Norm Derivatives and Characterizations of Inner Product Spaces

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Norm Derivatives and Characterizations of

Inner Product Spaces

This page intentionally left blank

Norm Derivatives and Characterizations of

Inner Product Spaces

Claudi Alsina

Universitat Politècnica de Catalunya, Spain

Justyna Sikorska Silesian University, Poland

M Santos Tomás

Universitat Politècnica de Catalunya, Spain

World Scientific NEW JERSEY

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LONDON

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Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office 57 Shelton Street, Covent Garden, London WC2H 9HE

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

NORM DERIVATIVES AND CHARACTERIZATIONS OF INNER PRODUCT SPACES Copyright © 2010 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

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ISBN-13 978-981-4287-26-5 ISBN-10 981-4287-26-1

Printed in Singapore.

Preface

The aim of this book is to provide a complete overview of characterizations of normed linear spaces as inner product spaces based on norm derivatives and generalizations of the most basic geometrical properties of triangles in normed linear spaces. Since the monograph by Amir that has appeared in 1986, with only a few results involving norm derivatives, a lot of papers have been published in this field, many of them by us and our collaborators. So we have decided to collect all these results and present them in a systematic way. In doing this, we have found new results and improved proofs which may be of interest for future researchers in this field. To develop this area, it has been necessary to find new techniques for solving functional equations and inequalities involving norm derivatives. Consequently, in addition to the characterizations of Banach spaces which are Hilbert spaces (and which have their own geometrical interest), we trust that the reader will benefit from learning how to deal with these questions requiring new functional tools. This book is divided into six chapters. Chapter 1 is introductory and includes some historical notes as well as the main preliminaries used in the different chapters. The bulk of this chapter concerns real normed linear spaces, inner product spaces and the classical orthogonal relations of James and Birkhoff and the Pythagorean relation. In presenting this, we also fix the terminology and notational conventions which are used in the sequel. Chapter 2 is devoted to the key concepts of the publication: norm derivatives. These functionals extend inner products, so many geometrical properties of Hilbert spaces may be formulated in normed linear spaces by means of the norm derivatives. We develop a complete description of their main properties, paying special attention to orthogonality relations associated to these norm derivatives and proving some interesting charv

vi

Norm Derivatives and Characterizations of Inner Product Spaces

acterizations on the derivability of the norm from inner products. New orthogonality relations are introduced and studied in detail, comparing these orthogonalities with the classical Pythagorean, Birkhoff and James orthogonalities. Chapters 3, 4 and 5 are devoted to studying heights, perpendicular bisectors and bisectrices in triangles located in normed linear spaces, respectively. In doing a detailed study of the basic geometrical properties of these lines and their associated points (orthocenters, circumcenters and incenters), we show a distinguished collection of characterizations of normed linear spaces as inner product spaces. Chapter 6 is devoted to areas of triangles in normed linear spaces. The book concludes with an appendix in which we present a series of open problems in these fields that may be of interest for further research. Finally, we list a comprehensive bibliography about this topic and a general index. This publication is primarily intended to be a reference book for those working on geometry in normed linear spaces, but it is also suitable for use as a textbook for an advanced undergraduate or beginning graduate course on norms and inner products and analytical techniques for solving functional equations characterizing norms associated to inner products. We are grateful to Prof. Roman Ger (Katowice, Poland) for his positive remarks, and to Ms Rosa Navarro (Barcelona, Spain) for her efficient typing of the various versions of our manuscript.

Special Notations

|·| \ A± (·, ·) b± (·, ·) BX Bx (r) ◦ dim h·, ·i f −1 inf i.p.s. max min PA ⊥P ⊥A ⊥B ⊥J ⊥ρ , ⊥ρ R R+ k·k

absolute value of a real number difference of two sets angle generalized bisectrix closed unit ball in a normed space closed ball of radious r centered at x composition of functions dimension inner product inverse function of f infimum inner product space maximum minimum metric projection on A Pythagorean orthogonality area orthogonality Birkhoff orthogonality James orthogonality ρ′± orthogonality real line positive half-line norm

vii

viii


ρ′+ , ρ′− ρ′′+ , ρ′′− sgn k · ke k · k+ k · k∨ k · kn X (X, h·, ·i) xy (X, k · k) [x, y] (x, y) hx, yi kxk Sx (r) SX Kx (r) w(·, ·) [x]B k·k [x]ρk·k

direction one-sided derivative of the square of the norm direction one-sided derivative of ρ′+ , ρ′− , respectively generalized sign function sgn(x) := x/kxk for x 6= 0, sgn(0) = 0 Euclidean norm taxi-cab norm diamond norm mixed norm generic vector space generic i.p.s. line determined by x, y generic real normed space closed segment determined by x, y open closed segment determined by x, y inner product of x, y norm of x sphere of radius r centered at x unit sphere centered at 0 open ball of radius r centered at x vectorial bisectrix segment the Birkhoff’s orthogonal set of x the ρ-orthogonal set of x

Contents

Preface

v

Special Notations

vii

1. Introduction

1

1.1 1.2 1.3 1.4 1.5

Historical notes . . . . . . . . . . . . . . Normed linear spaces . . . . . . . . . . . Strictly convex normed linear spaces . . Inner product spaces . . . . . . . . . . . Orthogonalities in normed linear spaces

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

2. Norm Derivatives

1 3 7 7 11 15

2.1 2.2 2.3 2.4 2.5 2.6

Norm derivatives: Definition and basic properties . . . . . . Orthogonality relations based on norm derivatives . . . . . ρ′± -orthogonal transformations . . . . . . . . . . . . . . . . On the equivalence of two norm derivatives . . . . . . . . . Norm derivatives and projections in normed linear spaces . Norm derivatives and Lagrange’s identity in normed linear spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 On some extensions of the norm derivatives . . . . . . . . . 2.8 ρ-orthogonal additivity . . . . . . . . . . . . . . . . . . . . . 3. Norm Derivatives and Heights

15 26 30 35 38 41 45 51 57

3.1 Definition and basic properties . . . . . . . . . . . . . . . . 3.2 Characterizations of inner product spaces involving geometrical properties of a height in a triangle . . . . . . . . . . . ix

57 60

x


3.3 3.4 3.5 3.6

Height functions and classical orthogonalities A new orthogonality relation . . . . . . . . . Orthocenters . . . . . . . . . . . . . . . . . . A characterization of inner product spaces isosceles trapezoid property . . . . . . . . . . 3.7 Functional equations of the height transform

. . . . . . . . . . . . . . . . . . involving . . . . . . . . . . . .

. . . . . . an . . . .

4. Perpendicular Bisectors in Normed Spaces

5. Bisectrices in Real Normed Spaces Bisectrices in real normed spaces . . . . . . . . . . . A new orthogonality relation . . . . . . . . . . . . . Functional equation of the bisectrix transform . . . . Generalized bisectrices in strictly convex real normed Incenters and generalized bisectrices . . . . . . . . .

91 94 103

4.1 Definitions and basic properties . . . . . . . . . . . . . . . . 4.2 A new orthogonality relation . . . . . . . . . . . . . . . . . 4.3 Relations between perpendicular bisectors and classical orthogonalities . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 On the radius of the circumscribed circumference of a triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Circumcenters in a triangle . . . . . . . . . . . . . . . . . . 4.6 Euler line in real normed space . . . . . . . . . . . . . . . . 4.7 Functional equation of the perpendicular bisector transform

5.1 5.2 5.3 5.4 5.5

74 81 85

103 106 111 115 117 124 125 131

. . . . . . . . . . . . spaces . . . .

6. Areas of Triangles in Normed Spaces 6.1 Definition of four areas of triangles . . . . . . . . . . . . 6.2 Classical properties of the areas and characterizations of inner product spaces . . . . . . . . . . . . . . . . . . . . 6.3 Equalities between different area functions . . . . . . . . 6.4 The area orthogonality . . . . . . . . . . . . . . . . . . .

131 136 144 149 156 163

. . 163 . . 164 . . 169 . . 172

Appendix A Open Problems

177

Bibliography

179

Index

187

Chapter 1

Introduction

1.1

Historical notes

Functional analysis arose from problems on mathematical physics and astronomy where the classical analytical methods were inadequate. For example, Jacob Bernoulli and Johann Bernoulli introduced the calculus of variations in which the value of an integral is considered as a function of the functions being integrated, so functions became variables. Indeed, the word “functional” was introduced by Hadamard in 1903, and derivatives of functionals were introduced by Fréchet in 1904 [Momma (1973); Dieudonné (1981)]. A key step in this historical development is precisely the contribution made in 1906 by Maurice Fréchet in formulating the general idea of metric spaces, extending the classical notion of the Euclidean spaces Rn , so distance measures could be associated to all kinds of abstract objects. This opened up the theory of metric spaces and their future generalizations, extending topological concepts, convergence criteria, etc. to sequence spaces or functional structures. In 1907, Fréchet himself, and Hilbert’s student, Schmidt, studied sequence spaces in analogy with the theory of square summable functions, and in 1910, Riesz founded operator theory. Motivated by problems on integral equations related to the ideas of Fourier series and new challenges in quantum mechanics, Hilbert used distances defined via inner products. In 1920, Banach moved further from inner product spaces to normed linear spaces, founding what we may call modern functional analysis. Indeed, the name “Banach spaces” is due to Fréchet and, independently, Wiener also introduced this notion. Banach’s research [Banach (1922); Banach (1932)] generalized all previous works on integral equations by

1

2


Volterra, Fredholm and Hilbert, and made it possible to prove strong results, such as the Hahn-Banach or Banach-Steinhaus theorems. Abstract Hilbert spaces were introduced by von Neumann in 1929 in an axiomatic way, and work on abstract normed linear spaces was done by Wiener, Hahn and Helly. In all these cases, the underlying structure of linear spaces followed the axiomatic approach made by Peano in 1888. The theory of Hilbert and Banach spaces was subsequently generalized to abstract topological sets and topological vector spaces by Weil, Kolmogorov and von Neumann. During the 20th century, a lot of attention was given to the problem of characterizing, by means of properties of the norms, when a Banach space is indeed a Hilbert space, i.e., when the norm derives from an inner product. While early characterizations of Euclidean structures were considered by Brunn in 1889 and Blaschke in 1916, the first and most popular characterization (the parallelogram law) was given by Jordan and von Neumann in 1935. In subsequent years, Kakutani, Birkhoff, Day and James proved many characterizations involving, among others, orthogonal relations and dual maps, and Day wrote a celebrated monograph on this subject [Day (1973)]. The topic became very active, as shown in [Amir (1986)], where 350 characterizations are presented, summarizing the main contributions up to 1986, such as those by Phelps, Hirschfeld, Rudin-Smith, Garkavi, Joly, Ben´ıtez, del R´ıo, Baronti, Senechalle, Oman, Kirˇcev-Troyanski, etc. A lot of work has been done in the field of functional equations [Aczél (1966); Aczél and Dhombres (1989)] to solve equations in normed linear spaces where the unknown is the norm, and in this way new characterizations have been found. It is also necessary to mention the interest in orthogonally additive mappings developed by Pinsker, Drewnovski, Orlicz, Sunderesan, Gudder, Strawther, Rätz, Szabó, etc. (and where in further studies, the second author also made many contributions) as well as solving functional equations in normed linear spaces. We will be making use of results and techniques arising in functional equations theory through this book. In a normed linear space (X, k · k), the norm derivatives are given for fixed x and y in X by the two expressions lim

λ→0±

kx + λyk − kxk . λ

The question [K¨ othe (1969)] of when a boundary point of the unit ball has a tangent hyperplane is connected with the differentiability of the norm

Introduction

3

at this point [Mazur (1933)], so norm derivatives have been considered in problems looking for smooth conditions (see [Köthe (1969)], §26), but very few characterizations of i.p.s. given in terms of norm derivatives were reported in [Amir (1986)]. Note that instead of considering the above norm derivatives, it is more convenient to introduce the functionals kx + λyk − kxk kx + λyk2 − kxk2 = kxk · lim λ→0± λ→0± 2λ λ

ρ′± (x, y) = lim

because when the norm comes from an inner product h·, ·i, we obtain ρ′± (x, y) = hx, yi, i.e., functionals ρ′± are perfect generalizations of inner products. Our chief concern in this publication is precisely to see how by virtue of these functionals ρ′± one can state natural generalizations of geometric properties of triangles, and how by introducing new functional techniques one can obtain a very large collection of new characterizations of norms derived from inner products. In doing this, we report the latest results in the field and also find new advances. 1.2

Normed linear spaces

We begin with the description of the well-known class of real normed linear spaces Definition 1.2.1 A pair (X, k · k) is called a real normed linear space provided that X is a vector space over the field of real numbers R and the function k · k from X into R satisfies the properties: (i) (ii) (iii) (iv)

kxk ≥ 0 for all x in X, kxk = 0 if and only if x = 0, kαxk = |α|kxk for all x in X and α in R, kx + yk ≤ kxk + kyk for all x and y in X.

The function k · k is called a norm and the real number kxk is said to be the norm of x. In the real line R the only norms are those of the form kxk = |x|, x ∈ R, where | · | denotes the absolute value |x| := max(x, −x), x ∈ R. In general, for all x, y in X we have (1.2.1) kxk − kyk ≤ kx − yk ≤ kxk + kyk,

4


so introducing the mapping d from X × X into R by d(x, y) := kx − yk, for all x, y in X, we infer that d is a metric induced by the norm k · k, so (X, d) is a metric space and therefore a topological space. With respect to the metric topology, by virtue of (1.2.1), the norm k · k is continuous and the topology induced by the norm is compatible with the vector space operations, i.e., R × X ∋ (α, x) 7→ αx ∈ X and X × X ∋ (x, y) 7→ x + y ∈ X are continuous in both variables together. The open ball Kx (r) of radius r centered at x consists of all points y in X such that ky − xk < r and can be obtained as the x-translation of the ball K0 (r) centered at the origin, i.e., Kx (r) = x + K0 (r). Analogously, one considers the closed ball Bx (r) = {y ∈ X : ky − xk ≤ r} = x + B0 (r). The sphere of radius r centered at x will be defined by Sx (r) = {y ∈ X : kx − yk = r}, and we denote the unit closed ball B0 (1) by BX and the corresponding unit sphere S0 (1) by SX . When all Cauchy sequences in (X, k · k) are convergent, i.e., the space is complete, then the real normed space is said to be a Banach space. Isometries in real normed spaces are characterized by Mazur and Ulam (see, e.g., [Mazur and Ulam (1932); Benz (1994)]). Theorem 1.2.1 Assume that (X, k · k), (Y, k · k) are real normed spaces and let f be a surjective mapping from X onto Y which is an isometry, i.e., kf (x) − f (y)k = kx − yk, for all x, y in X. Then the mapping T := f − f (0) is linear. Other interesting results on isometries on real normed spaces may be found in [Benz (1992); Benz (1994)]. Let us recall the most characteristic examples of real normed linear space.

5

Introduction

(i) The space Rn n ≥ 1 being fixed, and with the usual linear structure of Rn , one can consider the following norms: v u n uX x2i , kxke = t i=1

kxk+ =

n X i=1

|xi |,

kxk∨ = max{|x n |},  1 |, ..., |x!  1/2  n−1  X kxkm = max x2i , |xn |   i=1

n

for all x = (x1 , x2 , ..., xn ) in R . The norm k · ke is the classical Euclidean norm, k · k+ is the so-called taxi-cab norm, k · k∨ is the diamond norm and k · km is a mixed norm. (ii) The spaces c0 , c and l∞ If l∞ denotes the linear space of all bounded real sequences then its natural norm is defined by kxk∞ = sup{|xn | : n ≥ 1}, for any bounded real sequence x = (xn ). One can restrict this norm to the subspace c of all convergent real sequences and, in particular, to the subspace c0 of all real sequences convergent to 0. (iv) The space l1 In the linear space of real sequences (xn ) such that

n=1

considers the norm kxk1 = for all x = (xn ).

∞ P

∞ X

n=1

|xn |

|xn | < ∞ one

6


(v) The spaces lp , 1 < p < ∞ The classical inequality aα b1−α ≤ αa + (1 − α)b for α in (0,1) and a, b ≥ 0, implies H¨ older’s inequality !1/q ∞ !1/p ∞ ∞ X X X q p |xn yn | ≤ |xn | |yn | , (1.2.2) n=1

n=1

n=1

whenever 1/p + 1/q = 1, 1 < p < ∞, and real sequences (xn ) and (yn ) are such that the right-hand side of (1.2.1) converges. From (1.2.1) one easily derives Minkowski’s inequality !1/p !1/p !1/p ∞ ∞ ∞ X X X p p p ≤ + , |xn + yn | |xn | |yn | n=1

n=1

n=1

p

where 1 < p < +∞. Thus the space l (1 < p < ∞) of infinite sequences ∞ P (xn ) such that |xn |p < ∞ admits the norm defined by n=1

∞ X

kxkp =

n=1

p

|xn |

!1/p

for all x = (xn ). (vi) The space C(K) Given a compact space K, let C(K) be the vector space of all real-valued continuous functions f defined on K. Then one considers the norm kf k = sup{|f (x)| : x ∈ K}. (vii) The Lp -spaces, p ≥ 1 A closed real interval [a, b] being fixed with a < b, for p ≥ 1, let Lp denote the space of continuous real-valued functions defined on [a, b] and such that Z b |f (t)|p dt < ∞. a

Then one defines the norm kf kp =

Z

a

b

p

|f (t)| dt

!1/p

.

Introduction

1.3

7

Strictly convex normed linear spaces

An important class of real normed linear spaces is formed by the so-called strictly convex spaces. Precisely, we quote the following definition (see, e.g., [K¨ othe (1969)]): Definition 1.3.1 A real normed linear space (X, k·k) is said to be strictly convex if in its closed unit ball BX every boundary point in BX is an extreme point, i.e., any one of the following equivalent conditions holds: (i) (ii) (iii) (iv) (v) (vi)

SX contains no line segments; Every supporting hyperplane intersects SX in at most one point; Distinct boundary points have distinct supporting planes;

If kxk = kyk = 1 and x 6= y then 12 (x + y) < 1; If kx + yk = kxk + kyk and y 6= 0 then x = αy for some α ≥ 0; If x, y in X are linearly independent then kx + yk < kxk + kyk.

The spaces (Rn , k · ke ), lp and Lp are strictly convex for 1 < p < ∞ and for all n ∈ N, while (Rn , k · k+ ), (Rn , k · k∨ ), (Rn , k · km ), c0 , l1 , l∞ , L1 , for n > 1 are not. 1.4

Inner product spaces

Definition 1.4.1 A real vector space X is called an inner product space (briefly i.p.s.) if there is a real-valued function h·, ·i on X × X that satisfies the following four properties for all x, y, z in X and α in R: (i) (ii) (iii) (iv)

hx, xi is nonnegative and hx, xi = 0 if and only if x = 0, hx, y + zi = hx, yi + hx, zi, hx, αyi = αhx, yi, hx, yi = hy, xi.

An inner product h·, ·i defined on X × X induces the norm p kxk := hx, xi, x ∈ X

so all inner product spaces are normed linear spaces. When a norm is induced by an inner product one says that the norm derives from an inner product. In the case where (X, k · k) is a Banach space and k · k derives from an inner product h·, ·i, then (X, h·, ·i) is called a Hilbert space. One can show that any i.p.s. is strictly convex.

8


In the Hilbert space (Rn , k·ke ) one considers the standard inner product hx, yi =

n X

xj yj

j=1

for all vectors x = (x1 , ..., xn ), y = (y1 , ..., yn ) in Rn . In the space L2 of continuous real-valued functions on [a, b] such that Z b |f (t)|2 dt < ∞, a

one may consider the inner product Z b hf, gi = f (t)g(t)dt. a

2

In the space l one defines its inner product structure by means of the expression hx, yi =

∞ X

xn yn

n=1

for all infinite sequences x = (xn ) and y = (yn ) in l2 . A classical result (see, e.g. [Reed and Simon (1972)]) states that, indeed, l2 is in some sense the canonical example of a Hilbert space because any Hilbert space which contains a countable dense set and is not finite-dimensional is isomorphic to l2 . The classical result by Jordan and von Neumann [Jordan and von Neumann (1935)] states the following criterium for checking when the norm derives from an inner product. Theorem 1.4.1 Let (X, k · k) be a real normed linear space. Then k · k derives from an inner product if and only if the parallelogram law holds, i.e., kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2 for all x, y in X. The name of the law comes from its geometrical interpretation: the sum of the squares of the lengths of the diagonals in a parallelogram equals the sum of the squares of the lengths of the four sides.

Introduction

9

There exist some modifications of Theorem 1.4.1, where the sign of equality is replaced by one of the inequality signs, or where the condition is satisfied for unit vectors only [Day (1947); Schoenberg (1952)]. Theorem 1.4.2 Let (X, k · k) be a real normed linear space. Then k · k derives from an inner product if and only if kx + yk2 + kx − yk2 ∼ 2kxk2 + 2kyk2 for all x, y ∈ X, where ∼ stands either for ≤ or ≥. Theorem 1.4.3 Let (X, k · k) be a real normed linear space. Then k · k derives from an inner product if and only if ku + vk2 + ku − vk2 ∼ 4 for all u, v ∈ SX , where ∼ stands for one of the signs =, ≤ or ≥. Theorem 1.4.4 [Ben´ıtez and del Rio (1984)] Let (X, k · k) be a real normed linear space. Then k · k derives from an inner product if and only if for all u, v in SX there exist α, β 6= 0 such that kαµ + βvk2 + kαu − βvk2 ∼ 2(α2 + β 2 ), where ∼ stands for =, ≤ or ≥. When the norm k · k derives from an inner product h·, ·i then the inner product can be obtained from the norm by means of the polarization identity hx, yi =

1 (kx + yk2 − kx − yk2 ). 4

In our considerations, we will still require another classical characterization of inner product spaces Theorem 1.4.5 Let (X, k · k) be a real normed linear space of dimension greater than or equal to 2. The space X is an i.p.s. if and only if each two-dimensional subspace of X is an i.p.s. We now quote some additional geometrical notions which play a crucial role in an i.p.s. Definition 1.4.2 Two vectors, x and y, in an inner product space (X, h·, ·i) are said to be orthogonal if hx, yi = 0. A collection {xi } of vectors in X is called an orthonormal set if hxi , xi i = 1 and hxi , xj i = 0 and for all positive integers i, j and i 6= j.

10


By virtue of orthonormal sets, one can formulate the following version of the Pythagorean theorem. Theorem 1.4.6 Let {x1 , ..., xn } be an orthonormal set in an i.p.s. (X, h·, ·i). Then for all x in X and positive integers n, one has kxk2 =

n X i=1

|hx, xi i|2 + kx −

n X i=1

hx, xi ixi k2 .

Therefore Bessel’s inequality follows kxk2 ≥

n X i=1

|hx, xi i|2 ,

as well as the Cauchy-Schwarz inequality |hx, yi| ≤ kxk · kyk, where equality holds if and only if x and y are linearly dependent. Given a closed subspace H of a Hilbert space X, one considers the orthogonal complement H ⊥ of H as the set of vectors of X which are orthogonal to H. Then one obtains X = H ⊕ H ⊥ , and the fact that every x in X may be uniquely written in the form x = u + v with u ∈ H and v ∈ H ⊥ constitutes the projection theorem. If S is an orthonormal set in a Hilbert space X and no other orthonormal set contains S as a proper subset then S is called an orthonormal basis. Such bases always exist in Hilbert spaces, and if S = {xα | α ∈ A} for some index set A is one of them, then for all x in X we have X X x= hx, xα ixα and kxk2 = |hx, xα i|2 . α∈A

α∈A

This last equality is called Parseval’s relation and coefficients hx, xα i are often called the Fourier coefficients of x with respect to the basis. Indeed, it is easy to show (Gram-Schmidt orthogonalization) that an orthonormal set may be constructed from an arbitrary sequence of independent vectors. Finally, we recall the celebrated Riesz lemma which states that if X is a Hilbert space and X ∗ is its dual space of all bounded linear transformations from X into R, then for any T in X ∗ there is a unique vector yT in X such that T (x) = hx, yT i and, moreover, kyT k = sup{|T (x)| : kxk = 1}.

11

Introduction

1.5

Orthogonalities in normed linear spaces

In this section we recall the most classical orthogonality relations that have been studied in real normed spaces and whose properties have shown to give interesting characterizations of i.p.s. James orthogonality1 R.C. James (see [James (1945); James (1947a); James (1947b)]) introduced in a real normed space (X, k · k) the following orthogonality relation: x ⊥J y

whenever kx + yk = kx − yk,

using the idea that, in the plane, a practical way to examine the orthogonality between two vectors x and y is to check whether the two diagonals of the parallelogram determined by x and y are of equal length. When the norm k · k derives from an inner product h·, ·i, then, James orthogonality x ⊥J y reduces to the classical condition hx, yi = 0. Moreover the James orthogonality ⊥J is symmetric (i.e., x ⊥J y if and only if y ⊥J x) and partially homogeneous (i.e., if x ⊥J y, then ax ⊥J ay for all a in R). A large family of properties of ⊥J characterizes i.p.s. of dimension greater than or equal to 2 (see [Amir (1986)]). Theorem 1.5.1 Let (X, k · k) be a real normed linear space, dim X ≥ 2. Each of the following conditions characterizes X as an i.p.s.: (i) (ii) (iii) (iv) (v)

x ⊥J y implies x ⊥J ay for all x, y in X and for all a in R; tu ⊥J v implies u ⊥J v for all u, v in SX , t 6= 0; x ⊥J y implies x ⊥J λy for all x, y in X and for some λ in (0, 1); x ⊥J z and y ⊥J z implies (x + y) ⊥J z for all x, y, z in X; The set {x ∈ X : x ⊥J z} is convex for all z in X.

Birkhoff ’s orthogonality2 In a real normed linear space (X, k · k), one says that a vector x is orthogonal to y in the sense of Birkhoff [Birkhoff (1935)] if the following relation holds: x ⊥B y 1 In

whenever kxk ≤ kx + tyk

for all t in R.

the literature this orthogonality relation is also denoted by # instead of ⊥J (see e.g. [Amir (1986)] ). It can also be called isosceles orthogonality. 2 It is also called the Birkhoff-James orthogonality.

12


This relation means, geometrically, that the line through x in the ydirection supports the ball {z : kzk ≤ kxk} at x. Note that x ⊥B y yields αx ⊥B βy for all α, β in R (i.e., ⊥B is full homogeneous). In an i.p.s. the Birkhoff orthogonality x ⊥B y is reduced to the usual condition hx, yi = 0. Theorem 1.5.2 [Amir (1986)] Let (X, k · k) be a real normed linear space, dim X ≥ 2. Each of the following conditions characterizes X as an i.p.s.: (i) x ⊥J y implies x ⊥B y for all x, y in X; (ii) (u + v) ⊥B (u −v) for all u, v in SX ; kxk (iii) x + kxk kyk y ⊥B x − kyk y for all x, y in X, y 6= 0;

(iv) x ⊥B y implies x ⊥J y for all x, y in X; (v) x ⊥B y implies kx+ yk = F (kxk, kyk) for some F : R+ × R+ → R+ and for all x, y in X; (vi) If dim X ≥ 3, then x ⊥B z and y ⊥B z implies (x + y) ⊥B z for all x, y, z in X; (vii) If dim X ≥ 3, then x ⊥B y implies y ⊥B x for all x, y in X. Pythagorean orthogonality Taking into account the classical Pythagorean theorem, one can define the orthogonal relation in a normed space (X, k · k): x ⊥P y

whenever kx + yk2 = kxk2 + kyk2 .

