.4 implies that α = β = γ = hence,
a − c, b − c
= 9σ(αa, βb, γc) = 3. Therefore,
a, b
=
a, c
=
b, c
=

1 3

and

1
a − c, b − c
, 3

which implies that 0 is the center of T (a, b, c). Since c = −(a + b), the statement (3) makes this an impossibility. Thus, σ(αa, βb, γc) < 13 and the statement (4) holds.

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STRICT 2–CONVEXITY

173

(4) ⇒ (1): If the statement (4) holds and
a, b
=
a, c
=
b, c
=

1
a + c, b + c
= 1, 3

then letting α = β = γ = 13 , it follows immediately that −c = −(a + b) or c = a + b. Therefore, the statement (4) implies that (X,
·, ·
) is strictly 2–convex. 7.3. Strict 2–Convexity by (α, β, γ)–2–Norm Interior Points and (α, β, γ)–Algebraic Interior Points In this section, we show that some results of the section 7.1 are generalized by using the concepts of (α, β, γ)–2–norm and (α, β, γ)–algebraic interior points. The results in this sections are due to M.E. Newton in his doctoral dissertation [200]. Definition 7.3.1. In a linear 2–normed space (X,
·, ·
), let x and y be distinct points with y = 0. Then the set

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L(x, y) = {x + ty : t is a real number } is called the algebraic line determined by x and y. This makes it possible to define the unit cylinder with central axis L(0, c) for all c = 0. Definition 7.3.2. In a linear 2–normed space (X,
·, ·
), let c be any non–zero point. Then the set B(c) = {x :
x, c
≤ 1, x ∈ X} is called the unit cylinder with central axis L(0, c). Further, the set bdy B(c) = {x :
x, c
= 1, x ∈ X} is called the boundary of B(c). We will, shortly, examine the properties of B(c). To do so, we need some preliminary definitions and theorems.

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R. W. FREESE AND Y. J. CHO

Definition 7.3.3. (X,
·, ·
), the set

If x, y are elements of a linear 2–normed space

S(x, y) = {αx + βy : α, β ≥ 0, α + β = 1 for α, β real numbers} is called the algebraic line segment determined by x and y. Further, if x, y, z ∈ X, they are said to be collinear when z ∈ S(x, y) or y ∈ S(x, z) or x ∈ S(y, z). Otherwise, x, y, z are non–collinear. Definition 7.3.4. In a linear 2–normed space (X,
·, ·
), let x, y, z be three non–collinear points. The area of the triangle with vertices x, y, z is: A(x, y, z) =
x − z, y − z
. In the section 6.4 of Chapter VI, C.R. Diminnie, S. G¨ ahler and A. White [59] define the algebraic and 2–norm midpoints of three given points. Analogously, we make the following definitions:

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Definition 7.3.5. Let x, y, z be three points in a linear 2–normed space (X,
·, ·
). A point b ∈ X is said to be an (α, β, γ)–algebraic interior point of x, y, z if b = αx + βy + γz where α, β, γ are positive real numbers and α + β + γ = 1. Definition 7.3.6. If x, y, z are three non–collinear points of (X,
·, ·
), a point d ∈ X is an (α, β, γ)–2–norm interior points of x, y, z if
x − d, y − d
= γ
x − z, y − z
,
x − d, z − d
= β
x − z, y − z
,
y − d, z − d
= α
x − z, y − z
, where α, β, γ are positive real numbers and α + β + γ = 1. Note that if d is an (α, β, γ)–2–norm interior point of x, y, z, then A(x, y, z) = A(x, y, d) + A(x, z, d) + A(y, z, d). This brings us to the following generalization of some results in the section 6.4 of Chapter VI and the section 7.1.

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Theorem 7.3.1. Let (X,
·, ·
) be a linear 2–normed space, and let x, y, z be any three non–collinear points of X. If b is an (α, β, γ)–algebraic interior point of x, y, z, then it is also an (α, β, γ)–2–norm interior point of x, y, z for the same values of α, β, γ. Proof. Let b = αx + βy + γz, where α, β, γ are positive real numbers and α + β + γ = 1. Consider the following:
x − b, y − b
=
(x − y) + (y − b), y − b
=
x − y, y − b
=
x − y, y − (αx + βy + γz)
=
x − y, (1 − β)y − (αx + γz)
=
x − y, (α + γ)y − (αx + γz)
=
x − y, −α(x − y) + γ(y − z)
=
x − y, γ(y − z)
= γ
x − y, y − z
= γ
(x − z) − (y − z), y − z
= γ
x − z, y − z
. Analogously, we can show that

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x − b, z − b
= β
x − z, y − z
,
y − b, z − b
= α
x − z, y − z
. Thus, b is an (α, β, γ)–2–norm interior point of x, y, z, as desired. Under certain conditions, the converse of this theorem holds. We need to introduce two more definitions and a theorem. First, however, Definition 7.3.7. A linear 2–normed space (X,
·, ·
) is said to be strongly strictly convex if
x + y, z
=
x, z
+
y, z
and z ∈ / V (x, y) imply x = ky for some real number k. The next definition is a generalization of strict 2–convexity. Definition 7.3.8. A linear 2–normed space (X,
·, ·
) is said to be strongly strictly 2–convex if
x + z, y + z
= 1,
x, z
= β
x + z, y + z
= β,
y, z
= α
x+z, y+z
= α and
x, y
= γ
x+z, y+z
= γ, where α, β, γ are positive real numbers and α+β +γ = 1, imply that z = (α/γ)x+(β/γ)y. (We assume the labelling so that α ≥ β ≥ γ > 0.)

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Thus, we have the following theorem, which is a generalization of Theorem 7.1.3. Theorem 7.3.2. If (X,
·, ·
) is strongly strictly convex, then it is strongly strictly 2–convex. Proof. Let (X,
·, ·
) be strongly strictly convex, and let x, y, z be any three points of X satisfying the assumptions of Definition 7.3.8. Then, 1 =
x + z, y + z
≤
x, y
+
x, z
+
y, z
= α + β + γ = 1. Hence,
x + z, y + z
=
x, y + z
+
z, y
= 1. Therefore,
x, y + z
= 1 −
y, z
= 1 − α = β + γ =
x, y
+
x, z
.

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Further,
y, z
= 0 by hypothesis, so y is not a multiple of z. Thus, by Definition 7.3.5, x ∈ V (y, z). Consequently, there exist real numbers r1 , r2 such that x = r1 y + r2 z. Consider, on one hand, that
x, y
=
r1 y + r2 z, y
= |r2 |
y, z
= |r2 |α. On the other hand,
x, y
= γ by hypothesis. Thus r2 = ±(γ/α). Anlogously,
x, z
=
r1 y + r2 z, z
= |r1 |
y, z
= |r1 |α, on one hand. On the other,
x, z
= β by hypothesis. Therefore, we have r1 = ±(β/α). We consider the following three cases: Case 1. If r1 = β/α and r2 = γ/α, then x = (β/α)y + (γ/α)z. Hence z = (α/γ)x − (β/γ)y. Therefore, we have γ = γ
x + z, y + z
  α β α β x − y, x + 1 − y = γ 1 + γ γ γ γ

    α β αβ ≤γ 1+ ( − 1 + 2
x, y
γ γ γ

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STRICT 2–CONVEXITY

β

177

αβ αβ  α − 1 + −
x, y
γ γ2 γ γ2

β − α − γ 2αβ  + 2
x, y
=γ γ γ

2α(1 − α − γ) 
x, y
= 1 − 2α − 2γ + γ

 α − α2  = 1 − 2 2α + γ +
x, y
. γ

=γ

+

Now, 0 < α < 1, so α > α2 . Hence 2α + γ + (α − α2 )/γ > 0. Therefore, 1 − 2(2α + γ + (α − α2 )/γ) < 1. Hence, we have

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 α − α2 
x, y
<
x, y
. γ ≤ 1 − 2 2α + γ + γ But, by hypothesis, since γ =
x, y
, we have a contradiction. So x = (β/α)y + (γ/α)x. Case 2. If r1 = β/α and r2 = −(γ/α), then x = (β/α)y − (γ/α)z. Thus, we have β  γ z, y + z α = α
x + z, y + z
= α y + 1 − α α

β 

β + α − γ  γ  + 1− ≤α
y, z
= α
y, z
α α α = (β + α − γ)
y, z
= (1 − 2γ)
y, z
<
y, z
since 0 < γ < 1. But, by hypothesis, since α =
y, z
, we have a contradiction. Hence, x = (β/α)y − (γ/α)z. Case 3. If r1 = −(β/α) and r2 = −(γ/α), then x = −(β/α)y − (γ/α)z. In that case, we have β  γ z, y + z α = α
x + z, y + z
= α − y + 1 − α α

β 

α + β − γ  γ  + 1− ≤α
y, z
= α
y, z
α α α = (α + β − γ)
y, z
= (1 − 2γ)
y, z
<
y, z
. Thus, the contradiction is the same as in case 2. So x = −(β/α)y − (γ/α)z. Therefore, the only remaining possibility must hold. That is, x = −(β/α)y+ (γ/α)z. Equivalently, we have z = (α/γ)x + (β/γ)y, as desired.

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R. W. FREESE AND Y. J. CHO

In a strongly strictly convex space, the converse of Theorem 7.3.1 holds. This result is a generalization of Theorem 6.4.3 as follows: Theorem 7.3.3. If (X,
·, ·
) is strongly strictly convex, then an (α, β, γ)–2–norm interior point of three non–collinear points of X is also an (α, β, γ)–algebraic interior point of these points for the same values of α, β, γ. Proof. Let x, y, z be three non–conllinear points of X, and let d be an (α, β, γ)–2–norm interior point of x, y, z. Then by Definition 7.3.6,
x − d, y − d
= γ
x − z, y − z
,
x − d, z − d
= β
x − z, y − z
and
y − d, z − d
= α
x − z, y − z
,

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where α, β, γ are positive real numbers and α + β + γ = 1. We may assume the labelling so that α ≥ β ≥ γ > 0. Since x, y, z are non–collinear, then
x − z, y − z
= 0. Let t2 =
x − z, y − z
, hence t = 0. Thus, we have x − d y − d x − z y − z , , = γ = γ, t t t t x − z y − z x − d z − d , , = β =β t t t t and

x − z y − z y − d z − d , , = α = α. t t t t

Now since (X,
·, ·
) is strongly strictly convex, it follows, by Theorem 7.3.2, that (X,
·, ·
) is strongly strictly 2–convex. Hence, by Definition 7.3.8, z−d αx − d β y − d − = + . t γ t γ t Therefore, −γ(z − d) = α(x − d) + β(y − d). So we have αx + βy + γz = αd + βd + γd, αx + βy + γz = (α + β + γ)d, αx + βy + γz = d

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STRICT 2–CONVEXITY

179

and −γz + γd = αx − αd + βy − βd. Hence, d is an (α, β, γ)–algebraic interior point of x, y, z for the same values of α, β, γ, as desired. We now make the following definition: Definition 7.3.9. Let (X,
·, ·
) be a linear 2–normed space. (X,
·, ·
) is said to be strongly round if, for any non–collinear points x, y, z and non– zero c ∈ X,
x, c
=
y, c
=
z, c
= 0 imply
αx + βy + γz, c
<
x, c
, where α, β, γ are positive real numbers such that α + β + γ = 1.

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It is possible to show that if a linear 2–normed space (X,
·, ·
) is strongly strictly convex, then it is strongly round. Theorem 7.3.4. If (X,
·, ·
) is strongly strictly convex, then it is also strongly round. Proof. Let (X,
·, ·
) be strongly strictly convex. Also let x, y, z be non–collinear points of X and c = 0 be an element of X such that
x, c
=
y, c
=
z, c
= 0. Let α, β, γ be positive real numbers such that α+β+γ = 1. Consider two cases: c ∈ V (x, y) ∩ V (x, z) ∩ V (y, z) or not. Case 1. c ∈ / V (x, y) ∩ V (x, z) ∩ V (y, z). Then, either c ∈ / V (x, y) or c∈ / V (x, z) or c ∈ / V (y, z). We may assume without loss of generality that c∈ / V (x, y). Now since x, y, z are non–collinear by hypothesis, x and y are distinct. Also,
x, c
=
y, c
implies x = ky except possibly for k = ±1. Clearly, k = 1, since x = y. If k = −1, then y = −x. Consequently, we have
αx + βy + γz, c
=
(α − β)x + γz, c
≤ |α − β|
x, c
+ γ
z, c
= (|α − β| + γ)
x, c
. Since α, β > 0, |α − β| < α + β. Thus we have
αx + βy + γz, c
< (α + β + γ)
x, c
=
x, c
. Thus, if x is a multiple of y, we have our desired result. If x is not a multiple of y, consider the following:
αx + βy + γz, c
≤
αx + βy, c
+
γz, c
.

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R. W. FREESE AND Y. J. CHO

Since
z, c
> 0, we have αx αx + βy + γz αx + βy βy , c ≤ , c + γ = + , c + γ.
x, c

z, c

z, c

z, c
Now, since x is not a multiple of y, the points αx/
z, c
and βy/
z, c
are distinct, and not multiples of one another. The hypothesis c ∈ / V (x, y) implies c ∈ / V (αx/
z, c
, βy/
z, c
). Hence, by Definition 7.3.7, it follows that βy αx + βy αx , c = , c + , c ,
z, c

z, c

z, c
that is, αx + βy αx βy , c , c , c < + .
z, c

z, c

z, c
Hence we have βy αx + βy + γz αx , c < , c + , c +γ
z, c

z, c

z, c
x y = α , c +β , c +γ
x, c

y, c


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= α + β + γ = 1. Multiplying by
z, c
, we have
αx+βy +γz, c
<
z, c
=
x, c
, as desired. Case 2. c ∈ V (x, y) ∩ V (x, z) ∩ V (y, z). Since c ∈ V (x, y), there exist a, b ∈ R, not both zero, such that c = ax + by. Now,
x, c
=
y, c
by hypothesis. Further,
x, c
=
x, ax + by
= |b|
x, y
. Similarly,
y, c
= |a|
x, y
. If
x, y
= 0, then y = kx for some real number k. But then V (x, y) = V (y), so c ∈ V (y). Thus
y, c
= 0, which is contrary to hypothesis. Hence,
x, y
= 0. Therefore, we have
x, c
=
y, c
, |b|
x, y
= |a|
x, y
, |b| = |a|, b = ±a. If b = a, then c = a(x + y). In that case, we have
αx + βy + γz, c
=
αx + αy + (β − α)y + γz, c
=
α(x + y) + (β − α)y + γz, c
=
(β − α)y + γz, c
≤ |β − α|
y, c
+ γ
z, c
= (|β − α| + γ)
x, c


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STRICT 2–CONVEXITY

181

since
x, c
=
y, c
=
z, c
. Also, since α, β > 0, |β − α| < α + β. Thus,
αx + βy + γz, c
< (α + β + γ)
x, c
=
x, c
, as desired. If b = −a, then c = a(x − y). Now since c ∈ V (x, z) also, z ∈ V (x, y). There exist real numbers r, t, not both zero, such that z = rx + ty. Consider the following:
x, c
=
rx + ty, ax − ay
= |r(−a) − at|
x, y
= |r + t||a|
x, y
= |r + t|
x, ax − ay
= |r + t|
x, c
. Now
z, c
=
x, c
by hypothesis and so |r + t| = 1. Therefore, r + t = 1 or r + t = −1. In the case r + t = 1, we have
z, c
=
rx + ty, c
≤
rx, c
+
ty, c
= |r|
x, c
+ |t|
y, c
= (|r| + |t|)
x, c
.

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Hence, |r| + |t| ≤ 1. There are the following three possibilities: (i) r = 1 and t = 0, (ii) r = 0 and t = 1, (iii) 0 < r < 1 and 0 < t < 1. If (i) holds, then z = x. But x, y, z were non–collinear, and so all distinct. If (ii) holds, then z = y, and we have the same contradiction. If (iii) holds, then by Definition 7.3.3, z ∈ S(x, y), which also contradicts the hypothesis that x, y, z are non–collinear. Hence, r + t = 1. In the case r + t = −1, consider the following:
αx + βy + γz, c
=
αx + βy + γ(rx + ty), c
=
(α + γr)x + (β + γt)y, ax − ay
= |(α + γr)(−a) − a(β + γt)|
x, y
= |α + β + γ(r + t)||α|
x, y
= |(α + β) + γ(−1)||α|
x, y
= |(α + β) − γ||α|
x, y
= |(α + β) − γ|
x, c
.

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R. W. FREESE AND Y. J. CHO

Now (α + β) − γ = (1 − γ) − γ = 1 − 2γ. Since 0 < γ < 1, |1 − 2γ| < 1. Hence, we have
αx + βy + γz, c
= |1 − 2γ|
x, c
<
x, c
, as desired. Therefore, X is strongly round. Theorem 7.3.4 brings us, finally, to a description of unit cylinders B(c), c = 0, in the following theorem: Theorem 7.3.5. Let (X,
·, ·
) be a strongly strictly convex linear 2– normed space. Let x, y, z be three non–collinear points on bdy B(c) for some c = 0 in X. Then the (α, β, γ)–2–norm interior points of x, y, z do not lie on bdy B(c). Proof. Let x, y, z be non–collinear points on bdy B(c) for some c = 0. Then by Definition 7.3.2,
x, c
=
y, c
=
z, c
= 1. Let d be an (α, β, γ)– 2–norm interior point of x, y, z. Since (X,
·, ·
) is strongly strictly convex, then by Theorem 7.3.3, d is an (α, β, γ)–algebraic interior point of X also for the same α, β, γ. Thus, by Definition 7.3.5, d = αx + βy + γz. Also, since (X,
·, ·
) is strongly strictly convex, it follows from Theorem 7.3.4 that (X,
·, ·
) is strongly round. Hence,
d, c
=
αx + βy + γz, c
<
x, c
= 1. Therefore, d does not lie on the bdy B(c).

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7.4. Strict 2–Convexity by Duality Mappings ∗ be the space of all bounded linear 2–functionals on V (p, q, r) × Let Xpqr V (p, q, r) for every p, q, r in X. As in Chapter VI, we define the duality ∗ mapping types (A

) and (B

), I and Jφ : V (p, q, r) × V (p, q, r) → 2Xpqr as follows, respectively:

(A

)

∗ I(x, y) = {F ∈ Xpqr ; F (x, y) =
F

x, y
}

and (B

)

∗ Jφ (x, y) = {F ∈ Xpqr ; F (x, y) =
F

x, y
,
F
= φ(
x, y
)}

for every x, y in V (p, q, r), where φ is a semi–positive function. Note that by Theorem 6.6.3, if x and y are linearly independent, then I(x, y) = Φ, and some basic properties of I(x, y) and Jφ (x, y) are as in Chapter VI.

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Now, we give some characterizations of strict 2–convexity for linear 2– normed spaces by using the duality mapping types, which are due to Y.J. Cho, K.S. Park and B.H. Park [30]. Theorem 7.4.1. The following statements are equivalent: (1) (X,
·, ·
) is strictly 2–convex. (2) I(x, y) ∩ I(y, z) ∩ I(z, x) = Φ for x, y, z in V (p, q, r) implies that z = αx + βy for some α > 0 and β > 0. Proof. (1) implies (2): Assume (1). Let F ∈ I(x, y) ∩ I(y, z) ∩ I(z, x) and F = 0. Then we have F (x, y) =
F

x, y
, F (y, z) =
F

y, z
and F (z, x) =
F

z, x
and so we have
F

x + z, y + z
≥ F (x + z, y + z) = F (x, y) + F (x, z) + F (z, y) =
F

x, y
+
F

x, z
+
F

z, y
=
F
(
x, y
+
x, z
+
z, y
) and hence we have

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x + z, y + z
=
x, y
+
x, z
+
z, y
. Therefore, since (X,
·, ·
) is strictly 2–convex, by Theorem 7.1.1, z = αx + βy for some α > 0 and β > 0. ∗ , F = (2) implies (1): Let
x, y
=
y, z
=
z, x
= 1 and F ∈ Xpqr 0, with F (x, y) = F (y, z) = F (z, x) =
F
. Then, since F (x, y) =
F

x, y
, F (y, z) =
F

y, z
and F (z, x) =
F

z, x
, we have F ∈ I(x, y) ∩ I(y, z) ∩ I(z, x), that is, I(x, y) ∩ I(y, z) ∩ I(z, x) = Φ. Hence, by assumption, z = αx + βy for some α > 0 and β > 0. Since
x, y
=
y, z
=
z, x
= 1, we have α = β = 1. Therefore, by Theorem 7.1.1, (X,
·, ·
) is strictly 2–convex. Theorem 7.4.2. Let φ be an onto semi–positive function. Then the following statements are equivalent: (1) (X,
·, ·
) is strictly 2–convex.

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R. W. FREESE AND Y. J. CHO

(2) Jφ (x, y) ∩ Jφ (y, z) ∩ Jφ (z, x) = Φ for x, y, z in V (p, q, r) implies that z = αx + βy for some α > 0 and β > 0. Proof. (1) implies (2): Assume (1). Let Jφ (x, y)∩Jφ (y, z)∩Jφ (z, x) = Φ. Then, since Jφ (x, y) ⊂ I(x, y), Jφ (y, z) ⊂ I(y, z) and Jφ (z, x) ⊂ I(z, x), we have I(x, y) ∩ I(y, z) ∩ I(z, x) = Φ. Therefore, by Theorem 7.4.1, z = αx + βy for some α > 0 and β > 0. ∗ , F = 0, F (x, y) = F (y, z) = (2) implies (1): Assume (2). Let F ∈ Xpqr F (z, x) =
F
and
x, y
=
y, z
=
z, x
= 1. Since φ is an onto semipositive function, for
F
> 0, there exist a number a > 0 such that φ(a2 ) =
F
. Hence we have φ(
ax, ay
) =
F
, F (ax, ay) =
F

ax, ay
, φ(
ay, az
) =
F
, F (ay, az) =
F

ay, az
, φ(
az, ax
) =
F
, F (az, ax) =
F

az, ax
, that is, Jφ (ax, ay) ∩ Jφ (ay, az) ∩ Jφ (az, ax) = Φ.

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Hence, by assumption, az = αax+βay for some α > 0 and β > 0. Therefore, by Theorem 7.1.1, (X,
·, ·
) is strictly 2–convex. The following Corollary 7.4.3 and Corollary 7.4.4 follow immediately from Theorem 7.4.1 and Theorem 7.4.2, respectively: Corollary 7.4.3. Let (X,
·, ·
) be a strictly 2–convex linear 2–normed space and φ1 , φ2 and φ3 be semipositive functions. Then, if Jφ1 (x, y) ∩ Jφ2 (y, z) ∩ Jφ3 (z, x) = Φ for x, y, z in V (p, q, r), then z = αx + βy for some α > 0 and β > 0. If φ is the function φ(a) = a, we denote Jφ by J. Corollary 7.4.4. The following statements are equivalent: (1) (X,
·, ·
) is strictly 2–convex. (2) J(x, y) ∩ J(y, z) ∩ J(z, x) = Φ for x, y, z in V (p, q, r) implies that z = x + y. Corollary 7.4.5. Let φ be an onto pseudo–gauge function. Then the following statements are equivalent:

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(1) (X,
·, ·
) is strictly 2–convex. (2) Jφ (x, y) ∩ Jφ (y, z) ∩ Jφ (z, x) = Φ for x, y, z in V (p, q, r) implies that z = x + y. 7.5. Strict 2–Convexity by Extreme Points

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In [91], R.W. Freese, Y. J. Cho and S. S. Kim introduced the concept of extreme points in 2–metric spaces and gave some geometric properties and characterizations of 2–metric spaces and strict 2–convexity in linear 2–normed spaces. In a 2–metric space (X, σ), a point p is said to be a 2–metric extreme point of a set M in X provided, for arbitrary points x, y, z in M with σ(x, y, z) = σ(x, y, p) + σ(x, z, p) + σ(y, z, p), at least one of the terms on the right– hand side is zero. A point p is said to be 2–metric ultra–extreme point of a set M provided, for arbitrary points x, y, z in M with σ(x, y, z) = σ(x, y, p) + σ(x, z, p) + σ(y, z, p), at least two of the terms on the right–hand side are equal to zero. In a linear 2–normed space (X,
·, ·
), a point p is said to be an algebraic extreme point of a set M provided, for arbitrary points x, y, z in M with p = αx + βy + γz, α + β + γ = 1 and α, β, γ ∈ [0, 1], at least one of α, β, γ is zero. A point p is said to be an algebraic ultra–extreme point of a set M provided, for arbitrary points x, y, z in M with p = αx + βy + γz, α + β + γ = 1 and α, β, γ ∈ [0, 1], at least two of α, β, γ are equal to zero. We remark that: (1) A 2–metric(resp., algebraic) ultra–extreme point is a 2–metric (resp., algebraic) extreme point but not conversely. (2) In the Euclidean plane a 2–metric extreme point is an algebraic extreme point and conversely. We now present some characterizations of strict 2–convexity for linear 2–normed spaces by using the concept of extreme points: Theorem 7.5.1. A 2–metric extreme point of a set M in a 2–metric space (X, σ) is an algebraic extreme point of M . Proof. Suppose that there exists a point p in M such that p is a 2– metric extreme point of M but not an algebraic extreme point of M . This means that there exist three points x, y, z in M such that p = αx + βy + γz, α + β + γ = 1 and α, β, γ ∈ [0, 1], with each of α, β, γ are non–zero. Therefore z − p and y − p are linearly independent and hence we have σ(p, y, z) =
z − p, y − p
= 0.

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Similarly, we have σ(p, x, z) = 0 and σ(p, x, y) = 0, which is contrary to p being a 2–metric extreme point. By a similar argument used in Theorem 7.5.1, we have the following theorem: Theorem 7.5.2. A 2–metric ultra–extreme point of a set M in a 2– metric space (X, σ) is an algebraic ultra–extreme point of M . Theorem 7.5.3. An algebraic ultra–extreme point of a set M in a linear 2–normed space (X,
·, ·
) that is a 2–metric extreme point of M is a 2– metric ultra–extreme point of M . Proof. Suppose that there exists a point p in M such that p is an algebraic ultra–extreme point of M and a 2–metric extreme point but not a 2–metric ultra–extreme point. This means that there exist three distinct points x, y, z in M such that σ(p, x, y) = 0 but

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σ(x, y, z) = σ(x, y, p) + σ(x, z, p) + σ(y, z, p). That is, σ(x, y, z) = σ(x, z, p)+σ(y, z, p). Therefore, since p is not a 2–metric ultra–extreme point of M , then we have p = x, p = y, p = z and similarly, x, y, z are distinct. From σ(p, x, y) = 0, we have that p − x and y − x are linearly dependent, that is, p = λx + (1 − λ)y. Thus p = 0z + λx + (1 − λ)y which implies, since p is an algebric ultra–extreme point, that λ = 0 or λ = 1, thus implying p = x or p = y, which is a contradition. The proof is similar in the event that σ(p, x, z) = 0 or σ(p, y, z) = 0. Note that in the above theorem, if the hypothesis that a point is a 2– metric extreme point is deleted, the theorem is false as is shown by the following example: Example 7.5.1. Let (X,
·, ·
) be a linear 2–normed space that is not strictly 2–convex. Then by Theorem 7.2.5, there exists a triple p, q, r in X with a non–unique center. That is, we have σ(p, q, r) = σ(p, q, m1 ) + σ(p, r, m1 ) + σ(q, r, m1 ) and σ(p, q, r) = σ(p, q, m2 ) + σ(p, r, m2 ) + σ(q, r, m2 ) where it may be assumed without loss of generality that m1 is an element of the algebraic triangle T (p, q, r) with m1 = m2 . Therefore, since X is

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a linear space, it follows that m2 is an algebraic ultra–extreme point of M = T (p, q, r) ∪ T (p, q, m2 ) ∪ T (p, r, m2 ) ∪ T (q, r, m2 ). However, since none of the values σ(p, q, m2 ), σ(p, r, m2 ) and σ(q, r, m2 ) are zero, m2 is not a 2–metric ultra–extreme point. This leads naturally to the following theorem: Theorem 7.5.4. In a strictly 2–convex linear 2–normed space (X,
·, ·
), a point of a set M is an algebraic ultra–extreme point of M if and only if it is a 2–metric ultra–extreme point of M . Proof. Since every 2–metric ultra–extreme point is an algebraic ultra– extreme point, it suffices to show that every algebraic ultra–extreme point p of M is a 2–metric ultra–extreme point. By Theorems 2.6.2 and 2.6.3, we know that in a strictly 2–convex linear 2–normed space, a point is an algebraic between point if and only if it is a 2–metric between point. Therefore, p is a 2–metric ultra–extreme point of M . Theorem 7.5.4 gives directly the following:

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Corollary 7.5.5. In a strictly 2–convex linear 2–normed space (X,
·, ·
), a point is an algebraic extreme point of a set M if and only if it is a 2–metric extreme point of M . Furthermore, since the equivalence of algebraic and 2–metric interior points is equivalent to strict convexity and strict 2–convexity, the equivalence of algebraic and 2–metric extreme points implies strict 2–convexity and thus we have shown the following: Theorem 7.5.6. In a linear 2–normed space (X,
·, ·
), the set of algebraic extreme points of a set M is identical to the set of 2–metric extreme points if and only if (X,
·, ·
) is strictly 2–convex. Note that the circle in the Euclidean plane illustrates the fact that there exists a set M in a strictly convex linear 2–normed space with an infinite number of both 2–metric and algebraic ultra–extreme points. Also the set of all points (x, y) in the Euclidean plane such that 1 ≤ x ≤ 2 is a closed set M which has no ultra–extreme points. Theorem 7.5.7. In a strictly 2–convex linear 2–normed space (X,
·, ·
), the set of algebraic ultra–extreme points of a set M is identical to the set of 2–metric ultra–extreme points of M .

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Proof. By Theorem 7.5.2, since every 2–metric ultra–extreme point is an algebraic ultra–extreme point, it suffices to show that every algebraic ultra–extreme point of a set M is a 2–metric extreme point. Furthermore, by Theorem 7.5.3, if p is an algebraic ultra–extreme point and a 2–metric extreme point of a set M , then p is a 2–metric ultra–extreme point. Therefore, suppose that p is an algebraic extreme point but neither a 2–metric ultra–extreme point nor a 2–metric extreme point. Then there exist three points x, y, z in M such that σ(x, y, z) = σ(x, y, p) + σ(x, z, p) + σ(y, z, p) but p is not in T (x, y, z). Thus p is a 2–metric between point of x, y, z but not an algebraic between point of x, y, z. Therefore, by Theorems 2.3.1 and 2.3.2, (X,
·, ·
) is not strictly 2–convex, which is a contradiction.

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Theorem 7.5.8. In a linear 2–normed space (X,
·, ·
), if the set of algebraic ultra–extreme points of a set M is identical to the set of 2–metric ultra–extreme points of M , then (X,
·, ·
) is strictly 2–convex. Proof. Suppose that (X,
·, ·
) is not strictly 2–convex. Then, by Theorem 7.2.5, there exist a point p and a triple x, y, z in X such that p is a 2–metric center of x, y, z but p is not the algebraic center of x, y, z. Then the set M = T (x, y, p) ∪ T (x, z, p) ∪ T (y, z, p) ∪ T (x, y, z) has p as an algebraic ultra–extreme point but not a 2–metric extreme point. This contradiction proves this theorem. By Theorems 7.4.7 and 7.4.8, we have the following: Corollary 7.5.9. In a linear 2–normed space (X,
·, ·
), the set of algebraic ultra–extreme points of a set M is identical to the set of 2–metric ultra–extreme points of M if and only if the set of algebraic extreme points of a set M is identical to the set of 2–metric extreme points of M . Next, in Theorems 7.5.10 and 7.5.11, we give some characterizations of strict 2–convexity in linear 2–normed spaces by using some results in the section 2.6 of Chapter II. In fact, a linear 2–normed space in which the properties from Theorems 2.6.1 and 2.6.2 hold is said to be strictly 2–convex. By using Theorem 2.4.7 and Corollary 2.6.5, we hane the following:

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Theorem 7.5.10. Every semi–2–normed space with weakly continuous semi–2–norm such that the properties from Theorem 2.6.3 are valid is a strictly 2–convex linear 2–normed space. Proof. Suppose the contrary. Then by Theorem 2.4.7 and Corollary 2.6.5 to Theorem 2.6.4, there exist a, b, c, d in X such that σ(a, b, c) > σ(a, b, d) + σ(a, d, c) + σ(d, b, c) and d is not an element of the plane determined by a, b, c. From the properties of a semi–2–norm, it follows the existence of a point e = a + (1 − )b,  ∈ (0, ∞), such that σ(a, b, c) < σ(a, b, e) + σ(a, e, c) + σ(e, b, c). Because of the weak continuity of the semi–2–norm, there exists a point f = φe + (1 − φ)d, φ ∈ (0, 1) such that σ(a, b, c) = σ(a, b, f ) + σ(a, f, c) + σ(f, b, c).

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Hence f is a σ–between point of a, b, c and therefore by property (1) of Theorem 2.6.3 must be an algebraic between point of a, b, c, which is contrary to the fact. Theorem 7.5.11. The following properties hold in every linear 2– normed space (X,
·, ·
): (1) If the norm
·
b is strictly convex for every b ∈ X, b = 0, then (X,
·, ·
) is strictly 2–convex. (2) If there exists a strictly convex norm
·
on BX , so that for arbitrary a, b, in X,
a, b
=
b(a × b)
), then (X,
·, ·
) is strictly convex. Proof. Theorems 6.1.1 and 6.1.2, as well as, Theorem 7.1.3.

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CHAPTER 8. UNIFORM CONVEXITY In this chapter, we now consider the standard definition of uniform convexity for a normed linear space in light of the definition of strict convexity. By using the concept of uniform convexity in normed linear spaces, in [200], M.E. Newton constructed a parallel definition for uniform convexity in linear 2–normed spaces. We summarize some characterizations of uniform convexity in linear 2–normed space ontained by M.E. Newton [200] and C.S. Lin [166]. 8.1. Elementary Characterizations

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In 1936, the class of uniformly convex spaces was initially introduced by J.A. Clarkson [34]. Since 1936, this class and related classes of Banach spaces, etc., have been developed extensively by many mathematicians, B.J. Pettis [193], V. Klee [140]–[143], A. Lovaglia [162], V. Smulian [219]–[222], J. Lindenstrauss, M. Kadets, D.F. Cudia, M.M. Day, F. Sullivan [225], [226], K. Sundaresan [227]–[229], M.A. Smith [216]–[218], J.R. Giles [110], [111] and R.C. James [129]–[131], etc. First, we introduce the concept of uniform convexity in a normed linear space: Definition 8.1.1. A normed linear space (X,
·
) is said to be uniformly ∞ convex if for any sequences {xn }∞ n=1 and {yn }n=1 in X,
xn
≤ 1,
yn
≤ 1 1 and limn→∞
2 (xn + yn )
= 1 imply that limn→∞
xn − yn
= 0. A number of characterizations and properties of uniform convexity in normed linear spaces may be found in the books edited by B. Beauzamy [11], J. Diestel [48] and V.I. Istratescu [129]. The natural generalization of Definition 8.1.1 to linear 2–normed spaces is as follows: Definition 8.1.2. A linear 2–normed space (X,
·, ·
) is said to ∞ be uniformly convex if for any sequences {xn }∞ n=1 and {yn }n=1 in X, 1
xn , c

≤ 1,
yn , c
≤ 1, n = 1, 2, 3, · · · , limn→∞
2 (xn + yn ), c
= 1 and V (c) ∩ ∩∞ n=1 V (xn , yn ) = {0} imply that limn→∞
xn − yn , c
= 0. It is relatively simple to find examples of spaces which are either both strictly convex and uniformly convex, or neither. Typeset by AMS-TEX Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

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Example 8.1.1. Let X = R × R × R with 2–norm defined as follows: For x = (a1 , b1 , c1 ) and y = (a2 , b2 , c2 ) in X, let
x, y
=

 (b1 c2 − b2 c1 )2 + (a1 c2 − a2 c1 )2 + (a1 b2 − a2 b1 )2 .

Then the space (X,
·, ·
) is both strictly and uniformly convex linear 2– normed space. The proof of this will be evident after Section 8.3. Example 8.1.2. Let X = R × R × R with 2–norm defined as follows: For x = (a1 , b1 , c1 ) and y = (a2 , b2 , c2 ) in X, let
x, y
= |b1 c2 − b2 c1 | + |a1 c2 − a2 c1 | + |a1 b2 − a2 b1 |. Let vector addition and scalar multiplication be defined componentwise. We first verify that this is indeed a 2–norm and then show that (X,
·, ·
) is neither strictly convex nor uniformly convex. The 2–norm properties are satisfied: (1) if y = αx for some real number α, then we have

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x, y
=
x, αx
= |(b1 (αc1 ) − (αb1 )c1 )| + |(a1 (αc1 ) − (αa1 )c1 )| + |(a1 (αb1 ) − (αa1 )b1 )| = |α(b1 c1 − b1 c1 )| + |α(a1 c1 − a1 c1 )| + |α(a1 b1 − a1 b1 )| = 0. Conversely, if
x, y
= 0, we need to show that x and y are linearly dependent. If
x, y
= 0, then |b1 c2 −b2 c1 | +|a1 c2 −a2 c1 | +|a1 b2 −a2 b1 | = 0. Since absolute values are non-negative, then b1 c2 − b2 c1 = 0, a1 c2 − a2 c1 = 0 and a1 b2 − a2 b1 = 0. If c2 = 0, then a1 = (c1 /c2 )a2 , b1 = (c1 /c2 )b2 and c1 = (c1 /c2 )c2 and thus x = αy, where α = c1 /c2 . If c2 = 0, then algebraic manipulations yield that either x = (a1 , b1 , 0) and y = (0, 0, 0) or else x = ((b1 /b2 )a2 , (b1 /b2 )b2 , 0) and y = (a2 , b2 , 0). In either case, x and y are linearly dependent. (2)
x, y
=
y, x
, which is obvious. (3) for λ a real number,
x, λy
= |λ|
x, y
.

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In fact, we have
x, λy
= |b1 (λc2 ) − (λb2 )c1 | + |a1 (λc2 ) − (λa2 )c1 | + |a1 (λb2 ) − (λa2 )b1 | = |λ(b1 c2 − b2 c1 )| + |λ(a1 c2 − a2 c1 )| + |λ(a1 b2 − a2 b1 )| = |λ|(|b1 c2 − b2 c1 | + |a1 c2 − a2 c1 | + |a1 b2 − a2 b1 |) = |λ|
x, y
. (4) The fourth condition follows from the triangle inequality for absolute values. To see this, let x = (a1 , b1 , c1 ), y = (a2 , b2 , c2 ) and z = (a3 , b3 , c3 ) in X. Then we have
x, y + z
= |b1 (c2 + c3 ) − (b2 + b3 )c1 | + |a1 (c2 + c3 ) − (a2 + a3 )c1 | + |a1 (b2 + b3 ) − (a2 + a3 )b1 | = |(b1 c2 − b2 c1 ) + (b1 c3 − b3 c1 )| + |(a1 c2 − a2 c1 ) + (a1 c3 − a3 c1 )| + |(a1 b2 − a2 b1 ) + (a1 b3 − a3 b1 )| ≤ (|b1 c2 − b2 c1 | + |a1 c2 − a2 c1 | + |a1 b2 − a2 b1 |) + (|b1 c3 − b3 c1 | + |a1 c3 − a3 c1 | + |a1 b3 − a3 b1 |)

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=
x, z
+
y, z
. Next, we shall show that (X,
·, ·
) is not strictly convex. To see this, let x = (1, 0, 0), y = (0, 1, 0) and z = (0, 0, 1) in X. Then we have
x, z
=
(1, 0, 0), (0, 0, 1)
= 1,
y, z
=
(0, 1, 0), (0, 0, 1)
= 1 and V (z) ∩ V (x, y) = {(0, 0, 0)}. But clearly, we have x = y and so (X,
·, ·
) is not strictly convex. Finally, we show that (X,
·, ·
) is not uniformly convex. In fact, let ∞ ∞   1 1 ∞ , 0, 0) , 0) = (1 − , {y } = (0, 1 − {xn }∞ n n=1 n=1 n n n=1 n=1 be sequences in X and let c = (0, 0, 1). Then, we have, for n = 1, 2, · · · ,   1 1  
xn , c
= (1 − , 0, 0), (0, 0, 1) = 1 − ≤ 1 n n

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R. W. FREESE AND Y. J. CHO

and

  1 1  
yn , c
= (0, 1 − , 0), (0, 0, 1) = 1 − ≤ 1. n n Further, we have  1 1   lim
xn + yn , c
lim  (xn + yn ), c = n→∞ 2 2 n→∞   1 1 1   = lim (1 − , 1 − , 0), (0, 0, 1) n→∞ 2 n n  1 1 = lim 2 − = 1. 2 n→∞ n 
Clearly, V (c) ∩ ∩∞ n=1 V (xn , yn ) = {(0, 0, 0)}. Thus, the hypotheses for uniform convexity are satisfied. But the conclusion fails to hold:   1 1   lim
xn − yn , c
= lim (1 − , −1 + , 0), (0, 0, 1) n→∞ n→∞ n n  2 = lim 2 − = 2 = 0. n→∞ n

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Thus, the space (X,
·, ·
) is not uniformly convex. We have thus far produced examples of spaces which are both strictly and uniformly convex, or neither. We will also set forth a proof that uniform convexity of a linear 2–normed space implies its strict convexity. The converse is not true. Theorem 8.1.1. If (X,
·, ·
) is a uniformly convex linear 2–normed space, then it is also strictly convex. Proof. Let x, y, z ∈ X such that
x, z
=
y, z
= 1,
x + y, z
= ∞
x, z
+
y, z
and z ∈ / V (x, y). Let {xn }∞ n=1 and {yn }n=1 be sequences in X defined as follows: xn = x and yn = y for n = 1, 2, · · · . Then we have certainly
xn , z
≤ 1 and
yn , z
≤ 1 for n = 1, 2, · · · . Also, we have  1 1      lim  (xn + yn ), z  = lim  (x + y), z  n→∞ 2 n→∞ 2 1 lim
x + y, z
= 2 n→∞ 1 = lim (
x, z
+
y, z
) = 1. 2 n→∞

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UNIFORM CONVEXITY

195

Moreover, it follows / V (x,  y) implies V (z) ∩ V (x, y) = {0} and also
∞ that z ∈ we have V (z) ∩ ∩n=1 V (xn , yn ) = {0}. Thus, since (X,
·, ·
) is uniformly convex, limn→∞
xn − yn , z
= 0. Therefore, by Theorem 3.1.2 and the definition of the sequences, we have
x − y, z
= 0 and so x − y and z are linearly dependent. Thus, either z = α(x − y) for some real number α or else x − y = 0. If z = α(x − y), there are two possibilities: First, if α = 0, then z = 0, which contradicts the assumption that
x, z
= 1. Secondly, if α = 0 and x−y = 0, then z = α(x−y) = 0, which contradicts the hypothesis that z ∈ / V (x, y). Thus, x − y = 0, or equivalently, x = y. Hence, (X,
·, ·
) is strictly convex. To show that the converse is false, we need the following, which is from G. Kˆ othe [147], [148]. Theorem 8.1.2. Let (E,
·
1 ) be a normed linear space. Also let A be a continuous, one–to–one linear mapping from (E,
·
1 ) onto a strictly convex normed space (F,
·
F ). Then (E,
·
2 ) is a strictly convex normed space, where
x
2 =
x
1 +
Ax
F for all x ∈ E.