As in the case of our previous orthogonalities, if the space considered is an i.p.s., the above definition x ⊥P y is reduced to the standard condition hx, yi = 0. Moreover, ⊥P is symmetric, partially homogeneous and admits diagonals (i.e., for all x, y 6= 0 there exists a unique t ≥ 0 with (x + ty) ⊥P (x − ty)). Theorem 1.5.3 [Amir (1986)] Let (X, k·k) be a real normed linear space, dim X ≥ 2. Each of the following conditions characterizes X as an i.p.s.: (i) (ii) (iii) (iv) (v)

x ⊥P x ⊥P x ⊥J x ⊥P x ⊥P

y implies x ⊥P (−y) for all x, y in X; y implies x ⊥J y for all x, y in X; y implies x ⊥P y for all x, y in X; y implies αx ⊥P βy for all α, β in R and for all x, y in X; y implies x ⊥B y for all x, y in X;

Introduction

(vi) x ⊥B y implies x ⊥P y for all x, y in X; (vii) (u + v) ⊥P (u − v) for all u, v in SX ; (viii) x ⊥P z and y ⊥P z imply (x + y) ⊥P z for all x, y, z in X.

13

Chapter 2

Norm Derivatives

We start this chapter by defining functions ρ′+ and ρ′− , which play a crucial role in this book. Then we give several properties of these functions, which are used for different characterizations of inner product spaces.

2.1

Norm derivatives: Definition and basic properties

Let (X, k · k) be a real normed linear space of at least dimension two. We consider the functions ρ′+ , ρ′− : X × X → R defined as follows ρ′± (x, y) := lim

t→0±

kx + tyk2 − kxk2 , 2t

x, y ∈ X

(2.1.1)

and call them norm derivatives. In order to show that ρ′+ and ρ′− are well-defined, we recall a simple fact (cf., e.g., [Kuczma (1985)]). Lemma 2.1.1 Let I ⊂ R be an open interval and let f : I → R be convex. Then for every x ∈ I there exist the right and left side derivatives. Moreover, ′ ′ f− (x) ≤ f+ (x),

x ∈ I.

Now we are able to show Proposition 2.1.1 Proof.

Functions given by (2.1.1) are well-defined.

Fix x, y ∈ X and define f : R → R by f (t) := kx + tyk2 , 15

t ∈ R.

16


For arbitrary s, t ∈ R and α ∈ [0, 1], by the convexity of function x ∈ R → x2 ∈ R we have kx + (αt + (1 − α)s)yk2 = kα(x + ty) + (1 − α)(x + sy)k2

≤ (αkx + tyk + (1 − α)kx + syk)2 ≤ αkx + tyk2 + (1 − α)k(x + sy)k2

= αkx + tyk2 + (1 − α)kx + syk2 , so f is convex. Observe that ′ f± (0) = lim± t→0

kx + tyk2 − kxk2 , t

so ρ′+ (x, y) =

1 ′ f (0), 2 +

ρ′− (x, y) =

1 ′ f (0) 2 −

(2.1.2)

and on account of Lemma 2.1.1, for every x, y ∈ X limits ρ′± exist.

Proposition 2.1.2 If (X, h·, ·i) is a real inner product space, then both ρ′+ and ρ′− coincide with h·, ·i. Proof.

Take x, y ∈ X. Then hx + ty, x + tyi − hx, xi kx + tyk2 − kxk2 = lim± 2t 2t t→0 t→0 2 2thx, yi + t hy, yi = lim = hx, yi. 2t t→0±

ρ′± (x, y) = lim±

Lemma 2.1.2 by

Function h : R \ {0} → R defined for each fixed x, y ∈ X h(t) :=

kx + tyk − kxk , t

t ∈ R \ {0}

is bounded and increasing on each of the intervals (−∞, 0) and (0, +∞). Proof.

Fix x, y ∈ X. Since

kxk = kx + ty − tyk ≤ kx + tyk + |t|kyk and kx + tyk ≤ kxk + |t|kyk, we have −kyk ≤

kx + tyk − kxk ≤ kyk, |t|

17

Norm Derivatives

whence |h(t)| ≤ kyk,

t ∈ R\{0}.

(2.1.3)

Take t1 , t2 ∈ R such that 0 < t1 ≤ t2 . Then t2 kx + t1 yk = kt2 x + t1 t2 yk = kt1 (x + t2 y) + (t2 − t1 )xk ≤ t1 kx + t2 yk + (t2 − t1 )kxk,

whence kx + t2 yk − kxk kx + t1 yk − kxk ≤ , t1 t2 which means that h is increasing in (0, +∞). Now take t1 , t2 ∈ R such that t1 ≤ t2 < 0. Then −t1 kx + t2 yk = kt1 x + t1 t2 yk = kt2 (x + t1 y) + (t1 − t2 )xk ≤ −t2 kx + t1 yk + (t2 − t1 )kxk,

whence kx + t1 yk − kxk kx + t2 yk − kxk ≥ , t2 t1 which means again that h is increasing in (−∞, 0). Example 2.1.1 have

and

Consider (R2 , k · k+). Then for (x1 , x2 ), (y1 , y2 ) ∈ R2 we

 (x1 + x2 )(y1 + y2 )    (x1 − x2 )(y1 − y2 ) ρ′+ ((x1 , x2 ), (y1 , y2 )) =  x2 y2 + |x2 y1 |   x1 y1 + |x1 y2 |

if if if if

x1 x2 > 0, x1 x2 < 0, x1 = 0, x2 = 0.

 (x1 + x2 )(y1 + y2 )    (x1 − x2 )(y1 − y2 ) ′ ρ− ((x1 , x2 ), (y1 , y2 )) =  x y − |x2 y1 |   2 2 x1 y1 − |x1 y2 |

if if if if

x1 x2 > 0, x1 x2 < 0, x1 = 0, x2 = 0.

These functionals ρ′± can be written in the form (A. Monreal) ρ′± ((x1 , x2 ), (y1 , y2 )) = x1 y1 + sgn(x1 x2 )(x1 y2 + x2 y1 ) ± (1 − |sgn(x1 x2 )|)(|x1 y2 | + |x2 y1 |) + x2 y2 .

18


Still another form for ρ′± [James (1945)] is:   2 X   X xi yi ± |yi | . ρ′± ((x1 , x2 ), (y1 , y2 )) = k(x1 , x2 )k+  |xi | x =0 i=1 i i∈{1,2}

xi 6=0

Consider (R2 , k · k∨ ). Then for (x1 , x2 ), (y1 , y2 ) ∈ R2 we

Example 2.1.2 have

and

   x1 y1 ,    x2 y2 ,    x1 max{y1 , y2 } ρ′+ ((x1 , x2 ), (y1 , y2 )) =  x1 min{y1 , y2 }     x1 max{y1 , −y2 }    x1 min{y1 , −y2 }

if if if if if if

|x1 | > |x2 |, |x1 | < |x2 |, x1 = x2 ≥ 0, x1 = x2 < 0, x1 = −x2 > 0, x1 = −x2 < 0

  x1 y1 ,     x2 y2 ,    x1 min{y1 , y2 } ρ′− ((x1 , x2 ), (y1 , y2 )) =  x 1 max{y1 , y2 }     x1 min{y1 , −y2 }    x1 max{y1 , −y2 }

if if if if if if

|x1 | > |x2 |, |x1 | < |x2 |, x1 = x2 ≥ 0, x1 = x2 < 0, x1 = −x2 > 0, x1 = −x2 < 0.

Consider the Banach space (lp , k · k) with p > 1 of all ∞ P sequences x = (xn ) such that |xn |p is convergent and with the norm n=1 ∞ 1/p P kxk = |xn |p . Then ([James (1945)]) for all x 6= 0 we have Example 2.1.3

n=1

ρ′+ (x, y) = ρ′− (x, y) =

∞ X |xn |p−2 xn yn . kxkp−2 n=1

This shows that all spaces lp with p > 1 are examples of smooth spaces, which only in the case p = 2 are inner product spaces. Example 2.1.4 Let c0 be the space of all sequences convergent to 0 with supremum norm. Then for x = (xn ), y = (yn ) we have (see [Dragomir and Koliha (2000)]) ρ′+ (x, y) =

sup |xn |=kxk

xn yn

and

ρ′− (x, y) =

inf

|xn |=kxk

xn yn .

Norm Derivatives

19

Example 2.1.5 Let C([0, 1]) be the space of all continuous real-valued functions equipped with the norm kxk = sup{|x(u)| : u ∈ [0, 1]}. Then for functions x, y in C([0, 1]) we have (see [Dragomir and Koliha (1999)]): ρ′+ (x, y) = sup{x(u)y(u) : |x(u)| = kxk}, ρ′− (x, y) = inf{x(u)y(u) : |x(u)| = kxk}. The next theorem describes several properties of ρ′+ and ρ′− (cf. [Amir (1986); Lorch (1948); Tapia (1973)]). Theorem 2.1.1 Let (X, k · k) be a real normed linear space at least twodimensional and let ρ′+ , ρ′− : X × X → R be given by (2.1.1). Then ρ′± (0, y) = ρ′± (x, 0) = 0 for all x, y ∈ X; ρ′± (x, x) = kxk2 for all x ∈ X; ρ′± (αx, y) = ρ′± (x, αy) = αρ′± (x, y) for all x, y ∈ X and α ≥ 0; ρ′± (αx, y) = ρ′± (x, αy) = αρ′∓ (x, y) for all x, y ∈ X and α ≤ 0; ρ′± (x, αx + y) = αkxk2 + ρ′± (x, y) for all x, y ∈ X and α ∈ R; |ρ′± (x, y)| ≤ kxk kyk for all x, y ∈ X; If X is strictly convex, then |ρ′± (x, y)| < kxk kyk for all linearly independent x, y ∈ X; (viii) ρ′− (x, y) ≤ ρ′+ (x, y) for all x, y ∈ X; (ix) ρ′+ (x, y) = ρ′+ (y, x) for all x, y ∈ X or ρ′− (x, y) = ρ′− (y, x) for all x, y ∈ X if and only if the norm in X comes from an inner product; (x) ρ′+ (u, v) = ρ′+ (v, u) for all u, v ∈ SX or ρ′− (u, v) = ρ′− (v, u) for all u, v ∈ SX if and only if the norm in X comes from an inner product. (i) (ii) (iii) (iv) (v) (vi) (vii)

Proof.

(i) and (ii) follow directly from the definition of ρ′± .

(iii) Take x, y ∈ X and α ≥ 0. If α = 0 then ρ′± (0, y) = ρ′± (x, 0) = 0. Assume that α > 0. Then ρ′± (αx, y) = lim± t→0

kx + αt yk2 − kxk2 kαx + tyk2 − kαxk2 = lim± α 2t t→0 2 αt

= lim± α s→0

where s :=

t α.

Analogously,

ρ′± (x, αy) = lim± t→0

where s := αt.

kx + syk2 − kxk2 = αρ′± (x, y), 2s

kx+αtyk2 −kxk2 kx+syk2 −kxk2 = lim± α = αρ′± (x, y), 2t 2s s→0

20


(iv) The proof follows the same lines as in (iii) with the same substitutions. Now t → 0± if and only if s → 0∓ . (v) Take x, y ∈ X and α ∈ R. If t is small enough then 1 + αt > 0 and kx + t(αx + y)k2 − kxk2 2t t→0± k(1 + tα)x + tyk2 − kxk2 lim 2t t→0±

2

t 2 2 (1 + tα) x + 1+tα y − kxk + (1 + tα)2 kxk2 − kxk2 lim 2t t→0±

2

t 2 (1 + tα) x + 1+tα y − kxk (2αt + α2 t2 )kxk2 + lim lim t 2t t→0± t→0± 2 1+tα 1 2 2 kx + syk − kxk + αkxk2 = ρ′± (x, y) + αkxk2 , lim± 1−αs 2s s→0 ρ′± (x, αx + y) = lim

=

=

= =

where s :=

so

t 1+αt ,

whence 1 + αt =

1 1−αs ,

and t → 0± if and only if s → 0± .

(vi) Take x, y ∈ X. From (2.1.3) we have lim kx + tyk + kxk kx + tyk − kxk = kxk lim h(t) ≤ kxk kyk, t→0+ 2 t t→0+ ′ ρ (x, y) ≤ kxkkyk, +

x, y ∈ X.

(vii) Observe first that in a strictly convex space for all linearly independent vectors x and y, we have kx + yk < kxk + kyk and kx − yk < kxk + kyk.

(2.1.4)

From (v) and (iv) we have the following equalities: ρ′+ (x, x + y) = kxk2 + ρ′+ (x, y)

(2.1.5)

and ρ′+ (x, y − x) = −ρ′− (x, x − y) = −kxk2 + ρ′+ (x, y). On account of (vi), equality (2.1.5) leads to kxk2 + ρ′+ (x, y) ≤ kxk kx + yk,

(2.1.6)

21

Norm Derivatives

which from the linear independency of x and y and from (2.1.4) gives ρ′+ (x, y) ≤ kxk(kx + yk − kxk) < kxk kyk. Similarly, from (2.1.6) and (vi) we have −kxk2 + ρ′+ (x, y) ≥ −kxk kx − yk. Thus, the linear independency of x and y and (2.1.4) yield ρ′+ (x, y) ≥ kxk(kxk − kx − yk) > −kxk kyk. These inequalities allow us to write ′ ρ− (x, y) = −ρ′+ (−x, y) < kxk kyk

for all linearly independent vectors x and y, which completes the proof of this property. (viii) Follows from (2.1.2) and Lemma 2.1.1. (ix) By means of Proposition 2.1.2, it is enough to show only one implication. Assume that ρ′+ (x, y) = ρ′+ (y, x) for all x, y ∈ X, and observe that this condition implies ρ′− (x, y) = ρ′− (y, x) for all x, y ∈ X, and conversely. Indeed, if, for example, ρ′+ (x, y) = ρ′+ (y, x) for all x, y ∈ X, then on account of (iv) we obtain ρ′− (x, y) = −ρ′+ (−x, y) = −ρ′+ (y, −x) = ρ′− (y, x). Let P be any two-dimensional subspace of X. Define a mapping h·, ·i : P × P → R by the formula hx, yi :=

ρ′+ (x, y) + ρ′− (x, y) , 2

x, y ∈ P.

We will show that h·, ·i is an inner product in P . It is easy to see that the function is symmetric, homogeneous and nonnegative. The only condition which needs some explanation is the additivity of h·, ·i in each variable. However, by the symmetry, it is enough to show the additivity with respect to the second variable. Take x, y, z ∈ P . We consider two cases. Assume first that x and y are linearly dependent, so y = λx for some λ ∈ R. From

22


the earlier considerations and properties of ρ′± , we get ρ′+ (x, λx + z) + ρ′− (x, λx + z) 2 ′ ′ ρ (x, z) + ρ (x, z) − = λkxk2 + + 2 = hx, λxi + hx, zi = hx, yi + hx, zi.

hx, y + zi = hx, λx + zi =

Now let x and y be linearly independent, so z = αx + βy for some α, β ∈ R. Then hx, y + zi = hx, αx + (1 + β)yi ρ′+ (x, αx + (1 + β)y) + ρ′− (x, αx + (1 + β)y) 2 αkxk2 + (1 + β)ρ′+ (x, y) + αkxk2 + (1 + β)ρ′− (x, y) = 2 ρ′+ (x, y) + ρ′− (x, y) = 2 αkxk2 + ρ′+ (x, βy) + αkxk2 + ρ′− (x, βy) + 2 ρ′+ (x, αx + βy) + ρ′− (x, αx + βy) = hx, yi + 2 = hx, yi + hx, zi.

=

So, we have proved that h·, ·i defined in P is an inner product. By the free choice of P , on account of Theorem 1.4.5, the norm in X comes from an inner product. (x) Assume that ρ′+ (u, v) = ρ′+ (v, u) for all u, v ∈ SX . Take arbitrary y x and v := kyk . Of course, u, v ∈ SX , and we x, y ∈ X \ {0} and let u := kxk have y y x x ′ ′ , = ρ+ , . ρ+ kxk kyk kyk kxk From (iii) we obtain ρ′+ (x, y) = ρ′+ (y, x). Since ρ′+ (x, 0) = ρ′+ (0, x), the above equality holds for all x, y ∈ X, which together with (ix) shows that the norm in X comes from an inner product. Other properties of ρ′± that we will use in the sequel are the following.

23

Norm Derivatives

Proposition 2.1.3 The function ρ′+ is continuous and subadditive in the second variable, and the function ρ′− is continuous and superadditive in the second variable. Proof. Fix x0 ∈ X. From the convexity of function k·k2 , for all y1 , y2 ∈ X and t ∈ R, we have

2x0 + t(y1 + y2 ) 2

≤ 1 kx0 + ty1 k2 + 1 kx0 + ty2 k2 ,

2 2 2 whence

" #

2

t 2

2 x0 + (y1 + y2 ) − kx0 k ≤ kx0 + ty1 k2 − kx0 k2 2

+(kx0 + ty2 k2 − kx0 k2 ), (2.1.7)

and for t > 0 we get kx0 + 2t (y1 + y2 )k2 − kx0 k2 kx0 + ty1 k2 − kx0 k2 kx0 + ty2 k2 − kx0 k2 ≤ + , t 2t 2t 2· 2 from which we obtain the subadditivity of ρ′+ with respect to the second variable ρ′+ (x0 , y1 + y2 ) ≤ ρ′+ (x0 , y1 ) + ρ′+ (x0 , y2 ).

(2.1.8)

In order to prove its continuity, we combine (2.1.8) with condition (vi) of Theorem 2.1.1. Take h ∈ X. Then we have ρ′+ (x0 , y + h) − ρ′+ (x0 , y) ≤ ρ′+ (x0 , h) ≤ kx0 kkhk. From the other side − ρ′+ (x0 , y + h − h) − ρ′+ (x0 , y + h) ≥ −ρ′+ (x0 , −h) = ρ′− (x0 , h) ≥ −kx0 kkhk. This gives ′ ρ+ (x0 , y + h) − ρ′+ (x0 , y) ≤ kx0 kkhk,

which proves the continuity of ρ′+ in the second variable. The superadditivity in the second variable of function ρ′− is obtained from (2.1.7), after dividing it by 2t for t < 0 and taking the limit while t → 0− .

24


The continuity in the second variable of ρ′− is obtained immediately from the condition ρ′− (x0 , y) = −ρ′+ (−x0 , y). Proposition 2.1.4 [Amir (1986)] Let (X, k · k) be a real normed linear space. Then there exists a set F ⊂ X of Lebesgue measure zero such that for all x in X\F and y in X we have ρ′+ (x, y) = ρ′− (x, y), and X\F is dense in X. Definition 2.1.1 [K¨ othe (1969); Precupanu (1978)] Let (X, k · k) be a real normed linear space. We say that X is smooth if the norm k · k has the Gateaux (or weak) derivative in X, i.e., G(x, y) := lim

t→0

kx + tyk − kxk t

exists for each x, y in X. If G(x, y) exists for an x in X and for each y in X, we say that X is smooth at x or that x is a point of smoothness. Remark 2.1.1. Because k · k is a convex function, the existence of the above limit implies that G(x, ·) is a linear function for each fixed x in X. Moreover, we may state Definition 2.1.1 in an equivalent form, namely: X is smooth if and only if ρ′+ = ρ′− . Proposition 2.1.5 inequality

In a real normed linear space (X, k · k) the following

|ρ′+ (z, y + λx) − ρ′+ (z, y)| ≤ |λ|kxkkzk holds for all x, y, z in X and λ in R.

(2.1.9)

25

Norm Derivatives

Proof.

By the definition of the functional ρ′+ :

kz + µ(y + λx)k2 − kzk2 |ρ′+ (z, y + λx) − ρ′+ (z, y)| = lim+ 2µ µ→0 kz + µyk2 − kzk2 − lim 2µ µ→0+ kz + µ(y + λx)k2 − kz + µyk2 = lim 2µ µ→0+ kz + µ(y + λx)k + kz + µyk = lim + 2 µ→0 kz + µ(y + λx)k − kz + µyk × µ |kz + µy + µλxk − kz + µyk| = kzk lim µ µ→0+ |λ||µ|kxk ≤ kzk lim+ = |λ|kxkkzk, µ µ→0 so (2.1.8) follows.

From (11) in [Precupanu (1978)] we can derive the next property of ρ′+ . Proposition 2.1.6

Let u, v ∈ SX . Then lim ρ′+ (u + τ v, v) = ρ′+ (u, v).

τ →0+

Remark 2.1.2. Employing the same argument as that used by Precupanu, one can show that with the same assumptions lim ρ′+ (u + τ v, v) = ρ′+ (u, v),

τ →0−

so, in fact we have lim ρ′+ (u + τ v, v) = ρ′+ (u, v).

τ →0

Corollary 2.1.1

(2.1.10)

Let u, v ∈ SX . Then lim ρ′− (u + τ v, v) = ρ′− (u, v).

τ →0

(2.1.11)

26


Proof.

By the properties of ρ′± , we have lim ρ′− (u + τ v, v) = lim (−ρ′+ (u + τ v, −v))

τ →0

τ →0

= − lim ρ′+ (u + (−τ )(−v), −v) =

−τ →0 −ρ′+ (u, −v)

= ρ′− (u, v).

As a consequence of the above results and by properties of ρ′± , one has as even more general result. Corollary 2.1.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Functions R ∋ t 7→ ρ′± (x + ty, y) ∈ R are continuous at zero for every fixed x, y in X. Finally, the functions ρ′± characterize the Birkhoff orthogonality in the following sense. Proposition 2.1.7 [Amir (1986)] Let (X, k · k) be a real normed linear space. Then for all x, y in X and α in R, the condition x ⊥B y − αx is satisfied if and only if ρ′− (x, y) ≤ αkxk2 ≤ ρ′+ (x, y). 2.2

Orthogonality relations based on norm derivatives

In this section we introduce two orthogonality relations based on the norm derivatives and their connections with standard orthogonalities in normed linear spaces. We start with the generalization of the usual notion of the orthogonality in i.p.s. given by means of an inner product by defining in a normed space the ρ-orthogonal relation ⊥ρ by x ⊥ρ y

if and only if

ρ′+ (x, y) = 0.

Proposition 2.2.1 Let (X, k · k) be a real normed linear space. The relation ⊥ρ satisfies the following conditions: (i) For all x in X, 0 ⊥ρ x, x ⊥ρ 0; (ii) For all x in X, x ⊥ρ x if and only if x = 0; (iii) For all x, y in X and for all nonnegative a, if x ⊥ρ y then ax ⊥ρ y and x ⊥ρ ay;

27

Norm Derivatives

(iv) For all nonzero x, y in X, if x ⊥ρ y then x and y are linearly independent; (v) For all x in X and for all t > 0 there exists a y in X such that kyk = t and x ⊥ρ y; (vi) If the norm derives from an inner product, then the relation ⊥ρ is equivalent to the usual orthogonality in i.p.s. Proof. We will prove (v), the rest follows easily from the definition of ⊥ρ . Fixed x in X and t > 0, define f from {z ∈ X : k zk = t} into R by f (z) := ρ′+ (x, z). Then f is a continuous function with the property x x = −f −t f t kxk kxk and therefore it is immediate to show that there exists a y in X such that kyk = t and f (y) = 0, i.e., x ⊥ρ y. Proposition 2.2.2 Let (X, k · k) be a real normed linear space. The following conditions are satisfied: (i) For all x, y in X and α in R, x ⊥ρ y − αx if and only if αkxk2 = ρ′+ (x, y); (ii) (iii) (iv) (v)

For all x, y in X, if x ⊥ρ y, then x ⊥B y; If for all x, y in X, x ⊥ρ y implies x ⊥P y, then X is an i.p.s.; If for all x, y in X, x ⊥ρ y implies x ⊥J y, then X is an i.p.s.; The condition x ⊥B y implies that x ⊥ρ y holds for all x, y in X if and only if X is smooth; (vi) If for all x, y in X, x ⊥P y implies x ⊥ρ y, then X is an i.p.s.; (vii) If for all x, y in X, x ⊥J y implies x ⊥ρ y, then X is an i.p.s.; (viii) If for all u, v in SX , u + v ⊥ρ u − v, then X is an i.p.s. Proof. We will prove (iii) and (v). For (iii), consider w, v ∈ SX such that ρ′+ (w, v) = ρ′− (w, v) 6= 0. v v , and by hypothesis w ⊥P w − ρ′ (w,v) , i.e., Then w ⊥ρ w − ρ′ (w,v) +

+

1+

w −

Moreover, w ⊥ρ −w +

2

2

v v

= 2w −

. ′ ′

ρ+ (w, v) ρ+ (w, v)

v ρ′+ (w,v)

1+

−w +

and w ⊥P −w +

v ρ′+ (w,v) ,

2

v 1

= . ′ ρ+ (w, v) ρ′+ (w, v)2

(2.2.1)

i.e., (2.2.2)

28


By (2.2.1) and (2.2.2) we have k2wρ′+ (w, v) − vk = 1 for all w, v ∈ SX such that ρ′+ (w, v) = ρ′− (w, v) (when ρ′+ (w, v) = ρ′− (w, v) = 0 the last equation is obvious) and by (5.12) in [Amir (1986)], X is an i.p.s. In order to prove (v), assume first that X is smooth, so ρ′+ (x, y) = ′ ρ− (x, y) for all x, y in X, and let x ⊥B y. Then by Proposition 2.1.7, ρ′− (x, y) ≤ 0 ≤ ρ′+ (x, y) and, consequently, ρ′− (x, y) = ρ′+ (x, y) = 0, whence x ⊥ρ y. For the converse, fix x, y in X. There exists a t in R such that x ⊥B tx + y. Then ρ′+ (x, tx + y) = 0, whence tkxk2 + ρ′+ (x, y) = 0. But we also have −x ⊥B tx + y, whence −tkxk2 − ρ′− (x, y) = 0. Consequently, ρ′+ (x, y) = ρ′− (x, y) for all x, y in X. Let us now define the ρ-orthogonal relation ⊥ρ by the condition x ⊥ρ y

if and only if

ρ′+ (x, y) + ρ′− (x, y) = 0.

Then we have analogous results to Propositions 2.2.1 and 2.2.2. Proposition 2.2.3 Let (X, k · k) be a real normed linear space. The relation ⊥ρ satisfies the following conditions: For all x in X, 0 ⊥ρ x, x ⊥ρ 0; For all x in X, x ⊥ρ x if and only if x = 0; For all x, y in X and for all a, b in R, if x ⊥ρ y then ax ⊥ρ by; For all nonzero x, y in X, if x ⊥ρ y, then x and y are linearly independent; (v) For all x in X and for all t > 0 there exists y in X such that kyk = t and x ⊥ρ y; (vi) If the norm derives from an inner product, then the relation ⊥ρ is equivalent to the usual orthogonality in i.p.s.

(i) (ii) (iii) (iv)

Proposition 2.2.4

Let (X, k · k) be a real normed linear space. Then:

(i) For all x, y in X and α in R, x ⊥ρ y − αx if and only if 2αkxk2 = ρ′+ (x, y) + ρ′− (x, y); (ii) For all x, y in X and α in R, if x ⊥ρ y − αx, then ρ′− (x, y) ≤ αkxk2 ≤ ρ′+ (x, y); (iii) For all x, y in X, if x ⊥ρ y, then x ⊥B y; (iv) If for all x, y in X, x ⊥ρ y implies x ⊥P y, then X is an i.p.s.; (v) If for all x, y in X, x ⊥ρ y implies x ⊥J y, then X is an i.p.s.;

29

Norm Derivatives

(vi) The condition x ⊥B y implies that x ⊥ρ y holds for all x, y in X if and only if X is smooth; (vii) If for all x, y in X, x ⊥P y implies x ⊥ρ y, then X is an i.p.s.; (viii) If for all x, y in X, x ⊥J y implies x ⊥ρ y, then X is an i.p.s.; (ix) If for all u, v in SX , u + v ⊥ρ u − v, then X is an i.p.s. Proof. We will prove (iv) and (vi). For (iv), let w, v in SX be such that ρ′+ (w, v) = ρ′− (w, v) 6= 0 (i.e., X is smooth at w). Then it is easy to check that v v and w ⊥ρ −w + ′ w ⊥ρ w − ′ ρ+ (w, v) ρ+ (w, v) so, by the assumption, w ⊥P w −

v

and

ρ′+ (w, v)

w ⊥P −w +

v ρ′+ (w, v)

and

1+

w −

2

2

v v

= 2w −

, ′ ′

ρ+ (w, v) ρ+ (w, v)

1+

−w +

2

1 v

= . ′ ′

ρ+ (w, v) ρ+ (w, v)2

From the above two equalities, we deduce k2ρ′+ (w, v)w − vk2 = 1 whenever ρ′+ (w, v) 6= 0 (if ρ′+ (w, v) = 0, last equality is trivial) and by (5.12) in [Amir (1986)], X is an i.p.s. In order to prove (vi), take x, y in X. From the obvious inequality ρ′− (x, y) ≤

αρ′− (x, y) + (1 − α)ρ′+ (x, y) kxk2 ≤ ρ′+ (x, y) kxk2

for every α ∈ [0, 1], and by Proposition 2.1.7 we get x ⊥B tx + y with t = −

αρ′− (x,y)+(1−α)ρ′+ (x,y) . kxk2

From the assumption we have

x ⊥ρ tx + y, i.e., 0 = ρ′+ (x, tx + y) + ρ′− (x, tx + y) = 2tkxk2 + ρ′+ (x, y) + ρ′− (x, y),

30


whence (1 − 2α)(ρ′− (x, y) − ρ′+ (x, y)) = 0 for each α ∈ [0, 1], and consequently, ρ′+ (x, y) = ρ′− (x, y). The converse implication is verified in the same way as the analogous condition for ⊥ρ .

2.3

ρ′± -orthogonal transformations

In the context of inner product spaces, the functional equation hT (x), T (y)i = hx, yi corresponding to orthogonal transformations has received a lot of contributions in the literature. A natural generalization of such equation in real normed linear spaces (X, k · k) is to look for the solutions of ρ′+ (T (x), T (y)) = ρ′+ (x, y)

and

ρ′− (T (x), T (y)) = ρ′− (x, y),

(2.3.1)

for all x, y in X. In the Euclidean plane, we have the following examples. Example 2.3.1 Consider (R2 , k · k+ ), then T : R2 → R2 verifies ρ′+ (T (x), T (y)) = ρ′+ (x, y) if and only if T belongs to D4 , the Euclidean symmetry group of the square. This fact is referred to the result of [Schattschneider (1984)], where it was proved that the group of isometries of the plane with respect to the taxi-cab metric is the semi-direct product of the group D4 and the group of all translations of the plane. Example 2.3.2 Consider (R2 , k · k∨ ), then T : R2 → R2 verifies ρ′+ (T (x), T (y)) = ρ′+ (x, y) if and only if T belongs to D4 . The next result is a technical lemma that we need for the main theorem. Lemma 2.3.1 For any x in X, x 6= 0 there exists a nonzero vector y in X independent of x such that |ρ′+ (x, y)| < kxk kyk. Proof. Fix x ∈ X, x 6= 0. If for any vector z in X, z 6= 0, independent of x we would have |ρ′+ (x, z)| = kxk kzk, then for any such z, the vector x + z would also be independent of x and would satisfy

31

Norm Derivatives

|ρ′+ (x, x + z)| = kxk kx + zk, and consequently there would exist numbers ǫ(z), ǫ(x + z) ∈ {−1, 1} such that ǫ(x + z)kxk kx + zk = ρ′+ (x, x + z) = kxk2 + ρ′+ (x, z) = kxk2 + ǫ(z)kxk kzk, i.e., ǫ(x + z)kx + zk = kxk + ǫ(z)kzk. Therefore, for all vectors z independent of x we would have kx + zk2 = kxk2 + kzk2 + 2kxk kzkǫ(z). Let z be a nonzero vector independent of x and kx + zk2 = kxk2 + kzk2 (such a vector exists by the properties of Pythagorean orthogonality (see [James (1945); Amir (1986)])). Then kxk kzk = 0 and we obtain a contradiction. Next we prove the main result of this section [Alsina and Tomás (1991)]. Theorem 2.3.1 A continuous mapping T from a real normed linear space (X, k · k) of dimension two into itself verifies the equations: ρ′+ (T (x), T (y)) = g(x)ρ′+ (x, y)

ρ′− (T (x), T (y)) = g(x)ρ′− (x, y) (2.3.2) for a function g from X into R satisfying the condition and

g(x) = 0 implies x = 0 if and only if T is linear and there exists a positive constant k such that (x)k2 = k for all x in X\{0}. g(x) = kTkxk 2 Proof. First we will show that T preserves the linear independence of any couple of independent vectors x, y in X. In fact, if we had T (y) = λT (x) for some λ 6= 0, then kT (y)k = |λ|kT (x)k and by (2.3.2): λkT (x)k2 = ρ′+ (T (x), λT (x)) = ρ′+ (T (x), T (y)) = g(x)ρ′+ (x, y), λkT (x)k2 = ρ′− (T (x), λT (x)) = ρ′− (T (x), T (y)) = g(x)ρ′− (x, y), kT (x)k2 = ρ′+ (T (x), T (x)) = g(x)kxk2 , whence, since x 6= 0, we obtain that necessarily λ=

ρ′− (x, y) ρ′+ (x, y) = . kxk2 kxk2

(2.3.3)

32


Moreover, for all z in X, z 6= 0 we have

1 ′ 1 ′ ρ (T (z), T (y)) = ρ (T (z), λT (x)) g(z) + g(z) + ρ′ (x, y) λ ′ ρsgn(λ) (T (z), T (x)) = λρ′sgn(λ) (z, x) = + 2 ρ′sgn(λ) (z, x). = g(z) kxk (2.3.4) x x , y kxk , which cannot be zero because of the inLet z := y − ρ′+ kxk

ρ′+ (z, y) =

dependence of x and y. Then by (2.3.4) and the property (v) of ρ′+ we have:

2

x x x x x x ′ ′ ′

y − ρ′+ ,y

+ ρ+ y − ρ+ kxk , y kxk , ρ+ kxk , y kxk

kxk kxk x x ′ ′ = ρ+ y − ρ+ ,y ,y kxk kxk x x x x , y ρ′sgn(λ) y − ρ′+ ,y , . = ρ′+ kxk kxk kxk kxk x , y , we deduce that Since, by (2.3.3), sgn(λ) = sgn ρ′+ kxk

x x , y kxk

= 0, which gives a contradiction. Thus, T preserves

y − ρ′+ kxk the linear independence. Let F be the set of Lebesgue measure zero in X such that ρ′− (x, y) = ′ ρ+ (x, y) for all x in X\F and for all y in X (see Proposition 2.1.4). Next we will show that T (λx) = λT (x) for all λ ∈ R and x in X\F , x 6= 0. Take x in X\F . By Lemma 2.3.1 let us choose y independent of x and such that |ρ′+ (x, y)| < kxk kyk. Since T (x) and T (y) are also independent, for every λ in R there exist α, β in R (depending on λ) such that T (λx) = αT (x) + βT (y). Therefore, by (2.3.2) and the general properties of ρ′± , we obtain the equalities (in the case λ = 0 we assume ρ′sgn(λ) (·, ·) = 0): λg(x)kxk2 = g(x)ρ′+ (x, λx) = ρ′+ (T (x), T (λx)) = ρ′+ (T (x), αT (x) + βT (y)) = αkT (x)k2 + βρ′sgn(β) (T (x), T (y)) = αg(x)kxk2 + βg(x)ρ′sgn(β) (x, y) and, analogously, λg(y)ρ′sgn(λ) (y, x) = g(y)ρ′+ (y, λx) = ρ′+ (T (y), T (λx)) = ρ′+ (T (y), αT (x) + βT (y)) = βkT (y)k2 + αρ′sgn(α) (T (y), T (x))

33

Norm Derivatives

= βg(y)kyk2 + αg(y)ρ′sgn(α) (y, x). Consequently, α and β satisfy the system of linear equations: αkxk2 + βρ′sgn(β) (x, y) = λkxk2 , αρ′sgn(α) (y, x) + βkyk2 = λρ′sgn(λ) (y, x). The determinant of this system is ∆ = kxk2 kyk2 − ρ′sgn(α) (y, x)ρ′sgn(β) (x, y) > 0, because using Lemma 2.3.1 and properties of ρ′± , ρ′sgn(α) (y, x)ρ′sgn(β) (x, y) ≤ ρ′sgn(α) (y, x) ρ′sgn(β) (x, y) = ρ′sgn(α) (y, x) ρ′+ (x, y) < kxk2 kyk2 .

Then we have λ kxk2 kyk2 − ρ′sgn(λ) (y, x)ρ′sgn(β) (x, y) and sgn(α) = sgn(λ). α= ∆ Consequently, β=

kxk2 λ ′ (ρsgn(λ) (y, x) − ρ′sgn(α) (y, x)) = 0 ∆

and

α = λ.

Thus we obtain T (λx) = λT (x) for all λ in R and x in X\F . Since T is continuous and X\F is dense in X, we conclude that T (λx) = λT (x) for all x in X and λ in R. Take e1 ∈ X\F and using Lemma 2.3.1 choose e2 in X, linearly independent of x and such that |ρ′± (e1 , e2 )| < ke1 kke2 k. If we consider x = v1 e1 and y = v2 e2 for v1 , v2 in R\{0}, then T (x), T (y) are independent vectors and there exist a, b in R such that T (x + y) = aT (x) + bT (y). Therefore, we obtain g(x)(kxk2 + ρ′+ (x, y)) = = = =

g(x)ρ′+ (x, x + y) = ρ′+ (T (x), T (x + y)) ρ′+ (T (x), aT (x) + bT (y)) akT (x)k2 + bρ′sgn(b) (T (x), T (y)) ag(x)kxk2 + bg(x)ρ′sgn(b) (x, y)

and g(y)(kyk2 + ρ′+ (y, x)) = g(y)ρ′+ (y, x + y) = ρ′+ (T (y), T (x + y)) = ρ′+ (T (y), aT (x) + bT (y)) = bkT (y)k2 + aρ′sgn(a) (T (y), T (x))

34


= bg(y)kyk2 + ag(y)ρ′sgn(a) (y, x). Thus a, b satisfy the system of linear equations akxk2 + bρ′sgn(b) (x, y) = kxk2 + ρ′+ (x, y), aρ′sgn(a) (y, x) + bkyk2 = kyk2 + ρ′+ (y, x). Since the associated determinant is ∆ > 0, we have 1 kyk2 (kxk2 + ρ′+ (x, y)) − ρ′sgn(b) (x, y)(kyk2 + ρ′+ (y, x)) ≥ 0 a= ∆ and 1 kxk2 (kyk2 + ρ′+ (y, x)) − ρ′sgn(a) (y, x)(kxk2 + ρ′+ (x, y)) ≥ 0 b= ∆

and consequently, a = b = 1 and T (x + y) = T (x) + T (y). Finally, for all v in X there exist v1 , v2 in R such that v = v1 e1 + v2 e2 and T (v) = T (v1 e1 + v2 e2 ) = T (v1 e1 ) + T (v2 e2 ) = v1 T (e1 ) + v2 T (e2 ) and T is linear on X. Moreover, for all x, y in X, x 6= 0 and t in R, t 6= 0 such that x + ty 6= 0 (x)k2 , we have since g(x) = kTkxk 2 2 2 1 g(x + ty) − g(x) 2 kT (x) + tT (y)k − kT (x)k = kxk 2t kxk2 kx + tyk2 2t 2 2 2 kx + tyk − kxk − kT (x)k 2t and we have lim

t→0±

1 g(x + ty) − g(x) 2 ′ 2 ′ = kxk ρ (T (x), T (y)) − kT (x)k ρ (x, y) ± ± 2t kxk4 1 kxk2 g(x) − kT (x)k2 ρ′± (x, y) = 0 = kxk4

and g is a constant function.

The converse implication has a trivial verification.

In the particular case where g ≡ 1 is the constant function, we have the following result. Corollary 2.3.1 A continuous mapping T from X into itself satisfies (2.3.1) if and only if T is linear and kT (x)k = kxk for all x in X.

Norm Derivatives

Corollary 2.3.2 equations

35

A continuous mapping T from X into itself satisfies the

ρ′ (x, y) ρ′+ (T (x), T (y)) = + kT (x)k kT (y)k kxk kyk

and

ρ′− (T (x), T (y)) ρ′ (x, y) = − (2.3.5) kT (x)k kT (y)k kxk kyk

for all nonzero x, y in X such that T (x) 6= 0, T (y) 6= 0, if and only if there exists a function λ from X into R+ such that T (x) = λ(x)(N ◦ L)(x), where L : X → X is linear, kL(x)k = kxk for all x in X and N is a function x for all x in X, x 6= 0 and N (0) := 0. defined by N (x) := kxk (x)kxk Proof. Let L be the function L(x) = TkT (x)k , then L satisfy the hypothesis of Corollary 2.3.1: L is linear and kL(x)k = kxk for all x in X. Then N (T (x)) = (N ◦ L)(x) and this condition is equivalent to the equality T (x) = λ(x)(N ◦ L)(x) for a function λ from X into R+ .

From Theorem 2.3.1 and the last corollaries, we obtain the following results. Corollary 2.3.3 A continuous function T satisfies (2.3.1) and (2.3.5), if and only if T is linear and kT (x)k = kxk for all x in X. (In this case λ(x) = kxk.) Corollary 2.3.4 A continuous function T satisfies (2.3.2) and (2.3.5), if and only if there exist a constant k and a linear function L such that kL(x)k = kxk for all x in X and T = kL. (In this case λ(x) = kkxk.) Corollary 2.3.5 A continuous function T satisfies (2.3.5) and it is homogeneous of degree 1, if and only if T (x) = λ(x)(N ◦ L)(x), L is linear, kL(x)k = kxk and λ(ax) = |a|λ(x) for all x in X and a in R. Corollary 2.3.6 A continuous function T satisfies (2.3.5) and it is invertible, if and only if T (x) = λ(x)(N ◦ L)(x), L is linear, kL(x)k = kxk and λ(ax) 6= |a|λ(x) for all x in X and for all a in R, a 6= 1. Some extensions of the previous results to semi-norms in R2 can be found in [Alsina and Tom´ as (1991)]. 2.4

On the equivalence of two norm derivatives

If X is a real linear space endowed with two norms k · k1 and k · k2 , the two norms are said to be equivalent whenever their respective induced topologies are the same. However, such a condition may be formulated in terms of a

36


double inequality, namely, there exist two positive constants 0 < A ≤ B such that Akxk2 ≤ kxk1 ≤ Bkxk2

(2.4.1)

holds for all x in X. Since each norm k · ki , i ∈ {1, 2}, has the associated functional (ρ′+ )i , i ∈ {1, 2}, let us find out how condition (2.4.1) may be established in terms of the functionals (ρ′+ )i . Let us denote these functionals by ρi , i ∈ {1, 2}. Then we have the following. Theorem 2.4.1 In a real linear space, two norms k · k1 and k · k2 are equivalent if and only if there exists a positive constant α such that |ρ1 (x, y) − ρ2 (x, y)| ≤ α min {kxk1 kyk1 , kxk2 kyk2 }

(2.4.2)

for all x, y in X. Proof.

If (2.4.2) holds, the substitution y := x yields −αkxk21 ≤ kxk21 − kxk22 ≤ αkxk21 , −αkxk22 ≤ kxk21 − kxk22 ≤ αkxk22 ,

whence kxk22 ≤ (1 + α)kxk21

and

kxk21 ≤ (1 + α)kxk22 ,

i.e., √ 1 √ kxk2 ≤ kxk1 ≤ 1 + αkxk2 , 1+α √ √ so (2.4.1) holds with A = 1/ 1 + α ≤ 1 + α = B. Conversely, let us assume (2.4.1). Then we have ρ1 (x, y) − ρ2 (x, y) ≤ kxk1 kyk1 + kxk2 kyk2 ≤ (B 2 + 1)kxk2 kyk2 and ρ1 (x, y) − ρ2 (x, y) ≥ −kxk1 kyk1 − kxk2 kyk2 ≥ −(B 2 + 1)kxk2 kyk2 . Therefore, |ρ1 (x, y) − ρ2 (x, y)| ≤ (B 2 + 1)kxk2 kyk2 .

37

Norm Derivatives

In a similar way, one checks that 1 1 + 2 kxk1 kyk1, A 1 2 so (2.4.2) follows by taking α := max 1 + 2 , B + 1 > 1. A |ρ2 (x, y) − ρ1 (x, y)| ≤

Remark 2.4.1. In the case where both functionals ρ1 and ρ2 are inner products, by using the polarization identities ρi (x, y) = (kx + yk2i − kx − yk2i )/4 for i ∈ {1, 2}, one obtains an alternative to (2.4.2) by means of the inequality 2 2 2 2 ρ1 (x, y) − A + B ρ2 (x, y) ≤ B − A (kxk2 + kyk2 ), 2 2 2 4 where A and B are the positive constants of equation (2.4.1).

Definition 2.4.1 Let X be a real linear space endowed with two norms k · k1 and k · k2 , whose corresponding functionals (ρ′+ )1 and (ρ′+ )2 are denoted by ρ1 and ρ2 , respectively. Then ρ1 and ρ2 will be called equivalent functionals if there exist two constants A, B > 0, A ≤ B such that A|ρ1 (x, y)| ≤ |ρ2 (x, y)| ≤ B|ρ1 (x, y)|

(2.4.3)

for all x, y in X. The next theorem clarifies the relation between (2.4.1) and (2.4.3). Theorem 2.4.2 Under the above notations, ρ1 and ρ2 are equivalent if and only if the norms k · k1 and k · k2 are equivalent and ρ2 (x, y) ρ1 (x, y) = kxk21 kxk22

(2.4.4)

for all x, y in X, x 6= 0. Proof.

If (2.4.3) holds, then for all x, y in X, x 6= 0, and α ∈ R, we have A|ρ1 (x, αx + y)| ≤ |ρ2 (x, αx + y)| ≤ B|ρ1 (x, αx + y)|,

(2.4.5)

i.e., A αkxk21 + ρ1 (x, y) ≤ αkxk22 + ρ2 (x, y) ≤ B αkxk21 + ρ1 (x, y) .

38


The substitution α := −ρ1 (x, y)/kxk21 yields at once ρ1 (x, y) 2 ≤ 0, kxk + ρ (x, y) 0 ≤ − 2 2 2 kxk1 i.e.,

ρ2 (x, y) ρ1 (x, y) = , kxk21 kxk22 so (2.4.4) holds. Moreover, the substitution y := x into (2.4.5) yields the equivalence of the norms √ √ Akxk1 ≤ kxk2 ≤ Bkxk1 . Conversely, assume that (2.4.1) is satisfied and consider the set kxk21 : x ∈ X, ∆= x = 6 0 ⊂ [A2 , B 2 ]. kxk22 Since it is a non-empty bounded set, its infimum and supremum exist in this interval A2 ≤ inf ∆ ≤ sup ∆ ≤ B 2 . Consequently, we have for all x, y in X, x 6= 0 kxk21 |ρ2 (x, y)| = |ρ1 (x, y)| kxk22 ≤ sup ∆|ρ2 (x, y)| ≤ B 2 |ρ2 (x, y)|.

A2 |ρ2 (x, y)| ≤ inf ∆|ρ2 (x, y)| ≤

2.5

Norm derivatives and projections in normed linear spaces

Equation (2.4.4), considered in the previous section, has a geometrical interpretation. It is the generalized equality of the two projections hx, yi2 hx, yi1 = kxk21 kxk22 because in a real inner product space (X, h·, ·i) with associated norm k · k, given two vectors x, y in X\{0}, the projection P{x} (y) of y on the lin{x} is the vector P{x} (y) =

hx, yi x. kxk2

Norm Derivatives

39

For the following results, we first give a definition of orthogonal sets. Definition 2.5.1 Let (X, k · k) be a real normed linear space. We define [x]B k·k as being Birkhoff ’s orthogonal set of x, i.e., B ′ ′ [x]B k·k := y ∈ X : x ⊥ y = y ∈ X : ρ− (x, y) ≤ 0 ≤ ρ+ (x, y) .

The next theorem clarifies completely the relation between condition (2.4.4) and Birkhoff’s orthogonal sets.

Theorem 2.5.1 Let X be a real normed linear space endowed with two norms k ·ki , i ∈ {1, 2}. Then, condition (2.4.4) holds if and only if [x]B k·k1 = B [x]k·k2 for all x in X. B Proof. Assume that [x]B k·k1 = [x]k·k2 for all x in X. Consider for x 6= 0 and any y in X, the associated real numbers αi := (ρ′+ )i (x, y)/kxk2i , i ∈ {1, 2}. Then, using Theorem 2.1.1 we have

(ρ′− )1 (x, y) ≤ α1 kxk21 = (ρ′+ )1 (x, y), whence (ρ′− )1 (x, y − α1 x) ≤ 0 ≤ (ρ′+ )1 (x, y − α1 x), B therefore y − α1 x ∈ [x]B k·k1 , and consequently, y − α1 x ∈ [x]k·k2 . So, in ′ ′ (ρ )2 (x, y) (ρ )1 (x, y) ≤ + 2 . particular, α1 kxk22 ≤ (ρ′+ )2 (x, y) and + 2 kxk1 kxk2 B B Analogously one shows that y − α2 x ∈ [x]k·k2 = [x]k·k1 , and we obtain ′ (ρ′ )1 (x, y) (ρ+ )2 (x, y) ≤ + 2 . Thus condition (2.4.4) holds. 2 kxk2 kxk1 Conversely, if (2.4.4) is valid, then

−(ρ′+ )1 (x, − y) −(ρ′+ )2 (x, −y) (ρ′ )2 (x, y) (ρ′− )1 (x, y) = = = − 2 , 2 2 2 kxk1 kxk1 kxk2 kxk2 and consequently, if y ∈ [x]B k·k1 , then we have (ρ′− )2 (x, y)

kxk21 kxk21 = (ρ′− )1 (x, y) ≤ 0 ≤ (ρ′+ )1 (x, y) = (ρ′+ )2 (x, y) , 2 kxk2 kxk22

B B i.e., y ∈ [x]B k·k2 , so [x]k·k1 ⊂ [x]k·k2 . The other inclusion follows at once. The proof is complete.

In what follows, we define the ρ-orthogonal set.

40


Definition 2.5.2 Let (X, k · k) be a real normed linear space. We call [x]ρk·k the ρ-orthogonal set of X if it is described by the expression [x]ρk·k = y ∈ X : ρ′+ (x, y) = 0 . We will see that condition (2.4.4) is related to the orthogonal sets described by two norms, as in the case of the Birkhoff orthogonality.

Theorem 2.5.2 Let X be a real normed linear space endowed with two norms k · ki , i ∈ {1, 2}. Then condition (2.4.4) holds if and only if [x]ρk·k1 = [x]ρk·k2 for all x in X.

Proof. It is obvious that (2.4.4) implies [x]ρk·k1 = [x]ρk·k2 . Conversely, if we start with the assumption [x]ρk·k1 = [x]ρk·k2 for all x in X, let us consider any x, y in X, x 6= 0 and the real number α = −(ρ′+ )1 (x, y)/kxk21 . Then (ρ′+ )1 (x, αx + y) = αkxk21 + (ρ′+ )1 (x, y) = 0, so αx + y ∈ [x]ρk·k1 , and it must belong to [x]ρk·k2 , i.e., 0 = (ρ′+ )2 (x, αx + y) = αkxk22 + (ρ′+ )2 (x, y) kxk22 + (ρ′+ )2 (x, y), = −(ρ′+ )1 (x, y) kxk21 i.e., (2.4.4) follows.

Another interesting result combining (2.4.4), the ⊥ρ -orthogonality and Birkhoff orthogonality is given in the following. Theorem 2.5.3 Let X be a real normed linear space endowed with two norms k · ki , i ∈ {1, 2}. Assume that (ρ′+ )1 = (ρ′− )1 . Then condition (2.4.4) holds if and only if for all x in X we have [x]ρk·k1 = [x]B k·k2 . Proof. If (2.4.4) holds then it is obvious that [x]ρk·k1 ⊆ [x]B k·k2 . In the other B direction, if y ∈ [x]k·k2 , then by Theorem 2.5.1 we have y ∈ [x]B k·k1 , and by virtue of condition (ρ′+ )1 = (ρ′− )1 we obtain y ∈ [x]ρk·k1 . Conversely, assume ′ 2 that [x]ρk·k1 = [x]B k·k2 . Given x, y in X, x 6= 0, take α := −(ρ+ )2 (x, y)/kxk2 . Then (ρ′+ )2 (x, αx + y) = αkxk22 + (ρ′+ )2 (x, y) = 0 ≥ (ρ′− )2 (x, αx + y), ρ i.e., αx + y ∈ [x]B k·k2 , so αx + y ∈ [x]k·k1 and we have

0 = (ρ′+ )1 (x, αx + y) = αkxk21 + (ρ′+ )1 (x, y)

41

Norm Derivatives

= −(ρ′+ )2 (x, y)

kx1 k21 + (ρ′+ )1 (x, y), kx2 k22

and (2.4.4) holds.

2.6

Norm derivatives and Lagrange’s identity in normed linear spaces

In three-dimensional inner product space (R3 , h·, ·i) one has the cross product × satisfying, among others, the bi-additive conditions x × (y + z) = x × y + x × z and (x + y) × z = x × z + y × z and the well-known Lagrange identity hx × y, z × vi = hx, zihy, vi − hx, vihy, zi. Then, as a natural generalization, in a real normed linear space (X, k ·k) of dimension 3 we consider the following problem: determine functions F from X × X into X satisfying the following conditions for all x, y, z, v in X: F (x, y + z) = F (x, y) + F (x, z),

(2.6.1)

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

(2.6.2)

and Lagrange’s identity in normed linear spaces: ρ′+ (F (x, y), F (z, v)) = ρ′+ (x, z)ρ′+ (y, v) − ρ′+ (x, v)ρ′+ (y, z).

(2.6.3)

Note, in particular, that (2.6.3) implies by taking z := x and v := y that kF (x, y)k2 = kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x). Lemma 2.6.1

(2.6.4)

If F satisfies (2.6.1), (2.6.2) and (2.6.4), then

(i) F (x, x) = 0 for all x in X; (2.6.5) (ii) F (y, x) = −F (x, y) for all x, y in X; (2.6.6) (iii) F (x, ay + bz) = aF (x, y) + bF (x, z) for all real a, b and x, y, z in X. (2.6.7)

42


Proof. The substitution of y := x into (2.6.4) yields (i). Next, by (i), F (x + y, x + y) = 0 and by (2.6.1) and (2.6.2), one gets (ii). Finally, by (2.6.4) and the properties of ρ′+ , F (x, ·) is continuous at y = 0, and by (2.6.1) condition (iii) follows. Lemma 2.6.2

If F satisfies (2.6.1), (2.6.2) and (2.6.3), then ρ′+ (x, y) = ρ′− (x, y) for all x, y in X.