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Theorem 6.4.5 brings us to the counterexample of the converse of Theorem 8.1.1. Example 8.1.3. Let X = {q(x) : q is a polynomial on [0, 1]}. Define polynomial addition and scalar multiplication in the usual way. Further, define W (p, q) = |pq  − qp |, where polynomial multiplication is defined in the usual way. Let

1 |W (p, q)|2 dx
p, q
= sup |W (p, q)| + 0≤x≤1

0

for all p, q in X. Then (X,
·, ·
) is a strictly convex, but not uniformly convex linear 2–normed space. That X is a linear space is well–known. We will divide the rest of the discussion into several parts as follows: 1. We look briefly at known properties of the normed linear space X 2 and its subspace. 2. We define a space (E,
·
 ) and use Theorem 8.1.2 to show it is strictly convex.

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3. We show that (X,
·, ·
) is a linear 2–normed space. 4. We use the space (E,
·
 ) and Theorem 6.4.5 to show that (X,
·, ·
) is strictly convex. 5. We show that (X,
·, ·
) is not uniformly convex. Now, let us prove these steps: 1. (X 2 ,
·
) is defined as follows: Let X 2 = {f (x) : f is continuous function on [0,1]} with vector addition and scalar multiplication being function addition and  1 |f (x)|2 dx. Then (X 2 ,
·
) is a strictly scalar multiplication. Let
f
= 0 convex normed linear space. Further, let

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F = {q(x) : q is a polynomial on [0,1]} and define
q
F =
q
. Then (F,
·
F ) is a linear subspace of X 2 with the same norm. It is also strictly convex. 2. Let (E,
·
E ) be the normed linear space defined by E = F and
q
E = sup0≤x≤1 |q(x)|. Let A : E → F be the identity mapping. Then by Theorem 8.1.2, if
q
 =
q
E +
Aq
F , then (E,
·
 ) is strictly convex. Note that

1  |q(x)|2 dx.
q
= sup |q(x)| + 0≤x≤1

0

3. Recalling our original definition of (X,
·, ·
), we need to show that it is in fact a linear 2–normed space. To see this, consider the following: (1) Since p, q are polynomials, they are analytic functions. G. H. Meisters stated that W (p, q) = 0 identically on [0,1] if and only if p and q are linearly dependent on [0,1]. Thus,
p, q
= 0 if and only if p and q are linearly dependent. (2) It is clear that
p, q
=
q, p
. (3) Note that for any real number λ, W (p, λq) = |p(λq) − (λq)p | = |λ(pq  − qp )| = |λ||pq  − qp | = |λ|W (p, q). Therefore, we have
p, λq
= |λ|
p, q
.

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UNIFORM CONVEXITY

197

(4) For p, q, r ∈ X,
p + r, q
=
W (p + r, q)
E +
W (p + r, q)
F . Now, we have W (p + r, q) = |(p + r)q  − (p + r) q| = |(pq  − p q) + (rq  − qr  )| ≤ |pq  − qp | + |rq  − qr  | = W (p, q) + W (r, q), which, together with the norm properties of
·
E and
·
F , yields that
p + r, q
=
W (p + r), q)
E +
W (p + r), q)
F ≤
W (p, q)
E +
W (r, q)
E +
W (p, q)
F +
W (r, q)
F = (
W (p, q)
E +
W (p, q)
F ) + (
W (r, q)
E +
W (r, q)
F ) =
p, q
+
r, q
. 4. Let T (p, q) = pq  − qp for all p, q ∈ X. Then T is the mapping from X × X into (E,
·
) required to apply Theorem 6.4.5. It is easy to see that T satisfies the four conditions specified. In fact if a, b, c, d ∈ X, then we have for any real numbers α and β, T (a + c, b + d) = (a + c)(b + d ) − (b + d)(a + c ) Copyright © 2001. Nova Science Publishers, Incorporated. All rights reserved.

(1)

= (ab − ba ) + (ad − da ) + (cb − bc ) + (cd − dc ) = T (a, b) + T (a, d) + T (c, b) + T (c, d),

(2)

T (αa, βb) = (αa)(βb) − (βb)(αa) = αβab − αβba = αβ(ab − ba ) = αβT (a, b),

(3) T (a, b) = 0 if and only if a, b are linearly dependent, as in part 3 (1) of this example, (4)
a, b
=
T (a, b)
by the definition of T and the 2–norm on X. Thus, applying Theorem 6.4.5 yields that (X,
·, ·
) is strictly convex. ∞ 5. To see that (X,
·, ·
) is not uniformly convex, let {pn }∞ n=1 and {qn }n=1 be sequences in X defined by pn (x) =

1 x, 2

qn (x) =

1 xn+1  x− 2 n+1

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for n = 1, 2, · · · . Let c(x) = −1. Now, we have, for n = 1, 2, · · · ,  1    
pn , c
=
W (pn , c)
=   = 1. 2 Similarly, we have for n = 1, 2, · · · , 1    n 
qn , c
=
W (qn , c)
=  1 − x  2



1 1 sup |1 − xn | + = |1 − xn |2 dx 2 0≤x≤1 0 

  2 1 1  = 1+ 1− − 2 n + 1 2n + 1 1 (1 + 1) = 1. 2 
Certainly, we have V (c) ∩ ∩∞ n=1 V (pn , qn ) = {0} and  1 1 xn  1       (pn + qn ), c =
W (pn + qn , c)
 = 1 − 2 2 2 2



1  n n 2 x x 1     sup 1 − = +  dx 1 − 2 0≤x≤1 2 2 0 

  1 1 1 − . = 1+ 1− 2 n + 1 4(2n + 1)

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≤

Letting n → ∞, we have



  1  1 1 1 lim
(pn + qn ), c
= lim − 1+ 1− n→∞ 2 n→∞ 2 n + 1 4(2n + 1) =

1 (1 + 1) = 1. 2

However, we have xn
pn − qn , c
=
W (pn − qn , c)
 =

 2

 xn  1  n 2   x  = sup  +   dx 2 0≤x≤1 2 0  1 1 1 . = + 2 2 2n + 1

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UNIFORM CONVEXITY

199

Again, letting n → ∞, we have

lim
pn − qn , c
= lim

n→∞

n→∞

=

1 1 + 2 2



1 2n + 1



1 1 +0= , 2 2

which means that limn→∞
pn − qn , c
= 0. Therefore, (X,
·, ·
) is not uniformly convex. 8.2. Uniform Convexity In Quotient Spaces

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In this section, we will give some characterizations of uniform convexity in a quotient spaces (Xc ,
·
c ). First, we look at several examples of quotient spaces. The results in this section are due to M.E. Newton [189] and C.S. Lin [157]. Example 8.2.1. Let X = R × R × R with the 2–norm as in example 8.1.1. Let c = (0, 0, 1). Then V (c) is the “z–axis”. The equivalence classes which are the elements of Xc are lines parallel to the z–axis. We can denote each equivalence class by its representative whose z– coordinate is zero. That is, if x, y, z are real numbers, (x, y, z) ∈ ((x, y, 0))c. Vector addition and scalar multiplication are defined as follows: (1) ((x1 , y1 , 0))c + ((x2 , y2 , 0))c = ((x1 + x2 , y1 + y2 , 0))c , (2) α((x, y, 0))c = ((αx, αy, 0))c for any real number α. Note that, as a vector space, Xc is isomorphic to the Euclidean plane. Let us compute the norm of a vector ((x, y, 0))c in Xc . By definition of
·
c , we have
((x, y, 0))c
c =
(x, y, 0), c
=
(x, y, 0), (0, 0, 1)
 = (y − 0)2 + (x − 0)2 + (0 − 0)2  = x2 + y 2 . This quantity is independent of the value of z and so we could just as easily have used (x, y, z), z = 0, as a representative of ((x, y, 0))c. The result would have been the same. But note that the usual norm of a vector (x, y) in the

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R. W. FREESE AND Y. J. CHO

Euclidean plane is also plane is defined by



x2 + y 2 . Thus, a mapping m : Xc →Euclidean

m[((x, y, 0))c] = (x, y). Example 8.2.2. Let X =  {p(x) : p is a polynomial on [0, 1]} and let 1 |W (p, q)|2 dx. This is the space of Exam
p, q
= sup0≤x≤1 |W (p, q)| + 0 ple 8.1.3. Let c = xk . Then the polynomials comprising an equivalence class can characterized as follows: p(x) = an xn + an−1 xn−1 + · · · + ak xk + · · · + a1 x + a0 and

q(x) = bn xn + bn−1 xn−1 + · · · + bk xk + · · · + b1 x + b0

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are in the same equivalence class if and only if an = bn , an−1 = bn−1 , · · · , ak+1 = bk+1 , · · · , a1 = b1 and a0 = b0 . Note that ak need not equal bk . The norm of a vector in Xc is not affected by the particular choice of representative for that vector (equivalence class). To illustrate this, consider the case where n = 3. That is, let c = x3 . We will find the norm of the vector (x4 + rx3 + x2 )c and show it does not depend upon the value of r, where r is a real number. In fact, we have W (x4 + rx3 + x2 , c) = W (x4 + rx3 + x2 , x3 ) = |(x4 + rx3 + x2 )(3x2 ) − x3 (4x3 + 3rx2 + 2x)| = |3x6 + 3rx5 + 3x4 − 4x6 − 3rx5 − 2x4 | = | − x6 + x4 |. So we have
(x4 + rx3 + x2 )c
c =
x4 + rx3 + x2 , c
6

4

= sup | − x + x | + 0≤x≤1

=

4 + 27





0

1

| − x6 + x4 |2 dx.

8 . 1287

In any case, the terms involving the real number r vanish when W is computed. Thus we may vary the value of r, without affecting the norm.

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UNIFORM CONVEXITY

201

Our attempt to find a relationship between the uniform convexity of a linear 2–normed space (X,
·, ·
) and its quotient space will be facilitated by having alternative definitions of uniform convexity. G. Kˆ othe [150], [151] made such a definition for normed linear spaces: Definition 8.2.1. A normed linear space (X,
·
) is said to be uniformly convex if for any  ∈ (0, 2], there exists a δ() > 0 such that
x
≤ 1,
y
≤ 1 and
x − y
≥  imply that
12 (x + y)
≤ 1 − δ(). The number x + y      δ() = inf 1 −   :
x
≤ 1,
y
≤ 1,
x − y
≥  2 is called the modulus of convexity of (X,
·
). It is clear that (X,
·
) is uniformly convex if and only if, for every  > 0, δ() > 0. It is also clear that if 1 < 2 , δ(1 ) < δ(2 ), and that δ(0) = 0. The most immediate example of a uniformly convex space is the Hilbert space. In fact, the Parallelogram law

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 x − y 2  x + y 2 1       = (
x
2 +
y
2 ) −   2 2 2 shows that, if
x
≤ 1,
y
≤ 1 and
x − y
≥ , then
12 (x + y)
2 ≤  1 − 14 2 and hence δ() = 1 − 1 − 14 2 . He further has shown that Definition 8.1.1 implies Definition 8.2.1 and conversely. We will refer to these definitions as being “equivalent”. Analogously, we make the following definition in a linear 2–normed space : Definition 8.2.2. A linear 2–normed space (X,
·, ·
) is said to be uniformly convex if for every  in (0, 2] and c = 0 in X, there exists a δ(, c) > 0 such that
x, c
≤ 1,
y, c
≤ 1 and
x − y, c
≥  imply that  1   (x + y), c  ≤ 1 − δ(, c).  2 To prove the “equivalence” of Definitions 8.1.2 and 8.2.2, we first need the following:

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R. W. FREESE AND Y. J. CHO

Lemma 8.2.1. Let (X,
·, ·
) be a linear 2–normed space. And let ∞ {xn }∞ n=1 and {yn }n=1 be sequences in X such that
xn , c
≤ 1,
yn , c
≤ 1 for all n = 1, 2, · · · and limn→∞
21 (xn + yn ), c
= 1 for some c = 0 in X. Then limn→∞
xn , c
= 1 and limn→∞
yn , c
= 1. Proof. For all n = 1, 2, · · · , we have
xn + yn , c
≤
xn , c
+
yn , c
≤ 1 + 1 = 2. On the other hand, limn→∞
xn + yn , c
= 2. Thus, we have 2 = lim
xn + yn , c
≤ lim (
xn , c
+
yn , c
) ≤ 2, n→∞

n→∞

which implies that lim (
xn , c
+
yn , c
) = 2

n→∞

or lim
xn , c
+ lim
yn , c
= 2.

n→∞

n→∞

Since both
xn , c
≤ 1 and
yn , c
≤ 1 for all n = 1, 2, · · · , it follows that limn→∞
xn , c
= 1 and limn→∞
yn , c
= 1.

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This lemma enables us to prove the following theorem: Theorem 8.2.2. If a linear 2–normed space (X,
·, ·
) is uniformly convex in the sense of Definition 8.1.2, then it is uniformly convex in the sense of Definition 8.2.2, and conversely. Proof. We first show that if (X,
·, ·
) is uniformly convex in the sense of Definition 8.1.2, then it satisfies Definition 8.2.2. Assume that (X,
·, ·
) satisfies Definition 8.1.2, but not Definition 8.2.2. Thus, there exist an ∞  > 0, c(c = 0) in X and sequences {xn }∞ n=1 , {yn }n=1 in X such that for all n = 1, 2, · · · , (1)
xn , c
≤ 1 and
yn , c
≤ 1, (2)
xn − yn , c
≥  but (3)
12 (xn + yn ), c
> 1 − n1 . Then we have, for all n = 1, 2, · · · ,  1 1 1   1 − <  (xn + yn ), c ≤ (
xn , c
+
yn , c
) ≤ 1. n 2 2

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UNIFORM CONVEXITY

203

Letting n → ∞, we have  1   1 ≤ lim  (xn + yn ), c ≤ 1, n→∞ 2 which means that limn→∞
21 (xn + yn ), c
= 1. Further, since
xn − yn , c
≥  > 0, it follows that lim
n→∞
xn − yn ,c
= 0. Thus, by Definition 8.1.2, it must be that V (c) ∩ ∩∞ n=1 V (xn , yn ) = {0}. Therefore, for every n = 1, 2, · · · , there are real numbers αn and βn , not both zero, such that c = αn xn + βn yn . Then we have
xn − yn , c
=
xn − yn , αn xn + βn yn
=
xn − yn , αn (xn − yn ) + (αn + βn )yn
=
xn − yn , (αn + βn )yn
=
xn , (αn + βn )yn
= |αn + βn |
xn , yn
for n = 1, 2, · · · . Thus, for all n = 1, 2, · · · , we have |αn + βn |
xn , yn
=
xn − yn , c
≥  > 0.

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By Theorem 2.1.3, we have
xn , c
=
xn , αn xn + βn yn
= |βn |
xn , yn
and
yn , c
=
yn , αn xn + βn yn
= |αn |
xn , yn
. Since
xn , c
≤ 1,
yn , c
≤ 1 and limn→∞
21 (xn + yn ), c
= 1, it follows from Lemma 8.2.1 that limn→∞
xn , c
= 1 and limn→∞
yn , c
= 1. So we have lim |βn |
xn , yn
= 1. lim |αn |
xn , yn
= 1, n→∞

n→∞

Now,
xn , yn
> 0 for n = 1, 2, · · · . For if not, for each n, there exists a real number λn such that yn = λn xn . In that case,
xn − yn , c
=
(1 − λn )xn , αn xn + βn (λn xn )
=
(1 − λn )xn , (αn + βn λn )xn
= 0,

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R. W. FREESE AND Y. J. CHO

which contradicts
xn − yn , c
≥  > 0 for n = 1, 2, · · · . Therefore, since
xn , yn
> 0 for all n, it follows that α  1 |αn |
xn , yn
 n lim   = lim = = 1. n→∞ βn n→∞ |βn |
xn , yn
1 Also, we have
xn + yn , c
=
xn + yn , αn xn + βn yn
=
xn + yn , αn (xn + yn ) + (βn − αn )yn
=
xn + yn , (βn − αn )yn
, which means that limn→∞
xn +yn , c
= 2 implies limn→∞ |αn −βn |
xn , yn
= 2 also. Since
xn , yn
> 0 for all n, we have  αn  2 |βn − αn |
xn , yn
 lim 1 − = = 2.  = lim n→∞ n→∞ βn |βn |
xn , yn
1 Now since limn→∞ |1 − (αn /βn )| = 2 and limn→∞ |αn |/|βn | = 1, it follows that limn→∞ αn /βn = −1. Thus we have lim
xn − yn , c
= lim |αn + βn |
xn , yn
n→∞  βn   = lim 1 + |αn |
xn , yn
= 0, n→∞ αn

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n→∞

which contradicts the fact that
xn − yn , c
≥  > 0 for n = 1, 2, · · · . Thus, (X,
·, ·
) must be uniformly convex in the sense of Definition 8.2.2. Conversely, let us assume that (X,
·, ·
) is uniformly convex in the sense of Definition 8.2.2. We want to show that it satisfies Definition 8.1.2. If (X,
·, ·
) is not uniformly convex in the sense of Definition 8.1.2, then there ∞ exist sequences {xn }∞ n=1 , {yn }n=1 in X and a non–zero c in X such that (1)
xn , c
≤ 1 and
yn , c
≤ 1 for n = 1, 2, · · · , (2) limn→∞

12 (xn + yn ), c
 = 1, (3) V (c) ∩ ∩∞ n=1 V (xn , yn ) = {0}, (4) limn→∞
xn − yn , c
= 0. By (4), there exists an o > 0( o ≤ 2) such that, for any positive integer No , there is an N ≥ No with
xN − yN , c
≥ o . Also, by (2), for every δ > 0, there is some positive integer N1 such that n ≥ N1 implies  1    (xn + yn ), c > 1 − δ. 2

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UNIFORM CONVEXITY

205

Choose M to be any positive integer greater than N1 such that
xM − yM , c
≥ o , since this can be done for any positive integer. Thus, we have produced an o in (0, 2] and c = 0 in X such that, for any δ > 0, there are xM , yM in X with the following properties: (1)
xM , c
≤ 1 and
yM , c
≤ 1, (2)
xM − yM , c
≥ o , (3) 12
(xM + yM ), c
> 1 − δ, which contradicts our assumption that (X,
·, ·
) is uniformly convex in the sense of Definition 8.2.2. Hence (X,
·, ·
) must be uniformly convex in the sense of Definition 8.1.2. Next, we give some characterizations of uniform convexity in linear 2normed spaces obtained by C.S. Lin [166]. Theorem 8.2.3. For a linear 2-normed space (X,
· · ·
), the following four statements are equivalent: (1) (X,
· · ·
) is uniformly convex. (2) For any each  > 0, there corresponds δ() > 0 such that, if
x, z
=
y, z
, z ∈ / V (x, y) and
x − y, z
≥ 
x, z
, then

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1
x + y, z
≤ (1 − δ())
x, z
. 2 (3) For any  > 0, there corresponds δ  () > 0 such that, if
x − ay, z
= / V (x, y), then
x, z
, where a = x,z y,z and z ∈ 1
x + ay, z
≤ (1 − δ  ())
x, z
. 2 (4) For any  > 0, there corresponds δ  () > 0 such that, if
x − ay, z
≥ 
x, z
with the form (3) and z ∈ / V (x, y), then
x + y, z
≤
x, z
+
y, z
− δ  ()
x, z
or


x + y, z
≤ 3
x, z
−
y, z
− δ  ()
x, z
.

Proof. (1)⇒(2): Let
x, z
=
y, z
= c, c = 0 and z ∈ / V (x, y). Then, z z z by (1),
x, c
=
y, c
= 1 and
x − y, c
≥  imply that z 1   x + y,  ≤ 1 − δ(), 2 c

i.e.,

1
x + y, z
≤ (1 − δ())
x, z
. 2

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R. W. FREESE AND Y. J. CHO

(2)⇒(3):
x, z
=
ay, z
obviously. Replace y by ay in (2) to get (3). (3)⇒(4): Since 12
x + ay, z
≤ (1 − δ  ())
x, z
by (3), from
y, z
≥
x, z
, then it follows that
x + y, z
=
x + ay + y(1 − a), z
≤
x + ay, z
+ (1 − a)
y, z
≤ 2(1 − δ  ())
x, z
+
y, z
−
x, z
=
x, z
+
y, z
− 2δ  ()
x, z
. Let δ() = 2δ  (). Then the first inequality follows. In the case of
y, z
≤
x, z
, we have
x + y, z
≤
x + ay, z
+ (a − 1)
y, z
≤ 3
x, z
−
y, z
− 2δ  ()
x, z
.

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(4)⇒(1): If
x, z
=
y, z
= 1, then
x + y, z
≤ 2 − δ  () by (4). We may let δ() = 12 δ  (). Corollary 8.2.4. Let (X,
· · ·
) be a uniformly convex linear 2-normed  n  space. If xi ∈ X, i = 1, 2, · · · , n and y = x and
x , z
≤ y − k i=1 i  k  j=1 xj , z  for k = 1, 2, · · · , n − 1, then
y, z
≤

n  i=1


xi , z
−

n−1 

δ(k )
xk , z


k=1

for all z ∈ X with z ∈ / V (x1 , x2 , · · · , xn ), where   k     xk y − j=1 xj ,   , z − k= k  
xk , z
   y − j=1 xj , z   for k = 1, 2, · · · , n − 1 and δ is the function from Definition 8.2.1 with δ(0) = 0. Proof. Use the first inequality of (4) in Theorem 8.2.3 and induction.