Proof.

(2.6.8)

By (2.6.3) and Lemma 2.6.1, we have

0 = ρ′+ (F (x, −y), F (y, −y)) = ρ′+ (x, y)ρ′+ (−y, −y) − ρ′+ (x, −y)ρ′+ (−y, y) = ρ′+ (x, y)kyk2 − (−ρ′− (x, y))(−ρ′− (y, y)) = (ρ′+ (x, y) − ρ′− (x, y))kyk2 ,

whence for y 6= 0, ρ′+ (x, y) = ρ′− (x, y), and since this last equality is obvious for y = 0, we can conclude (2.6.8). Theorem 2.6.1 If (X, k · k) is a real normed linear space of dimension 3, and there exists a function F from X × X into X satisfying (2.6.1), (2.6.2) and (2.6.3), then necessarily the norm k · k is induced by an inner product. Proof. Assume that F from X × X into X satisfies (2.6.1), (2.6.2) and (2.6.3). By the previous lemmas for all x, z in X and a, b in R, we have kF (z, x + bz)k2 = kF (z, x + az + bz)k2 , i.e., kzk2 kx + bzk2 − ρ′+ (z, x + bz)ρ′+ (x + bz, z) = kzk2kx + (a + b)zk2 − ρ′+ (z, x + (a + b)z)ρ′+ (x + (a + b)z, z).

(2.6.9)

We can rewrite (2.6.9) in the form kzk2 kx + (a + b)zk2 − kx + bzk2 = ρ′+ (z, x + (a + b)z)ρ′+ (x + (a + b)z, z) − ρ′+ (z, x + bz)ρ′+ (x + bz, z).

(2.6.10)

Let us fix x and z, two independent vectors in X, and let us introduce the function f from R into R defined by f (t) := ρ′+ (x + tz, z) = ρ′− (x + tz, z). Thus, by means of (2.6.10) and (2.6.11) we can write kzk2 kx + (a + b)zk2 − kx + bzk2 = (a + b)kzk2 + ρ′+ (z, x) f (a + b) − bkzk2 + ρ′+ (z, x) f (b)

(2.6.11)

Norm Derivatives

= bkzk2 + ρ′+ (z, x) [f (a + b) − f (b)] + akzk2 f (a + b).

43

(2.6.12)

Since the norm is continuous and |f (a + b)| ≤ kx + (a + b)zkkzk, taking limits in (2.6.12) when a → 0± , we obtain bkzk2 + ρ′+ (z, x) lim (f (a + b) − f (b)) = 0, a→0±

i.e., for any real b, b 6= b0 := −ρ′+ (z, x)/kzk2 , we obtain lim f (b + a) = f (b).

a→0±

(2.6.13)

Note that by (2.6.12) and (2.6.11) at point b0 we have kx + b0 z + azk2 − kx + b0 zk2 a a→0± = 2ρ′± (x + b0 z, z) = 2f (b0 ).

lim f (b0 + a) = lim

a→0±

(2.6.14)

We claim that f (b0 ) = 0. To verifies this, consider the following chain of equalities for any real λ and for our fixed x, z: λ2 kF (x, z)k2 = kF (x, λz)k2 = kF (x + b0 z, λz)k2 = kF (x + b0 z, x + b0 z + λz)k2 = kx + b0 zk2 kx + b0 z + λzk2 −ρ′+ (x + b0 z, x + b0 z + λz)ρ′+ (x + b0 z + λz, x + b0 z + λz − λz) = kx + b0 zk2 kx + b0 z + λzk2 − kx + b0 zk2 + ρ′+ (x + b0 z, λz) kx + b0 z + λzk2

+ ρ′+ (x + b0 z + λz, −λz)

= −kx + b0 zk2 ρ′+ (x + (b0 + λ)z, −λz) − kx + b0 z + λzk2 ρ′+ (x + b0 z, λz) −ρ′+ (x + b0 z, λz)ρ′+ (x + (b0 + λ)z, −λz).

(2.6.15)

Since by Lemma 2.6.2 we have ρ′+ = ρ′− , division by λ < 0 in (2.6.15) yields λkF (x, z)k2 = kx + b0 zk2 ρ′+ (x + (b0 + λ)z, z)

44


− kx + b0 z + λzk2 ρ′+ (x + b0 z, z) + λρ′+ (x + b0 z, z)ρ′+ (x + (b0 + λ)z, z)

(2.6.16)

and taking limits when λ → 0− , by using (2.6.14) we obtain 0 = kx + b0 zk2 2f (b0 ) − kx + b0 zk2 f (b0 ), and since x and z are independent, f (b0 ) = 0. Therefore, by (2.6.16), for any λ < 0 it is λkF (x, z)k2 = kx + b0 zk2 f (b0 + λ), i.e., for any t < b0 : f (t) = f (b0 + (t − b0 )) =

kF (x, z)k2 (t − b0 ), kx + b0 zk2

so f is an affine function on (−∞, b0 ]. Since f (b0 ) = 0, by (2.6.15) we also have for λ > 0 λ2 kF (x, z)k2 = λkx + b0 zk2 f (b0 + λ), i.e., for any t > b0 : f (t) = f (b0 + (t − b0 )) =

kF (x, z)k2 (t − b0 ), kx + b0 zk2

so f is an affine function on R vanishing at b0 . Thus for all real t ρ′+ (z, x) kF (x, z)k2 ′ , ρ+ (x + tz, z) =

t+ ρ′ (z,x) 2 kzk2

x − +kzk2 z and for t = 0 we obtain: ρ′+ (x, z) =

kxk2 kzk2 − ρ′+ (x, z)ρ′+ (z, x) ρ′+ (z, x) · ,

ρ′ (z,x) 2

kzk2

x − +kzk2 z

i.e., ρ′+ (x, z) = 0 if and only if ρ′+ (z, x) = 0 and the symmetry of the orthogonality relation ρ′+ (x, z) = 0 yields that necessarily (in dimension 3) the norm derives from an inner product (see (19.6) in [Amir (1986)]). This completes the proof.

Norm Derivatives

2.7

45

On some extensions of the norm derivatives

Our aim in this section is to consider orthogonal relations in situations where we use functionals satisfying weaker conditions than those quoted above for ρ′+ , and also to check in which cases the required properties force the derivability of the norm from an inner product. Precisely, let (X, k · k) be a real normed linear space, and let F be a function from X × X into R such that (i) F (x, x) 6= 0 = F (0, 0) whenever x 6= 0; (ii) There exist ε > 0 and a function f : R+ → R+ such that for all x, y in X and λ > 0: |F (x, x + λy) − F (x, x) − f (λ)F (x, y)| ≤ ε; (iii) There exists δ > 0 such that for all x, y in X and λ > 0 |F (λx, y) − F (x, λy)| ≤ δ. The above set of conditions (i), (ii) and (iii) is a weaker requirement than that satisfying the corresponding properties of ρ′± from Theorem 2.1.1. The following results will in fact show a kind of stability of generalization of conditions (v) and (iii) for Theorem 2.1.1. Remark 2.7.1. Conditions (i), (ii) and (iii) are independent even in dimension 1. To verify this, consider the following examples: (a) The function F (x, y) = 0 for all x, y satisfies (ii) and (iii), but not (i). (b) Given δ > 0, the function F : R × R → R defined by F (x, y) := x2 , if x = y and F (x, y) :=min(xy, δ) if x 6= y, satisfies (i) and (iii), but does not satisfy (ii) because there is no function f such that | min(x2 + λxy, δ) − x2 − f (λ) min(xy, δ)| ≤ ε, since the term on the left-hand side tends to infinity when x tends to infinity. (c) The function F : R×R → R, F (x, y) = y satisfies (i), (ii) with f (λ) = λ, for all λ, but (iii) fails. We begin with a technical but crucial result. Lemma 2.7.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If a functional F : X × X → R satisfies (i), (ii), (iii) and

46


the condition F (x, λx)(F (x, λx) − λkxk2 ) = 0,

(2.7.1)

for all x in X and λ > 0, then necessarily F (x, λy) = F (λx, y) = λF (x, y),

(2.7.2)

F (x, x + λy) = kxk2 + λF (x, y),

(2.7.3)

and

for all x, y in X and λ > 0. Proof. Assume that F satisfies (i), (ii), (iii) as well as condition (2.7.1). By (2.7.1) with λ = 1 and using (i), we deduce F (x, x) = kxk2 for all x in X. Therefore, by (ii) with x = y, for any λ > 0 we have |F (x, (1 + λ)x) − (1 + f (λ))kxk2 | ≤ ε and hence, F (x, (1 + λ)x) = 1 + f (λ) kxk2 kxk→∞ lim

(2.7.4)

for every x ∈ X, x 6= 0. It follows that the inequality F (x, (1 + λ)x) > 0 is satisfied for kxk large enough, thus by (2.7.1) we also have (1 + λ)kxk2 F (x, (1 + λ)x) = = 1 + λ. lim kxk2 kxk2 kxk→∞ kxk→∞ lim

(2.7.5)

From (2.7.4) and (2.7.5) we get f (λ) = λ for every λ > 0. Next, for b > 0 the substitution y := bz into (ii) implies |λbF (x, z) − λF (x, bz)| ≤ |λbF (x, z) + kxk2 − F (x, x + (λb)z)| + |F (x, x + λ(bz)) − kxk2 − λF (x, bz)| ≤ 2ε, i.e., |bF (x, z) − F (x, bz)| ≤ 2ε/λ, and letting λ tend to infinity F (x, bz) = bF (x, z), for all b > 0. Consequently, by (iii) we obtain |F (bx, z) − bF (x, z)| = |F (bx, z) − F (x, bz)| ≤ δ.

(2.7.6)

Norm Derivatives

47

Dividing by b and taking limit when b tends to infinity we get F (bx, z) = F (x, z), b→∞ b lim

i.e., for all a > 0 F (ax, z) = lim

b→∞

F (bax, z) F (bax, z) = lim a = aF (x, z). b→∞ b ba

(2.7.7)

Thus, (2.7.6) and (2.7.7) prove (2.7.2). Finally, taking x := nu and y := nv in (ii): |F (nu, nu + λnv) − n2 ||uk2 − λF (nu, nv)| ≤ ε, but using (2.7.6) and (2.7.7): |F (u, u + λv) − kuk2 − λF (u, v)| ≤ ε/n2 , so letting n tend to infinity, we obtain (2.7.3).

Theorem 2.7.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. A functional F : X × X → R satisfies (i), (ii), (iii) and the Pythagorean identity

2

F (y, x) 2 F (y, x) 2

y + x − y (iv) kxk = kyk2 kyk2

for all x, y in X, y 6= 0, if and only if X is an inner product space whose inner product associated with the norm k · k is precisely F . Proof.

Assume that F satisfies (i), (ii), (iii) and (iv). Then

2 kxk2 kyk4 = [F (y, x)]2 kyk2 + kyk2 x − F (y, x)y ,

and for x := λy, λ > 0 we obtain

F (y, λy)(F (y, λy) − λkyk2 ) = 0, that is (2.7.1), so we can apply the previous lemma and induce the validity of both (2.7.2) and (2.7.3). Bearing this in mind, and substituting y := u and x := u + λv, λ > 0 into (iv), we get ku + λvk2 kuk4 = [F (u, u + λv)]2 kuk2 + kkuk2(u + λv) − F (u, u + λv)uk2 = (kuk2 + λF (u, v))2 kuk2 + kkuk2(u + λv) − kuk2 u − λF (u, v)uk2 ,

48


whence kuk4 (ku + λvk2 − kuk2 ) = λ2 [F (u, v)]2 kuk2 + 2λF (u, v)kuk4 + λ2 kkuk2 v − F (u, v)uk2 ,

(2.7.8)

and dividing by 2λ and taking limits when λ → 0+ , we obtain ρ′+ (u, v) = F (u, v).

(2.7.9)

Finally, using (2.7.9), (2.7.8) with λ = 1 and (iv), kuk4 (ku + vk2 − kuk2 ) = [ρ′+ (u, v)]2 kuk2 + 2ρ′+ (u, v)kuk4 + kkuk2v − ρ′+ (u, v)uk2 = [ρ′+ (u, v)]2 kuk2 + 2ρ′+ (u, v)kuk4 + kvk2 kuk4 − [ρ′+ (u, v)]2 kuk2

i.e.,

= 2ρ′+ (u, v) + kvk2 kuk4 , ku + vk2 = kuk2 + kvk2 + 2ρ′+ (u, v),

so ρ′+ is symmetric, and therefore it must be an inner product. The converse is immediate. Remark 2.7.2. If (i) or (ii) do not hold, we may still have (iii) and (iv) for a norm derivable from an inner product, but different from F . The following examples show why this is possible. Example 2.7.1

F ≡ 0 satisfies (ii), (iii) and (iv) but does not verify (i).

Example 2.7.2 In the Euclidean space (R2 , k · k), if χ denotes the characteristic function of the real set [0, 1] ∩ Q, define F : R2 × R2 → R by means of F (x, y) = χ(|x · y|/kxkkyk)(x · y). Then it is easy to see that F satisfies (i), (iii) for any δ > 0 and (iv). If F satisfied (ii) then x = y would yield f (λ) = λ for all λ > 0, and

Norm Derivatives

49

substituting x = (3n, 4n), y = (1, 0) one would arrive at an inequality, where the right-hand side is ε and on the left-hand side the term may tend to infinity when n grows to infinity. Now we turn our attention to the James orthogonality. Theorem 2.7.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. A functional F : X × X → R satisfies (i), (ii), (iii) and the James identity

F (y, x)

y for all x, y in X, y 6= 0, (v) kxk = x − 2 kyk2 if and only if X is an i.p.s. whose inner product is F .

Proof. Assume that F satisfies the above mentioned conditions. Using (v) with x = λy, λ > 0, and bearing in mind (i), we obtain F (y, λy) = 0

or

F (y, λy) = λkyk2 ,

i.e., (2.7.1) holds, and by Lemma 2.7.1 we have (2.7.2) and (2.7.3). Substituting y := u and x := u + λv, λ > 0, into (v) and using (2.7.3), we get

F (u, v)

, (2.7.10) u − v ku + λvk = u + λ 2

2 kuk whence

2

F (u,v)

u + λ 2 kuk2 u − v − kuk2 ku + λvk2 − kuk2 = . 2λ 2λ

So, taking limit when λ tends to zero from the right, we have F (u, v) F (u, v) ′ ′ u−v =2 kuk2 − ρ′− (u, v) ρ+ (u, v) = ρ+ u, 2 kuk2 kuk2 and therefore F (u, v) = (ρ′+ (u, v) + ρ′− (u, v))/2. Substituting the last expression in (2.7.10) dividing by λ > 0 and taking limit when λ tends to infinity, one obtains

ρ′+ (u, v) + ρ′− (u, v)

u kvk = v −

, u, v ∈ X, u 6= 0. kuk2

50


So in all nonzero points of smoothness we have ρ′ (u, v) := ρ′+ (u, v) = ρ′− (u, v) and

2ρ′ (u, v)

u kvk = v −

, kuk2

which means that the property (i) of Theorem 1 from [Precupanu (1978)] ′ (u,v) ) for all nonzero points of smoothness u is satisfied (with η ′ (u, v) := ρ kuk in X, and on account of Theorem 2 from that paper the norm in X derives from an inner product. The converse is obvious. Note. Theorems 2.7.1 and 2.7.2 may also be useful for showing that a real normed space is not an i.p.s. Consider X = c0 (i.e., the space of all real sequences convergent to zero) and take F = ρ′+ , so (i), (ii) and (iii) are obviously satisfied, and consider x = (1/n) and y = (1/n2 ) in c0 , then kxk = kyk = 1 and ρ′+ (y, x) = 1, so (iv) would imply kx − yk = 0, i.e., x = y, which is impossible. Therefore c0 is not an i.p.s. Analogously, condition (v) may be used to reach the same conclusion. Theorem 2.7.3 In a real normed linear space (X, k · k) with dim X ≥ 2, let F : X × X → R be a functional satisfying (i), (ii), (iii). Then, the Birkhoff ’s inequalities F (y, x) F (y, x) F (y, x) F (y, x) ′ ′ y, x − y ≤ 0 ≤ ρ+ y, x − y (vi) ρ− kyk2 kyk2 kyk2 kyk2

hold for all x, y in X, y 6= 0, if and only if ρ′− (x, y) ≤ F (x, y) ≤ ρ′+ (x, y) for all x, y in X. Proof. Assume that F satisfies (i), (ii), (iii) and (vi). By the substitution x := λy, λ > 0, into (vi), using the properties of ρ′± we obtain precisely condition (2.7.1) F (y, λy) = 0, F (y, λy) λ − kyk2

so F satisfies the hypothesis of Lemma 2.7.1, i.e., (2.7.2) and (2.7.3) hold. Setting y := u and x := u + λv, λ > 0 in (vi), and using (2.7.3), we have F (u, v) F (u, v) u, u ≤0 ρ′− 1+λ λv − λ kuk2 kuk2 F (u, v) F (u, v) ′ ≤ ρ+ u, λv − λ u 1+λ kuk2 kuk2

Norm Derivatives

51

(u,v) is a positive and hence, assuming that λ is small enough, so that 1+λ Fkuk 2 number, we get F (u, v) −F (u, v) + ρ′− (u, v) ≤ 0 λ 1+λ 2 kuk F (u, v) −F (u, v) + ρ′+ (u, v) , ≤λ 1+λ 2 kuk

whence it follows that the inequalities ρ′− (u, v) ≤ F (u, v) ≤ ρ′+ (u, v) hold for all u, v in X, u 6= 0. On the other hand it is easy to verify that the above inequalities still hold for u = 0. Reciprocally, assume that F satisfies (i), (ii), (iii) and ρ′− (x, y) ≤ F (x, y) ≤ ρ′+ (x, y) for all x, y in X. Let x, y be in X, y 6= 0. If F (y, x) = 0, then (vi) is trivially satisfied. If F (y, x) 6= 0, then using the properties of ρ′± from Theorem 2.1.1, one concludes that (vi) is equivalent to F (y, x)ρ′− (y, x) ≤ [F (y, x)]2 ≤ F (y, x)ρ′+ (y, x)

in the case F (y, x) > 0, and also to the inequality with opposite signs in the other case. Both cases give finally ρ′− (y, x) ≤ F (y, x) ≤ ρ′+ (y, x). 2.8

ρ-orthogonal additivity

The Cauchy functional equation, i.e., the equation of additivity, has been widely investigated (see e.g. [Aczél (1966); Kuczma (1985); Aczél and Dhombres (1989)]). Its conditional form described below deserves further study. Definition 2.8.1 A mapping f from a linear space X into a group (G, +) is called orthogonally additive provided that for every x, y ∈ X one has x⊥y

implies

f (x + y) = f (x) + f (y),

(2.8.1)

where ⊥ denotes an orthogonal relation defined on X. For instance, in an inner product space (X, h·, ·i) the functional X ∋ x 7→ hx, xi ∈ R is orthogonally additive (Pythagoras theorem). The notion of orthogonal additivity has been intensively studied by many authors; see e.g.

52


[Drewnowski and Orlicz (1968); Sundaresan (1972); Gudder and Strawther (1975); R¨ atz (1985); Szab´ o (1986); Szabó (1990); Szabó (1991); Szabó (1993); Szab´ o (1995); Szab´ o (1997)] and others. Let us state the notion of an abstract orthogonality space (see [Rätz (1985)]). Definition 2.8.2 An ordered pair (X, ⊥) is called an orthogonality space in the sense of R¨ atz whenever X is a real linear space with dim X ≥ 2 and ⊥ is a binary relation on X such that (i) (ii) (iii) (iv)

x ⊥ 0 and 0 ⊥ x for all x in X; if x, y ∈ X \ {0} and x ⊥ y, then x and y are linearly independent; if x, y ∈ X and x ⊥ y, then for all α, β in R we have αx ⊥ βy; for any two-dimensional subspace P of X and for every x ∈ P , λ ∈ [0, ∞), there exists a y ∈ P such that x ⊥ y and x + y ⊥ λx − y.

A normed linear space with Birkhoff orthogonality is a typical example of an orthogonality space (see [Rätz (1985); Szabó (1990); Szabó (1991)]). According to the results from the papers [Rätz (1985); Baron and Volkmann (1998)], we state the following theorem concerning the orthogonal additivity for a function defined on an orthogonality space. Theorem 2.8.1 Let (X, ⊥) be an orthogonality space and let (G, +) be an Abelian group. A mapping f : X → G satisfies condition (2.8.1) if and only if there exist an additive mapping a : X → G and a biadditive and symmetric mapping b : X × X → G such that f (x) = a(x) + b(x, x) for all x ∈ X

(2.8.2)

b(x, y) = 0 for all x, y ∈ X with x ⊥ y.

(2.8.3)

and

As an immediate consequence of the above result, we deduce that each Birkhoff orthogonally additive mapping has the form (2.8.2). And finally, on account of Proposition 2.2.4 (iii) and (vi), since in smooth spaces the relations ⊥ρ and ⊥B are equivalent, as a corollary we get the following result. Corollary 2.8.1 Let (X, k · k) be a smooth normed linear space with dim X ≥ 2, and let (G, +) be an Abelian group. A mapping f : X → G

Norm Derivatives

53

satisfies condition x ⊥ρ y

implies

f (x + y) = f (x) + f (y),

(2.8.4)

if and only if there exist an additive mapping a : X → G and a biadditive and symmetric mapping b : X × X → G such that f has the form (2.8.2) and condition (2.8.3) is satisfied with ⊥ := ⊥ρ . The question is: What about spaces which are not smooth? Assume that (X, k · k) is a normed linear space with dim X ≥ 2. We will show that the relation ⊥ρ satisfies the four properties of the orthogonality space (see [Alsina, Sikorska and Tomás (2007)]. The first three are easy to check. In order to check the fourth, we need some auxiliary results. Lemma 2.8.1

For any two vectors x and w in X, we have lim ρ′± (x + tw, w) = ρ′± (x + t0 w, w).

t→t0

Proof.

By Corollary 2.1.2, we can write

lim ρ′± (x + tw, w) = lim ρ′± (x + t0 w + sw, w) = ρ′± (x + t0 w, w).

t→t0

s→0

Lemma 2.8.2

For any two vectors x and w in X, we have lim ρ′± (x + tw, x) = ρ′± (x + t0 w, x).

t→t0

Proof.

By Theorem 2.1.1 (iii)-(v) for t 6= 0, we have ρ′± (x + tw, x) = kx + twk2 − tρ′∓ sgn (t) (x + tw, w),

and by Lemma 2.8.1 we obtain: if t0 = 0, immediately lim ρ′± (x + tw, x) = kxk2 ,

t→0

and if t0 6= 0, then lim ρ′± (x+tw, x) = kx+t0 wk2 −t0 ρ′∓ sgn (t0 ) (x+t0 w, w) = ρ′± (x+t0 w, x).

t→t0

Lemma 2.8.3 For any x in X \ {0} and λ in [0, ∞) there exists z in X \ {0} such that h ih i 4kxk2 kzk2 . (2.8.5) ρ′+ (x, z) + ρ′− (x, z) ρ′+ (z, x) + ρ′− (z, x) = λ+1

54


Proof. If λ = 0, it is enough to take z := x. Assume that λ > 0. Let w ∈ X be linearly independent of x and such that ρ′+ (x, w) + ρ′− (x, w) 6= 0. Define ϕ : R → R by ϕ(t) :=

i 4kxk2 kx + twk2 h ′ − ρ+ (x, x + tw) + ρ′− (x, x + tw) λ+1 h i ′ × ρ+ (x + tw, x) + ρ′− (x + tw, x) .

We have ϕ(0) = 4kxk

If t1 := −

4

1 −1 λ+1

< 0.

2kxk2 , then ρ′+ (x, w) + ρ′− (x, w)

ρ′+ (x, x + t1 w) + ρ′− (x, x + t1 w) = 2kxk2 + t1 ρ′+ (x, w) + ρ′− (x, w) = 0

and

ϕ(t1 ) =

4kxk2 kx + t1 wk2 > 0. λ+1

On account of Proposition 2.1.3 and Lemma 2.8.2, function ϕ is continuous, and consequently between t1 and 0 there exists t0 such that ϕ(t0 ) = 0, i.e., condition (2.8.5) is satisfied with z := x + t0 w. Now we are able to prove Proposition 2.8.1 For any two-dimensional subspace P of X and for every x in P , λ in [0, ∞), there exists a y in P such that x ⊥ρ y and x + y ⊥ρ λx − y. Proof. Fix x in X. If x = 0 then take y := 0. For x 6= 0 take nonzero z in X such that (2.8.5) is satisfied. Define y := −x +

i λ+1 h ′ ′ (z, ρ x) + ρ (z, x) z. + − 2kzk2

Norm Derivatives

55

We have ρ′+ (x, y) + ρ′− (x, y) λ+1 ′ ′ ρ (z, x) + ρ (z, x) z = ρ′+ x, −x + − 2kzk2 + λ+1 ′ ′ ρ (z, x) + ρ (z, x) z + ρ′− x, −x + − 2kzk2 + h ih i λ+1 ′ ′ ′ ′ ρ (z, = − 2kxk2 + x) + ρ (z, x) ρ (x, z) + ρ (x, z) = 0, − + − 2kzk2 +

and

ρ′+ (x + y, λx − y) + ρ′− (x + y, λx − y) λ+1 ′ ′ ρ (z, x) + ρ′− (z, x) z, = ρ+ 2kzk2 + λ+1 ′ ′ ρ (z, x) + ρ− (z, x) z (λ + 1)x − 2kzk2 + λ+1 ′ ρ (z, x) + ρ′− (z, x) z, + ρ′− 2kzk2 + λ+1 ′ ′ ρ (z, x) + ρ− (z, x) z (λ + 1)x − 2kzk2 + =−2 +

i2 (λ + 1)2 h ′ ′ ρ (z, x) + ρ (z, x) kzk2 + − 4kzk4

ih i (λ + 1)2 h ′ ′ ′ ′ ρ (z, x) + ρ (z, x) ρ (z, x) + ρ (z, x) = 0, + − + − 2kzk2

which concludes the proof.

Relation ⊥ρ satisfies (iv), so the main result of this section follows immediately. Theorem 2.8.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2, and let (G, +) be an Abelian group. A mapping f : X → G satisfies condition (2.8.4) if and only if there exist an additive mapping a : X → G and a biadditive and symmetric mapping b : X × X → G such that f has the form (2.8.2) and condition (2.8.3) holds with ⊥:=⊥ρ .

Chapter 3

Norm Derivatives and Heights

Our aim in this chapter is to consider heights in triangles located in real normed spaces and defined in terms of the norm derivatives, as well as to study in detail which generalizations of the usual properties of heights in the Euclidean plane give characterizations of inner product spaces.

3.1

Definition and basic properties

In the Euclidean plane R2 (or, in general, in a real inner product space), given the triangle determined by two linearly independent vectors x and y with the initial point at the origin, one can compute the height vector with the initial point at the origin and orthogonal to x − y using the formula h=y+

kyk2 − hx, yi hy − x, yi (x − y) = y + (x − y). 2 kx − yk kx − yk2

x−

h=

0

y

(x +α

−y

x Figure 3.1.1 57

)

y

y

58


Using the function ρ′+ as a generalization of an inner product, in a real normed space (X, k · k) we consider the following collection of height functions: h1 (x, y) := y + h2 (x, y) := y + h3 (x, y) := y +

kyk2 − ρ′+ (x, y) (x − y), kx − yk2 kyk2 − ρ′+ (y, x) (x − y), kx − yk2 ρ′+ (y − x, y) (x − y) kx − yk2

defined for x, y ∈ X, x 6= y and hi (x, x) := x for x ∈ X, i ∈ {1, 2, 3}. Remark 3.1.1. Observe that for all x in X, we have h(x, λx) = 0 if λ 6= 1, but h(x, λx) = x if λ = 1. Analogous definitions can be given by replacing the role of ρ′+ by ρ′− or by changing in the given height functions the order of the arguments appearing in ρ′+ . Since all results obtained in such cases are similar to those exhibit in our study on h1 , h2 and h3 , we restrict ourselves to this triplet of functions. If X is an inner product space, then for all x, y in X, h1 (x, y) = h2 (x, y) = h3 (x, y), and in the triangle determined by vectors x and y with the initial point at the origin, this is the usual height h described above. In general, if X is a real normed space, where the norm does not come from an inner product, these three height functions are different. Example 3.1.1 Consider (R2 , k · k+ ). Then, it is very easy to see that 7 1 , for x =(2, 1) and y = (1, −2), we obtain: h (x, y) = 1 4 4 , h2 (x, y) = 7 5 11 5 , − , − , . h (x, y) = 3 8 8 4 4

However, the equality of arbitrary two height functions characterizes already an inner product space. Namely, we have the following. Theorem 3.1.1 Let (X, k·k) be a real normed linear space. The following conditions are equivalent: (i) The norm in X comes from an inner product; (ii) h1 = h2 ; (iii) h1 = h3 ;


59

(iv) h2 = h3 . Moreover, we have the following characterization Theorem 3.1.2 Let (X, k · k) be a real normed linear space, dim X ≥ 2. Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)

The norm in X comes from an inner product; ρ′+ (x − y, h1 (y, x)) ∼ 0 for all x, y in X; ρ′+ (x − y, h1 (x, y)) ≤ 0 for all x, y in X; ρ′+ (x − y, h2 (x, y)) ∼ 0 for all x, y in X; ρ′+ (x − y, h2 (y, x)) ≤ 0 for all x, y in X,

where ∼ denotes either ≤ or ≥. Proof. If X is an inner product space, it is obvious that the other conditions hold. For the converse implications, it is enough to consider only x, y ∈ X such that x 6= y. If we assume condition (ii), then for all x, y in X, x 6= y, we have 0 ∼ ρ′+ (x − y, x) + ρ′+ (y, x) − kxk2 . Substituting y := tu, with u linearly independent of x, and z := x − tu, where t belongs to R+ , we obtain kz + tuk2 − kzk2 ∼ ρ′+ (z, u) + ρ′+ (u, z) + tkuk2 , t and if t → 0+ , then ρ′+ (u, z) ∼ ρ′+ (z, u) for all z, u ∈ X (also for linearly dependent u and z), whence from Theorem 2.1.1. (viii) and (ix), X is an inner product space. If we assume (iii), we have also ρ′+ (y − x, h1 (y, x)) ≤ 0 for all x, y in X. Then by the properties of ρ′± , we get ρ′+ (x − y, h1 (y, x)) ≥ ρ′− (x − y, h1 (y, x)) = −ρ′+ (y − x, h1 (y, x)) ≥ 0 for all x, y in X, which means that (ii) is satisfied and we have our assertion. If we assume (iv), then kyk2 − ρ′+ (y, x) + ρ′+ (x − y, y) ∼ 0 for all x, y in X. By the substitution z := x − y, we obtain ρ′+ (z, y) ∼ ρ′+ (y, z) for all z, y in X, whence the norm in X can be derived from an inner product. Finally, if we assume (v), then also ρ′+ (y − x, h2 (x, y)) ≤ 0 for all x, y in X. Using the properties of ρ′± , we obtain ρ′+ (x − y, h2 (x, y)) ≥ ρ′− (x − y, h2 (x, y)) = −ρ′+ (y − x, h2 (x, y) ≥ 0 for all x, y in X, which gives (iv) and completes the proof. Using the definition of h3 , immediately we get the following result.

60


Theorem 3.1.3 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then the following conditions hold for all x, y in X: (i) ρ′+ (x − y, h3 (y, x)) = 0; (ii) ρ′+ (x − y, h3 (x, y)) ≥ 0; (iii) ρ′+ (x − y, h3 (x, y)) = 0 if and only if ρ′+ (x, y) = ρ′− (x, y). If we assume that the height functions are symmetric, we obtain Theorem 3.1.4 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then, for all x, y in X: (i) h1 (x, y) = h1 (y, x) if and only if X is an i.p.s.; (ii) h2 (x, y) = h2 (y, x) if and only if X is an i.p.s.; (iii) h3 (x, y) = h3 (y, x) if and only if ρ′− (x, y) = ρ′+ (x, y). Proof. We prove only case (ii). The other cases are proved similarly. If h2 is symmetric, then it is immediate to get for all x, y ∈ X kx − yk2 = kxk2 + kyk2 − ρ′+ (y, x) − ρ′+ (x, y). If we change y by −y, we obtain kx + yk2 = kxk2 + kyk2 + ρ′− (y, x) + ρ′− (x, y). Adding the last two equalities yields kx − yk2 + kx + yk2 = 2kxk2 + 2kyk2 + ρ′− (y, x) − ρ′+ (y, x) + ρ′− (x, y) − ρ′+ (x, y) ≤ 2kxk2 + 2kyk2 for all x, y in X. From Theorem 1.4.2, the norm in X comes from an inner product. The converse implication is obvious. 3.2

Characterizations of inner product spaces involving geometrical properties of a height in a triangle

If (X, h·, ·i) is an inner product space and x and y are two linearly independent vectors in X, in the triangle determined by vectors x and y with the initial point at the origin, by using Heron’s formula to calculate the area, the length of the height over the side x − y is given by the formula p 2 s(s − kxk)(s − kyk)(s − kx − yk), kx − yk

(3.2.1)


61

where s is the semiperimeter s = (kxk + kyk + kx − yk)/2 and the norm is derived from the inner product. For our next results in a real normed linear space, we consider h1 as the height function. Among characterizations of i.p.s. given by means of h1 we start with one, where we compare the norm of h1 (x, y) and (3.2.1). Theorem 3.2.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then the norm k · k comes from an inner product if and only if p (3.2.2) kx − ykkh1 (x, y)k = 2 s(s − kxk)(s − kyk)(s − kx − yk) for all x, y in X.

Proof. Observe that if x and y are linearly dependent, both sides of (3.2.2) are equal to zero, so in fact it is enough to consider linearly independent vectors. If we substitute y by tz with t > 0 in the definition of h1 (x, y), we divide it by t and take the limit when t tends to zero, then we obtain

tkzk2 − ρ′+ (x, z) kh1 (x, tz)k

= lim+ z + (x − tz) lim+

2 t kx − tzk t→0 t→0

ρ′+ (x, z)

x . = z − kxk2

(3.2.3)

On the other hand, by (3.2.2) it is easy to see that kh1 (x, tz)k = lim + t t→0 =

p (kzkkxk + ρ′− (x, z))(kzkkxk − ρ′− (x, z)) kxk p kzk2kxk2 − ρ′− (x, z)2 . kxk

(3.2.4)

Then, by virtue of (3.2.3) and (3.2.4), we have the equality

and therefore,

p

′ ′ 2 2 2

ρ (x, z) +

= kzk kxk − ρ− (x, z)

z − x

2 kxk kxk

2

2

z x z x z x ′ ′

kzk − ρ+ kxk , kzk kxk = 1 − ρ− kxk , kzk .

(3.2.5)

62


If we consider u = by (3.2.5) we have

x z and v = , then for all u, v in SX , v 6= ±u, kxk kzk

kv − ρ′+ (u, v)uk2 = 1 − ρ′− (u, v)2 ,

(3.2.6)

u+v in (3.2.6), we obtain ku + vk

2 2

u+v u+v u+v ′ ′

u, − ρ u = . 1 − ρ u, + −

ku + vk ku + vk ku + vk

and if we replace v by

Bearing in mind that kuk = 1, and using the properties of ρ′± , we get kv − ρ′+ (u, v)uk2 = ku + vk2 − 1 − ρ′− (u, v)2 − 2ρ′− (u, v).

(3.2.7)

Thus (3.2.6) and (3.2.7) yield 2 + 2ρ′− (u, v) = ku + vk2 = kv + uk2 = 2 + 2ρ′− (v, u), for linearly independent u and v in SX , whence, in fact, ρ′− (u, v) = ρ′− (v, u) for all u, v in SX and X is an inner product space. The converse implication results from an easy computation. Formula (3.2.1) represents a function of kxk, kyk, kx − yk, so a generalization of (3.2.2) is to consider the existence of a function L : (0, ∞)3 → (0, ∞) for which kh1 (x, y)k = L (kxk, kyk, kx − yk) ,

x, y ∈ X\{0}, x 6= y.

We now use differentiability conditions to establish the following characterization of i.p.s. Theorem 3.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2, and let H : R3 → R be a function from class C 2 in an open set containing the set of points X = {P = (kxk, 0, kxk) : x ∈ X}. Assume that kh1 (x, y)k2 = H (kxk, kyk, kx − yk) for all x, y in X\{0}, x 6= y (3.2.8) holds. Then X is an inner product space if and only if for all P ∈ X : (i) H(P ) = 0; (ii) D2 H(P ) = D3 H(P ) = 0; (iii) D23 H(P ) = 0, D22 H(P ) = 2, D33 H(P ) = −2.


63

Proof. Assume that function H satisfies conditions (i), (ii) and (iii) for all P ∈ X . Substituting y := tz with t > 0 in (3.2.8), dividing the result by t2 and taking the limit when t tends to zero from the right, we obtain

2 ρ′+ (x, z) kh1 (x, tz)k2 H (kxk, tkzk, kx − tzk)

= lim = z − x . lim t2 t2 kxk2 t→0+ t→0+ (3.2.9) On the other hand, using (i), (ii) and (iii) and applying l’Hˆ opital’s rule twice, by a straightforward computation, if g(t) = kx − tzk, we obtain H (kxk, tkzk, kx − tzk) t2 t→0 D2 H(kxk, tkzk, kx − tzk)kzk + D3 H(kxk, tkzk, kx − tzk)g ′ (t) = lim+ 2t t→0 ρ′− (x, z) ρ′− (x, z)2 kzk2 = D22 H(P ) − 2D23 H(P ) + D33 H(P ) 2 kxkkzk kxk2 kzk2 ′ 2 ρ (x, z) = kzk2 1 − − 2 . (3.2.10) kxk kzk2 lim+

By (3.2.9) and (3.2.10), we have

2

2

z z x z x x ′ ′

kzk − ρ+ kxk , kzk kxk = 1 − ρ− kxk , kzk ,

and if u :=

x z , v := we obtain kxk kzk

kv − ρ′+ (u, v)uk2 = 1 − ρ′+ (u, v)2 , Then, if we substitute v by

u, v ∈ SX .

(3.2.11)

u+v , we have ku + vk

kv − ρ′+ (u, v)k2 = ku + vk2 − 1 − 3ρ′− (u, v), and by (3.2.11), 2 + 2ρ′− (u, v) = ku + vk2 . Thus ρ′− (u, v) = ρ′− (v, u), and X is an inner product space. Conversely, if X is an inner product space, it is immediate from (3.2.2) that the function H : R3 → R is given by (x2 + y 2 − z 2 )2 1 2 2 H(x, y, z) = 2 x y − z 4

64


for all (x, y, z) in R × R × R\{0}, and by a straightforward computation H verifies conditions (i), (ii) and (iii). Assuming first order differentiability conditions, in a way similar to that in Theorem 3.2.2, we obtain the following result. Theorem 3.2.3 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The norm in X derives from an inner product if and only if for all x, y ∈ X\{0}, x 6= y, we have kxk − kyk kxk + kyk H , (3.2.12) kh1 (x, y)k2 = kx − yk2 G kx − yk kx − yk where G, H are two functions in C 1 (R+ ) satisfying G(1) = H(1) = 0 and G′ (1)H ′ (1) = −1. Proof. gives

If X is an i.p.s. then a straightforward computation using (3.2.2)

kh1 (x, y)k2 = kx − yk2

kxk + kyk kx − yk 2

2

−1 1− ·

kxk − kyk kx − yk 2

2

,

and kh1 (x, y)k2 admits a representation of the form (3.2.12) with 1 − t2 t2 − 1 and H(t) = . G(t) = 2 2 Assume now (3.2.12), substitute y := tz with t > 0 into (3.2.12), divide by t2 and take the limit as t tends to zero from the right. Then, using the definition of h1 , we obtain

2

2 ′

kh1 (x, tz)k2

z + tkzk − ρ+ (x, z) (x − tz) = lim+ lim

2 2 t kx − tzk t→0 t→0+

2

′

ρ+ (x, z)

=

z − kxk2 x . On the other hand, by (3.2.12) and applying l’Hôpital’s rule we get with g(t) = kx − tzk: kxk−tkzk kxk+tkzk H 2 G kx−tzk kx−tzk kh1 (x, tz)k = kxk2 lim+ lim t2 t t t→0+ t→0 ′ kxk + tkzk kzkkx − tzk − g (t)(kxk + tkzk) = kxk2 lim G′ kx − tzk kx − tzk2 t→0+


65

kxk − tkzk −kzkkx − tzk − g ′ (t)(kxk − tkzk) ×H kx − tzk kx − tzk2 ′ ′ G (1)H (1) kzkkxk + ρ′− (x, z) −kzkkxk + ρ′− (x, z) = kxk2 2 x ,z . = kzk2 − ρ′− kxk ′

Then 2

kzk −

ρ′−

x ,z kxk

2

2

ρ′+ (x, z)

x , = z − kxk2

and dividing by kzk2 , we obtain

2 2

z z z x x x ′ ′

.

, = − ρ+ , 1 − ρ− kxk kzk kzk kxk kzk kxk x z By the substitution u := , v := we have, for all u, v in SX kxk kzk 1 − ρ′− (u, v)2 = kv − ρ′+ (u, v)uk2 , and, consequently, (see [Alsina, Guijarro and Tomás (1997)]) X is an i.p.s. In the next theorem we have other characterizations of i.p.s. Theorem 3.2.4 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The following conditions are equivalent: (i) The norm k · k comes from an inner product; (ii) h1 (x, y) + h1 (y, x) = x + y for all x, y in X with kxk = kyk; 2 −kxk2 for all x, y in X, x 6= y; (iii) h1 (x, y) + h1 (y, x) = x + y + (x − y) kyk kx−yk2 (iv) h1 (x, y) = (v) h1 (x, y) =

x+y 2 x+y 2

for all x, y in X with kxk = kyk; 2

2

−kxk + (x − y) kyk 2kx−yk2 for all x, y in X, x 6= y.

In an i.p.s., if we consider an isosceles triangle, then the height vector divides the third side into two segments of equal length. We will see that this property characterizes inner product spaces. Theorem 3.2.5 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then the following conditions are equivalent:

66

(i) (ii) (iii) (iv)


The norm k · k comes from an inner product; kh1 (x, y) − yk = kh1 (x, y) − xk for all x, y in X with kxk = kyk; kh1 (x, y) − yk = kh1 (y, x) − xk for all x, y in X with kxk = kyk; kh1 (x, y)−yk+kh1 (y, x)−xk = kx−yk for all x, y in X with kxk = kyk.

Proof. The implications i ⇒ ii, i ⇒ iii and i ⇒ iv are immediate. If condition (ii) holds, then take x, y ∈ X with kxk = kyk, x 6= ±y, then

2 ′ 2 ′

y + kyk − ρ+ (x, y) (x − y) − y = y + kyk − ρ+ (x, y) (x − y) − x

kx − yk2 kx − yk2 Since x 6= y,

kyk2 − ρ′ (x, y) = kyk2 − ρ′ (x, y) − kx − yk2 . + + −kyk2 + ρ′+ (x, y) = kyk2 − ρ′+ (x, y) − kx − yk2 kxk2 + kyk2 − kx − yk2 = 2ρ′+ (x, y).

Replacing y by −y, we get kxk2 + kyk2 − kx + yk2 = −2ρ′− (x, y). From the above, for all u, v ∈ SX , u 6= ±v we have 2kuk2 + 2kvk2 − ku + vk2 − ku − vk2 = 2(ρ′+ (u, v) − ρ′− (u, v)) ≥ 0, whence ku + vk2 + ku − vk2 ≤ 4 for all u, v ∈ SX (also for u = ±v), and by Theorem 1.4.3 the norm in X derives from an inner product. If condition (iii) holds, we have kyk2 − ρ′ (x, y) = kxk2 − ρ′ (y, x) (3.2.13) + + for all vectors x, y in X with kxk = kyk, and bearing in mind that by Theorem 2.1.1 (vi) ρ′+ (x, y) ≤ kxkkyk = kyk2 and ρ′+ (y, x) ≤ kxkkyk = kxk2 ,

67


we deduce from (3.2.13) that ρ′+ (x, y) = ρ′+ (y, x), and by Theorem 2.1.1 the norm on X derives from an inner product. If we assume condition (iv), then for all vectors x, y with kxk = kyk we get kyk2 − ρ′+ (x, y) + kxk2 − ρ′+ (y, x) = kx − yk2 ,

so

kyk2 − ρ′+ (x, y) + kxk2 − ρ′+ (y, x) = kx − yk2 . Consequently, for all unit vectors u, v, we have 2 − ρ′+ (u, v) − ρ′+ (v, u) = ku − vk2 ,

(3.2.14)

and replacing v by −v, from Theorem 2.1.1 (iv) we obtain 2 + ρ′− (u, v) + ρ′− (v, u) = ku + vk2 .

(3.2.15)

Finally, we deduce from (3.2.14) and (3.2.15) that ku + vk2 + ku − vk2 = 4 + ρ′− (u, v) − ρ′+ (u, v) + ρ′− (v, u) − ρ′+ (v, u) ≤ 4, and from Theorem 1.4.3, the norm in X comes from an inner product.

In the next two theorems, we give two characterizations of inner product space based on known geometrical properties using the definition and properties of the cosine of an angle in a triangle.

y

y)

x−

, h(x

h(−y, x − y)

y

x Figure 3.2.1 Theorem 3.2.6 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The norm in X comes from an inner product if and only

68


if for all vectors x, y in X, we have kx − ykkx − h1 (x, y)k = kx − y − h1 (−y, x − y)kkxk.

(3.2.16)

Proof. If we assume (see Figure 3.2.1) condition (3.2.16), by the definition of h1 , we have for all x, y in X, x 6= y and x 6= 0

2 ′

x − y − kyk − ρ+ (x, y) (x − y) kx − yk

2 kx − yk

2 ′

kx − yk − ρ+ (−y, x − y)

= kxk (−x)

, 2 kxk whence, for all x, y in X, kx − yk2 − kyk2 + ρ′ (x, y) = kx − yk2 − kyk2 + ρ′ (y, x) . + −

Substituting −tx, t > 0 in the place of x, dividing by t and taking the limit when t → 0+ , we obtain ′ 2ρ+ (y, x) − ρ′− (x, y) = ρ′+ (y, x) .

So, for all x, y in X,

ρ′+ (y, x) = ρ′− (x, y)

or

3ρ′+ (y, x) = ρ′− (x, y).

Replacing x by −x for all x, y in X, we get ρ′− (y, x) = ρ′+ (x, y)

or

3ρ′− (y, x) = ρ′+ (x, y),

and changing the roles of x and y, for all x, y in X, we obtain ρ′− (x, y) = ρ′+ (y, x)

or

3ρ′− (x, y) = ρ′+ (y, x).

Assume that for some x and y in X we have 3ρ′+ (y, x) = ρ′− (x, y), then from the last condition 3ρ′+ (y, x) = ρ′− (x, y) = aρ′+ (y, x), where a ∈ {1, 13 }. Hence, ρ′+ (y, x) = ρ′− (x, y) = 0. So, in fact, for all x, y in X we have ρ′+ (y, x) = ρ′− (x, y) ≤ ρ′+ (x, y). But this also means that ρ′+ (x, y) ≤ ρ′+ (y, x)

69


for all x, y in X, which gives the equality, and by means of Theorem 2.1.1 (ix) the proof of this implication is complete. The converse implication is trivial. Theorem 3.2.7 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The norm k · k comes from an inner product if and only if for all x, y in X, we have kx − h1 (y, x)kkx − yk = |ρ′+ (x, x − y)|. Proof.

whence

(3.2.17)

Assume (3.2.17), i.e., kx − h1 (y, x)kkx − yk = kxk2 − ρ′− (x, y) , kxk2 − ρ′+ (y, x) = kxk2 − ρ′− (x, y)

for all x, y in X. Take now x, y in SX . Then, on account of Theorem 2.1.1 (vi) and (viii) ρ′+ (y, x) = ρ′− (x, y) ≤ ρ′+ (x, y) for all x, y in SX , which characterizes X as an i.p.s. (cf. the proof of Theorem 3.2.6). The converse implication is trivial. Finally, we show that only in inner product spaces, given a triangle determined by vectors x, y, the vector h1 (x, y) − x is ρ-orthogonal to h1 (x, y). Theorem 3.2.8 Let (X, k · k) be a real normed linear space with dim X ≥ 2, then the norm k · k derives from an inner product if and only if for all vectors x, y in X h1 (x, y) − x ⊥ρ h1 (x, y). Proof.

By the hypothesis and the properties of ρ′+ 0 = kh1 (x, y) − xk2 + ρ′+ (h1 (x, y) − x, x) ρ′+

kyk2 − ρ′+ (x, y) − kx − yk2 (x − y), x

= − kyk2 − ρ′+ (x, y) − kx − yk2

2

.

Define the function F : X × X → R by the formula F (w, z) := kzk2 − ρ′+ (w + z, z) − kwk2 ,

w, z ∈ X,

(3.2.18)

70


then (3.2.18) has the form ρ′+ (F (x − y, y)(x − y), x) = −(F (x − y, y))2 . Now, if we make substitutions: i) x := tu + v, y := v with t > 0, then we obtain F (tu, v)t2 kuk2 + tρ′+ (F (tu, v)u, v) = −(F (tu, v))2 .

(3.2.19)

From the definition of F , we have F (tu, v) = −(kv + tuk2 − kvk2 ) + tρ′− (v + tu, u) − t2 kuk2 . Without loss of generality, we may assume that X = R2 , and then lim

t→0+

F (tu, v) = ρ′− (v, u) − 2ρ′+ (v, u) t

(see Proposition 2.1.6), and by (3.2.19) 2 F (tu, v) ′ u, v = − ρ′− (v, u) − 2ρ′+ (v, u) . lim ρ+ t t→0+

(3.2.20)

(3.2.21)

ii) x := tu − v, y := −v with t > 0, then analogously to i) we have lim+ −

t→0

F (tu − v) = ρ′+ (v, u) − 2ρ′− (v, u), t

and 2 F (tu, −v) u, v = − ρ′+ (v, u) − 2ρ′− (v, u) . lim ρ′+ − t t→0+

iii) x := −tv, y := u with t > 0, then defining a new function k : X×X → R, k(w, z) := kzk2 − ρ′+ (w, z) − kz − wk2 ,

w, z ∈ X

from (3.2.18) we get 2 k(−tv, u) k(−tv, u) ′ (u + tv), v = − , ρ+ t t and since lim

t→0+

k(−tv, u) = ρ′− (v, u) − 2ρ′+ (u, v), t

then lim

t→0+

ρ′+

k(−tv, u) (u + tv), v t

2 = − ρ′− (v, u) − 2ρ′+ (u, v) .


71

Let A := ρ′− (v, u) − 2ρ′+ (v, u),

B := ρ′+ (v, u) − 2ρ′− (v, u),

C := ρ′− (v, u) − 2ρ′+ (u, v).

If A 6= 0, by (3.2.20) for t > 0 in a neighborhood of 0, sgn(A) = , and consequently sgn F (tu,v) t ρ′+

F (tu, v) u, v t

=

F (tu, v) ′ ρsgn(A) (u, v). t

Then, by (3.2.20) and (3.2.21) ρ′sgn(A) (u, v) = −A.

(3.2.22)

ρ′sgn(B) (u, v) = −B,

(3.2.23)

Analogously, if B 6= 0, then and if C 6= 0, then (from Proposition 2.1.6) ρ′sgn(C) (u, v) = −C,

(3.2.24)

and we can consider three cases: Case 1: A > 0. If A = ρ′− (v, u) − 2ρ′+ (v, u) > 0, then by property (viii) of ρ′+ and ρ′− (see Theorem 2.1.1), we deduce ρ′+ (v, u) < 0, and consequently B = (ρ′+ (v, u) − ρ′− (v, u)) + (−ρ′− (v, u)) > 0. Then, by (3.2.22) and (3.2.23) ρ′+ (u, v) = −A = −ρ′− (v, u) + 2ρ′+ (v, u), and ρ′+ (u, v) = −B = −ρ′+ (v, u) + 2ρ′− (v, u), respectively, from which we deduce ρ′+ (v, u) = ρ′− (v, u) = ρ′+ (u, v)

72


Case 2: A < 0. In this case, by (3.2.22) 0 < −A = ρ′− (u, v) ≤ ρ′+ (u, v). Thus, C = (ρ′− (v, u) − ρ′+ (u, v)) + (−ρ′+ (u, v)) < 0, and by (3.2.24) ρ′− (u, v) = −C = −ρ′− (v, u) + 2ρ′+ (u, v), but by (3.2.22) ρ′− (u, v) = −A = −ρ′− (v, u) + 2ρ′+ (v, u), so ρ′+ (u, v) = ρ′+ (v, u). Case 3: A = 0. If A = ρ′− (v, u) − 2ρ′+ (v, u) = 0, then 2ρ′+ (v, u) = ρ′− (v, u) ≤ ρ′+ (v, u).

(3.2.25)

Therefore, ρ′+ (v, u) ≤ 0, and B = ρ′+ (v, u) − 2ρ′− (v, u) = −3ρ′+ (v, u) ≥ 0. If B = −3ρ′+ (v, u) > 0, by (3.2.23) ρ′+ (u, v) = −B = 3ρ′+ (v, u) < 0.

(3.2.26)

Then from (3.2.25) and (3.2.26), C = ρ′− (v, u) − 2ρ′+ (u, v) = −4ρ′+ (v, u), and by (3.2.24) ρ′+ (u, v) = −C = 4ρ′+ (v, u). Finally, from (3.2.26) and (3.2.27), we deduce the contradiction: ρ′+ (v, u) = 0. Hence, B = −3ρ′+ (v, u) = 0, and C = ρ′− (v, u) − 2ρ′+ (u, v) = −2ρ′+ (u, v). Now, by (3.2.24), if C > 0, ρ′+ (u, v) = −C = 2ρ′+ (u, v) < 0 and we have a contradiction. Analogously, for C < 0, since ρ′+ (u, v) ≥ ρ′− (u, v) = −C = 2ρ′+ (u, v) > 0.