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UNIFORM CONVEXITY

207

Corollary 8.2.5. Let (X,
· · ·
) be a uniformly convex linear 2-normed n space. If xi ∈ X, i = 1, 2, · · · , n, and y = i=1 xi , then
y, z
≤

n 

(1 − 2δ(i ))
xi , z


i=1

for all z ∈ X with z ∈ / V (x1 , x2 , · · · , xn ), where    xi  y  i =  
xi , z
−
y, z
, z  and δ is the function from Definition 8.2.1 with δ(0) = 0. Proof. Clearly, we have      xi   y      
xi , z
, z  = 
y, z
, z  = 1 for i = 1, 2, · · · , n. By the definition, there exists δ(i ) > 0 such that

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    xi y   
xi , z
+ |
y, z
, z  ≤ 2(1 − δ(i )) or
xi
y, z
+ y
xi , z
, z
≤ 2(1 − δ(i ))
xi , z

y, z
. Hence, it follows that   n   n      =  
y, z
y + y
x , z
, z (x
y, z
+ y
x , z
), z i i i     i=1

≤ ≤

i=1 n 

   xi
y, z
+ y
xi , z
, z 

i=1 n 

2(1 − δ(i ))
xi , z

y, z


i=1

or

n n    
xi , z
≤ 2(1 − δ(i ))
xi , z

y, z
.
y, z

y, z
+ i=1

i=1

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Thus, we have
y, z
≤

n 

(1 − 2δ(i ))
xi , z
.

i=1

In Theorem 8.2.2, we have shown that Definition 8.1.2 implies Definition 8.2.2, and conversely, it is a relatively simple matter to prove the next theorem. Theorem 8.2.4. Let (X,
·, ·
) be a linear 2–normed space. Then the following statements are equivalent: (1) (X,
·, ·
) is uniformly convex. (2) For every c = 0 in X, (Xc ,
·
c ) is uniformly convex. Proof. To see that condition (1) implies condition (2), let (X,
·, ·
) be uniformly convex. Let  in (0, 2] and c = 0 in X be given. Then by Definition 8.2.2, there exists a δ(, c) > 0 such that
x, c
≤ 1,
y, c
≤ 1 and
x − y, c
≥  imply that
12 (x + y), c
≤ 1 − δ. Let (a)c , (b)c be any elements of (Xc ,
·
c ) such that
(a)c
c ≤ 1,
(b)c
c ≤ 1 and
(a−b)c
c ≥ . Then we have
a, c
=
(a)c
c ≤ 1,


b, c
=
(b)c
c ≤ 1

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and
a − b, c
=
(a − b)c
c ≥ . Hence, for the δ previously found, we have  1    (a + b), c ≤ 1 − δ. 2 But then
12 (a + b)c
c ≤ 1 − δ, as desired. Thus, for arbitrary c (c = 0) in X, (Xc ,
·
c ) is uniformly convex. To see that the condition (2) implies the condition (1), we prove the contrapositive. Let us assume that (X,
·, ·
) is not uniformly convex. By ∞ Definition 8.1.2, there are sequences {xn }∞ n=1 , {yn }n=1 in X and an element c (c = 0) in X with the following properties: (1)
xn , c
≤ 1 and
yn , c
≤ 1 for n = 1, 2, · · · , (2) limn→∞

12 (xn + yn ), c
 = 1,  (3) V (c) ∩∞ n=1 V (xn , yn ) = {0}, (4) limn→∞
xn − yn , c
= 0.

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UNIFORM CONVEXITY

209

∞ Hence the sequences {(xn )c }∞ n=1 and {(yn )c }n=1 have
(xn )c
c =
xn , c
≤ 1 and
(yn )c
c =
yn , c
≤ 1. Further, we have

 1  1     lim  (xn + yn )c  = lim  (xn + yn ), c = 1. n→∞ 2 n→∞ 2 c But limn→∞
(xn − yn )c
c = lim→∞
xn − yn , c
= 0. Thus, by Definition 8.1.1, (Xc ,
·
c ) is not uniformly convex. 8.3. Uniform Convexity by 2–inner Products In this section, we show that every 2–inner product space is uniformly convex and an example is given. Theorem 8.3.1. If (X, (·, ·|·)) is a 2–inner product space, then  (X,
·, ·
) is a uniformly convex linear 2-normed space, where
x, y
= (x, x|y). Proof. That
·, ·
is a 2–norm is given in Theorem 5.1.10. Further, by Theorem 5.1.12 and the remarks which follow it, for every c = 0 in X, (Xc ,
·
c ) is an inner product space and its norm satisfies

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(a + b)c
2c +
(a − b)c
2c = 2(
(a)c
2c +
(b)c
2c ) for all a, b in X. Let  in (0, 2] be given and let x, y ∈ X such that
(x)c
c ≤ 1,
(y)c
c ≤ 1 and
(x − y)c
c ≥ . Thus, we have
(x + y)c
2c +
(x − y)c
2c ≤ 2(1 + 1) = 4. Also, since
(x − y)c
c ≥ , we have
(x + y)c
2c + 2 ≤
(x + y)c
2c +
(x − y)c
2c . Combining these inequalities yields
(x + y)c
2c ≤ 4 − 2 or
(x + y)c
2c + 2 ≤ 4. Thus, for some δ > 0, we have  2 1  1 2   = 1 − δ. 4 − 2 = 1 −  (x + y)c  ≤ 2 2 4 c

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R. W. FREESE AND Y. J. CHO

Hence, by Definition 8.2.1, (Xc ,
·
c ) is uniformly convex. Therefore, it follows from Theorem 8.2.2 that (X,
·, ·
) is also uniformly convex, as desired. We now will develop a 2-normed analogue of Hilbert space. Corollary 5.1.5 brings us to the following example: Example 8.3.1. Let H = {x : x = (x1 , x2 , · · · , xn , · · · ) and ∞ 2 xn is a real number for n = 1, 2, · · · . The usual n=1 |xn | < ∞}, where ∞ 2 norm on H is
·
H = n=1 |xn | . Vector addition and scalar multiplication are defined componentwise. G. K¨ othe [147], [148] has shown that (H,
·
H ) is a uniformly convex normed linear space. Further, (H, (·|·)) is an inner product space, where (x|y) = x1 y1 + x2 y2 + · · · + xn yn + · · · =

∞ 

xn y n .

n=1

 Note that (x|x) =
x
. Now by Corollary 5.1.5, we can define a 2–inner product on H as follows:

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(x, y|z) = (x, y)
z
2 − (x|z)(y|z) for any x, y, z in H. Hence, by Theorem 5.1.10, we can define a 2–norm for H. 8.4. Uniform Convexity by Bivectors In this section, by using the concept of bisectors, we give some characterizations of uniform convexity in linear 2–normed spaces. Theorem 8.4.1. Let (X,
·, ·
) be a linear 2–normed space whose dimension is less than or equal to three. Let (BX ,
·
) denote the space of bivectors over X with
b(a × b)
=
a, b
. Then the following statements are equivalent: (1) (BX ,
·
) is uniformly convex. (2) (X,
·, ·
) is uniformly convex. Proof. We first show that the condition (1) implies the condition (2). ∞ Let (BX ,
·
) be uniformly convex. Let {xn }∞ n=1 and {yn }n=1 be sequences in X and c = 0 be an element in X such that
xn , c
≤ 1 and
yn , c
≤ 1,

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UNIFORM CONVEXITY

n = 1, 2, · · · , limn→∞
12 (xn + yn ), c
= 1 and V (c) {0}. By the definition of
·
, we therefore have:
b(xn × c)
=
xn , c
≤ 1,

211




 ∩∞ n=1 V (xn , yn ) =


b(yn × c)
=
yn , c
≤ 1.

Further, we have   1 1     lim  [b(xn × c) + b(yn × c)] = lim  (b(xn + yn ) × c) n→∞ 2 n→∞ 2 1    = lim  (xn + yn ), c = 1. n→∞ 2 Since BX is uniformly convex, it follows that lim
b(xn × c) − b(yn × c)
= 0.

n→∞

Hence, by the definition of
·
, we have lim
xn − yn , c
= lim
b((xn − yn ) × c)


n→∞

n→∞

= lim
b(xn × c) − b(yn × c)
n→∞

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= 0. Conversely, we next show that the condition (2) implies condition (1). Let (X,
·, ·
) be uniformly convex. Let  ∈ (0, 2] be given. Let b1 , b2 be  , bivectors such that
b1
≤ 1,
b2
≤ 1 and
b1 − b2
≥ . Since BX = BX it follows, from the section 2.3 of Chapter II, that there are a, b, c in X such that b1 = b(a × c) and b2 = b(b × c). Therefore, we have
a, c
=
b(a × c)
=
b1
≤ 1,
b, c
=
b(b × c)
=
b2
≤ 1,
a − b, c
=
b((a − b) × c)
=
b(a × c) − b(b × c)
=
b1 − b2
≥ . Hence, by Definition 8.2.2, the uniform convexity of (X,
·, ·
) implies that there exists a δ(, c) > 0 such that
21 (a + b), c
≤ 1 − δ. But then we have, for the same value of δ,   1 1     (b [b(a × c) + b(b × c)] + b ) +    1 2  2 2  1  1     =  [b(a + b) × c) =  (a + b), c ≤ 1 − δ. 2 2

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R. W. FREESE AND Y. J. CHO

Therefore, by Definition 8.2.1, (BX ,
·
) is uniformly convex. From Theorem 8.4.1, we have also two corollaries: Corollary 8.4.2. Every 2–dimensional linear 2–normed space (X,
·, ·
) is uniformly convex. Proof. In Chapter II, it is stated that, when X is two–dimensional, then the dimension of BX is one. Every one–dimensional normed space is trivially uniformly convex. Hence, by Theorem 8.4.1, (X,
·, ·
) is also uniformly convex. Corollary 8.4.3. Let (X,
·, ·
) be a linear 2–normed space whose dimension is less than or equal to three. Then the uniform convexity of any one of (X,
·, ·
), (BX ,
·
), and (Xc ,
·
c ) for all non–zero c in X implies the uniform convexity of the two others. Proof. By Theorem 8.2.2 and Theorem 8.4.1, we can prove this corollary.

8.5. Uniform 2–Convexity in Linear 2–Normed Spaces

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In [166], C.S. Lin introduced the concept of uniform 2-convexity in linear 2-normed spaces and gave some relations between uniform 2-convexity and strict 2-convexity in linear 2-normed spaces. Definition 8.5.1. A linear 2-normed space (X,
·, ·
) is said to be uniformly 2-convex if, for any  ∈ (0, 2], there corresponds δ() > 0 such that, if
x, y
=
y, z
=
z, x
= 1 and
x + y, z
≥ , then 1
x + z, y + z
≤ 1 − δ(). 3 Theorem 8.5.1. For a linear 2-normed space (X,
·, ·
) the following six statements are equivalent: (1) (X,
·, ·
) is uniformly 2-convex. (2) For any  ∈ (0, 2], there corresponds δ() > 0 such that, if
x, y
=
y, z
=
z, x
= 0 and
x + y, z
≥ 
x, z
, then 1
x + y, y + z
≤ (1 − δ())
x, z
. 3

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UNIFORM CONVEXITY

213

(3) For any  ∈ (0, 2], there corresponds δ() > 0 such that, if
x, y
=
y, z
= 0 and
x + ay, z
≥
x, z
, where a = x,z y,z , then 1
x + z, ay + z
≤ (1 − δ())
x, z
. 3 (4) For any  ∈ (0, 2], there corresponds δ() > 0 such that, if
y, z

x, z
= 0,
x, y
=
y, z
and
x + ay, z
≥ 
x, z
, where a = x,z y,z , then
x + y, y + z
≤
x, z
+ 2
y, z
− δ()
x, z
or
x + y, y + z
≤ 5
x, z
− 2
y, z
− δ()
x, z
. (5) For any  ∈ (0, 2], there corresponds δ() > 0 such that, if
x, y

y, z

z, x
= 0 where b =

y,z x,y

and c =

x,z x,y ,

and
bx + cy, z
≥ 
bx, z
,

then

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1
bx + z, cy + z
≤ (1 − δ())
bx, z
. 3 (6) For any  ∈ (0, 2], there corresponds δ() > 0 such that, if
x, y

y, z

z, x
= 0 where b =

y,z x,y

and c =

x,z , x,y

and
bx + cy, z
≥ 
bx, z
,

then


x + z, y + z
≤
x, y
+
y, z
+
z, x
− δ()
bx, z
or
x + z, y + z
≤
x, y
(8bc + 1) − 3(
y, z
+
z, x
) − δ()
bx, z
or
x + z, y + z
≤
x, y
(2bc − 1) + 3
x, z
−
y, z
− δ()
bx, z


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R. W. FREESE AND Y. J. CHO

or
x + z, y + z
≤
x, y
(2bc − 1) + 3
y, z
−
x, z
− δ()
bx, z
. Proof. (1)⇒(2): Let
x, z
= d2 for some x y  y z  z x        ,  =  ,  =  ,  = 1, d d d d d d

d = 0. Then we have, by (2), x y z     + ,  ≥ . d d d

Hence, it follows from (1) that  1 x z y z  + , +   ≤ 1 − δ(), 3 d d d d i.e., 1
x + z, y + z
≤ (1 − δ())
x, z
. 3 (2)⇒(3):
x, ay
=
ay, z
=
x, z
= 0 obviously. Replace y by ay in (2) to get(3). (3)⇒(4): Assume that
y, z
≥
x, z
. Then, by (3), we have
x + z, y + z
=
x, z + ay + z + y(1 − a)
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≤
x + z, ay + z
+
x + z, y
(1 − a) ≤ 3(1 − δ  ())
x, z
+ 2(1 − a)
y, z
=
x, z
+ 2
y, z
− 3δ  ()
x, z
. Let δ() = 3δ  (). Then the first inequality follows. If
y, z
≤
x, z
, then we have
x + z, +z
≤ 3(1 − δ  ())
x, z
+ 2(a − 1)
y, z
x, z
= 5
x, z
− 2
y, z
− 3δ  ()
x, z
. (4)⇒(1): If
x, y
=
y, x
=
z, x
= 1, then
x + z, y + z
≤ 3 − δ  () by (4). We may let δ() = 13 δ  (). (3)⇒(5): Since
b, y
=
y, z
= 0 and
bx + cy, cy + z
≥
bx, z
, where x,z = bx,z c = x,y y,z , we have, by (3), 1
bx + z, cy + z
≤ (1 − δ())
bx, z
. 3

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UNIFORM CONVEXITY

215

(5)⇒(6): We must consider the following four cases: Case 1. If
x, y
≥
x, z
and
x, y
≥
y, z
, then
x + z, y + z
=
bx + z − x(b − 1), cy + z + y(1 − c)
≤
bx + z + cy + z
+
bx + z + y
(1 − c) +
x + +cy + z
(1 − b) + (1 − b)(1 − c)
x, y
≤ 3(1 − δ  ())
bx, z
+ 2(1 − c)
y, z
+ 2(1 − b)
x, z
+
x, y
−
y, z
−
x, z
+
bx, z
=
x, y
+
y, z
+
z, x
− 3δ  ()
bx, z
. Let δ() = 3δ  (), so that the first inequality follows. Case 2. If
x, y
≤
x, z
and
x, y
≤
y, z
, then we have
x + z, y + z
≤ 3(1 − δ  ())
bx, z
+ 2(c − 1)
y, z
+ 2(b − 1)
x, z
+
x, y
−
y, z
−
x, z
+
bx, z
=
x, y
(8bc + 1) − 3(
y, z
+
z, x
) − 3δ  ()
bx, z
.

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Case 3. If
x, z
≥
x, y
≥
y, z
, then we have
x + z, y + z
≤ 3(1 − δ  ())
bx, z
+ 2(c − 1)
y, z
+ 2(1 − b)
x, z
−
x, y
+
y, z
+
x, z
−
bx, z
=
x, y
(2bc − 1) + 3
x, z
−
y, z
− 3δ  ()
bx, z
. Case 4. If
y, z
≥
x, y
≥
x, z
, then we should obtain the result directly by interchanging x and y in Case 3. Of course, b and c are interchanged, too. (6)⇒(1): If
x, y
=
y, z
=
z, x
, then b = c = 1 in (6). Thus, we have, by (6),
x + z, y + z
≤ 3 − δ  (). Let δ() = 13 δ  (). The proof is now completed. Theorem 8.5.2. strictly convex.

Every uniformly convex linear 2-normed space is

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216

R. W. FREESE AND Y. J. CHO

Proof. Let 12
x + y, z
=
x, z
=
y, z
= 1 and z ∈ / V (x, y). If  =
x − y, z
> 0, there exists δ() > 0 such that 2 =
x + y, z
≤ 2 − 2δ(), as the space is uniformly convex. This is impossible unless δ() = 0. It follows that  = 0. Therefore, since z ∈ / V (x, y), we have x = y. Remark 8.5.1. We include here an alternative proof of Theorem 8.5.2. From Corollary 8.2.4, we see that, if x, y ∈ X, then
x + y, z
≤
x, z
+
y, z
− δ()
x, z
y x for al z ∈ X with z ∈ / V (x, y), where  =
x,z − y,z , z
. If
x + y, z
= y x
x, z
+
y, z
, then δ() = 0 and so  = 0. Hence, x,z = y,z . By Definition 6.1.2, X is strictly convex.

Theorem 8.5.3. Every uniformly 2-convex 2-normed space is strictly 2-convex. Proof. Let 13
x + z, y + z
=
x, y
=
y, z
=
z, x
= 1 and also let
x + y, z
= . Then, there exists δ() > 0 such that

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3 =
x + z, y + z
≤ 3 − 3δ(). This is impossible unless δ() = 0. It follws that  = 0, i.e., z = a(x + y) for some real a and so 1 =
x, z
=
x, a(x + y)
= |a|
x, y
= |a|. a = −1 leads to a contradiction as 1 = 13
x + z, y + z
= 13
− y, −x
= 13 . Therefore a = 1 implies z = x + y. We mention that a different proof of the theorem is possible by using (6) in Theorem 8.5.1 and Theorem 5.1.12. The arguments are similar to that of the alternative proof of Theorem 8.5.2. Theorem 8.5.4. Every uniformly convex linear 2-normed space is uniformly 2-convex. Proof. For any  ∈ (0, 2], there exists δ  () > 0 such that, if
x, y
=
y, z
=
z, x
= 1, z ∈ / V (x, y) and
x + y, z
≥ , then
x − y, z
≤ 2 − 2δ  (),

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UNIFORM CONVEXITY

217

since the space is uniformly convex. But we have
x + z, y + z
=
y + z + (x − y), y + z
=
x − y, y + z
≤
x − y, y
+
x − y, z
=
x, y
+
x − y, z
≤ 1 + 2 − 2δ  () = 3 − 2δ  (). The proof is finished if we let δ() = 23 δ  ().

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Remark 8.5.2. The following implications hold for a linear 2-normed space: (1) Uniformly convex ⇒ Strictly convex ⇒ (
) Strictly 2-convex. (2) Uniformly convex ⇒ Uniformly 2-convex ⇒ Strictly 2-convex.

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CHAPTER 9. ISOMETRY CONDITIONS IN LINEAR 2–NORMED SPACES The purpose of this chapter is to deal with isometries between a linear 2–normed space and strictly convex and strictly 2–convex linear 2–normed spaces and to obtain some results under weaker conditions than isometries in linear 2–normed spaces and an example of a non–linear isometry. We also show the existence of fixed points of non–expansive mappings on linear 2–normed spaces. The results in this chapter are due to J. Baker [9], C.R. Diminnie [50], C.R. Diminnie, S. G¨ ahler and A. White [56], S. G¨ ahler [96], Y.J. Cho and R.W. Freese [32].