(3.2.27)


73

Therefore, C = −2ρ′+ (u, v) = 0, and then ρ′+ (u, v) = ρ′+ (v, u) = 0. In any case, ρ′+ (u, v) = ρ′+ (v, u) for all u and v in X, so X is an inner product space. Theorem 3.2.9 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then we have the following properties: (i) x − y ⊥ρ h1 (x, y) for all x, y in X if and only if X is an i.p.s.; (ii) x − y ⊥ρ h2 (x, y) for all x, y in X if and only if X is an i.p.s.; (iii) If x − y ⊥ρ h3 (x, y) for all x, y in X then ρ′+ = ρ′− . Proof.

We will show (i). By hypothesis ρ′+ (x − y, h1 (x, y)) = 0, so ρ′+ (x − y, y) + kyk2 − ρ′+ (x, y) = 0.

If x := u and y := u − v, then for all u, v in X we have kuk2 + kvk2 − ku − vk2 = ρ′− (u, v) + ρ′+ (v, u), and replacing v by −v kuk2 + kvk2 − ku + vk2 = −ρ′+ (u, v) − ρ′− (v, u). By interchanging the roles of u and v and adding all four equalities, we obtain 2(kuk2 + kvk2 ) − ku − vk2 − ku + vk2 = 0, which is a characterization of inner product spaces. The converse implication is trivial. Theorem 3.2.10 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then we have the following properties: (i) x − y ⊥ρ h1 (x, y) for all x, y in X if and only if X is an i.p.s.; (ii) If x − y ⊥ρ h2 (x, y) for all x, y in X if and only if X is an i.p.s.; (iii) If x − y ⊥ρ h3 (x, y) for all x, y in X then ρ′+ = ρ′− . Proof.

(i) By hypothesis ρ′+ (x − y, h1 (x, y)) = −ρ′− (x − y, h1 (x, y)), so

ρ′+ (x − y, y) + kyk2 − ρ′+ (x, y) = −ρ′− (x − y, y) − kyk2 + ρ′+ (x, y). If x := u and y := u − v, then for all u, v in X we have 2(ku − vk2 − kuk2 − kvk2 ) = −ρ′− (v, u) − ρ′+ (v, u) − 2ρ′− (u, v),

74


and replacing v by −v 2(ku + vk2 − kuk2 − kvk2 ) = ρ′+ (v, u) + ρ′− (v, u) + 2ρ′+ (u, v). By adding last two equalities, we obtain ku − vk2 + ku + vk2 − 2(kuk2 + kvk2 ) = ρ′+ (u, v) − ρ′− (u, v) ≥ 0, and from Theorem 1.4.2 we deduce that X is an inner product space. The converse implication is trivial. (ii) By the hypothesis ρ′+ (x − y, h2 (x, y)) = −ρ′− (x − y, h2 (x, y)), so ρ′+ (x − y, y) + kyk2 − ρ′+ (y, x) = −ρ′− (x − y, y) − kyk2 + ρ′+ (y, x). If x := u + v and y := v, then for all u, v in X we have ρ′+ (u, v) + ρ′− (u, v) = 2ρ′+ (v, u), and replacing v by −v ρ′− (u, v) + ρ′+ (u, v) = 2ρ′− (v, u). Comparing the last two equalities, we have ρ′+ (v, u) = ρ′− (v, u) for all u, v in X and, consequently, ρ′+ (u, v) = ρ′+ (v, u) for all u, v in X and X is an i.p.s. The converse implication is trivial. (iii) By the hypothesis ρ′+ (x − y, h3 (x, y)) = −ρ′− (x − y, h3 (x, y)), so ρ′+ (x − y, y) + ρ′+ (y − x, y) = −ρ′− (x − y, y) − ρ′+ (y − x, y). If x := u + v and y := v, then for all u, v in X we have ρ′+ (u, v) = ρ′− (u, v). 3.3

Height functions and classical orthogonalities

By combining some well-known facts about orthogonalities and the height functions h1 , h2 , h3 , we obtain several new characterizations of i.p.s. Theorem 3.3.1 dim X ≥ 2.

Let (X, k · k) be a real normed linear space with

(i) If x − y ⊥J h2 (x, y) for all x, y in X then X is an i.p.s.. (ii) If x − y ⊥J h1 (x, y) for all x, y in X then ρ′+ = ρ′− . (iii) If x − y ⊥J h3 (x, y) for all x, y in X then ρ′+ = ρ′− .

75


Proof. We will show case (i). By hypothesis, for all x, y in X, x 6= y we have

2 ′ 2 ′

x − 2y − kyk − ρ+ (y, x) (x − y) = x + kyk − ρ+ (y, x) (x − y) .

2 2 kx − yk kx − yk If y := u and x := u + v, then for all u, v in X, v 6= 0,

′ ′

v − u + ρ+ (u, v) v = v + u − ρ+ (u, v) v .

2 2 kvk kvk

By applying this last equality (with u replaced by it tu, t′ > 0), ρ (u,v) is a straightforward verification to show that ρ′+ v, u − +kvk2 v = ρ′ (u,v) −ρ′− v, u − +kvk2 v , and therefore 2ρ′+ (u, v) = ρ′+ (v, u) + ρ′− (v, u) ≤ 2ρ′+ (v, u) for all u, v in X, v 6= 0, which implies ρ′+ (u, v) = ρ′+ (v, u) for all u, v in X and X is an i.p.s. Another interesting result is given in: Let (X, k · k) be a real normed linear space with

Theorem 3.3.2 dim X ≥ 2.

(i) If y − h3 (x, y) ⊥J h3 (x, y) for all x, y in X, then ρ′+ = ρ′− . (ii) The norm k·k comes from an inner product if and only if y−h1 (x, y) ⊥J h1 (x, y) for all x, y in X. Proof.

(i) By hypothesis, we have for all x, y in X, x 6= y

ρ′+ (y − x, y)

. (x − y) kyk = y − 2 y +

2 kx − yk

Take u, v in X, u 6= 0 and substitute x := v − u, y := v in the above

ρ′+ (u, v)

u . kvk = v − 2 kuk2

Replacing v by u + tv, where t > 0, we obtain

ρ′+ (u, v)

u . ku + tvk = −u + tv − 2t kuk2 By using the last equality, we have 2

2

ku + tvk − kuk = 2t

ρ′ (u,v) ku + t −v + 2 +kuk2 u k2 − kuk2 2t

,

76


and taking limit when t tends to 0, we deduce that ρ′+ (u, v) = ρ′− (u, v) for all u, v in X. (ii) Assume y − h1 (x, y) ⊥J h1 (x, y) for all x, y in X. Then for all x, y in X, x 6= y

kyk2 − ρ′+ (x, y)

(x − y) kyk = −y − 2

. 2 kx − yk If we substitute y by x + λy, λ > 0, for all x, y in X, y 6= 0 we obtain

kx + λyk2 − kxk2 − λρ′+ (x, y)

, y kx + λyk = x + λy − 2

λkyk2

and taking limit when t tends to zero,

y ′ ′

(2ρ+ (x, y) − ρ+ (x, y)) kxk = x − 2

. kyk2

If we take x, y in SX then

1 = kx − 2ρ′+ (x, y)yk, and by (5.12) in [Amir (1986)] (see also [Precupanu (1978)]), the norm k · k comes from an inner product. The converse is trivial. In what follows, we have some similar results for the Birkhoff orthogonality. Theorem 3.3.3 Then

Let (X, k · k) be a real normed linear space, dim X ≥ 2.

(i) x − y ⊥B h1 (x, y) for all x, y in X if and only if X is an i.p.s.; (ii) x − y ⊥B h2 (x, y) for all x, y in X if and only if X is an i.p.s.; (iii) For all x, y in X we have x − y ⊥B h3 (x, y). Proof.

In order to prove (i), assume that x − y ⊥B h1 (x, y) for all x, y in ρ′ (x,y)−kyk2

X. If x 6= y, then x − y ⊥B y − α(x − y) with α = + kx−yk2 . This from Proposition 2.1.6 means that ρ′− (x − y, y) ≤ αkx − yk2 ≤ ρ′+ (x − y, y), ρ′− (x − y, y) ≤ ρ′+ (x, y) − kyk2 ≤ ρ′+ (x − y, y), which is valid for all x, y in X. Replacing y by x + y in the above inequality, we get ρ′− (−y, x + y) ≤ ρ′+ (x, x + y) − kx + yk2 ≤ ρ′+ (−y, x + y),

77


whence, ρ′+ (y, x) + kyk2 ≥ kx + yk2 − ρ′+ (x, y) − kxk2 . Replacing y by ty, where t > 0, and dividing by 2t, we obtain ρ′ (x, y) ρ′+ (y, x) tkyk2 kx + tyk2 − kxk2 + ≥ − + . 2 2 2t 2 Taking the limit when t tends to 0+ yields ρ′+ (y, x) ≥ ρ′+ (x, y) for all x, y in X, so X is an i.p.s. In order to prove (ii), observe that analogously to the previous case we have ρ′− (x, y) ≤ ρ′+ (y, x + y) − kyk2 ≤ ρ′+ (x, y) for all x, y in X. Then ρ′− (x, y) ≤ ρ′+ (y, x) ≤ ρ′+ (x, y) and X is an i.p.s. Condition (iii) has a straightforward verification. There are also connections between height functions and the Pythagorean orthogonality. Theorem 3.3.4 Let (X, k · k) be a real normed linear space, dim X ≥ 2. Then X is an i.p.s. if and only if x − y ⊥P h2 (x, y) for all x, y in X. Proof.

By hypothesis, for all x, y in X, kx − y + h2 (x, y)k2 = kx − yk2 + kh2 (x, y)k2 .

If y := u and x := u + v, we have for all u, v in X, v 6= 0

2

2

′ ′

u + v − ρ+ (u, v) v = kvk2 + u − ρ+ (u, v) v .

kvk2 kvk2 Then

′ + t u−

v ρ (u, v) ρ′+ v, u − + 2 v = lim+ kvk t→0 = lim

t→0+

ρ′+ (u,v) kvk2 v

2t

ρ′ (u,v) 2

tu − t +kvk2 v 2t

2

− kvk2

= 0.

Therefore, ρ′+ (v, u) = ρ′+ (u, v) for all u, v in X and X is an i.p.s. The converse implication is trivial.

Theorem 3.3.5 Let (X, k · k) be a real normed linear space, dim X ≥ 2. If x − y ⊥P h3 (x, y) for all x, y in X, then ρ′+ = ρ′− . If y − h3 (x, y) ⊥P h3 (x, y) for all x, y in X then X is an i.p.s.

78


Proof. (i) Assume that x − y ⊥P h3 (x, y) for all x, y in X and substitute x := v − u, y := v. Then for all u, v in X u 6= 0, we have

2

2

′ ′

v − u − ρ+ (u, v) u = kuk2 + v − ρ+ (u, v) u .

2 2 kuk kuk In this case, it is very easy to show that ρ′+ (u, v) ′ u =0 ρ+ u, −v + kuk2

and

ρ′+ = ρ′− .

(ii) If y−h3 (x, y) ⊥P h3 (x, y) for all x, y in X, then kyk2 = kh3 (x, y)k2 +

ρ′+ (y−x,y)2 kx−yk2

for all x, y in X, x 6= y. Substituting x := v − u, y := v, for all u, v in X, u 6= 0, we get

2

′ 2

ρ′+ (u, v)

+ ρ+ (u, v) , u kvk2 = v −

kuk2 kuk2

and for all t > 0, u, v in X, u 6= 0,

2

′ 2

ρ′+ (u, v + tu) 2

+ ρ+ (u, v + tu) .

u kv + tuk = v + tu −

kuk2 kuk2 By applying the last two equalities, we obtain

tkuk2 + 2ρ′+ (u, v) kv + tuk2 − kvk2 = , 2t 2 and taking limit when t tends to 0, we get ρ′+ (v, u) = ρ′+ (u, v), which is valid for all u, v in X and X is an i.p.s. Theorem 3.3.6 If (X, k · k) is a real normed linear space with dim X ≥ 2, then for all the previously defined orthogonalities ⊥P , ⊥J , ⊥B , ⊥ρ , ⊥ρ the following statements are equivalent: (i) X is an inner product space; (ii) x − h2 (x, y) is orthogonal to h2 (x, y) for all vectors x, y in X. Proof. If we assume condition (ii) for the Pythagorean orthogonality ⊥P , then for all x, y in X, x 6= y we have

2

2

kyk2 − ρ′+ (y, x) kyk2 − ρ′+ (y, x) 2

(x − y) + kx − yk 1 − , kxk = y + kx − yk2 kx − yk2 2

79


and replacing x by −tx, with t > 0, dividing by (2t)2 , and taking limits when t → 0+ , for all x, y in X, y 6= 0, we obtain

1

2ρ′ (y, x) − ρ′ (y, x) y − kyk2 x 2 + − 2 kyk 2 = kxk2 kyk2 − 2ρ′+ (y, x) − ρ′− (y, x) .

In particular, if kxk = kyk = 1 this equality becomes

k(2ρ′+ (y, x) − ρ′− (y, x))y − xk2 = 1 − (2ρ′+ (y, x) − ρ′− (y, x))2 . Then, if we replace x by

x+y kx+yk ,

(3.3.1)

for x 6= −y,

′

(2ρ (y, x) − ρ′ (y, x))y − x 2 = kx + yk2 − (2ρ′ (y, x) − ρ′ (y, x) + 1)2 , + − + −

and using (3.3.1), for all x, y in X, we get

kx + yk2 = 2 + 2(2ρ′+ (y, x) − ρ′− (y, x)).

(3.3.2)

Now, if we replace y by −y: kx − yk2 = 2 + 2(−2ρ′− (y, x) + ρ′+ (y, x)).

(3.3.3)

From (3.3.2) and (3.3.3), we have kx + yk2 + kx − yk2 = 4 + 6(ρ′+ (y, x) − ρ′− (y, x)) ≥ 4, and on account of Theorem 1.4.3, X is an inner product space. If we assume (ii) for the James orthogonality ⊥J , then for all x, y in X, x 6= y, we have

kyk2 − ρ′+ (y, x)

, (x − y) kxk = x − 2 y +

2 kx − yk

and by the substitution x := v + tu, y := tu, for all v, u in X, v = 6 0 and t > 0, we obtain

ρ′+ (tu, v)

. v (3.3.4) kv + tuk = v − tu + 2

kvk2 On the other hand, by the definition of ρ′+

′

v − tu + 2ρ+ (u, v) ′ v = lim ρ+ v, −u + kvk2 t→0+

2ρ′+ (tu,v) 2 v

2 kvk 2t

− kvk2

,

80


and by (3.3.4) and by the properties of ρ′+ , −ρ′− (v, u) + 2ρ′+ (u, v) = lim

t→0+

kv + tuk2 − kvk2 = ρ′+ (v, u). 2t

Therefore, for all u, v in X, 2ρ′+ (u, v) = ρ′+ (v, u) + ρ′− (v, u) ≤ 2ρ′+ (v, u), and X is an i.p.s. For the Birkhoff orthogonality ⊥B , if x − h2 (x, y) ⊥B h2 (x, y) for all vectors x, y in X, x 6= y, then kyk2 − ρ′+ (y, x) (x − y) ⊥B h2 (x, y); 1− kx − yk2 i.e., for all x, y in X x − y ⊥B h2 (x, y), and by Theorem 3.3.3 this is a characterization of inner product spaces. For the orthogonality ⊥ρ , if ρ′+ (x − h2 (x, y), h2 (x, y)) = 0, then for x 6= y, 2 ′ ′ ρ+ (y, x − y) ρ+ (y, x − y) 2 ′ + + 1 kx − yk − ρ 1 (x − y), x = 0. + kx − yk2 kx − yk2 By the substitution x := u + v and y := v, we have 2 ρ′+ (v, u) + kuk2 − ρ′+ (ρ′+ (v, u) + kuk2 )u, u + v = 0,

valid for all u, v in X. Then, for all u, v in SX

Analogously,

(ρ′+ (v, u) + 1) ρ′+ (v, u) − ρ′+ (u, v) = 0.

(3.3.5)

(ρ′+ (u, v) + 1) ρ′+ (u, v) − ρ′+ (v, u) = 0,

(3.3.6)

and from (3.3.5) and (3.3.6) for all u, v ∈ SX , we have ρ′+ (v, u) = ρ′+ (u, v),

so X is an inner product space. For the last orthogonality ⊥ρ the considerations are similar.


3.4

81

A new orthogonality relation

In the Euclidean plane R2 , if we consider orthogonal vectors x, y and the triangle determined by x and y with the initial point at the origin, the height for sides x and x − y over y is x. Then, in a real normed linear space (X, k · k) with dim X ≥ 2 we can consider the height orthogonality: x ⊥h y

if

h1 (x, x − y) = x,

which is equivalent to the condition kx − yk2 = kxk2 + kyk2 − ρ′− (x, y). It is immediate to see in the following examples that ⊥h may have very special behaviours. Example 3.4.1 If we consider (R2 , k · k+ ), then for x 6= 0, xy1 . In particular, (x, 0) ⊥h (y1 , y2 ) if and only if |y2 | = 2|x−y1 |−2|y 1 |−|x| (x, 0) ⊥h (0, y) if and only if y = 0. Example 3.4.2 If we consider (R2 , k · k∨ ), then for x 6= 0, (x, 0) ⊥h (y1 , y2 ) if and only if x(x − y1 ) = max((x − y1 )2 , y22 ) − max(y12 , y22 ). In particular, (x, 0) ⊥h (0, y) if and only if y = 0. (See also Example 5.2.1.) Proposition 3.4.1 Let (X, k · k) be a real normed linear space and dim X ≥ 2. The relation ⊥h satisfies the following conditions for all x, y in X and for all a in R: (i) (ii) (iii) (iv) (v) (vi) (vii)

0 ⊥h x and x ⊥h 0; x ⊥h x if and only if x = 0; If x ⊥h y, then ax ⊥h ay; If x ⊥h y and x, y 6= 0, then x and y are linearly independent; For all x in X and t > 0 there exists y in X with kyk = t and x ⊥h y; For all x, y in X\{0} there exists t0 > 0 such that (x+t0 y) ⊥h (x−t0 y); If the norm k · k comes from an inner product, then the relation ⊥h is equivalent to the usual orthogonality.

Theorem 3.4.1 Let (X, k · k) be a real normed linear space, dim X ≥ 2, and let for all x, y in X\{0} the following implication holds x ⊥h y

implies

Then X is an inner product space.

x ⊥h −y.

82


Proof. If x ⊥h y then kx−yk2 = kxk2 +kyk2 −ρ′− (x, y). Since this implies x ⊥h −y, kx + yk2 = kxk2 + kyk2 + ρ′+ (x, y), we have kx − yk2 + kx + yk2 = 2(kxk2 + kyk2) + ρ′+ (x, y) − ρ′− (x, y) ≥ 2(kxk2 + kyk2 ) for all x, y in X such that x ⊥h y. By (vi) of the previous proposition and Theorem 1.4.4 (with α := 1, β := t0 ), we infer that X must be an inner product space. In the rest of this section, we study the relations between ⊥h and the orthogonality of x − y and h1 (x, y). Proposition 3.4.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If x − y ⊥J h1 (x, y) for all x, y in X, then x ⊥h y implies x ⊥J y. Proof. If x ⊥h y, we have h1 (x, x − y) = x. However, by hypothesis, y ⊥J h1 (x, x − y), then y ⊥J x and x ⊥J y. Theorem 3.4.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If x − y ⊥J h1 (x, y) for all x, y in X, then the following conditions are equivalent: (i) X is an i.p.s.; (ii) If x ⊥h y then ρ′− (x, y) ∼ 0; (iii) The inequality kh1 (x, y)k ≤ kλx + (1 − λ)yk holds for all λ ∈ R. Proof. Condition (i) implies both (ii) and (iii). Assume now condition (ii) and let x, y be two vectors in X such that x ⊥h y. Then kx − yk2 = kxk2 + kyk2 − ρ′− (x, y) and kx − yk = kx + yk. Therefore, kx+yk2 +kx−yk2 = 2kx−yk2 = 2(kxk2 +kyk2 )−2ρ′− (x, y) ∼ 2(kxk2 +kyk2 ) for all x, y in X with x ⊥h y. Thus, from Proposition 3.4.1 (vi) and Theorem 1.4.4, X is an i.p.s. Assume condition (iii) and take x, y in X, x 6= y, and α in R. Let λ :=

kyk2 − ρ′+ (x, y) + α. kx − yk2

Then from (iii)

so

kyk2 − ρ′+ (x, y)

, (x − y) kh1 (x, y)k ≤ y + + α(x − y)

kx − yk2 kh1 (x, y)k ≤ kh1 (x, y) + α(x − y)k,

83


which means that h1 (x, y) ⊥B x − y, and on account of Proposition 2.1.7 ρ′− (h1 (x, y), x − y) ≤ 0. Let x, y be two vectors in X such that x ⊥h x − y; then h1 (x, y) = x and ρ′− (x, x − y) = ρ′− (h1 (x, y), x − y) ≤ 0, i.e., for all u, v in X with u ⊥h v we have ρ′− (u, v) ≤ 0 and from (ii), X is an i.p.s. Theorem 3.4.3 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If x − y ⊥P h1 (x, y) for all x, y in X, then the following conditions hold: (i) If x ⊥h y then x ⊥P y; (ii) If x ⊥h y implies ρ′− (x, y) ∼ 0, then X is an i.p.s. Proof. (i) In the condition x − y ⊥P h1 (x, y) for all x, y in X, substitute x := u and y := u − v. Then it is very easy to show that

2

2 2 ′

u + ku − vk − kuk + ρ− (u, v) v

2 kvk

2 2

ku − vk − kuk2 − ρ′− (u, v)

= kvk2 + u − v + v

kvk2

for all u, v in X, v 6= 0. If u ⊥h v, then ρ′− (u, v) = kuk2 + kvk2 − ku − vk2 and ku + vk2 = kvk2 + kuk2, i.e., u ⊥P v. (ii) Assume that x ⊥h y implies ρ′− (x, y) ∼ 0. In this case, if u ⊥h v, then using (i) we obtain ku + vk2 + ku − vk2 = 2(kuk2 + kvk2 ) − ρ′− (u, v) ∼ 2(kuk2 + kvk2 ), and we deduce that X is an i.p.s.

If X is an inner product space, then the length of the height from the origin to the side x − y is the minimal distance from 0 to the point λx + (1 − λ)y for all λ in R, i.e., for all x, y in X and for all λ in R we have the inequality kh1 (x, y)k ≤ kλx + (1 − λ)yk.

(3.4.1)

We now study (3.4.1) when x and y are vectors in a real normed linear space (X, k · k), and by using the orthogonality relation ⊥h , we obtain a characterization of inner product spaces. To this end, we state the following result.

84


Lemma 3.4.1 Let (X, k · k) be a real normed linear space, then the following conditions hold: (i) If inequality (3.4.1) holds, then kh1 (x, y)k = kh1 (y, x)k for all x, y in X; (ii) Inequality (3.4.1) holds if and only if h1 (x, y) ⊥B x − y and h1 (y, x) ⊥B x − y for all x, y in X; (iii) Inequality (3.4.1) holds if and only if ρ′− (h1 (x, y), x) ≤ kh1 (x, y)k2 ≤ ρ′+ (h1 (x, y), x), and ρ′− (h1 (x, y), y) ≤ kh1 (x, y)k2 ≤ ρ′+ (h1 (x, y), y) for all x, y in X. Proof.

kx−yk2 −kxk2 +ρ′ (y,x)

+ , then by (i) If h1 (y, x) = λx + (1 − λ)y, where λ = kx−yk2 (3.4.1), kh1 (x, y)k ≤ kh1 (y, x)k for all x, y in X, x 6= y. Replacing x by y, we get kh1 (y, x)k ≤ kh1 (x, y)k, and consequently kh1 (x, y)k = kh1 (y, x)k for all x, y in X.

kyk2 −ρ′ (x,y)+tkx−yk2

+ (ii) If inequality (3.4.1) holds, for all t in R, take λ := kx−yk2 and then, kh1 (x, y)k ≤ kh1 (x, y) + t(x − y)k and h1 (x, y) ⊥B x − y.

kyk2 −ρ′ (x,y)−tkx−yk2

+ , we obtain h1 (x, y) ⊥B y − x. If λ = kx−yk2 Now, assume h1 (x, y) ⊥B x − y, then for all λ in R take t :=

λkx−yk2 −kyk2 +ρ′+ (x,y) , kx−yk2

then, using kh1 (x, y)k ≤ kh1 (x, y) + t(x − y)k we obtain inequality (3.4.1). (iii) By using (ii), it is immediate to deduce that h1 (x, y) ⊥B x − y if and only if ρ′− (h1 (x, y), x − y) ≤ 0 ≤ ρ′+ (h1 (x, y), x − y), and as h1 (x, y) = y + λ(x − y), if λ > 0 ρ′− (h1 (x, y), λ(x − y)) ≤ 0 ≤ ρ′+ (h1 (x, y), λ(x − y)), kh1 (x, y)k2 − ρ′+ (h1 (x, y), y) ≤ 0 ≤ kh1 (x, y)k2 − ρ′− (h1 (x, y), y) and if λ < 0, we have a similar inequality. An analogous method works for the proof of the second inequality.

85


Theorem 3.4.4 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If inequality (3.4.1) holds, then X is an inner product space if and only if x − y ⊥h h1 (x, y) for all x, y in X. Proof. Assume x − y ⊥h h1 (x, y). By (ii) in the last lemma and the properties of ⊥B , for all x, y in X, ρ′− (h1 (x, y), x − y) ≤ 0. Take z, u in X, and let y := z − u − h1 (z, u) and x := z − u. By hypothesis z − u ⊥h h1 (z, u), and therefore, h1 (z − u, z − u − h1 (z, u)) = z − u

and

h1 (x, y) = x,

i.e., ρ′− (z − u, h1 (z, u)) = ρ′− (h1 (x, y), x − y) ≤ 0, and by (iii) in Theorem 3.1.2 the norm in X comes from an inner product. We conclude this section with the following collection of characterizations. Theorem 3.4.5 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If inequality (3.4.1) holds, then the following conditions are equivalent (i) X is an inner product space;

≤ kh1 (u, v)k; (ii) For all u, v in SX , we have u+v 2 (iii) For all vectors u, v in SX , vectors h1 (u, v) and u + v are linearly dependent; y y x x , kyk + kyk = 12 kxk ; (iv) For all x, y in X\0, we have h1 kxk

2

x y (v) For all x, y in X\{0}, ρ′+ (x, y) = kxkkyk 1 − 12 kxk − kyk

. Proof. Condition (ii) implies (i) by (3.8) of [Amir (1986)] and (3.4.1). The remaining implications can be easily proved.

3.5

Orthocenters

Let (X, k · k) be a real normed linear space with dim X ≥ 2. If we consider for all x, y, z in X, x, y, z, linearly dependent but not collinear (= affinely independent), then we can consider the triangle ∆xyz

86


with sides x − y, y − z, x − z, and the three height lines: h1 := x − Rh(x − z, x − y), h2 := y − Rh(y − x, y − z),

(3.5.1)

h3 := z − Rh(z − y, z − x),

where h = h1 (see Section 3.1), and in a natural way we introduce the following Definition 3.5.1 The triangle ∆xyz has an orthocenter H if and only if the three height lines (3.5.1) intersect at the point H. Now we can state a new characterization of inner product spaces. Theorem 3.5.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2 and let α 6= 0 be a fixed real number. Then X is an i.p.s. if and only there exists the orthocenter of ∆xy(αx + y) for all linearly independent x, y in X . Proof. Of course, the orthocenter condition is necessary for X to be an i.p.s. We prove the converse implication. If we consider ∆xy(αx + y), it is a straightforward computation to show that the three height lines (3.5.1) meet at a point if and only if the following equality holds ABC = (1 − A)(1 − B)(1 − C),

(3.5.2)

where A=

kαxk2 − ρ′+ (y − x, −αx) , k(α − 1)x + yk2

B=

kx − yk2 − ρ′+ ((1 − α)x − y, x − y)) , kαxk2

and C=

k(α − 1)x + yk2 − ρ′+ (αx, (α − 1)x + y) . ky − xk2

If we substitute in (3.5.2) z = y − x, replace αx by x, and for all t > 0 replace z by tz, dividing by t2 , using Remark 2.1.2, (iii), (iv) and (v) of Thereom 2.1.1, taking limit when t tends to zero, by a straightforward computation we have, for all x, z in X ρ′+ (x, z)(ρ′+ (x, z) − ρ′− (z, x)) = 0.


87

By interchanging x and z, we obtain ρ′+ (z, x)(ρ′+ (z, x) − ρ′− (x, z)) = 0

(3.5.3)

and changing x by −x in (3.5.3): ρ′− (z, x)(ρ′− (z, x) − ρ′+ (x, z)) = 0.

(3.5.4)

If ρ′+ (x, z) = ρ′− (z, x), using (viii) of Theorem 2.1.1 we have ρ′+ (x, z) ≤ ρ′+ (z, x). If ρ′+ (x, z) = 0, by (3.5.3) we have two possibilities, either ρ′+ (z, x) = 0 and in this case ρ′+ (x, z) = ρ′+ (z, x) or ρ′+ (z, x) = ρ′− (x, z). In the last case, by (3.5.4) we have ρ′− (z, x) = 0 (and therefore ρ′+ (z, x) ≥ 0 = ρ′+ (x, z)) or ρ′+ (x, z) = ρ′− (z, x) (and ρ′+ (x, z) = 0 = ρ′− (z, x) ≤ ρ′+ (z, x)). Therefore, for all x, z in X we deduce ρ′+ (x, z) ≤ ρ′+ (z, x) and X must be an i.p.s. Note. Observe that Theorem 3.5.1 implies the corresponding theorem in triangles ∆xyz where z is in the plane determined by x and y. In a general real normed space (not i.p.s.) X, the intersection by pairs of the three height lines of ∆xyz yield (if they exist) three different points B (y − z + A(z − x)) , {H12 } := h1 ∩ h2 = y − 1 − A + AB C {H31 } := h3 ∩ h1 = x − (x − y + B(y − z)) , 1 − B + BC A {H23 } := h2 ∩ h3 = z − (z − x + C(x − y)) , 1 − C + CA where ky − zk2 − ρ′+ (y − x, y − z) , kz − xk2 2 ′ kx − yk − ρ+ (x − z, x − y) B= , ky − zk2 2 ′ kz − xk − ρ+ (z − y, z − x) C= . ky − xk2

A=

(3.5.5)

Definition 3.5.2 The points H12 , H31 , H23 defined above are called the orthocenter points of ∆xyz. We now consider the following classical property in the Euclidean plane concerning the orthocenter: in a triangle ∆xyz, where x, y, z have the

88


same norm, the orthocenter is the point x + y + z (see [Hofmann (1958)], p. 178/179). We can translate this property into a real normed space by considering the orthocenter point H12 (we have by symmetry the same result for H31 and H23 ), and as can be immediately seen we obtain a new characterization of i.p.s. Theorem 3.5.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an i.p.s. if and only if the orthocenter point H12 of ∆xyz is x + y + z, where x, y, z are vectors in X with kxk = kyk = kzk. Proof. One implication of the theorem can be found in [Hofmann (1958)]. Conversely, assume that x, y, z in X satisfy kxk = kyk = kzk, x, y are linearly independent and z is in the plane determined by x and y. Consider the triangle with vertices x, y and z and the orthocenter point H12 . By hypothesis y−

B (y − z + A(z − x)) = x + y + z, 1 − A + AB

where A and B are defined in (3.5.5). However, z = z1 x + z2 y, where z1 , z2 belong to R, so, by using the linear independence of x and y, it is very easy to show that necessarily 2 +1 1 −1 and B = z1 +z . A = z2 −z 2z2 2z1 Substituting the values A, B in (3.5.5), it is a straightforward computation to by verify that, if z1 + z2 6= 1, and z1 z2 6= 0 z1 − z2 + 1 kz − xk2 + ky − zk2 , 2z2 z1 + z2 + 1 ρ′+ (x − z, x − y) = − ky − zk2 + kx − yk2 . 2z1 ρ′+ (y − x, y − z) =

(3.5.6)

Now, consider u, v unit vectors in X, then if a, b in R satisfy kau + bvk = 1, ab 6= 0 and a + b 6= 1 by (3.5.6), we have ρ′+ (v − u, v − au − bv) =

a−b+1 kau + bv − uk2 + kv − au − bvk2 , (3.5.7) 2b

which yields kv − uk2 + ρ′+ (v − u, u − au − bv) =

a−b+1 kau + bv − uk2 + kv − au − bvk2 , 2b

89


and replacing u by v and a by b, we get b−a+1 kau + bv − vk2 + ku − au − bvk2 . 2a (3.5.8) ′ ′ Then by (3.5.7) and (3.5.8), and using ρ− ≤ ρ+

kv − uk2 − ρ′− (v − u, v − au − bv) =

ku − vk2 ≤ ku − bv − auk2 If a =

−1 ku+λvk

and b =

−λ ku+λvk ,

1+a+b 1+a+b + kv − bv − auk2 . 2b 2a

with λ > 0, then

2

2 !

u + u + λv + λ v + u + λv ,

ku + λvk ku + λvk

ku + λvk − 1 − λ ku − vk ≤ −2λ 2

and taking limit when λ tends to 0 ku − vk2 ≤ 2 − 2ρ′+ (u, v)

and

ku + vk2 ≤ 2 + 2ρ′− (u, v),

so on adding the last two equations, we obtain ku − vk2 + ku + vk2 ≤ 4 and X is an i.p.s. Another well-know property in the real space is the generalization to the height Theorem: The product of a whole height AD of a triangle and the part of this height between its foot and the orthocenter HD is equal to the product of the two segments into which the height divides the side. In Figure 3.5.1, we have BD · DC = AD · HD (see [Honsberger (1995)], p. 19). A

H

B

D Figure 3.5.1

C

90


We can translate this property into a real normed space to obtain the following characterization of inner product spaces. Theorem 3.5.3 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an i.p.s. if and only if the orthocenter point H23 of the triangle ∆yxo satisfies the equation kx − h(x, y)kky − h(x, y)k = kh(x, y)k · kh(x, y) − H23 k

(3.5.9)

for linearly independent vectors x, y in X and h = h1 . Proof. The direct part is proved in [Honsberger (1995)]. Conversely, if the triangle ∆yxo satisfies equation (3.5.9), then on substituting the expressions of h(x, y) and H23 , we have kx − yk2 − kyk2 + ρ′+ (x, y) kyk2 − ρ′+ (x, y) kx − yk kx − yk

2 2 ′

kyk − ρ+ (x, y) = (x − y)

y +

kx − yk2 kx − yk2 (kxk2 − ρ′+ (x − y, x)) , × 1 − kx − yk2 kyk2 + (kyk2 − ρ′ (x, y))(kxk2 − kyk2 − ρ′ (x − y, x) +

+

and replacing x by tx for t > 0 and dividing by t2 , we obtain for all x, y in X, ktx − yk2 − kyk2 + tρ′+ (x, y) kyk2 − tρ′+ (x, y) ktx − yk2 t ktx − yk2 kyk2 + (kyk2 − ρ′+ (tx, y))(t2 kxk2 − kyk2 − tρ′+ (tx − y, x)) × t

yktx − yk2 + (kyk2 − ρ′+ (tx, y))(tx − y) 2

=

t × ktx − yk2 kyk2 + (kyk2 − tρ′+ (x, y))(t2 kxk2 − kyk2 − tρ′+ (tx − y, x)) −ktx − yk2 t(tkxk2 − ρ′ (tx − y, x)) . +

Taking limit when t tends to zero, using Proposition 2.1.6 and the definition of ρ′± , we arrive at the following equation for all x, y in X ′ ρ+ (x, y) − 2ρ′− (y, x) ρ′+ (x, y) − ρ′− (y, x) = 0.

Then, ρ′+ (x, y) = ρ′− (y, x) or ρ′+ (x, y) = 2ρ′− (y, x), but analogously ρ′− (x, y) = ρ′+ (y, x) or ρ′− (x, y) = 2ρ′+ (y, x).

91


Therefore, if ρ′+ (x, y) = 2ρ′− (y, x), then ρ′+ (x, y) = 2ρ′+ (x, y) or = 4ρ′+ (x, y), and ρ′+ (x, y) = 0 = ρ′− (y, x), and for all x, y in X, we have ρ′+ (x, y) = ρ′− (y, x) ≤ ρ′+ (y, x), so X is an i.p.s.

ρ′+ (x, y)

3.6

A characterization of inner product spaces involving an isosceles trapezoid property

Suzuki’s property (see [Suzuki (1995)]) states that, considering an isosceles trapezoid ABCD in the Euclidean plane R2 , for any point S in R2 we have (see Figure 3.6.1) SB 2 − SC 2 SA2 − SD2 = . AD BC B

(3.6.1)

C

A

D

S Figure 3.6.1

Using a height function defined in the first section of this chapter, we translate Suzuki’s property into a real normed linear space and study how the condition obtained characterizes the norm as a norm derivable from an inner product (see [Alsina, Cruells and Tomás (1999a); Alsina, Cruells and Tom´ as (1999b)]). Let (X, k · k) be a real normed linear space with dim X ≥ 2, and let x, y in X be linearly independent. Using the height function h := h1 for the triangle formed by two vectors −y, x − y (see Figure 3.6.2)

92


y

h(−y, x − y)

x−

y

x z Figure 3.6.2 we can write equation (3.6.1) in the form: 2

2

kzk − kx − zk = kxk

2

x kz − yk2 − y − z + (kxk − 2ky + h(−y, x − y)k) kxk

kxk − 2ky + h(−y, x − y)k

.

(3.6.2)

Let f : X\{0} × X → R be given by the formula kxk2 + kyk2 − kx − yk2 − ρ′− (y, x) . f (x, y) = 1 − 2 kxk2

(3.6.3)

(kzk2 − kx − zk2 ) · f (x, y) = ky − zk2 − ky − z + f (x, y)xk2 .

(3.6.4)

Then, equation (3.6.2) in the real normed space becomes

Theorem 3.6.1 Let (X, k · k) be a real normed linear space and let Af := {(x, y) ∈ X\{0} × X : f (x, y) 6= 0, ρ′+ (x, y) ≥ 0}. If for all (x, y) ∈ Af and z ∈ X equation (3.6.4) is satisfied, then the norm in X derives from an inner product. Note: The equality f (x, y) = 0 corresponds to the case where the side of the trapezoid which is parallel to x vanishes. In this case, both sides of (3.6.4) also vanish. The condition ρ′+ (x, y) ≥ 0 is related to the “angle” between x and y. Proof. Observe first that for each x ∈ X\{0}, the pair (x, 0) is in Af , so Af is not empty. Take x, y in X such that (x, y) ∈ Af and substitute z := x + y in (3.6.4). Then, (kx + yk2 − kyk2 )f (x, y) = f (x, y)(2 − f (x, y))kxk2 .

(3.6.5)


93

On dividing by f (x, y) and using the expression (3.6.3) of f , the above equation becomes: kx + yk2 − kxk2 − kyk2 = 2 kxk2 + kyk2 − kx − yk2 − ρ′+ (y, x) . (3.6.6) Observe that lim f (x, ty) = 1, so we can state that for any t > 0 small t→0+

enough, the pair (x, ty) is in Af and satisfies equation (3.6.6), too. Therefore, we can exchange y for ty under the same conditions, and equation (3.6.6) can be rewritten, after being divided by 2t, as kx − tyk2 − kxk2 ρ′− (y, x) kyk2 tkyk2 kx + tyk2 − kxk2 −t = 2 + − . 2t 2 −2t 2 2 By letting t decrease to 0, we have ′ 1 ′ ′ ρ+ (x, y) = 2 ρ− (x, y) − ρ− (y, x) . 2

Similarly, if we rewrite equation (3.6.4) for z := y, with the same remarks as before, replacing y by ty, with t > 0, dividing by 2t and calculating the limit when t → 0+ , we obtain ′ 1 ′ ′ ρ− (x, y) = 2 ρ− (x, y) − ρ− (y, x) . 2

We arrive at the conclusion that for all x, y in X, x 6= 0 such that (x, y) ∈ Af , we have 1 ′ ′ ′ ′ ρ+ (x, y) = ρ− (x, y) ∈ ρ− (y, x), ρ− (y, x) . (3.6.7) 3 Consider x, y in X such that ρ′+ (x, y) ≥ 0. Using the same reasoning as above, we can state that for any t > 0 small enough f (x, ty) 6= 0, so equation (3.6.7) becomes 1 ρ′+ (x, ty) = ρ′− (x, ty) ∈ ρ′− (ty, x), ρ′− (ty, x) . (3.6.8) 3 Dividing by t, we infer that for all x, y in X such that ρ′+ (x, y) ≥ 0 equation (3.6.7) is also satisfied. Using the above result, we prove that for all x, y in X such that ρ′+ (x, y) ≥ 0, ρ′+ is a symmetric map. Let x, y in X be such that ρ′+ (x, y) ≥ 0. By (3.6.7) we also have ρ′− (y, x) ≥ 0, and because of ρ′− ≤ ρ′+

94


we deduce that ρ′+ (y, x) ≥ 0; so using (3.6.7) again for the couple y, x ∈ X, we have 1 ′ ′ ′ ′ (3.6.9) ρ+ (y, x) = ρ− (y, x) ∈ ρ− (x, y), ρ− (x, y) . 3 Thus from (3.6.7) and (3.6.9), we can state that for all x, y in X if ≥ 0, then ρ′+ (x, y) = ρ′+ (y, x). Finally, for all x, y in X such that ρ′+ (x, y) ≤ 0, we have ρ′+ (−x, y) ≥ 0, and using (3.6.7) and (3.6.9) once more we obtain the symmetry of ρ′+ for all x, y in X, which is a characterization of i.p.s. ρ′+ (x, y)

For a new orthogonality relation defined by means of the trapezoid rule, the reader is refered to [Alsina, Cruells and Tomás (1999b)] (see also [Ger (2000)]). 3.7

Functional equations of the height transform

Let (X, k·k) be a real normed linear space of dimension greater than or equal to two. In this section we will study functions f : X → X that transform the height of the triangle with sides x, y, x− y into the corresponding height of the triangle determined by sides f (x), f (y) and f (x) − f (y). Namely, we study the condition f (h3 (y − x, y)) = h1 (f (y) − f (x), f (y)), which leads to the functional equation of the height transform kf (y)k2 − ρ′+ (f (y) − f (x), f (y)) ρ′+ (x, y) x = f (y) − f (x) (3.7.1) f y− kxk2 kf (x)k2 for all x, y in X, x 6= 0. We begin with the following (see [Alsina and Garcia Roig (1991)]). Lemma 3.7.1 If f : X → X is a surjective, continuous solution of ρ′+ (f (y), f (x)) ρ′+ (x, y) x = f (y) − f (x), (3.7.2) f y− kxk2 kf (x)k2 and f (x) 6= 0 whenever x 6= 0, then f must be a linear automorphism. Proof. First we show that f preserves the linear independence of any couple of vectors x, y in X, x, y 6= 0. In fact, if we had f (y) = λf (x) for

95


some λ 6= 0, then by (3.7.2) and the properties of ρ′+ we would immediately obtain ρ′+ (λf (x), f (x)) ρ′+ (x, y) x = λf f (x) = 0, (x) − f y− kxk2 kf (x)k2 and this would yield y =

ρ′+ (x, y) x, which contradicts the linear indepenkxk2

dence of x and y. Next, consider two independent vectors u, v in X. The substitutions x := u and y := u + v into (3.7.2) yield ρ′ (f (u + v), f (u)) f (u) f (u + v) − + kf (u)k2 ρ′ (u, u + v) ρ′+ (u, v) =f u+v− + u = f v − u kuk2 kuk2 ′ ρ (f (v), f (u)) = f (v) − + f (u). kf (u)k2 Therefore, f (u + v) = f (v) +

ρ′+ (f (u + v), f (u)) − ρ′+ (f (v), f (u)) f (u), (3.7.3) kf (u)k2

and interchanging the roles of u and v, f (u + v) = f (u) +

ρ′+ (f (v + u), f (v)) − ρ′+ (f (u), f (v)) f (v). (3.7.4) kf (v)k2

By (3.7.3) and (3.7.4), and bearing in mind that f (u) and f (v) are independent, we can conclude that f must satisfy ρ′+ (f (u + v), f (u)) − ρ′+ (f (v), f (u)) = 1, kf (u)k2 i.e., by (3.7.4) f (u + v) = f (u) + f (v),

(3.7.5)

where (3.7.5) holds for all couples of independent vectors u and v. Moreover, if v = λu we can choose in a plane containing u a sequence (vn ) of vectors independent of u but with lim vn = λu. By (3.7.5) and the continuity of n→∞ f , we obtain f (u + v) = f (u + λu) = lim f (u + vn ) = lim (f (u) + f (vn )) n→∞

= f (u) + f (λu) = f (u) + f (v).

n→∞

96


Thus, (3.7.5) holds for all u and v in X\ {0} and obviously also if either u = 0 or v = 0. Consequently, f satisfies the classical Cauchy functional equation on X and f is a linear transformation (see [Aczél (1966)]). Since f −1 ({0}) = {0} and f is onto, f is an automorphism. Remark 3.7.1. Observe that the hypothesis of f being onto, used at the end of the preceding proof, is only needed in the infinite dimensional case. Now we will prove Lemma 3.7.2 A continuous function f from a real normed linear space (X, k · k) with dim X ≥ 2, onto itself satisfies (3.7.2) and vanishes only at 0 if and only if (X, k · k) is an inner product space and f is a similitude. Proof. It is a straightforward verification to show that any similitude satisfies (3.7.2) whenever X is an inner product space. Conversely, assume that f is a solution of (3.7.2). By virtue of Lemma 3.7.1, f must necessarily be a linear automorphism and must satisfy for all x 6= 0 the equation ρ′+ (f (y), f (x)) ρ′+ (x, y) = . kxk2 kf (x)k2

(3.7.6)

Consider any couple of linearly independent vectors u, v in X, and any real t > 0. The substitution x := f −1 (u) and y := f −1 (u + tv) in (3.7.6) implies ρ′+ ((f ◦ f −1 )(u + tv), (f ◦ f −1 )(u)) ρ′+ (u + tv, u) = kuk2 k(f ◦ f −1 )(u)k2 kf −1 (u)k2 + ρ′+ (f −1 (u), f −1 (tv)) ρ′+ (f −1 (u), f −1 (u + tv)) = = kf −1 (u)k2 kf −1 (u)k2 ′ ′ −1 −1 ρ (f (u), f (tv)) ρ (tv, u) tρ′ (v, u) =1+ + =1+ + 2 =1+ + 2 , −1 2 kf (u)k kuk kuk i.e., ρ′+ (u + tv, u) = kuk2 + tρ′+ (v, u), whence ρ′+ (v, u) = lim+ t→0

ρ′+ (u + tv, u) − kuk2 . t

(3.7.7)

Now we compute the limit (3.7.7) using (3.7.6), the linearity of f , the


97

definition and the properties of ρ′+ and the fact that u + tv 6= 0: ρ′+ (u + tv, u) − kuk2 t t→0+ ρ′+ (f (u), f (u) + tf (v))ku + tvk2 − kuk2 kf (u + tv)k2 = lim+ tkf (u + tv)k2 t→0 kf (u)k2 ku + tvk2 + tρ′+ (f (u), f (v))ku + tvk2 − kuk2kf (u + tv)k2 = lim tkf (u + tv)k2 t→0+ n 2kf (u)k2 ku + tvk2 − kuk2 ku + tvk2 = lim+ ρ′+ (f (u), f (v)) + 2 kf (u + tv)k kf (u + tv)k2 2t t→0 2 2 kuk kf (u) + tf (v)k − kf (u)k2 o −2 · kf (u + tv)k2 2t 2 kuk kuk2 ′ ′ = ρ′+ (f (u), f (v)) + ρ (f (u), f (v)), 2ρ (u, v) − 2 + kf (u)k2 kf (u)k2 + ρ′+ (v, u) = lim

whence kuk2 ′ ρ (f (u), (f (v)) kf (u)k2 + kuk2 kf (v)k2 ′ ρ (v, u), = 2ρ′+ (u, v) − kf (u)k2 kvk2 +

ρ′+ (v, u) = 2ρ′+ (u, v) −

therefore we obtain 2ρ′+ (u, v) = 1 +

kuk2 kf (v)k2 ′ ρ+ (v, u). kf (u)k2 kvk2

(3.7.8)

By (3.7.8), ρ′+ (u, v) = ρ′+ (v, u) if one of these values is zero. Otherwise, interchanging the roles of u and v, we obtain kvk2 kf (u)k2 ′ ρ+ (u, v). 2ρ′+ (v, u) = 1 + (3.7.9) kf (v)k2 kuk2 Combining (3.7.8) and (3.7.9), we immediately get kuk2 kf (v)k2 = 1, kf (u)k2 kvk2

(3.7.10)

ρ′+ (u, v) = ρ′+ (v, u),

(3.7.11)

and therefore,

for independent u and v. In the case when u and v are linearly dependent, we immediately see that (3.7.11) also holds. Thus, we have the derivability of the norm from an inner product, and (3.7.10) forces f to be a similitude.

98


For the study of (3.7.1), we also need the following lemma: Lemma 3.7.3 If f is a continuous function from X onto X satisfying (3.7.1) and vanishing only at zero, then f must be a linear bijection. Proof. First we show that f preserves the linear independence. In fact, if x and y are two linearly independent vectors in X and if f (y) = λf (x) for some λ 6= 0, then kλf (x)k2 − ρ′+ (λf (x) − f (x), λf (x)) ρ′ (x, y) f (x) f y − + 2 x = λf (x) − kxk kf (x)k2 λ2 kf (x)k2 − (λ − 1)λkf (x)k2 = λf (x) − f (x) = 0, kf (x)k2 ρ′ (x,y)

and since f vanishes only at 0, this would yield y = +kxk2 x, in contradiction to the independence of x and y. Next we prove that if u, v in X are two independent vectors, then f (u + v) = f (u) + f (v). To that effect, if we substitute x := u and y := u + v in (3.7.1), we have ρ′ (u, u + v) u f u+v− + kuk2 kf (u + v)k2 − ρ′+ (f (u + v) − f (u), f (u + v)) = f (u + v) − f (u). kf (u)k2 This, together with the properties of ρ′+ , yield ρ′+ (u, v) u f (u + v) = f v − kuk2 kf (u + v)k2 − ρ′+ (f (u + v) − f (u), f (u + v)) + f (u) kf (u)k2 kf (v)k2 − ρ′+ (f (v) − f (u), f (v)) = f (v) − f (u) kf (u)k2 2 ′ kf (u + v)k − ρ+ (f (u + v) − f (u), f (u + v)) + f (u) = f (v) + A(u, v)f (u), kf (u)k2 where A(u, v) =

kf (u + v)k2 − ρ′+ (f (u + v) − f (u), f (u + v)) kf (u)k2 2 ′ kf (v)k + ρ+ (f (v) − f (u), f (v)) − , kf (u)k2


99

and interchanging the roles of u and v, f (u + v) = f (u) + A(v, u)f (v). But since f (u) and f (v) are independent, from the previous equalities we can conclude A(u, v) = A(v, u) = 1, and consequently, f (u + v) = f (u) + f (v) for all couples of independent vectors u and v. However, if u, v are dependent, this formula also holds. To that effect, we can consider in a plane containing u, a sequence (vn ) of vectors independent of u and with lim vn = v, and by the additivity of f on independent vectors and the n→∞

continuity of f , we have f (u + v) = lim f (u + vn ) = lim (f (u) + f (vn )) = f (u) + f (v). n→∞

n→∞

Therefore, f satisfies the classical Cauchy functional equation f (u + v) = f (u) + f (v)

for all u, v in X,

and in consequence (see [Aczél and Dhombres (1989)], p. 36) f is a linear transformation. Since f is onto and vanishes only at zero, f is a linear bijection. Note that in the above result the continuity assumption may be replaced by other mild regularity conditions which force an additive function to be linear (e.g., local boundedness). Now we can solve (3.7.1) completely Theorem 3.7.1 Let (X, k · k) be a real normed linear space and f a continuous function from X onto X. Then, f satisfies (3.7.1) and vanishes only at zero, if and only if, (X, k · k) is an inner product space and f is a similitude. Proof. If we assume that f is a solution of (3.7.1), and vanishes only at zero, by Lemma 3.7.3 f is a linear bijection and equation (3.7.1) is equivalent to kf (y)k2 − ρ′+ (f (y) − f (x), f (y)) ρ′+ (x, y) = . kxk2 kf (x)k2

100


Putting z := y − x, we obtain 1+

kf (x + z)k2 − kf (z)k2 − ρ′+ (f (z), f (x)) ρ′+ (x, z) = . kxk2 kf (x)k2

(3.7.12)

Now, if we take linearly independent u, v in X, the substitution x := f −1 (u) and z := f −1 (u + tv), with t ∈ R, t > 0, into (3.7.12) implies ku + u + tvk2 − ku + tvk2 − ρ′+ (u + tv, u) kuk2 ′ −1 ρ (f (u), f −1 (u + tv)) ρ′+ (f −1 (u), f −1 (tv)) = =1+ + 1 + 1 + kf −1 (u)k2 kf −1 (u)k2 ku + tvk2 − ktvk2 − ρ′+ (tv, u) =1+ , kuk2 i.e., k2u + tvk2 − ku + tvk2 − ρ′+ (u + tv, u) = kuk2 + ku + tvk2 − ktvk2 − tρ′+ (v, u), whence ρ′+ (v, u) = lim

t→0+

ρ′+ (u + tv, u) − k2u + tvk2 + 2ku + tvk2 + kuk2 − ktvk2 . t

On the other hand, by (3.7.12) we also have ρ′+ (u + tv, u) kf (2u + tv)k2 − kf (u)k2 − ρ′+ (f (u), f (u + tv)) − kf (u + tv)k2 ku + tvk2 kf (u + tv)k2 kf (2u + tv)k2 − 2kf (u)k2 − tρ′+ (f (u), f (v)) − kf (u + tv)k2 ku + tvk2 , = kf (u + tv)k2

=

therefore, we obtain kf (2u + tv)k2 − 2kf (u)k2 − tρ′+ (f (u), f (v)) ′ ku + tvk2 ρ+ (v, u) = lim+ tkf (u + tv)k2 t→0 −k2u + tvk2 + ku + tvk2 + kuk2 + − tkvk2 , t and consequently, ku + tvk2 ρ′+ (v, u) = lim+ −ρ′+ (f (u), f (v)) kf (u + tv)k2 t→0 2 2 2ku + tvk kf (2u + tv)k − kf (2u)k2 + kf (u + tv)k2 2t


101

4kuk2 kf (u + tv)k2 − kf (u)k2 4kf (u)k2 ku + tvk2 − kuk2 − kf (u + tv)k2 2t kf (u + tv)k2 2t k2u + tvk2 − k2uk2 ku + tvk2 − kuk2 +2 − tkvk2 −2 2t 2t 4kuk2 ′ kuk2 + ρ (f (u), f (v)) + 4ρ′+ (u, v) = −ρ′+ (f (u), f (v)) 2 kf (u)k kf (u)k2 + kuk2 ′ ρ (f (u), f (v)) − 4ρ′+ (u, v) + 2ρ′+ (u, v) −4 kf (u)k2 + kuk2 ′ ρ (f (u), f (v)), = 2ρ′+ (u, v) − kf (u)k2 + +

whence using (3.7.12) again we deduce kuk2 kf (v)k2 ′ ρ+ (v, u) 1 − kf (u)k2 kvk2 kuk2 kf (u + v)k2 − kf (u)k2 − kf (v)k2 .(3.7.13) = 2ρ′+ (u, v) − 2 kf (u)k Reversing the roles of u and v, we have kvk2 kf (u)k2 ρ′+ (u, v) 1 − kf (v)k2 kuk2 kvk2 = 2ρ′+ (v, u) − kf (u + v)k2 − kf (u)k2 − kf (v)k2 (3.7.14) kf (v)k2

and combining (3.7.13) and (3.7.14), we immediately get ρ′+ (v, u) = ρ′+ (u, v)

for all linearly independent u, v in X, which characterizes X as an inner product space. Hence, the equation (3.7.1) is equivalent to equation (3.7.2), and applying Lemma 3.7.2, f is a similitude. The converse statement is immediate from Lemma 3.7.2. A similar methodology can be applied to studying function f from X into itself, satisfying the functional equation of the height transform f (h3 (y − x, y)) = h3 (f (y) − f (x), f (y)), i.e., in explicit form, for all x, y in X with x and f (x) different from 0 ρ′+ (f (x), f (y)) ρ′+ (x, y) x = f (y) − f (x). (3.7.15) f y− kxk2 kf (x)k2

102


The following result holds: Theorem 3.7.2 If f : X → X is a continuous function and (X, k · k) is a real normed linear space, then f is a solution of (3.7.15) and vanishes only at zero if and only if, f is a linear transformation and a similitude. Proof. Following “mutatis mutandis” the proof of Lemma 3.7.3, it can be seen that if f is a solution of (3.7.15) and f (x) = 0 only for x = 0, then f is linear. Therefore equation (3.7.15) is equivalent to ρ′ (f (x), f (y)) ρ′+ (x, y) = + . 2 kxk kf (x)k2

(3.7.16)

Let g : X\{0} → R be defined by g(x) =

kf (x)k . kxk

If x, y are a couple in X\{0}, and t is in R, t 6= 0, then kf (x + ty)k2 kf (x)k2 − 2 g(x + ty) − g(x) kx + tyk kxk2 = 2t 2t kxk2 kf (x + ty)k2 − kx + tyk2 kf (x)k2 = kx + tyk2 kxk2 2t 2 2 1 2 kf (x) + tf (y)k − kf (x)k = kxk kx + tyk2 kxk2 2t 2 2 kx + tyk − kxk −kf (x)k2 , 2t

and using (3.7.16) lim

t→0+

1 g(x + ty) − g(x) = kxk2 ρ′+ (f (x), f (y)) − kf (x)k2 ρ′+ (x, y) = 0. 2t kxk4

Consequently, lim

t→0−

g(x + ty) − g(x) g(x + (−t)(−y)) − g(x) = − lim 2t 2(−t) t→0− g(x + t(−y)) − g(x) = 0, = − lim+ 2t t→0

so g is a constant function and f is a similitude. The “if” part is established by a straightforward verification.