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9.1. Isometry Conditions In this section, we will show that the functions from linear 2–normed spaces into strictly convex normed linear spaces which satisfy certain isometry conditions are affine and we will deal with isometries between linear 2– normed spacs and strictly 2–convex (resp. strictly convex) linear 2–normed spaces. S. Mazur and S. Ulam [168] have shown that an isometry of a real normed linear space onto another is necessarily affine (i.e.,x → f (x) − f (0) is linear). It is natural to ask if the result holds without the onto assumption. In this section, we prove that an isometry from a real normed linear space into a strictly convex real normed linear space is affine. We also show that, given a real normed linear space Y which is not strictly convex, there exists an isometry from R into Y which is not affine. By using these facts, we also prove that an isometry from a linear 2–normed space into strictly convex and strictly 2–convex linear 2–normed spaces is linear under some conditions. By an isometry from a normed linear space X into a normed linear space Y we mean a function f : X → Y such that f (x) − f (y) = x − y for all x, y ∈ X. The following lemmas and Theorem 9.1.3 are due to J. Baker [9] and we need these results for our main theorems: Lemma 9.1.1. If (Y, ·) is a normed linear space, a, b ∈ Y and a+b = a + b, then sa + tb = sa + tb for all s, t ≥ 0. Proof. Suppose, without loss of generality, that 0 ≤ s ≤ t. Then we Typeset by AMS-TEX Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

220

R. W. FREESE AND Y. J. CHO

have sa + tb ≤ sa + tb and, on the other hand, sa + tb = t(a + b) − (t − s)a ≥ ta + b − (t − s)a = sa + tb. Lemma 9.1.2. Let (Y, ·) be a real normed linear space which is strictly convex and let a, b ∈ Y . Then 12 (a + b) is the unique member of Y , which is a distance 12 a − b from both a and b. Proof. The result clearly holds if a = b. It is also easy to see that 1 1 2 (a + b) is a distance 2 a − b from both a and b. Thus it suffices to prove the uniqueness. Suppose that a = b and u, v ∈ Y with a − u = b − u = a − v = b − v = Then we have

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(9.1)

1 a − b. 2




1
1 1



(u + v) (a − u) + (a − v) a − =



2 2 2 1 1 ≤ a − u + a − v 2 2 1 = a − b. 2

Similarly, we have (9.2)


1
1


b − (u + v)
≤ a − b. 2 2

If either of these inequalities were strict we would have



1 1



a − b ≤
a − (u + v)
+
b − (u + v)
< a − b. 2 2 Thus the equality holds in (9.1) and (9.2), so that
1
1 1

(a − u) + (a − v)
= a − b
2 2 2
1

1




=
(a − u)
+
(a − v)
. 2 2

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ISOMETRY CONDITIONS IN LINEAR 2-NORMED SPACES

221

Since Y is strictly convex, a = u and a = v, it follows that a − u = t(a − v) for some t > 0. But, since a − u = a − v, t = 1 and thus u = v. By using Lemmas 9.1.1 and 9.1.2, we have the following: Theorem 9.1.3. Let (X,  · ) and (Y,  · ) be real normed linear spaces and suppose that (Y,  · ) is strictly convex. If f : X → Y is such that f (x) − f (y) = x − y

(9.3)

for all x, y ∈ X, then f is affine. Proof. If f (0) = 0, let g(x) = f (x) − f (0). Then g is an isometry and g(0) = 0. Thus we may assume that f (0) = 0. Putting y = 0 in (9.3), we have f (x) = x and hence f (−x) = x for all x ∈ X. Replacing y by −x in (9.3), we obtain f (x) + (−f (−x)) = 2x = f (x) +  − f (−x). Since (Y,  · ) is strictly convex, we must have f (x) = −tf (−x) for some t > 0. But f (x) = f (−x) so t = 1 . Thus f (−x) = −f (x) for all x ∈ X. Now, for all x, y ∈ X, we have

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f (x + y) = f (x + y) − f (0) = x + y = x − (−y) = f (x) − f (−y) = f (x) + f (y). Hence, we have

x − y
1
x + y  1



− f (x)
=

= x − y = f (x) − f (y)
f 2 2 2 2 and similarly,


x + y 
1

− f (y)
= f (x) − f (y)
f 2 2 for all x, y ∈ X. It follows from Lemma 9.1.2 that (9.4)

f

x + y  2

=

f (x) + f (y) 2

for all x, y ∈ X. Since f (0) = 0 we easily find from (9.4) that f is additive. But, being an isometry, f is continuous. Hence f is linear.

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Let (Y,  · ) be a real normed linear space which is not strictly convex. Following P. Fischer and Gy. Musz´ely [76], we give an example of an isometry from R into Y which is not affine. To this end, choose a, b ∈ Y such that a, b are linearly independent, a = b = 1,and a + b = a + b . This is possible according to Lemma 9.1.1 and the definition of strict convexity. For all x ∈ R, define  xa if x ≤ 1, f (x) = a + (x − 1)b if x > 1.

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It is easy to verify, using Lemma 9.1.1, that f is an isometry but is not affine. As a further example, let us construct a homogeneous isometry which is not linear. This answers a problem raised by P.R. Chernoff [26]. Let g : R2 → R be defined by    y if 0 ≤ y ≤ x or x ≤ y ≤ 0, x if 0 ≤ x ≤ y or y ≤ x ≤ 0, g(x, y) =   0 otherwise. It is not difficult to see that (1) g is homogeneous, i.e., g(tx, ty) = tg(x, y) for all t, x, y ∈ R , (2) |g(x, y) − g(u, v)| ≤ (x − u)2 + (y − v)2 for all x, y, u, v ∈ R, (3) g is not linear. Let X denote R2 with the usual normed linear space structure, and let Y denote R3 with the usual vector space structure but with (x, y, z) = max( x2 + y 2 , |z|). Then, with this norm, Y is a normed linear space. If f : X → Y is defined by f (x, y) = (x, y, g(x, y)), then it follows from (1), (2) and (3), that f is a homogeneous isometry which is not linear. By using Theorem 9.1.3. we have the following: Theorem 9.1.4. Let (X, ·, ·) be a linear 2–normed space and (X
, ·) be a normed linear space. (1) If (X,  · ) is strictly convex, c is a fixed non–zero element of X, and f is a function from X
into X which satisfies f (a) − f (b), c = a − b

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for every a, b ∈ X, then the function gc from X
into Xc defined by gc (a) = (f (a) − f (0))c is linear. (2) If (X
,  · ) is strictly convex, c is a fixed non–zero element of X, and f is a function from X into X
satisfying f (a) − f (b) = a − b, c for every a, b ∈ X, then the function g from X into X
defined by g(a) = f (a) − f (0) is linear. Proof. (1) Let (X, ·, ·) be strictly convex and f, c, gc be as given in statement (1). By Theorem 6.1.2, (Xc ,  · c ) is strictly convex. For every a, b ∈ X
, gc (a) − gc (b)c = (f (a) − f (b))c c = f (a) − f (b), c = a − b. By Theorem 9.1.3, it follows that gc is linear. (2) Let (X
,  · ) be strictly convex and f, c, g be as given in statement (2). Then, g(0) = 0 and for every a, b ∈ X, we have

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g(a) − g(b) = f (a) − f (b) = a − b, c. Thus, for a, b ∈ X which satisfy (a)c = (b)c , it follows that g(a) − g(b) = 0 and the function gc from Xc into X
given by gc ((a)c ) = g(a) is well–defined. For any a, b ∈ X, we have gc ((a)c ) − gc ((b)c ) = g(a) − g(b) = a − b, c = (a)c − (b)c c . Since gc ((0)c ) = g(0) = 0, Theorem 9.1.3 implies that gc is linear. For any a, b ∈ X and any real number α, we have g(αa) = gc ((αa)c ) = αgc ((a)c ) = αg(a) and g(a + b) = gc ((a)c + (b)c ) = gc ((a)c ) + gc ((b)c ) = g(a) + g(b).

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Therefore, g is linear. By Theorem 9.1.4, we have the following: Corollary 9.1.5. Let (X, ·, ·) be a strictly convex linear 2–normed space, c be a fixed non–zero element of X, and f be a function from X into X which satisfies f (0) = 0 and f (a) − f (b), c = a − b, c for every a, b ∈ X. Then the function gc from X into Xc defined by gc (a) = (f (a))c is linear. Proof. Since (X, ·, ·) is strictly convex, Theorem 6.1.2 implies that (Xc ,  · c ) is strictly convex. For any a, b ∈ X, we have gc (a) − gc (b)c = (f (a) − f (b))c c = f (a) − f (b), c = a − b, c. From (2) of Theorem 9.1.4, it follows that gc is linear.

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The following theorem was proved by S. G¨ ahler [96] and used to prove Theorem 9.1.7: Theorem 9.1.6. Let (X, ·, ·) and (X
, ·, ·) be linear 2–normed spaces. If f is a mapping from X into X
for which f (0) = 0 and f (a) − f (c), f (b) − f (c) = a − c, b − c for every a, b, c ∈ X, then f is linear. Proof. Let a, b ∈ X be linearly independent and α∗ , β ∗ , α, β, α
, β
be non–zero real numbers. Then we have f (α∗ a), f (β ∗ b) = σ(0, f (α∗a), f (β ∗ b)) = σ(0, α∗ a, β ∗ b) = |α∗ β ∗ |a, b = 0 and hence f (α∗ a) and f (α∗ b) are linearly independent. Since the following identies hold : f (2βb) − f (2αa), f (αa + βb) − f (2αa) = σ(f (2αa), f (2βb), f (αa + βb)) = σ(2αa, 2βb, αa + βb) = 2(βb − αa), βb − αa = 0,

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f (2αa), f (a) = σ(0, f (2αa), f (a)) = σ(0, 2αa, a) = 2αa, a = 0. Similarly, we have f (2βb), f (b) = 0. Hence, we get f (αa + βb) = α
f (a) + β
f (b). And from the following: |α|σ(0, a, b) = σ(0, αa + βb, b) = σ(0, α
f (a) + β
f (b), f (b)) = |a
|σ(0, f (a), f (b)) = |α
|σ(0, a, b), it follows that |α| = |α
|. Similarly, we have also |β| = |β
|. Moreover, from |α + β − 1|σ(0, a, b) = σ(0, a − b, αa + βb − b) = σ(b, a, αa + βb) = σ(f (b), f (a), α
f (a) + β
f (b)) = |α
+ β
− 1|σ(0, a, b), it follows that |α + β − 1| = |α
+ β
− 1|. Similarly, by using f (−a) = −f (a) and f (−b) = −f (b), we have

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|α + β − 1| = |α
+ β
− 1|,

|α − β − 1| = |α
− β
− 1|.

Therefore, by the above identities, we have α = α
and β = β
and hence f is linear. The following theorem given by C.R. Diminnie, S. G¨ ahler and A. White [56] deals with isometries between linear 2–normed spaces and strictly 2– convex linear 2–normed spaces. Using Theorem 9. 1. 6, we have the following: Theorem 9.1.7. Let (X, ·, ·) be a linear 2–normed space and (X
, ·, ·) be a strictly 2–convex linear 2–normed space. If f is a mapping from X into X
for which f (0) = 0 and f (a) − f (c), f (b) − f (c) = a − c, b − c for every a, b, c ∈ X, then f is linear and one–to–one and (X, ·, ·) is strictly 2–convex. If (X
, ·, ·) is strictly convex, then (X, ·, ·) is strictly convex also.

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Proof. The linearity of f follows by Theorem 9.1.6. For every a = 0, since there exists a point b ∈ X with f (a), f (b) = a, b = 0, f (a) = 0 and hence f is one–to–one. The remainder of the conclusions follows easily. The following result given by C.R. Diminnie and A. White [61] deals with isometries between a linear 2–normed space and strictly 2–convex 2–normed spaces, that is, 2–functionals, on the cross product of linear 2–normed spaces, which satisfy certain isometry conditions, are affine. Theorem 9.1.8. Let (X, ·, ·) be a linear 2–normed space. If F is a 2–functional with domain X × X such that |F (a, c) − F (b, c)| = ka − b, c for all a, b ∈ X, c is a fixed non–zero element in X and k > 0, then G(x, c) = F (x, c) − F (0, c) is linear in x. Proof. Note that |G(a, c) − G(b, c)| = |F (a, c) − F (b, c)| = ka − b, c and G(0, c) = 0. Thus, setting b = 0, we have

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(9.5)

|G(a, c)| = ka, c

for all a ∈ X. Therefore, G(a, c) = 0 if and only if a and c are linearly dependent. Also, we have (9.6)

G(a, c) = G(b, c) if and only if a − b and c are linearly dependent.

Now, |G(2a, c)| = 2ka, c = 2|G(a, c)| implies that either G(2a, c) = 2G(a, c) or G(2a, c) = 0. If G(2a, c) = 0, then a and c are linearly dependent and G(2a, c) = 0 = 2G(a, c) and so we have (9.7)

G(2a, c) = 2G(a, c)

for all a ∈ X. Let α be any real numbr. Then we have |G(αa, c) − G(a, c)| = k|α − 1|a, c = |α − 1||G(a, c)| = |(α − 1)G(a, c)|. Therefore, it follows that either G(αa, c) − G(a, c) = (α − 1)G(a, c) or G(αa, c) − G(a, c) = (1 − α)G(a, c), that is, either G(αa, c) = αG(a, c) or G(αa, c) = (2 − α)G(a, c).

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If G(αa, c) = (2 − α)G(a, c), then αG(a, c) = 2G(a, c) − G(αa, c). Hence, using (9.5) and (9.7), we have |αG(a, c)| = |2G(a, c) − G(αa, c)| = |G(2a, c) − G(αa, c)| = k(2 − α)a, c = |(2 − α)G(a, c)|. Therefore, either G(a, c) = 0 or |α| = |2 − α|. If G(a, c) = 0, then a and c are linearly dependent and G(αa, c) = 0 = αG(a, c). If |α| = |2 − α|, then α = 1 and hence G(αa, c) = αG(a, c), which implies that G(αa, c) = αG(a, c)

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(9.8)

for all a, b ∈ X and all real numbers α. Since the following equalities hold:
a − b


 a + b 







, c = k
, c
,

G(a, c) − G 2 2
a − b


 a + b 







, c = k
, c
,

G(b, c) − G 2 2 by (9.8), we have either G(a, c) = G(b, c) or a + b  , c = G(a + b, c). G(a, c) + G(b, c) = 2G 2 Consider the case G(a, c) = G(b, c). By (9.6), a − b and c are linearly dependent. Also, we have |G(a + b, c) − G(b, c)| = ka, c = |G(a, c)|. Therefore either G(a + b, c) = G(a, c) + G(b, c) or G(a + b, c) = G(b, c) − G(a, c). Since G(a, c) = G(b, c), G(a + b, c) = 0 and hence a + b and c are linearly dependent. Hence we have a=

1 [(a + b) + (a − b)], 2

b=

1 [(a + b) − (a − b)] 2

are linearly dependent on c. Thus, we get G(a, c) + G(b, c) = 0 = G(a + b, c) and so (9.9)

G(a + b, c) = G(a, c) + G(b, c)

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for all a, b ∈ X. By (9.8) and (9.9), G(x, c) is linear in x. Corollary 9.1.9. Let H be a 2–functional with domain X × X such that |H(c, a) − H(c, b)| = c, a − b for all a, b ∈ X and c is a fixed non–zero element in X. Then H(c, y) − H(c, 0) is linear in y. Corollary 9.1.10. Let I(x, y) = F (x, c) − F (0, c) + H(c, y) − H(c, 0), where F is defined in Theorem 9.1.8 and H is defined in Corollary 9.1.9. Then I is a bilinear 2–functional. Example 9.1.1. S. G¨ ahler [96] has shown that if (X, ·, ·) is a linear 2–normed space with basis {e1 , e2 } and a, b ∈ X with a = a1 e1 + a2 e2 , b = b1 e1 + b2 e2 , then a, b = |a1 b2 − a2 b1 |e1 , e2 . Define a functional F on X ×X by F (a1 e1 +a2 e2 , b1 e1 +b2 e2 ) = (a1 b2 −a2 b1 )e1 , e2 . Then |F (a, c)− F (b, c)| = a − b, c and F (x, c) − F (0, c) is linear in x. 9.2. An Example of a Non–linear Isometry

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In Theorem 9.1.4, it was demonstrated that functions from linear 2– normed spaces into strictly convex normed spaces, which satisfied certain isometry conditions, are affine. Here, an example given in C.R. Diminnie [50] shows that this result is false if the property of strict convexity is deleted. Example 9.2.1. Let (X, ·, ·) be a 2–dimensional linear 2–normed space with basis {e1 , e2 }. Since e1 and e2 are linearly independent, it may be assumed that e1 , e2  = 1. Let c = e1 + e2 be the fixed element of X as in Theorem 9.1.4. Then, by Theorem 3.1.1, we have a1 e1 + a2 e2 , c = |a1 − a2 |. Let (N,  · ) be any normed linear space which is not strictly convex. Then there exist two elements x, y ∈ N, x = y, such that x + y = x + y, Define f : X → N by f (a) =



x = y = 1.

(a1 − a2 )x

if a1 ≤ a2 + 1,

x + (a1 − a2 − 1)y

if a1 > a2 + 1,

where a = a1 e1 + a2 e2 . Then, for a = a1 e1 + a2 e2 and b = b1 e1 + b2 e2 , consider the following cases:

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Case 1. If a1 ≤ a2 + 1 and b1 ≤ b2 + 1, then we have f (a) − f (b) = (a1 − a2 )x − (b1 − b2 )x = |(a1 − b1 ) − (a2 − b2 )| = a − b, c. Case 2. If a1 ≤ a2 + 1 and b1 > b2 + 1, by Lemma 9.1.1, then we have f (a) − f (b) = (a1 − a2 )x − x − (b1 − b2 − 1)y = (a1 − a2 − 1)x − (b1 − b2 − 1)y = −(a1 − a2 − 1)x + (b1 − b2 − 1)y = |(a1 − b1 ) − (a2 − b2 )| = a − b, c. Case 3. If a1 > a2 + 1 and b1 > b2 + 1, then we have f (a) − f (b) = x + (a1 − a2 − 1)y − x − (b1 − b2 − 1)y = |(a1 − b1 ) − (a2 − b2 )| = a − b, c. Therefore, for every a, b ∈ X, we obtain

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f (a) − f (b) = a − b, c. Further, since f (0) = 0, the function g defined in Theorem 9.1.4 is equal to f . Finally, let a∗ = e1 + 12 e2 and b∗ = e1 + 14 e2 .Then f (a∗ ) = 12 x, f (b∗ ) = 34 x, and f (a∗ +b∗ ) = x+ 41 y. Since x = y, it follows that f (a∗ +b∗ ) = f (a∗ )+f (b∗ ) and hence f is not linear. 9.3. Weak Conditions of Isometries In this section, we introduce some weak conditions for isometrices in linear 2–normed spaces given in Y.J. Cho and R.W. Freese [32]. Let R+ , Q+ and Z + denote the sets of positive real numbers, positive rational numbers, and natural numbers, respectively. Definition 9.3.1. A linear 2–normed space (X, ·, ·) is said to have Property (P) if for an arbitrary element x in X and a fixed non–zero element z in X with x, z < 1, there exists an element y ∈ X such that x − y, z = 1 = x + y, z.

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Let ρ ∈ R+ and N > 1, N ∈ Z + , be fixed numbers. Let (X, ·, ·) be a linear 2–normed space with Property (P), and (Y,  · ) be a strictly convex normed linear space. If f is a function from X into Y such that (9.10) if x − y, z = ρ for arbitrary elements x, y in X and a fixed non–zero element z in X, then f (x) − f (y) ≤ ρ, and (9.11) if x − y, z = N ρ, then f (x) − f (y) ≥ N ρ, then we have the following seven propositions: Proposition 9.3.1. For arbitrary elements x, y in X and a fixed non– zero element z in X with x − y, z = ρ, we have f (x) − f (y) = ρ. Proof. Putting pn = y + n(x − y) for n = 0, 1, 2, . . . , N, we have pN − y, z = y + N (x − y) − y, z = N x − y, z = N ρ and pn − pn−1 , z = y + n(x − y) − y − (n − 1)(x − y), z = x − y, z = ρ for n = 1, 2, · · · , N , and so we have

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f (pN ) − f (y) ≥ N ρ,

f (pn ) − f (pn−1 ) ≤ ρ

for n = 1, 2, · · · , N . Therefore, since N ρ ≤ f (pN ) − f (y) ≤

N 

f (pN+1−n ) − f (pN−n ) ≤ N ρ,

n=1

f (x) − f (y) = f (p1 ) − f (p0 ) = ρ. Proposition 9.3.2. For arbitrary elements x, y in X and a fixed non– zero element z in X with x − y, z = 2ρ we have f (x) − f (y) = 2ρ. Proof. Putting pn = y + 12 n(x − y) for n = 0, 1, 2, · · · , N , we have
N
N (x − y)

− y, z
= x − y, z = N ρ, pN − y, z =
y + 2 2

x − y, z n−1 n(x − y)

pn − pn−1 , z =
y + −y− (x − y), z
= =ρ 2 2 2

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for n = 1, 2, · · · , N . Hence, by Proposition 9.3.1, we have N ρ ≤ f (pN ) − f (y) ≤

N 

f (pN+1−n ) − f (pN−n ) ≤ N ρ,

n=1

N that is, f (pN ) − f (y) = n=1 f (pN+1−n ) − f (pN−n ). Now we prove that f (p2 ) − f (p0 ) = f (p2 ) − f (p1 ) + f (p1 ) − f (p0 ). In fact, assume that f (p2 ) − f (p0 ) < f (p2 ) − f (p1 ) + f (p1 ) − f (p0 ). It follows that f (pN ) − f (y) ≤

N−2 

f (pN+1−n ) − f (pN−n ) + f (p2 ) − f (p0 )

n=1


1, putting pm = x + m(y − x), since pm − pm−1 , z = ρ = pm+1 − pm , z and pm+1 − pm−1 , z = 2ρ, by Propositions 9.3.1 and 9.3.2, we have f (pm ) − f (pm−1 ) = ρ = f (pm+1 ) − f (pm ), f (pm+1 ) − f (pm−1 ) = 2ρ. Hence, by Proposition 9.3.3, f (pm+1 ) = 2f (pm ) − f (pm−1 ). By mathematical induction, assume that f (x + k(y − x)) = f (x) + k(f (y) − f (x)). Then we have 2f (pk ) = 2f (x + k(y − x)) = 2(f (x) + k(f (y) − f (x))) = f (x + (k + 1)(y − x)) + f (x) + (k − 1)(f (y) − f (x)), that is,

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f (x + (k + 1)(y − x)) = f (x) + (k + 1)(f (y) − f (x)). Therefore, for every non–negative integer m, f (x + m(y − x)) = f (x) + m(f (y) − f (x)). Proposition 9.3.5. For all numbers α, β ∈ R+ with 2β ≥ α, arbitrary elements x, y in X and a fixed non–zero element z in X with x − y, z = α there exists an element u in X such that u − x, z = β = u − y, z. Proof. In case 2β = α, u = (x + y)/2. Let 2β > α. For a = (y − x)/2β, a, z = (y − x)/2β, z = (1/2β)y − x, z = (α/2β) < 1, and so, by Property (P), there exists an element b in X such that a − b, z = 1 = a + b, z. Putting u = (x + y)/2 + βb, we have u − x, z = (y + x)/2 + βb − x, z = βa + b, z = β,

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u − y, z = β − a + b, z = β. Proposition 9.3.6. For all numbers m, n ∈ Z, arbitrary elements x, y in X and a fixed non–zero element z in X with x − y, z = (m/n)ρ, we have f (x) − f (y) = (m/n)ρ. Proof. By Proposition 9.3.5, there exists an element u in X such that u − x, z = mρ = u − y, z. Let a, b ∈ X be such that x = u + m(a − u) and y = u + m(b − u). If we put p = u + n(a − u) and q = u + n(b − u), then we have a − u, z = b − u, z = p − q, z = ρ. Therefore, by Proposition 9.3.4, it follows that p − q, z = ρ = f (p) − f (q) = nf (a) − f (b) and f (x) − f (y) = mf (a) − f (b),

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that is, we have f (x) − f (y) = (m/n)ρ. Proposition 9.3.7. If γ1 ρ < x − y, z < γ2 ρ for all γ1 , γ2 ∈ Q+ with γ1 < γ2 , then we have γ1 ρ ≤ f (x) − f (y) < γ2 ρ. Proof. By Proposition 9.3.5, there exists an element u in X such that u − x, z = (γ2 /2)ρ = u − y, z . Hence, by Proposition 9.3.6, it follows that f (u) − f (x) = (γ2 /2)ρ = f (u) − f (y) and so

f (x) − f (y) ≤ f (x) − f (u) + f (u) − f (y) = (γ2 /2)ρ + (γ2 /2)ρ = γ2 ρ.

Assume that there exist two elements x, y in X such that γ1 ρ < x − y, z < γ2 ρ,

f (x) − f (y) < γ1 ρ.

Then we have γ1 ρ − x − y, z < (γ2 − γ1 )ρ < γ2 ρ − f (x) − f (y)

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and so, for u = x + λ(y − x) with λ = γ2 ρ/x − y, z > 1, we have u − x, z = γ2 ρ,

u − y, z = (λ − 1)x − y, z < (γ2 − γ1 )ρ.

It follows from Proposition 9.3.6. that f (u) − f (x) = γ2 ρ and f (u) − f (y) ≤ f (u) − f (x) + f (x) − f (y) ≤ γ2 ρ − γ1 ρ = (γ2 − γ1 )ρ. But we have

γ2 ρ = f (u) − f (x) ≤ f (u) − f (y) + f (y) − f (x) < (γ2 − γ1 )ρ + γ1 ρ = γ2 ρ,

which contradicts f (u) − f (x) = γ2 ρ. Therefore, for all elements γ1 , γ2 ∈ Q+ with γ1 < γ2 , if γ1 ρ < x − y, z < γ2 ρ, then we have γ1 ρ ≤ f (x) − f (y) < γ2 ρ.

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By Propositions 9.3.1∼9.3.7, we have the following: Theorem 9.3.8. Let ρ ∈ R+ and N > 1, N ∈ Z, be fixed numbers, (X, ·, ·) be a linear 2–normed space with Property (P) and (Y, ·, ·) be a strictly convex normed linear space. If f is a function from X into Y with the properties (9.10) and (9.11), then x − y, z = f (x) − f (y) for arbitrary elements x, y ∈ X and a fixed non–zero element z ∈ X and so the function g from X into Y defined by g(x) = f (x) − f (0) is linear. Proof. By Propositions 9.3.1∼9.3.7, we have x − y, z = f (x) − f (y) and hence, by Theorem 9.1.4, g is a linear function from X into Y . As a consequence of Theorem 9.3.8, we have the following: Corollary 9.3.8. Let ρ ∈ R+ and N > 1, N ∈ N , be fixed numbers and (X, ·, ·) be a strictly convex linear 2–normed space with Property (P). If f is a function from X into itself such that (9.12) if x − y, z = ρ for arbitrary elements x, y ∈ X and a fixed non– zero element z ∈ X, then f (x) − f (y) ≤ ρ, (9.13) if x − y, z = N ρ, then f (x) − f (y) ≥ N ρ, (9.14) if f (0) = 0, then the function gz from X into Xz defined by gz (x) = (f (x))z is linear.

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Proof. Since (X, ·, ·) is strictly convex, by Theorem 6.1.2, (Xz ,  · z ) is also strictly convex. If x − y, z = ρ , then we have f (x) − f (y), z = (f (x) − f (y))z z = (f (x))z − (f (y))z z = gz (x) − gz (y)z ≤ ρ. If x − y, z = N ρ , then we have f (x) − f (y), z = gz (x) − gz (y) ≥ N ρ. Therefore, by Theorem 9.3.8, we have x − y, z = g(x) − g(y) and hence the function gz from X into Xz defined by gz (x) = (f (x))z − (f (0))z = (f (x))z is linear. 9.4. Non–expansive Mappings in Linear 2–Normed Spaces In this section, the investigation was concerned with the structure of the fixed point set F (T ) of a non–expansive mapping T on linear 2–normed spaces. In [126], K. Is´eki introduced the concept of non–expansive mapping in linear 2–normed spaces. If G is a convex subset of X, a mapping T : G → X is said to be non–expansive if, for every x, y ∈ G and z ∈ X,

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(9.15)

T (x) − T (y), z ≤ x − y, z.

Non–expansive mappings are characterized in this section. In addition, a theorem of K. Is´eki [126] concerning fixed points of non–expansive mappings will be obtained as a corollary to the main results, which are due to C.R. Diminnie and A. White [63]. In the following, a subset L of X of the form {x1 + αx2 : α ∈ R} , where x1 and x2 are in X and x2 is a non–zero element, will be called a line. Theorem 9.4.1. Let G be a convex set which contains at least 2 elements and is not a subset of a line. Then T is non–expansive if and only if there is a point c ∈ R and there is a point zo ∈ X such that |c| ≤ 1 and T (x) = cx + z0 for every x ∈ G. Proof. Since all functions of the above type are non–expansive, we need show only that all non–expansive maps are of this type. 1. Assume first that 0 ∈ G and T (0) = 0. Then, for every x ∈ G and z∈X (9.16)

T (x), z ≤ x, z.

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Therefore, for each x ∈ G, there is a real number g(x) such that T (x) = g(x)x. If x and y are linearly independent elements of G, then 12 (x + y) ∈ G also, and by (9.15), we have
x + y

x − y




− T (x), x − y
≤
, x − y
= 0.
T 2 2 Therefore, there is a k ∈ R such that x + y  − T (x) = k(x − y) T 2 or g

 x + y  x + y  2

2

− g(x)x = k(x − y)

Then, we have

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  1 x + y  1  x + y  g − g(x) − k x = − k + g y, 2 2 2 2 which implies that g(x) = g( 21 (x + y)) by the linear independence of x and y. Since a similar argument shows that g(y) = g( 21 (x + y)), it follows that g(x) = g(y) whenever x and y are linearly independent. If x and y are non–zero, linearly dependent elements of G, then, since G is not a subset of a line, there is a z ∈ G such that x, z and y, z are linearly independent, respectively. By the arguments used above, we have g(x) = g(z) = g(y). Therefore, g(x) = g(y) for all non–zero x, y ∈ G. Since T (0) = 0, there is a real number c such that T (x) = cx for every x ∈ G. Finally, (9.16) implies that |c| ≤ 1. 2. For arbitrary T and G which satisfy the hypothesis, choose any x0 ∈ G and let G
= {x − x0 : x ∈ G}. Then, G
is convex, G
is not contained in a line since G is not a subset of a line, and 0 ∈ G
. Define S : G
→ X by S(x − x0 ) = T (x) − T (x0 ) for every x ∈ G. For x, y ∈ G and z ∈ X, we have S(x − x0 ) − S(y − x0 ), z = T (x) − T (y), z ≤ x − y, z = (x − x0 ) − (y − x0 ), z. Hence, S is non–expansive on G
and S(0) = S(x0 − x0 ) = T (x0 ) − T (x0 ) = 0.

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By part 1, there is a c ∈ R such that |c| ≤ 1 and for every x ∈ G, S(x − x0 ) = c(x − x0 ). Therefore, for every x ∈ G, we have T (x) = cx + T (x0 ) − cx0 . The following example given by C.R. Diminnie and A White [63] shows that the characterization fails if G is contained in a line. Example 9.4.1. Suppose that G is a subset of the line L = {x1 + αx2 : α ∈ R}. Define T : G → X by T (x1 + αx2 ) = (sin α)x2 . Then, if x1 + αx2 and x1 + γx2 are in G and z ∈ X, we have T (x1 + αx2 ) − T (x1 + γx2 ), z = | sin α − sin γ|x2 , z ≤ |α − γ|x2 , z = (x1 + αx2 ) − (x1 + γx2 ), z. Therefore, T is a non–expansive mapping which does not satisfy Theorem 9.4.1.

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For convex sets which are subsets of lines, we have the following characterization of non–expansive mappings. Theorem 9.4.2. Suppose G is a convex subset of a line L = {x1 + αx2 : α ∈ R}, where x1 ∈ G, and let A = {α : x1 +αx2 ∈ G}. Then, T : G → X is non–expansive if and only if there is a function g : A → R such that, for every α, γ ∈ A, |g(α)−g(γ)| ≤ |α−γ|, g(0) = 0 and T (x1 +αx2 ) = g(α)x2 +T (x1 ). Proof. Again, since the sufficiency of the above conditions is clear, we need prove only the necessity. 1. Assume that x1 = 0 and T (0) = 0. Then, for every α ∈ A and z ∈ X, (9.17)

T (αx2 ), z ≤ αx2 , z.

Therefore, for every non–zero α ∈ A, there is a real number g(α) such that (9.18)

T (αx2 ) = g(α)x2 .

If we define g(0) = 0, then (9.18) holds for every α ∈ A. By (9.15), it follows that |g(α) − g(γ)| ≤ |α − γ| for every α, γ ∈ A.

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2. If x1 = 0 or T (x1 ) = 0, let G
= {αx2 : α ∈ A}. Then, G
is convex, 0 ∈ G
. Define S : G
→ X by S(αx2 ) = T (x1 + αx2 ) − T (x1 ) for every α ∈ A. Note that S(0) = 0 and for α, γ ∈ A and z ∈ X, S(αx2 ) − S(γx2 ), z = T (x1 + αx2 ) − T (x1 + γx2 ), z ≤ αx2 − γx2 , z. Therefore, since S and G
satisfy the assumptions made in part 1, it follows that there is a function g : A → R such that g(0) = 0, |g(α) − g(γ)| ≤ |α − γ| for every α, γ ∈ A, and S(αx2 ) = g(α)x2 . Hence, for every α ∈ A. T (x1 + αx2 ) = g(α)x2 + T (x1 ).

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K. Is´eki [126] has shown that in a strictly convex linear 2–normed space, the set F (T ) of fixed points of a non–expansive T is always a convex set. This result can now be proven for any linear 2–normed space. Corollary 9.4.3. If G is a convex subset of a linear 2–normed space (X, ·, ·) and if T : G → X is non–expansive, then the set F (T ) of fixed points of T is convex. Proof. If G is not a subset of a line, then by Theorem 9.4.1, F (T ) is the empty set, or a single point, or all of G. In any case, F (T ) is convex. If G is a subset of a line and F (T ) is not empty, then the point x1 of Theorem 9.4.2 may be chosen to be an element of F (T ). With the same notation as in Theorem 9.4.2, T (x1 + αx2 ) = g(α)x2 + x1 for every α ∈ A. In this case, x1 + αx2 ∈ F (T ) if and only if g(α) = α. Note that since G is convex, A is a convex set of real numbers. Then, since g : A → R is non–expansive, it is well–known that the set F (g) of fixed points of g is convex. Therefore, since F (T ) = {x1 + αx2 : α ∈ F (g)}, F (T ) is convex also. Next, in [65], C.R. Diminnie and A. White [65] gave an alternate definition of non–expansive mappings and investigated the fixed point sets for such mappings [64]. A mapping S : X → X is said to be 2–metrically non– expansive if σ(S(a), S(b), S(c)) ≤ σ(a, b, c)

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ISOMETRY CONDITIONS IN LINEAR 2–NORMED SPACES

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for every a, b, c ∈ X. Theorem 9.4.4. If S is 2–metrically non–expansive, (X, ·, ·) is a strictly 2–convex linear 2–normed space, and a, b, c are non–collinear points in F (S), then T (a, b, c) ⊂ F (S). Proof. If x ∈ T (a, b, c), then since T (a, b, c) = C(a, b, c) (cf. Theorem 7.2.5), there are real numbers α, β, γ ≥ 0 such that α + β + γ = 1 and x = αa + βb + γc. Then, σ(a, b, c) = σ(S(a), S(b), S(c)) ≤ σ(S(a), S(b), S(x)) + σ(S(a), S(x), S(c)) + σ(S(x), S(b), S(c)) ≤ σ(a, b, x) + σ(a, x, c) + σ(x, b, c) = σ(a, b, c) Therefore, we have S(x) ∈ T (S(a), S(b), S(c)) = T (a, b, c), σ(a, b, S(x)) = σ(a, b, x),

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σ(a, S(x), c) = σ(a, x, c), σ(S(x), b, c) = σ(x, b, c). Since S(x) ∈ T (a, b, c), there are α1 , β1 , γ1 ≥ 0 such that α1 + β1 + γ1 = 1 and S(x) = α1 a + β1 b + γ1 c. Then, we have σ(x, b, c) = αa + βb + γc − c, b − c = αa + (β + γ − 1)c, b − c = αa − c, b − c = ασ(a, b, c) and similarly, we have σ(S(x), b, c) = α1 σ(a, b, c). Since σ(x, b, c) = σ(S(x), b, c) and σ(a, b, c) > 0, it follows that α = α1 . Similar arguments show that β = β1 and γ = γ1 . Therefore, S(x) = x and x ∈ F (S).

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Corollary 9.4.5. If S is 2–metrically non–expansive, (X, ·, ·) is strictly 2–convex and F (S) contains at least 3 non–collinear points, then F (S) is convex. Proof. Let a, b ∈ F (S) and x = αa + (1 − α)b for some α ∈ (0, 1).If c ∈ F (S) is chosen so that a, b, c are non–collinear, then x ∈ T (a, b, c). By Theorem 9.4.4, T (a, b, c) ⊂ F (S) and hence, we have x ∈ F (S). We finish with an example [65] to show that Corollary 9.4.5 is not true if F (S) fails to contain at least 3 non–collinear points. Example 9.4.2. Let (X, ·, ·) be a 2–dimensional linear 2–normed space with basis {e1 , e2 }. Then (X, ·, ·) is strictly 2–convex. Define S : X → X by S(α1 e1 + α2 e2 ) = α21 e1 .

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Then S is 2–metrically non–expansive since for any a, b, c ∈ X, S(a), S(b), S(c) are collinear and thus σ(S(a), S(b), S(c)) = 0 ≤ σ(a, b, c). However, it is easily seen that F (S) = {0, e1 } and hence, F (S) is not convex.

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CHAPTER 10. ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS In [51], C.R. Diminnie introduced a new orthogonality relation for normed linear spaces using a concept of area of a parallelogram given by E. Silverman [214]. Comparisons are drawn between this relation and an earlier relation used by G. Birkhoff [16]. In addition, we know that this new relation is utilized to obtain new characterizations of inner product spaces. Many the results in this chapter are due to Y.J. Cho, C.R. Diminnie, R.W. Freese, E.Z. Andalafte [33] and C.R. Diminnie [51] 10.1. Elementary Properties of Orthogonalities

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Now, we give some properties of orthogonalities and some characterizations of inner product spaces. Let (X,
·
) be a normed linear space of dimension greater than 1. If (X,
·
) is an inner product space with inner product (·|·), then the most obvious definition of orthogonality is: x⊥y if and only if (x|y) = 0. More general definitions involving only normed properties were introduced by G. Birkhoff [16] and R.C. James [130]. Research in this area usually concentrates on studying interesting properties of such relations and using these properties to find characterizations of inner product spaces. The most immediate desirable property of an orthogonality relation ⊥ is that it should agree with the condition (x|y) = 0 in inner product spaces. In addition, other properties usually studied are as follows: (A) Symmetry : if x⊥y, then y⊥x. (B) Homogeneity : if x⊥y, then αx⊥βy for all real α and β. (C) Additivity : if x⊥y and x⊥z, then x⊥(y + z). (D) If x, y ∈ X, then there is a real number α such that x⊥(αx + y). The last property is extremely useful in working with an orthogonality relation. Among other things, it guarantees that the relation is not vacuous. To formulate our new orthogonality relation, we will rely on a concept studied by E. Silverman [210]. Let F denote the set of linear functionals on X whose norms are less than or equal to 1. Then, define 
   f (x) f (y)    (10.1)
x, y
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Intuitively,
x, y
may be visualized as the area of the parallelogram with vertices at 0, x, y and x + y. This quantity was used by F. Sullivan [225], [226] to examine convexity properties of normed linear spaces. In Chapter II, this object was studied as an example of a more general notion, which S. G¨ ahler [96] called a “2–norm”. He demonstrated that
·, ·
satisfies the conditions of 2–norm. For our purposes, Corollary 5.1.5 in Chapter 5 states that if (X,
·
) is an inner product space, then we have (10.2)


x, y
2 =
x
2
y
2 − (x|y)2 .

Considering (10.2) and the intuitive interpretation of 2–norm
·, ·
given above, we introduce the following Diminnie orthogonality and Birkhoff orthogonality, which are denoted by (x⊥y)(D) and (x⊥y)(B), respectively: Definition 10.1.1. In a normed linear space (X,
·
), (10.3)

x⊥y(D) if and only if
x, y
=
x

y
.

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Definition 10.1.2. In a normed linear space (X,
·
), (10.4)

(x⊥y)(B) if and only if
x + αy
≥
x
for all real α.

If H is a subset of X, then (x⊥H)(B) if and only if (x⊥y)(B) for all y ∈ H. Certain properties of this relation are immediate. By property (N1 ) of 2–norm
·, ·
, it follows that x⊥x if and only if x = 0. Also, relation (10.2) shows that if (X,
·
) is an inner product space, then x⊥y if and only if (x|y) = 0. The usual properties of
·
and properties (N2 ) and (N3 ) of 2– norm
·, ·
imply that ⊥ is symmetric and homogeneous. Further, in normed linear spaces X of dimension three or greater, the following are equivalent: (1) X is an inner product space. (2) Diminnie orthogonality is additive. (3) Diminnie orthogonality is equivalent to Birkhoff orthogonality. Note that Birkhoff orthogonality is homogeneous. Further, a normed linear space X of dimension three or greater in which Birkhoff orthogonality is symmetric or additive must be an inner product space.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS243

Several properties of Birkhoff orthogonality may be found in the works of R.C. James [130]∼[131], J.L. Joly [132], D. D. Rio and C. Benitez [200]. In [129], R.C. James defined isosceles orthogonality by x⊥y provided
x + y
=
x − y
. It may be observed that, under this definition, x⊥y is equivalent to y⊥x and that this is equivalent to αx⊥αy for all real α. Then it can be shown that (X,
·
) is an inner product space if x⊥y implies x⊥αy for all real α. That is, James’ “isosceles” orthogonality possessing the above property of homogeneity in a normed linear space X implies that the space X is an inner product space. Likewise, it can be shown that if a normed linear space X possesses the above property of additivity with respect to isosceles orthogonality, X is an inner product space. We say that x and y is a normed linear space X are Pythagorean orthogonal ([129]) provided
x − y
2 =
x
2 +
y
2 .

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Then, although it is clear that x⊥y implies y⊥x and even αx⊥αy, hypothesizing that Pythagorean orthogonality is homogeneous forxes X to be an inner product space. Similarly, additivity of Pythagorean orthogonality in X implies X is an inner product space. A generalization of both Pythagorean and isosceles orthogonality is αorthogonality, which is defined by x⊥y provided (1 + α2 )
x + y
2 =
x + αy
2 +
y + αx
2 . Note that α-orthogonality becomes Pythogorean orthogonality when α = 0 and becomes isosceles orthogonality for α = −1. In a normed linear space X, homogeneity of α-orthogonality for some α = −1 implies X is an inner product space, as does additivity of α-orthogonality for some α = −1. A further generalization is the concept of (α, β)-orthogonality ([5]), which is defined by x⊥y provided
x − y
2 +
αx − βy
2 =
x − βy
2 +
y − αx
2 . It may be noted that (0, 0)-orthogonality is pythagorean orthogonality, while (0, −1)−, (−1, 0)− and (−1, −1)–orthogonality are isosceles orthogonality.

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The properties of homogeneity, symmetry and left and right additivity of (α, β)-orthogonality are defined in the usual way. In particular, (α, β)orthogonality is left additive if and only if for all x, y, z in X, x⊥y and y⊥z implies x+y⊥z. Likewise, (α, β)-orthogonality is right additive if and only if for all x, y, z in X, x⊥y and x⊥z imply x⊥y +z. Also, (α, β)-orthogonality is homogeneous provided for all x, y in X, a, b real, x⊥y implies ax⊥by. It has been shown that for a normed linear space X over the reals, the following are equivalent: (1) (α, β)-orthogonality is homogeneous. (2) (α, β)-orthogonality is symmetric and left additive. (3) (α, β)-orthogonality is symmetric and right additive. (4) (α, β)-orthogonality is left and right additive. (5) (X,
·
) is a real inner product space. Carlsson orthogonality generalizes this still further. In [24], S.O. Carlsson defined x orthogonal to y provided m 

ak
bk x + ck y
2 = 0,

k=1

where m ≥ 2 and ak , bk , ck are real numbers such that

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m 

ak bk ck = 1,

k=1

m  k=1

ak b2k

=

m 

ak c2k = 0.

k=1

Note that Carlsson orthogonality is symmetric for some cases, such as pythagorean and isosceles orthogonality but not for other values of the parameters. Boussouis orthogonality includes all of the preceding. In [20], B. Boussouis defined his orthogonality by x⊥y provided  α(ω)
β(ω)x + γ(ω)y
2dµ(ω) = 0, Ω

where (Ω, µ) is a positive measure space and α, β, γ are µ-measurable functions from Ω to the reals such that α(ω) = 0µ almost everywhere, αβ 2 and αγ 2 are µ-integrable and    2 2 α(ω)β (ω)dµ(ω) = α(ω)γ (ω)dµ(ω) = 0, α(ω)β(ω)γ(ω)dµ(ω) = 1. Ω

Ω

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Ω

ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS245

Area orthogonality ([4]) is defined by x⊥y provided either
x

y
= 0 or x and y are linearly independent and the lines spanned by them divide the unit ball of the plane determined by them into four parts of equal area. It may be noted that area orthogonality is homogeneous. Singer orthogonality ([214]) is defined by x⊥y provided  x   y   x   y          either
x

y
= 0 or  + = − .
x

y

x

y
Clearly Singer orthogonality is symmetric and homogeneous. It is known that the following are equivalent to Singer orthogonality in a normed linear space X: (1) X is an inner product space. (2) If x, y are elements of the unit sphere in X and x is orthogonal to αx + y, then |α| ≤ 1. (3) If x, y are elements of the unit sphere in X, then x + y is orthogonal to x − y. Note that Singer orthogonality is simply normalized isosceles orthogonality and is additive in a normed linear space of two dimensions.

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Some authors attribute to Roberts orthogonality x⊥y provided
x − λy
=
x + λy
for all real λ, appearing in [202] in 1935 earlier than the others. Another orthogonality appearing in [5] and [6] the height orthogonality defined by x⊥y provided h1 (x, y) = y + where ρ+ (x, y) = lim

t→0+


y
2 − ρ+ (x, y) ,
x − y
2 (x − y)
x + ty
2 −
x
. 2t

For this orthogonality in a normed linear space X, x⊥y implies αx⊥αy. The following properties are equivalent: (1) The normed linear space X is an inner product space.

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(2) (3) (4) (5)

For For For For

all all all all

x, y x, y x, y x, y

in in in in

X, h1 (x, y) = h1 (y, x). X with
x
=
y
= 1, h1 (x, y) + h1 (y, x) = x + y. X, with
x
=
y
= 1, h1 (x, y) = x+y 2 . X, x⊥y implies x⊥ − y.

Perpendicular bisector orthogonality ([117]) is defined by x⊥y provided ρ+ (x + y, x) = ρ+ (x − y, x), where ρ+ (x, y) is difined as in the height orthogonality. We note that perpendicular bisector orthogonality satisfies the properties that x⊥y implies αx⊥αy, x⊥y implies x⊥ − y and x⊥y implies −x⊥y. The following properties are equivalent: (1) A normed linear space X is an inner product space. (2) For all x, y in X with
x
=
y
= 1, x + y⊥x − y. (3) For all real α, x + αy⊥x − αy implies
x
=
αy
.

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Below are selected consequences of relationships between the various orthogonality relations. The following statements are equivalent in a normed linear space X: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17)

A normed linear space X is an inner product space. Birkhoff orthogonality implies Isosceles orthogonality. Diminnie orthogonality implies Isosceles orthogonality. Perpendicular bisector orthogonality implies Isosceles orthogonality. Singer orthogonality implies Isosceles orthogonality. Birkhoff orthogonality implies Carlsson orthogonality. Birkhoff orthogonality implies Pythagorean orthogonality. Diminnie orthogonality implies Pythagorean orthogonality. Singer orthogonality implies Pythagorean orthogonality. Perpendicular bisector orthogonality implies Pythagorean orthogonality. Isosceles orthogonality implies Birkhoff orthogonality. Isosceles orthogonality implies Pythagorean orthogonality. Carlsson orthogonality implies Birkhoff orthogonality. Pythagorean orthogonality implies Birkhoff orthogonality. Pythagorean orthogonality implies Isosceles orthogonality. Birkhoff orthogonality implies area orthogonality. Area orthogonality implies Birkhoff orthogonality.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS247

(18) (19) (20) (21) (22) (23) (24)

Boussouis orthogonality implies area orthogonality. Area orthogonality implies Boussouis orthogonality. Diminnie orthogonality implies area orthogonality. Carlsson orthogonality implies area orthogonality. Carlsson orthogonality is equivalent to Diminnie orthogonality. Diminnie orthogonality is equivalent to Birkhoff orthogonality. Carlsson orthogonality is equivalent to Birkhoff orthogonality.

Other properties of orthogonality relations have been studied by many authors. For example, we define an orthogonality relation in a normed linear space X to be convex provided for x, y such that x⊥y, there exists a c in (0, 1) such that x⊥cy. It can be shown that if Pythagorean orthogonality is convex, then X is an inner product space.

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Definition 10.1.3. A hyperplane of a normed linear space (X,
·
) is any proper closed linear subset M of X which is not properly contained in a proper linear subset of X, or any translation x + M of such a linear subset M. The following theorems were proven by R.C. James [130] and they are basic to much that follows: Theorem 10.1.1. If f (f = 0) is a linear functional on a normed linear space (X,
·
), then |f (x)| =
f

x
if and only if (x⊥H)(B), where H is the hyperplane of all h for which f (h) = 0. If x is orthogonal to a linear subset L of T , then there exists a linear functional f for which f (x) =
f

x
and f (y) = 0 if y ∈ L, and there is a hyperplane H through the origin with (x⊥H)(B) and L. The following theorem is a direct consequence of Theorem 10.1.1 and the fact that for any element x of a normed linear space (X,
·
), there is a linear functional f such that f (x) =
f

x
. Theorem 10.1.2. Any element of a normed linear space is Birkhoff orthogonal to some hyperplane through the origin. However, for a given hyperplane H through the origin there does not necessarily exist an element x with (x⊥H)(B). This problem is equivalent to finding an element with f (x) =
f

x
for a given linear functional f . Corollary 10.1.3. If x and y are any two elements of a normed linear

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R. W. FREESE AND Y. J. CHO

space (X,
·
), then there exists a number α such that (x⊥(αx + y))(B). Also, (x⊥(αx + y))(B) if and only if there is a linear functional f with |f (x)| =
f

x
and α = −f (y)/f (x). If (x⊥(αx + y))(B), then |α| ≤
y
/
x
. 10.2. Properties of Orthogonalities and the Characterization Theorem We begin this section with two lemmas involving Birkhoff orthogonality. The first was proved by R.C. James [130]. Lemma 10.2.1. (x⊥y)(B) if and only if there is a functional f ∈ F such that f (x) =
x
and f (y) = 0. Lemma 10.2.2. If (x⊥y)(B), then
x, y
≥
x

y
. Proof. If (x⊥y)(B), then, by Lemma 10.2.1 and the relation (10.1),
x, y
≥
x
g(y) for all g ∈ F . The result follows easily by the Hahn–Banach Theorem.

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Note that Lemma 10.2.2 and the properties of Birkhoff orthogonality enable us to show that Diminnie orthogonality ⊥ has the property (D). Theorem 10.2.3. If (x⊥(βx + y))(B), then there are real numbers α1 and α2 such that α1 ≤ β ≤ α2 , (x⊥α1 x + y)(D) and (x⊥α2 x + y)(D). In particular, for each x, y ∈ X, there is a real number α such that (x⊥αx + y)(D). Proof. If x = 0, the results are obvious since (0⊥y)(D) for all y ∈ X. If x = 0 and (x⊥(βx + y))(B), then by Lemma 10.2.2, we have
x, y
=
x, βx + y
≥
x

βx + y
. Since limt→±∞
tx + y
= ∞ and
tx + y
is a continuous function of t, there are real numbers α1 , α2 such that α1 ≤ β ≤ α2 ,
x

α1 x + y
=
x, y
=
x, α1 x + y
and
x

α2 x + y
=
x, y
=
x, α2 x + y
.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS249

The remainder follows from Corollary 10.1.3. In general, for x, y ∈ X and x = 0, the number α of Theorem 10.2.3 need not be unique. Following the convention used by R.C. James [130], we make the following definition: Definition 10.2.1. Diminnie orthogonality ⊥ is unique if for each x, y ∈ X with x = 0, there is a unique real number α such that (x⊥αx + y)(D). Lemma 10.2.4. If ⊥ is additive, then ⊥ is unique. Proof. Let x, y ∈ X with x = 0 and suppose that (x⊥α1 x + y)(D) and (x⊥α2 x + y)(D). Then, by the additivity and homogeneity of ⊥, it follows (x⊥α1 − α2 )(D)x. Using the homogeneity of Diminnie orthogonality ⊥ and the basic properties discussed after Definition 10.1.1, it follows that α1 = α2 . If (X,
·
) is an inner product space, then relation (10.2) implies that
x, y
≤
x

y
for all x, y ∈ X. We will see shortly that for spaces of dimension 3 or larger, this condition characterizes inner product spaces. The next lemma shows that this condition forces ⊥(D) and ⊥(B) to be equivalent.

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Lemma 10.2.5. If
x, y
≤
x

y
for all x, y ∈ X, then Diminnie orthogonality and Birkhoff orthogonality are equivalent. Proof. We may assume that x, y = 0 since in either case (x⊥y)(D) and (x⊥y)(B). If x⊥y, then for each real number α , we have
x + αy

y
≥
x + αy, y
=
x, y
=
x

y
. Therefore,
x+αy
≥
x
for all real α and (x⊥y)(B). The converse follows from Lemma 10.2.2 and the hypothesis. Similarly, the uniqueness of Diminnie orthogonality ⊥ forces the same result. Lemma 10.2.6. If Diminnie orthogonality ⊥ is unique, then Diminnie orthogonality and Birkhoff othogonality are equivalent. Proof. Again, we may assume that x, y = 0. Assume that (x⊥y)(D). By Corollary 10.1.3 there is a real number α such that (x⊥(αx + y))(B). By Theorem 10.2.3 and the uniqueness of Diminnie orthogonality ⊥, it follows that α = 0 and hence we have (x⊥y)(B).

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R. W. FREESE AND Y. J. CHO

Conversely, if (x⊥y)(B), then Theorem 10.2.3 implies that there are real numbers α1 , α2 such that α1 ≤ 0 ≤ α2 , (x⊥α1 x+y)(D) and (x⊥α2 x+y)(D). By the uniqueness of ⊥(D), α1 = α2 = 0 and thus we have (x⊥y)(D). R.C. James [132] proved that, for spaces of dimension 3 or larger, symmetry of Birkhoff orthogonality characterizes inner product spaces. Using this, Lemmas 10.2.4, 10.2.5, 10.2.6 and certain obvious implications, we obtain the main result of this section. Theorem 10.2.7. Let (X,
·
) be a normed linear space of dimension greater than or equal to 3. Then the following are equivalent: (1) (X,
·
) is an inner product space. (2) Birkhoff orthogonality is symmetric. (3) Birkhoff orthogonality and orthogonality are equivalent. (4)
x, y
≤
x

y
for all x, y ∈ X. (5) Diminnie orthogonality is unique. (6) Diminnie orthogonality is additive.

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10.3. The 2–Dimensional Case and Examples Throughout this section, we will assume that X is a 2–dimensional vector space unless otherwise stated. It is well–known that symmetry of Birkhoff orthogonality does not imply that (X,
·
) is an inner product space in this case. We will show that, in this setting, none of the other statements in Theorem 10.2.7 implies that (X,
·
) is an inner product space. Due to Lemmas 10.2.4 and 10.2.5 we need only show that statements (4) and (6) of Theorem 10.2.7 do not imply this result. Before doing so, we will begin with some examples which show that orthogonality and Birkhoff orthogonality are not equivalent in general. Example 10.3.1. Let X = R2 with
(x1 , x2 )
= max{|x1 |, |x2 |} and let e1 = (1, 0), e2 = (0, 1). Then, F = {f : |f (e1 )| + |f (e2 )| ≤ 1}, which implies that
e1 , e2
= 1 and hence, by Theorem 2.2.7, we have
(a1 , a2 ), (b1 , b2 )
= |a1 b2 − a2 b1 |. Now, if x = e1 and y = (1, 2), then x⊥y since
x, y
= 2 =
x

y
. On the other hand, x is not Birkhoff orthogonal to y since
x − 14 y
= 34 <
x
. Therefore, orthogonality does not imply Birkhoff orthogonality.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS251

Example 10.3.2. Let X = R2 with
(x1 , x2 )
= |x1 | + |x2 | and again let e1 = (1, 0), e2 = (0, 1). In this case, F = {f : max(|(f (e1 )|, |f (e2)|) ≤ 1}, which implies that
e1 , e2
= 2 and
(a1 , a2 ), (b1 , b2 )
= 2|a1 b2 − a2 b1 |. Then e1 is not orthogonal to e2 since
e1 , e2
= 2 >
e1

e2
. However, for all real α,
e1 + αe2
= 1 + |α| ≥
e1
and hence (e1 ⊥e2 )(B). Therefore, Birkhoff orthogonality does not imply orthogonality. Before proceeding to our final example, we will need two technical lemmas: Lemma 10.3.1. Diminnie orthogonality ⊥ is additive if and only if there do not exist linearly independent vectors x, y ∈ X with the properties
x
=
y
= 1 and
x, y
=
x − y
.

(10.5)

Proof. Let Diminnie orthogonality ⊥ be additive but suppose that there are linearly independent vectors x, y ∈ X which satisfy (10.5). Then we have
x − y, x
=
x, y
=
x − y
=
x − y

x
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and, similarly,
x − y, y
=
x − y

y
, which imply that (x − y⊥x)(D) and (x − y⊥y)(D). By additivity and homogeneity of Diminnie orthogonality ⊥, (x − y⊥x)(D) and hence x = y, which is impossible. Conversely, assume that there are no linearly independent vectors x, y ∈ X which satisfy (10.5). If (u⊥v)(D) and (u⊥w)(D) but u is not orthogonal to v + w, then, by the homogeneity of Diminnie orthogonality ⊥, we may assume that
v
=
w
= 1 and that v and w are linearly independent. Since X is 2–dimensional, there are real numbers α1 , α2 for which u = α1 v + α2 w. Then, the conditions
u, v
=
u

v
=
u
and
u, w
=
u

w
=
u
imply that |α1 | = |α2 | =


u
.
v, w


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R. W. FREESE AND Y. J. CHO

Therefore, replacing w by −w if necessary, we have found linearly independent vectors v, w ∈ X for which
v
=
w
= 1 and
u
=


u

v − w
or
v, w
=
v − w
.
v, w


Since this contradicts our assumption, the proof is complete. Note that a normed linear space (X,
·
) is strictly convex if and only if the unit sphere S = {x ∈ X :
x
= 1} has no line segment (See Chapter VI). Lemma 10.3.2. If (X,
·
) is strictly convex and
x, y
≤
x

y
for all x, y ∈ X, then Diminnie orthogonality ⊥ is additive. Proof. Suppose that there are linearly independent vectors x, y ∈ X which satisfy (10.5). Then, for 0 ≤ t ≤ 1 and z = tx + (1 − t)y, we note that
z
≤ 1 and
z

x − y
≥
z, x − y
=
y, x − y
=
x, y
=
x − y
≥
z

x − y
.

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Therefore, we have
z

x − y
=
x − y
, which implies that
z
= 1. Since this violates the strict convexity of (X,
·
), no such x and y can exist and Diminnie orthogonality ⊥ is additive by Lemma 10.3.1. The last example illustrates a strictly convex 2–dimensional space which satisfies statement (4) of Theorem 10.2.7 (and hence, statement (6) also) but is not an inner product space. Example 10.3.3. Let X = R2 with
·
defined by

(x1 , x2 )
=

(|x1 |3 + |x2 |3 )1/3 (|x1 |

3/2

+ |x2 |

if x1 x2 ≥ 0,

3/2 2/3

)

if x1 x2 ≤ 0.

(X,
·
) is strictly convex since its unit sphere contains no line segment. If e1 = (1, 0) and e2 = (0, 1), then F is the set of linear functionals f satisfying |f (e1 )|3/2 + |f (e2 )|3/2 ≤ 1 if f (e1 )f (e2 ) ≥ 0 or

|f (e1 )|3 + |f (e2 )|3 ≤ 1 if f (e1 )f (e2 ) ≤ 0.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS253

Note that in either case, |f (e1 )| ≤ 1 and |f (e2 )| ≤ 1. To evaluate
·, ·
, we need to calculate
e1 , e2
and then use Theorem 2.1.1. Let f, g ∈ F and consider the following two main cases: Case 1. If f (e1 )f (e2 ) ≥ 0 and g(e1 )g(e2 ) ≥ 0, then we have f (e1 )g(e2 ) − g(e1 )f (e2 ) ≤ max{|f (e1 )||g(e2 )|, |g(e1 )||f (e2 )|} ≤ 1. older’s Inequality, Case 2. If f (e1 )f (e2 ) ≥ 0 and g(e1 )g(e2 ) ≤ 0, then by Hˆ f (e1 )g(e2 ) − g(e1 )f (e2 ) ≤ (|f (e1 )|3/2 + |f (e2 )3/2 |)2/3 (|g(e1 )|3 + |g(e2 )|3 )1/3 ≤ 1. Since the remaining cases are similar, we see that
e1 , e2
≤ 1. Further, since the functionals f (x1 , x2 ) = x1 and g(x1 , x2 ) = x2 are in F, it follows that
e1 , e2
= 1 and hence, by Theorem 2.1.1, that
(a1 , a2 ), (b1 , b2 )
= |a1 b2 − a2 b1 |. If x = (a1 , a2 ) and y = (b1 , b2 ), consider the following cases:

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Case 1. If a1 a2 ≥ 0 and b1 b2 ≥ 0, then
x, y
= |a1 b2 − a2 b1 | ≤ max(|a1 ||b2 |, |a2 ||b1 |) ≤
x

y
. Case 2. If a1 a2 ≥ 0 and b1 b2 ≤ 0, then by Hˆ older’s Inequality,
x, y
= |a1 b2 − a2 b1 | ≤ (|a1 |3 + |a2 |3 )1/3 (|b1 |3/2 + |b2 |3/2 )2/3 =
x

y
. Since the remaining cases are similar, it follows that statement (4) of Theorem 10.2.7 is valid for (X,
·
). The strict convexity of (X,
·
) and Lemma 10.3.2 imply that statement (6) holds also. However. if we note that
e1 + e2
2 +
e1 − e2
2 = 22/3 + 24/3 while

2(
e1
2 +
e2
2 ) = 4.

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We see that the “Parallelogram law” is not valid in (X,
·
) and therefore, (X,
·
) is not an inner product space. 10.4. Orthogonal Triples in Linear 2–Normed Spaces In [33], Y.J. Cho, C.R. Diminnie, R.W. Freese and E.Z. Andalaffe defined an isosceles orthogonal triple I(·, ·, ·) in linear 2–normed spaces, gave some basic properties of I(·, ·, ·), and showed that under the assumption of strict convexity, every subspace of X of dimension ≤ 3 contains an isosceles orthogonal triple. Further, if (X,
·, ·
) is strictly convex and I(·, ·, ·) is either homogeneous or additive, then (X,
·, ·
) is a 2–inner product space. The concept of isosceles orthogonality was introduced by R.C. James [129]. In a real normed linear space (X,
·
), x and y are said to be isosceles orthogonal if
x + y
=
x − y
. The term “isosceles” refers to the fact that x + y and x − y form the congruent sides of an isosceles triangle. In an inner product space, this is equivalent to the usual meaning of orthogonality, that is, that the inner product of x and y is 0. Motivatived by these considerations, we offer the following analogous definition for linear 2–normed spaces.

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Definition 10.4.1. In a linear 2–normed space (X,
·, ·
), (x, y, z) is an isosceles orthogonal triple, denoted by I(x, y, z), provided that
x + y, z
=
x − y, z
,
x + z, y
=
x − z, y
and
y + z, x
=
y − z, x
. Some of the elementary properties of the relation I(·, ·, ·) are summarized in our first result. Theorem 10.4.1. Let (X,
·, ·
) be a linear 2–normed space. (1) (Symmetry) If I(x, y, z), then I(p(x), p(y), p(z)) for any permutation p of the set {x, y, z}. (2) If l(x, y, z) and |a| = |b| = |c| = 1, then I(ax, by, cz). (3) If V (x, y, z) is 0 or 1–dimensional, then I(x, y, z). (4) For any x, y in X, I(x, y, 0). (5) If V (x, y, z) is 2–dimensional, then I(x, y, z) if and only if two of the elements x, y, z are linearly independent and the third is 0.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS255

(6) (Consistency) If (X,
·, ·
) is a 2–inner product space with 2–inner product (·, ·|·), then I(x, y, z) if and only if (x, y|z) = (x, z|y) = (y, z|x) = 0. Proof. Statements (1) and (2) are direct consequences of the definition of I(·, ·, ·). Also, statements (3) and (4) follow immediately from the definition of
·, ·
. Half of statement (5) follows from statement (4) and the fact that V (x, y, z) is 2–dimensional. To prove the other half, we may assume that y and z are linearly independent and x = ay + bz for some real numbers a and b. If I(x, y, z), then from
x, ax + y
=
x, y
, we see that the condition
x+y, z
=
x−y, z
implies that |1+a|
y, z
= |1−a|
y, z
. Since
y, z
> 0, we must have a = 0. Similarly, the condition
x+z, y
=
x−z, y
implies that b = 0 and hence x = 0. Finially, statement (6) follows from the easily proven relation (See Theorem 5.1.9) (x, y|z) =

1 (
x + y, z
2 −
x − y, z
2 ). 4

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To prove the next main result, we need an additional concept and some technical lemmas: Our first lemma is a direct consequence of the definition of strict convexity. By C.R. Diminnie, S. G¨ ahler and A. White [56], it is proved that every 2– inner product space is strictly convex (cf. Corollary 6.1.3). Lemma 10.4.2. If (X,
·, ·
) is strictly convex and {x, y, z} is linearly independent, then
x + y, z
<
x, z
+
y, z
. Lemma 10.4.3. Let {x, y, z} be linearly independent and a1 , a2 be real numbers with a1 < a2 . (1) If a1 < b < a2 and
y + a1 x, z
=
y + a2 x, z
, then
y + bx, z
≤
y + a1 x, z
. (2) If
y + a1 x, z
=
y + a2 x, z
and a1 < a2 < b or b < a1 < a2 , then
y + a1 x, z
≤
y + bx, z
. (3) If
y + a1 x, z
=
y + a2 x, z
= min{
y + a x, z
: a ∈ R} and a1 < b < a2 , then
y + bx, z
= min{
y + a x, z
: a ∈ R} also. (4) If a1 < a2 < a3 and
y + a1 x, z
=
y + a2 x, z
=
y + a3 x, z
, then
y + ax, z
= min{
y + a x, z
: a ∈ R} for all a in [a1 , a3 ]. Proof. (1) If a1 < b < a2 , then there is a number t in (0,1) such that

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R. W. FREESE AND Y. J. CHO

b = ta1 + (1 − t)a2 . Therefore, we have
y + bx, z
=
t(y + a1 x) + (1 − t)(y + a2 x), z
≤ t
y + a1 x, z
+ (1 − t)
y + a2 x, z
=
y + a1 x, z
. (2) If a1 < a2 < b, then there is number t in (0,1) such that a2 = ta1 + (1 − t)b. Then, as in part (1), we have
y + a1 x, z
=
y + a2 x, z
≤ t
y + a1 x, z
+ (1 − t)
y + bx, z
, which implies that
y + a1 x, z
≤
y + bx, z
. The proof of remaining case is similar. (3) Note first that since it is easily shown that
y + a x, z
is a continuous function of a and
y + a x, z
→ ∞ as a → ±∞,
y + ax, z
attains a minimum value on R. The result is then a consequence of statement (1). (4) Statements (1) and (2) imply that the minimum value of
y + a x, z
is
y + a1 x, z
. The result then follows from statement (3).

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The next lemma is an adaptation of a similar result for normed linear spaces presented by R.C. James [131]. Lemma 10.4.4. If {r, s, t} is a linearly independent set and u(a) =
(a + 1)s + r, t
−
(a − 1)s + r, t
, then u(a) → ±2
s, t
as a → ±∞. Further, u(a) > 0 whenever a ≥ 1 + 2
r, t
/
s, t
and u(a) < 0 whenever a ≤ −(1 + 2
r, t
/
s, t
). Proof. For a > 1, since (a − 1)/(a + 1) + 2/(a + 1) = 1, we may write u(a) =
(a − 1)s + (a − 1)/(a + 1)r, t
−
(a − 1)s + r, t
+ 2
s  + 1/(a + 1)r, t
. Since 
(a − 1)s + (a − 1)/(a + 1)r, t
−
(a − 1)s + r, t
 ≤ 2/(a + 1)
r, t
, it is obvious that u(a) → 2
s, t
as a → ∞. Now, if we assume that a ≥ 1 + 2
r, t
/
s, t
, then all the above steps still hold. Further, it follows that   2 a−1      r, t− (a − 1)s + r, t <
r, t
<
s, t
(a − 1)s + a+1 a while we have

  2s +

 2 1  r, t ≥ 2
s, t
−
r, t
a+1 a+1 2 > 2
s, t
−
r, t
>
s, t
. a

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS257

Therefore, u(a) > 0 when a ≥ 1 + 2
r, t
/
s, t
. The remainder may be proved by substituting −a for a and using the properties of
·, ·
. Our next main result shows that in strictly convex linear 2–normed spaces, it is possible to prove a version of the Gram–Schmidt procedure. Whether this is true without the condition of strict convexity is an open question. Theorem 10.4.5. If (X,
·, ·
) is strictly convex, then for any x, y, z in X, there are real numbers a , b , c such that I(x, a x + y, b x + c y + z). Proof. Since the proof is quite lengthy, we will break it up into several smaller steps: Step 1. Suppose {x, y, z} is dependent, that is, there are real numbers p, q, r, not all 0, such that px + qy + rz = 0. If r = 0, then (p/r)x + (q/r)y + z = 0,

I(x, y, (p/r)x + (q/r)y + z).

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If r = 0, then px + qy = 0, with p and q not both 0. If q = 0, then (p/q)x + y = 0 and hence, I(x, (p/q)x + y, z). Finally, if q = 0, then p = 0 and px = 0 imply that x = 0 and we have I(x, y, z). For the remaining steps, we will assume that {x, y, z} is linearly independent. Step 2. By Lemma 10.4.4, (
(c + 1)y + z, x
−
(c − 1)y + z, x
) → ±2
y, x
as c → ±∞. By the usual continuity arguments, it follows that there is a real number c such that
(c + 1)y + z, x
=
(c − 1)y + z, x
. Throughout the remainder of this proof, let w = c y +z and note that {x, y, w} is linearly independent and
y+w, x
=
y−w, x
. By the strict convexity of (X,
·, ·
), we have 2
y, x
<
y + w, x
+
y − w, x
= 2
y − w, x
, which implies that
x, y
<
x, y + w
and
x, y
<
x, y − w
. Similar steps show that
x, w
<
x, w − y
and
x, w
<
x, w + y
. Step 3. Define f (a, b) =
(b + 1)x + w, ax + y
−
(b − 1)x + w, ax + y
and g(a, b) =
(a + 1)x + y, bx + w
−
(a − 1)x + y, bx + w
.

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258

R. W. FREESE AND Y. J. CHO

If we consider h(e, d) =
ex + y, dx + w
, then we have |h(e1 , d1 ) − h(e2 , d2 )| ≤ |h(e1 , d1 ) − h(e1 , d2 )| + |h(e1 , d2 ) − h(e2 , d2 )|   = 
e1 x + y, d1 x + w
−
e1 x + y, d2 x + w
   + 
e1 x + y, d2 x + w
−
e2 x + y, d2 x + w
 ≤
e1 x + y, (d1 − d2 )x
+
(e1 − e2 )x, d2 x + w


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= |e1 − e2 |
x, w
+ |d1 − d2 |
x, y
. Therefore, h is uniformly continuous and it follows easily that f and g are uniformly continuous as well. Step 4. If we apply Lemma 10.4.4 to the functions f and g, we have the following: A. f (a, b) → ±2
x, ax + y
= ±2
x, y
as b → ±∞. B. f (a, b) > 0 when b ≥ 1 + (2/
x, y
)
ax + y, w
and f (a, b) < 0 when b ≤ −(1 + (2/
x, y
)
ax + y, w
). C. g(a, b) → ±2
x, bx + w
= ±2
x, w
as a → ±∞. D. g(a, b) > 0 when a ≥ 1 + (2/
x, w
)
bx + w, y
and g(a, b) < 0 when a ≤ −(1 + (2/
x, w
)
bx + w, y
). As a consequence, if Z = {(a, b) : f (a, b) = 0} and L(a) = {(a, b) : b ∈ R}, we see that Z is a non–empty closed set such that Z ∩ L(a) = Φ for all real a. Also, Z is bounded above and below by the continuous curves b = ±(1 + (2/
x, y
)
ax + y, w
). The next two steps develop further properties of Z and Z ∩ L(a). Step 5. By our discussion at the end of Step 4, each set Z ∩ L(a) is a non–empty compact set. We will show that this set is also connected. Suppose (a, b1 ) and (a, b2 ) are elements of Z ∩ L(a) with b1 < b2 . Since b1 − 1 < b2 − 1 and b1 + 1 < b2 + 1, we need consider only two cases:

(10.6) (10.7)

b1 − 1 < b1 + 1 ≤ b2 − 1 < b2 + 1, b1 − 1 < b2 − 1 < b1 + 1 < b2 + 1.

If (10.6) holds, then by Lemma 10.4.3, the conditions b1 + 1 ≤ b2 − 1 and
(b1 − 1)x + w, ax + y
=
(b1 + 1)x + w, ax + y
imply that
(b1 + 1)x + w, ax + y
≤
(b2 − 1)x + w, ax + y
.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS259

Also, b1 + 1 ≤ b2 − 1 and
(b2 − 1)x + w, ax + y
=
(b2 + 1)x + w, ax + y
imply that
(b2 − 1)x + w, ax + y
≤
(b1 + 1)x + w, ax + y
. Therefore, we have
(b1 − 1)x + w, ax + y
=
(b1 + 1)x + w, ax + y
=
(b2 − 1)x + w, ax + y
=
(b2 + 1)x + w, ax + y
, which implies, by Lemma 10.4.3, that
dx + w, ax + y
= min{
b x + w, ax + y
: b ∈ R} for all d in [b1 −1, b2 +1]. If b1 < b < b2 , then we have b1 −1 < b +1 < b2 +1, which implies that
(b1 − 1)x + w, ax + y
=
(b + 1)x + w, ax + y


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= min{
b x + w, ax + y
: b ∈ R}. Therefore, (a, b) is in Z ∩ L(a) and Z ∩ L(a) is connected. If (10.7) holds, then b1 − 1 < b2 − 1 < b1 + 1 and
(b1 − 1)x + w, ax + y
=
(b1 + 1)x + w, ax + y
imply, by Lemma 10.4.3, that
(b2 − 1)x + w, ax + y
≤
(b1 + 1)x + w, ax + y
. Similarly, it follows that b2 −1 < b1 +1 < b2 +1 and
(b2 −1)x+w, ax+y
=
(b2 + 1)x + w, ax + y
imply
(b1 + 1)x + w, ax + y
≤
(b2 + 1)x + w, ax + y
. Then we have
(b1 − 1)x + w, ax + y
=
(b2 − 1)x + w, ax + y
=
(b1 + 1)x + w, ax + y
=
(b2 + 1)x + w, ax + y
and the proof proceeds as in Case (10.5) above.

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260

R. W. FREESE AND Y. J. CHO

Note that for each real a, since Z ∩L(a) is connected and f (a, b) → ±2
x, y
as b → ±∞, we see that if f (a, b1) < 0 then f (a, b) < 0 for all b ≤ b1 and if f (a, b2 ) > 0, then f (a, b) > 0 for all b ≥ b2 . Step 6. In this step, we show that Z is connected. To do so, it suffices to show that for each n ≥ 1, the set Zn = Z ∩ {(a, b) : −n ≤ a ≤ n} is connected. Suppose that for some n, Zn is disconnected, that is, there are disjoint, non–empty, closed subsets A1 , A2 of Zn such that Zn = A1 ∪ A2 . Since Zn is compact, A1 and A2 must be compact also. If p1 denotes the projection function p1 (a, b) = a, then since p1 is continuous, p1 (A1 ) and p1 (A2 ) are non–empty compact subsets of [−n, n]. Also, by Step 4, [−n, n] = p1 (A1 ) ∪ p1 (A2 ). Then, the connectedness of [−n, n] implies that p1 (A1 ) ∩ p1 (A2 ) = Φ. If a is a number common to p1 (A1 ) and p1 (A2 ), then A1 ∩ L(a ) and A2 ∩ L(a ) are disjoint, non–empty, closed subsets of Z ∩ L(a ) such that (A1 ∩ L(a )) ∪ (A2 ∩ L(a )) = (A1 ∪ A2 ) ∩ L(a )

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= Zn ∩ L(a ) = Z ∩ L(a ), which violates Step 5. Therefore, each Zn is connected and Z must also be connected. Step 7. Note that f (a, a) =
(a + 1)x + w, ax + y
−
(a − 1)x + w, ax + y
=
(a + 1)x + w, (a + 1)x + w − (x + w − y)
−
(a − 1)x + w, (a − 1)x + w + (x + y − w)
=
(a + 1)x + w, x + w − y
−
(a − 1)x + w, x + y − w
=
(a + 1)x + w, x + w − y
−
(a − 1)x + w, x + w − y
+
(a − 1)x + w, x + w − y
−
(a − 1)x + w, x + y − w
. By Lemma 10.4.4, (
(a + 1)x + w, x + w − y
−
(a − 1)x + w, x + w − y
) → 2
x, y − w
as a → ∞. Also, by the remarks at the end of Step 2,   
(a − 1)x + w, x + w − y
−
(a − 1)x + w, x + y − w
 ≤ 2
(a − 1)x + w, x
= 2
x, w
< 2
x, y − w
.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS261

Therefore, there is an a1 > 0 such that f (a, a) > 0 when a ≥ a1 . Similarly, there are real numbers a2 < 0, a3 > 0 and a4 < 0 such that    f (a, −a) > 0 if a ≤ a2 , f (a, −a) < 0 if a ≥ a3 ,   f (a, a) < 0 if a ≤ a4 ,

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and there real numbers b1 > 0, b2 < 0, b3  g(b, b) > 0     g(−b, b) < 0  g(−b, b) > 0    g(b, b) < 0

> 0 and b4 < 0 such that if b ≥ b1 , if b ≤ b2 , if b ≥ b3 , if b ≤ b4 .

Step 8. By Step 7 and the remarks at the end of Step 5, we have f (a, b) > 0 when b ≥ a ≥ a1 or b ≥ −a ≥ −a2 . Also, by Step 4, f (a, b) > 0 when b ≥ 1 + (2/
x, y
)
ax + y, w
. Since b = 1 + (2/
x, y
)
ax + y, w
is continuous as function of a, it is bounded on the interval [a2 , a1 ]. Therefore, there is a real number d > 0 such that f (a, b) > 0 when b ≥ d and a2 ≤ a ≤ a1 . If k1 = max{d, −a2 , a1 }, then f (a, b) > 0 when b ≥ k1 and |a| ≤ k1 or b ≥ |a| ≥ k1 . Similar arguments lead to the existence of positive real numbers k2 , k3 , k4 such that f (a, b) < 0 when b ≤ −k2 and |a| ≤ k2 or b ≤ −|a| ≤ −k2 , g(a, b) > 0 when a ≥ k3 and |b| ≤ k3 or a ≥ |b| ≥ k3 , and g(a, b) < 0 when a ≤ −k4 and |b| ≤ k4 or a ≤ −|b| ≤ −k4 . If we choose k = max{k1 , k2 , k3 , k4 }, then k > 0 and  f (a, b) > 0 if b = k and |a| ≤ k,     f (a, b) < 0 if b = −k and |a| ≤ k,  g(a, b) > 0 if a = k and |b| ≤ k,    g(a, b) < 0 if a = −k and |b| ≤ k. Since f (k, k) > 0 and f (k, −k) < 0, there is a number b5 in [−k, k] such that f (k, b5 ) = 0, that is, (k, b5 ) is in Z. Further, since |b5 | ≤ k, g(k, b5) > 0. Similarly, there is a real number b6 in [−k, k] such that (−k, b6 ) is in Z and g(−k, b6 ) < 0. Since Z is connected and g is continuous, it follows that there is a point (a , b ) in Z such that g(a , b ) = 0. Then, f (a , b ) = g(a , b ) = 0. Step 9. Recall from Step 2 that w = c y + z. If we let u = a x + y and v = w + b x = b x + c y + z, then we have 0 =
(c + 1)y + z, x
−
(c − 1)y + z, x
=
v + u, x
−
v − u, x
,

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262

R. W. FREESE AND Y. J. CHO

0 = f (a , b ) =
(b + 1)x + w, a x + y
−
(b − 1)x + w, a x + y
=
v + x, u
−
v − x, u
, 0 = g(a , b ) =
(a + 1)x + y, b + w
−
(a − 1)x + y, b x + w
=
u + x.v
−
u − x, v
. Therefore, I(x, u, v) and the proof is complete. Remark. Note that in the proof of Theorem 10.4.5, for linearly independent x, y, z, the number c was chosen first and it depended on x, y, z while the choices of a and b depended on c , x, y, z. We conclude with two results concerning some additional properties of I(·, ·, ·):

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Definition 10.4.2. (1) I(·, ·, ·) is said to be homogeneous if I(x, y, z) implies that I(ax, y, z) for all real a. (2) I(·, ·, ·) is said to be additive if I(x1 , y, z) and I(x2 , y, z) imply that I(x1 + x2 , y, z). By the symmetry of I(·, ·, ·), if I(·, ·, ·) is homogeneous, then I(ax, by, cz) for all a, b, c. Theorem 10.4.6. If (X,
·, ·
) is strictly convex and I(·, ·, ·) is homogeneous, then (X,
·, ·
) is a 2–inner product space. Proof. By a theorem of C.R. Diminnie, S. G¨ ahler and A. White, it suffices to show that
x + y, z
=
x − y, z
implies that
x + 2y, z
=
x − 2y, z
for all x, y, z in X. We consider the following cases: Case 1. If x, y and z are linearly independent, then since
x + y, z
=
x − y, z
, the proof of Theorem 10.4.5 implies that there are real numbers a and b such that I(z, a z + y, b z + x). By the homogeneity of I(·, ·, ·), it follows that I(z, 2(a z + y), b z + x) holds also. Therefore, we have
2(a z + y) + (b z + x), z
=
2(a z + y) − (b z + x), z
which reduces to
x + 2y, z
=
x − 2y, z
.

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ORTHOGONALITY RELATIONS BETWEEN THE NORMS AND 2-NORMS263

Case 2. If x, y, z are linearly dependent but y and z are linearly independent, then there are real numbers d and e such that x = d y + e z. Since
x + y, z
=
x − y, z
, we have
(d + 1)y + e z, z
=
(d − 1)y + e z, z
or

|d + 1|
y, z
= |d − 1|
y, z
,

which implies that d = 0 since
y, z
> 0. Therefore, we have x = e z and
x + 2y, z
=
2y, z
=
x − 2y, z
. Case 3. If y and z are linearly dependent, then
x − 2y, z
=
x, z
=
x − 2y, z
.

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Theorem 10.4.7. If (X,
·, ·
) is strictly convex and I(·, ·, ·) is additive, then (X,
·, ·
) is a 2–inner product space. Proof. It suffices to show that additivity of I(·, ·, ·) implies homogeneity of I(·, ·, ·). Assume I(·, ·, ·) is additive and let I(x, y, z) hold. Then, it follows that I(nx, y, z) holds for every positive integer n. Further, by Theorem 10.4.1, I(0, y, z) and I(−x, y, z) hold also. Therefore, we may extend this to obtain I(nx, y, z) for all integers n. By using the symmetry of I(·, ·, ·), we may repeat this process to get I(nx, my, kz) for all integers n, m, k. Thus, for all integer n, m, k, we have
nx + my, z
=
nx − my, z
,
my + kz, x
=
my − kz, x
and
nx + kz, y
=
nx − kz, y
, which imply that for every rational number r,
rx + y, z
=
rx − y, z
,
y + z, rx
=
y − z, rx
and
rx + z, y
=
rx − z, y
. By the usual continuity arguments, these in turn imply that I(ax, y, z) for all real a and the proof is complete.

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R. W. FREESE AND Y. J. CHO

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As mentioned previously, it remains an open question whether Theorem 10.4.5 is true without the assumption of strict convexity. Since Theorems 10.4.6 and 10.4.7 depend on Theorem 10.4.5, the same is true in their cases. Additionally, the reader might be interested in developing a similar treatment for other well–known concepts of orthogonality, such as Pythagorean orthogonality or Birkhoff orthogonality.

Geometry of Linear 2-Normed Spaces, Nova Science Publishers, Incorporated, 2001. ProQuest Ebook Central,

CHAPTER 11. QUADRATIC FORMS ON MODULES In this chapter, some properties of three new quadratic forms on a module over a *–algebra defined by C.S. Lin [155] are given. Also, he showed that there exists a sesquilinear form such that both forms are equal to each other. So far as application is concerned these results enable us to form new characterization formulas for an inner product space if we restrict attention to normed linear spaces. Finally, we introduce a new generalization of a real 2–inner product space among real linear 2–normed spaces by using the concept of quadratic forms.

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11.1. Generalized A–Quadratic Forms of Type (P) Recently, J. Vukman [239], [240] has shown that given an A–quadratic form on a unitary left A–module, there exists an A–sequilinear form with relations between them. In this section, we introduce the concept and some properties of a generalized A–quadratic form defined by C.S. Lin [155]. These results characterize those of J. Vukman [239], [240] which in turn generalized Kurepa’s extensions of Jordan–Neuman’s generalization of inner product spaces among normed linear spaces. A Banach *–algebra is a *–algebra (an algebra with involution) which also is a Banach algebra. Let A be a *–algebra with a unity element e and let X be a linear space which is also a left A–module. A mapping B : X × X → A is called an A–sesquilinear form if B is additive in both arguments, B(ax, y) = aB(x, y) and B(x, ay) = B(x, y)a∗ for all x, y ∈ X and a ∈ A. A left A–module X is said to be unitary if ex = x for all x ∈ X. Definition 11.1.1. A mapping Q : X → A is called a generalized A– quadratic form of type (P) if Q(ax) = aQ(x)a∗ ,

(11.1) cQ

n
 i=1

(11.2)

v1



n n
   Q vi − (n + c − 1)vj + j=2

i=1

n n n−1 
   Q(vi ) + Q(vi − vj ) = (n + c − 1) Q(v1 ) + c i=2

i