2

Chapter 4

Perpendicular Bisectors in Normed Spaces

The center of the circumscribed circumference of a triangle is the intersection point of the three perpendicular bisectors of its three sides. This is a classical situation in an i.p.s. What happens in a general normed linear space where various notions of orthogonality lead to alternative expressions for perpendicular bisectors? We explore such questions in this chapter.

4.1

Definitions and basic properties

In a real i.p.s. (X, h·, ·i) with dimension greater than or equal to 2, given two linearly independent vectors x, y ∈ X, we define the perpendicular bisector of the linear segment [x, y] = {αx + (1 − α)y : 0 ≤ α ≤ 1} as the set

x+y + λu 2

:

λ∈R ,

where u 6= 0 is a vector in the subspace generated by x and y (denoted by span(x, y)) and orthogonal to x − y. If u := αx + βy, then by the requirement hx − y, ui = 0, we have

and we may take

α kxk2 − hy, xi = β(kyk2 − hx, yi), u = (kyk2 − hx, yi)x + (kxk2 − hy, xi)y. 103

104


x+y 2

y

+ λu

u

x

Figure 4.1.1

In a real normed linear space (X, k · k), we can replace the inner product hx, yi by its generalizations ρ′± (x, y). Namely, we consider the vectors u± (x, y) = (kyk2 − ρ′± (x, y))x + (kxk2 − ρ′± (y, x))y

(4.1.1)

for all x, y in X, and define M± (x, y) :=

x+y + λu± (x, y) : λ ∈ R . 2

(4.1.2)

It follows from the above that if x and y are linearly dependent, e.g. y = λx for some λ ∈ R, then M± (x, λx) =

x + λx . 2

We call the lines (4.1.2) perpendicular bisectors of the linear segment [x, y], and later we justify their associated orthogonal relations. In what follows, we restrict ourselves to considerations related to M+ . Furthermore, in the real normed linear space (X, k·k), we can also define the perpendicular bisector of the linear segment [x, y] as the set of metric nature: M ∗ (x, y) = {z ∈ X : z ∈ span (x, y), kz − xk = kz − yk} . We have M ∗ (x, y) = M ∗ (y, x) and M+ (x, y) = M+ (y, x), but in general M+ (x, y) 6= M ∗ (x, y); for example, in (R2 , k·k+ ) if x = (3, 1) and y = (1, 1),


105

then M+ (x, y) = {(2, 1) + λ(−4, 4) : λ ∈ R}, but M ∗ (x, y) = {(2, 1) + µ(0, 1) : µ ∈ R}. The equality of the sets M+ and M ∗ is a characteristic property of inner product spaces. Theorem 4.1.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an inner product space if and only if for all x, y in X M+ (x, y) = M ∗ (x, y). For the proof of this theorem, we need the following lemma based on the James orthogonality. Lemma 4.1.1 In a real normed linear space (X, k · k) with dim X ≥ 2, for all vectors x and y in X, we have x+y 1 + w : w ∈ X, w ⊥J x − y . {z ∈ X : kz − xk = kz − yk} = 2 2 , Proof. If z ∈ X and kz − xk = kz − yk, and we take w = 2 z − x+y 2 1 + w and then z = x+y 2 2 kw − (x − y)k = 2kz − xk = 2kz − yk = kw + (x − y)k. Therefore, w ⊥J x − y. Conversely, if z = kx − zk =

x+y 2

+ 21 w with w ⊥J x − y, then

1 1 kx − y − wk = kx − y + wk = ky − zk. 2 2

Proof of Theorem 4.1.1. Assume M+ (x, y) = M ∗ (x, y) for all x, y in X. Fix v ∈ X and take ς ∈ X such that v ⊥J ς. We can consider linearly

106


independent vectors x, y ∈ span (v, ς) and such that ς = x − y. Then by the previous lemma x+y 1 x+y 1 + v ∈ + w : w ∈ X, v ⊥J x − y ∩ span (v, ς) 2 2 2 2 = {z ∈ X : kz − xk = kz − yk} ∩ span (x, y) = M ∗ (x, y) = M+ (x, y). Therefore, the vectors u+ (x, y) and v are linearly dependent, and x+y + λv : λ ∈ R = M+ (x, y). 2 Now, if λ ∈ R, using M+ (x, y) = M ∗ (x, y), we have

x + y

+ λv − x kς − 2λvk = kx − y − 2λvk = ky − x + 2λvk = 2

2

x + y

= 2

2 + λv − y = kx − y + 2λvk = kς + 2λvk.

Therefore,

2λv ⊥J ς

for all

λ ∈ R,

i.e., the James orthogonality is homogeneous, which is a characterization of inner product space [James (1945); Amir (1986)]. The converse implication is obvious. 4.2


In an inner product space, the perpendicular bisectors of the linear segments determined by the sides of a rhombus are its diagonals. Indeed, this property characterizes inner products spaces. Theorem 4.2.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an inner product space if and only if for all x, y in X with kxk = kyk one has M+ (x, y) = span (x + y). Proof. If we assume that M+ (x, y) defined by (4.1.2) is the subspace generated by x + y, for all x, y in SX , the vector u+ (x, y) ∈ span (x + y),


107

i.e., kyk2 − ρ′+ (x, y) = kxk2 − ρ′+ (y, x), so ρ′+ (x, y) = ρ′+ (y, x), and by the properties of ρ′+ , X is an inner product space. Since in an inner product space, two vectors x and y are orthogonal if and only if they are orthogonal in the sense of James, if we take the rhombus with sides x + y and x − y, then by the last theorem M+ (x + y, x − y) = span (x). Based on this property, we define a new orthogonality relation Definition 4.2.1 In a real normed linear space (X, k·k) with dim X ≥ 2, we consider the perpendicular bisector orthogonality x ⊥M y

if

M+ (x + y, x − y) = span (x),

or, equivalently, after using the definition of M+ and the properties of ρ′+ x ⊥M y Example 4.2.1

if and only if

ρ′+ (x + y, x) = ρ′+ (x − y, x).

(4.2.1)

In the space (R2 , k · k+ ) for all x 6= 0, we have

(x, 0) ⊥M (y1 , y2 ) if and only if y1 = 0 and xy2 < 0. The next lemma contains some basic properties of this orthogonality relation. Proposition 4.2.1 The perpendicular bisector orthogonality ⊥M satisfies the following properties for all x, y in X and for all a, t in R: (i) (ii) (iii) (iv) (v) (vi) (vii)

x ⊥M 0, 0 ⊥M x; x ⊥M x if and only if x = 0; x ⊥M tx if and only if tx = 0; If x ⊥M y, then ax ⊥M ay; If x ⊥M y, then x ⊥M −y and −x ⊥M y; If x ⊥M y and x, y 6= 0, then x and y are linearly independent; If the norm derives from an inner product h·, ·i, the relation x ⊥M y is equivalent to the usual orthogonality hx, yi = 0.

108


Proposition 4.2.2 If (X, k · k) is a real normed linear space with dim X ≥ 2, then for all pairs x, y in X\{0} there exist two unique real numbers t1 > 0 and t2 < 0 such that x + ti y ⊥M x − ti y,

i ∈ {1, 2}

(i.e., the ⊥M orthogonality admits diagonals). Proof. Fix x, y in X\{0}. For any t, the condition x + ty ⊥M x − ty is equivalent to kyk2 t2 + (ρ′sgn(t) (y, x) − ρ′sgn(t) (x, y))t − kxk2 = 0.

(4.2.2)

So taking t1 =

ρ′+ (x, y) − ρ′+ (y, x) +

p ′ [ρ+ (x, y) − ρ′+ (y, x)]2 + 4kxk2 kyk2 >0 2kyk2

ρ′− (x, y) − ρ′− (y, x) −

p ′ [ρ− (x, y) − ρ′− (y, x)]2 + 4kxk2 kyk2 0 be such that x + ty ⊥M x − ty. If we assume (4.2.3), then x + ty ⊥J x − ty, i.e., kxk = tkyk and by (4.2.2) we also have ρ′+ (x, y) = ρ′+ (y, x) for x = 0 or y = 0, which on the account of Theorem 2.1.1 means that X is an inner product space. Corollary 4.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2 such that ρ′+ = ρ′− . Then X is an inner product space if and only if the ⊥M -orthogonality is symmetric.


109

Proof. Assume that ⊥M -orthogonality is symmetric. For each pair of vectors x and y in X, let t > 0 be such that x + ty ⊥M x − ty, so (4.2.2) is satisfied, i.e., kyk2 t2 + (ρ′+ (y, x) − ρ′+ (x, y))t − kxk2 = 0. Analogously, by the symmetry of ⊥M , x − ty ⊥M x + ty and kyk2 t2 − (ρ′− (y, x) − ρ′− (x, y))t − kxk2 = 0, then from last equations and the equality of ρ′+ and ρ′− , we deduce ρ′+ (x, y) = ρ′+ (y, x), and X is an inner product space. The converse is immediate. In general, the perpendicular bisector orthogonality is not symmetric; for example, in (R2 , k · k+ ), if x = (1, 2) and y = (−1, 1), then x ⊥M y, but y 6⊥M x. Based on the properties of the ⊥M orthogonality and those of the perpendicular bisectors, we obtain a new geometrical characterizations of inner product spaces. Theorem 4.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The following conditions are equivalent: (i) X is an inner product space; (ii) For all x, y in X\{0}, if x ⊥M y, then M (x, x + y) =

y + span (x), 2

M (y, x + y) =

x + span (y); 2

and

(iii) For all u, v in SX , we have, u + v ⊥M u − v; (iv) For all x, y ∈ X\{0}, if for some α in R we have x + αy ⊥M x − αy, then kxk = kαyk.

(4.2.4)

Note that (ii) gives a characterization of the perpendicular bisectors of the rectangle sides (see Figure 4.2.1), (iii) states that in a rhombus the diagonals are orthogonal, and (iv) establishes that if the diagonals of a parallelogram are orthogonal then it is a rhombus.

110


M (x, x + y)

x+

x

y

y

Figure 4.2.1 Proof. If we assume condition (ii), let x, y be two vectors in X\{0} such that x ⊥M y, then x y M (x, x + y) = + span (x), M (y, x + y) = + span (y), 2 2 i.e., kxk2 = ρ′+ (x + y, x),

kyk2 = ρ′+ (x + y, y),

then kyk2 = ρ′+ (x + y, y) = kx + yk2 − ρ′− (x + y, x) and therefore, kx + yk2 = kyk2 + ρ′− (x + y, x) ≤ kyk2 + ρ′+ (x + y, x) = kyk2 + kxk2 . Furthermore, by (v) of Proposition 4.2.1, x ⊥M −y, and we also obtain kx − yk2 ≤ kyk2 + kxk2 . Finally, from last two inequalities, we deduce kx + yk2 + kx − yk2 ≤ 2(kxk2 + kyk2 ), which together with the fact that the ⊥M -orthogonality admits diagonals and by Theorem 1.4.4, we infer that X is an inner product space [Amir (1986); Ben´ıtez and del Rio (1984)].


111

If we assume condition (iii), then for all u, v in SX if u + v ⊥M u − v, by a straightforward computation ρ′+ (u, v) = ρ′+ (v, u) and X is an i.p.s. For condition (iv), given x, y in X\{0}, if ρ′+ (x, y) ≥ ρ′+ (y, x), let p ρ′+ (x, y) − ρ′+ (y, x) + (ρ′+ (y, x) − ρ′+ (x, y))2 + 4kxk2 kyk2 α := 2kyk2 be the unique positive root of (4.2.2) such that x + αy ⊥ x − αy. Then, by (4.2.4) α=

kxk , kyk

and consequently, 4kxkkyk(ρ′+(x, y) − ρ′+ (y, x)) = 0, so ρ′+ (x, y) = ρ′+ (y, x) and X is an i.p.s.

Corollary 4.2.3 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an inner product space if and only if for all u, v in SX , u + v ⊥M u − v. 4.3

Relations between perpendicular bisectors and classical orthogonalities

If (X, h·, ·i) is an i.p.s. and x, y are two vectors in X, we can consider the x+y + λu(x, y) with the vector u(x, y) orthogonal perpendicular bisector 2 to x − y. Now we show how this property characterizes i.p.s., whenever we consider in a real normed space (X, k · k) the classical orthogonal relations and u := u+ , where u+ is defined by (4.1.1). Proposition 4.3.1 Let (X, k · k) be a strictly convex space with dim X ≥ 2. Then the following conditions are equivalent: (i) (ii) (iii) (iv) (v)

X is an inner product space; x − y ⊥ρ u(x, y) for all x, y in X; x − y ⊥ρ u(x, y) for all x, y in X; x − y ⊥P u(x, y) for all x, y in X; x − y ⊥J u(x, y) for all x, y in X;

112


(vi) x − y ⊥B u(x, y) for all x, y in X. Proof. To prove this theorem, it is only necessary to check that each of properties (ii), (iii), (iv), (v) and (vii) implies (i). If we assume condition (ii), i.e., for all x, y in X ρ′+ (x − y, u(x, y)) = 0,

(4.3.1)

replacing y by x − z and using Theorem 2.1.1, we obtain for all x, z in X (4.3.2) ρ′+ z, (ρ′+ (x − z, x) − kxk2 )z + F (z, x)x = 0, where F (z, x) := kx − zk2 + ρ′− (x, z) − ρ′+ (x − z, x). However, lim F (z, λx) = kzk2 > 0.

λ→0+

(4.3.3)

So, for λ > 0 in a neighborhood of zero, F (z, λx) > 0, and replacing x by λx and using Theorem 2.1.1 in (4.3.2), we have (ρ′+ (λx − z, x) − λkxk2 )kzk2 + F (z, λx)ρ′+ (z, x) = 0. Taking limit when λ tends to zero in the last equality, by using Proposition 2.1.6 and (4.3.3), we obtain ρ′+ = ρ′− . On the other hand, by the substitutions z := x − y, y := y in (4.3.1), by Theorem 2.1.1 and using ρ′+ = ρ′− , we have for all z, y in X, 0 = (kyk2 − ρ′+ (z + y, y))kzk2 + (kz + yk2 − ρ′+ (z + y, y) − ρ′+ (y, z))ρ′+ (z, y). Now take linearly independent unit vectors w, v in X, and substitute z := w, y := v in the above equality: ρ′+ (w + v, v)(1 + ρ′+ (w, v)) = 1 + kw + vk2 ρ′+ (w, v) − ρ′+ (v, w)ρ′+ (w, v). (4.3.4) By interchanging the roles of w and v in (4.3.1), using Theorem 2.1.1, and the equality ρ′+ (w+v, w) = ρ′+ (w+v, w+v−v) = kw+vk2 −ρ′− (w+v, v) as well as the fact that ρ′+ = ρ′− , we obtain ρ′+ (w + v, v)(1 + ρ′+ (v, w)) = −1 + kw + vk2 + ρ′+ (v, w)ρ′+ (w, v). (4.3.5) By comparing ρ′+ (w + v, v) in (4.3.4) and (4.3.5), removing common factor 1 − ρ′+ (w, v)ρ′+ (v, w) (which is different from zero because X is strictly

113


convex and w, v are linearly independent) and dividing by the common factor, we obtain kw + vk2 = 2 + ρ′+ (v, w) + ρ′+ (w, v). Replacing v by −v and adding, we get kw + vk2 + kw − vk2 = 4, which on account of Theorem 1.4.3 gives a characterization of an i.p.s. To prove that (iii) implies (i), we first use the same argument as that used for orthogonality ⊥ρ in order to prove that ρ′+ = ρ′− (considering F is not necessary). The rest follows from the previous implication. If condition (iv) holds, then for all x, y in X ku(x, y)k2 + kx − yk2 = ku(x, y) + x − yk2 , and replacing x by tx and y by ty for all t > 0, using u(tx, ty) = t3 u(x, y), we have t4 ku(x, y)k2 + kx − yk2 = kt2 u(x, y) + x − yk2 . Then for all γ > 0, γ 2 ku(x, y)k2 = kγu(x, y)+x−yk2 −kx−yk2, dividing by 2γ and taking limit when γ tends to zero, finally using the definition of ρ′+ , we obtain condition (ii) and X is an i.p.s. If we assume condition (v), i.e., for all x, y in X ku(x, y) − x + yk = ku(x, y) + x − yk, replacing x by tx and y by ty for all t > 0, kt2 u(x, y) − x + yk2 = kt2 u(x, y) + x − yk2 , and for all γ > 0 kγu(x, y) − x + yk2 = kγu(x, y) + x − yk2 , and subtracting kx − yk2 from the both sides of the last equality, dividing by 2γ and taking limit when γ tends to zero, using the definition of ρ′± we obtain, ρ′+ (x − y, u(x, y)) + ρ′− (x − y, u(x, y)) = 0.

(4.3.6)

By the substitutions, z := x − y, x := x in (4.3.6) and using (v) from Theorem 2.1.1 2(ρ′+ (x − z, x) − kxk2 )kzk2 + ρ′+ (z, F (z, x)x) + ρ′− (z, F (z, x)x) = 0,

114


where F (z, x) is defined in (4.3.2). Replacing x by λx, λ > 0, dividing by λ, taking limit when λ tends to zero, and using (4.3.3) and Proposition 2.1.6, we get ρ′+ = ρ′− . Then by (4.3.6) ρ′+ (x − y, u(x, y)) = 0, and therefore we have condition (ii) and X is an i.p.s. Finally, if condition (vi) holds, then by Proposition 2.1.7 for all x, y in X ρ′− (x − y, u(x, y)) ≤ 0 ≤ ρ′+ (x − y, u(x, y)).

(4.3.7)

If we first make the substitution z := x−y, x := x and then z := −ty, t > 0, using (iv) from Theorem 2.1.1, ρ′+ (x + ty, x) = kx + tyk2 − tρ′− (x + ty, y) and dividing by t2 , we obtain ρ′− (y, Ft (x, y)x) ≤ (kxk2 − ρ′+ (x + ty, x))kyk2 ≤ ρ′+ (y, Ft (x, y)x), (4.3.8) where Ft (x, y) = ρ′+ (x, y) − ρ′− (x + ty, y). Now we consider two cases: ρ′+ (x, y) = ρ′− (x, y) or ρ′+ (x, y) > ρ′− (x, y). In this last case, by using Proposition 2.1.6 and Corollary 2.1.1, we get lim Ft (x, y) > 0 and for t > 0 in a neighborhood of zero Ft (x, y) > 0, so t→0+

by (4.3.8) and Theorem 2.1.1, Ft (x, y)ρ′− (y, x) ≤ (kxk2 − ρ′+ (x + ty, x))kyk2 ≤ Ft (x, y)ρ′+ (y, x). Taking limit when t tends to zero, we obtain ρ′− (y, x) ≤ 0 ≤ ρ′+ (y, x). Then for all x, y in X, ρ′+ (x, y) = ρ′− (x, y) or ρ′− (y, x) ≤ 0 ≤ ρ′+ (y, x). Now, for all t > 0, if we consider x and x + ty using (iii) and (v) from Theorem 2.1.1, we have ρ′+ (x, y) = ρ′− (x, y) or ρ′− (x + ty, x) ≤ 0 ≤ ρ′+ (x + ty, x). If ρ′+ (x, y) > ρ′− (x, y), for all t > 0, ρ′− (x + ty, x) ≤ 0 ≤ ρ′+ (x + ty, x) and taking limit when t tends to zero, we obtain the contradiction kxk = 0. Then ρ′+ = ρ′− and by (4.3.7) we have ρ′+ (x − y, u(x, y)) = 0, i.e., condition (ii) holds and X is an i.p.s. Now we generalize another property of perpendicular bisectors. Corollary 4.3.1 Let (X, k ·k) be a strictly convex space with dim X ≥ 2. Then X is an i.p.s. if and only if for all x, y in X and for all z in M+ (x, y), we have kx − zk = ky − zk.


Proof.

115

By hypothesis for all λ > 0

x + y

x + y

=

,

+ λu(x, + λu(x, y) − x y) − y

2

2

i.e., x − y ⊥J 2λu(x, y), and if λ = 12 , by our last proposition X is an i.p.s. 4.4

On the radius of the circumscribed circumference of a triangle

If (X, h·, ·i) is an i.p.s. and x, y are two linearly independent vectors in X, in the triangle of sides x, y and x − y, the radius R of the circumscribed circumference is given by the formula kxkkykkx − yk p , 4 s(s − kxk)(s − kyk)(s − kx − yk)

where s = (kxk + kyk + kx − yk) /2 is the semiperimeter of the triangle. Moreover, another equivalent expression for R is

kyk2 kxk2 − hx, yi x + kxk2 kyk2 − hy, xi y , 2kxk2 kyk2 − hx, yi2 − hy, xi2

which we get by finding R as kαx+βyk, and determining α and β by means of the condition that αx+βy− 12 x must be orthogonal to x and αx+βy− 12 y must be orthogonal to y. Then, in a real normed space (X, k · k), given two linearly independent vectors x and y, we can define the radius of the circumscribed circumference in the triangle of sides x, y and x − y as:

kyk2 kxk2 − ρ′− (x, y) x + kxk2 kyk2 − ρ′− (y, x) y R(x, y) := . 2kxk2 kyk2 − ρ′− (x, y)2 − ρ′− (y, x)2 This definition is only possible if

ρ′− (x, y)2 + ρ′− (y, x)2 < 2kxk2 kyk2 , i.e., ρ′− (x, y) < kxkkyk or ρ′− (y, x) < kxkkyk. For this reason we assume that X is strictly convex. Theorem 4.4.1 Let (X, k · k) be a strictly convex space with dim X ≥ 3. Then X is an i.p.s. if and only if for all linearly independent vectors x, y

116


in X, kxkkykkx − yk , R(x, y) = p 4 s(s − kxk)(s − kyk)(s − kx − yk)

where s = (kxk + kyk + kx − yk) /2.

Proof. If we let y := tz with t > 0 and we take limit when t tends to zero, we obtain

kxk2 kzk2x − ρ′− (z, x)z = lim R(x, tz) 2kxk2 kzk2 − ρ′− (x, z)2 − ρ′− (z, x)2 t→0+ 1/2 kxk2 t2 kzk2kx − tzk2 = lim t→0+ [kx − tzk2 − (tkzk − kxk)2 ] [(kxk + tkzk2 ) − kx − tzk2 ] 1/2 tkx − tzk = kxkkzk lim t→0+ kx − tzk2 − (tkzk − kxk)2 !1/2 tkx − tzk × lim+ 2 t→0 (kxk + tkzk) − kx − tzk2 s s kxk kxk = kxkkzk −2ρ′− (x, z) + 2kxkkzk 2kxkkzk + 2ρ′− (x, z) kzkkxk2 , = p 2 kxk2 kzk2 − ρ′− (x, z)2

and therefore,

2 2

2 − ρ′

x z z x , kxk − ρ′− kzk

x − kxk , kzk x z z ′

r .

kxk − ρ− kzk , kxk kzk = 2 x z ′ 2 1 − ρ− kxk , kzk If u := v in SX ,

x z , v := , then for all linearly independent vectors u and kxk kzk ′

(u, v)2 − ρ′− (v, u)2

u − ρ′− (v, u)v = 2 − ρ− p . 2 1 − ρ′− (u, v)2

Now, if ρ′− (v, u) = 0, then q 2 1 − ρ′− (u, v)2 = 2 − ρ′− (u, v)2 .

Therefore, ρ′− (u, v) = 0, and consequently, (see (19.6) in [Amir (1986)]) X is an i.p.s.


117

Analogous definitions of R(x, y) can be given by replacing the role of ρ′− by ρ′+ or by changing the order of the arguments appearing in ρ′− . For example, assume that the radius of the circumscribed circumference is given by

kyk2 kxk2 − ρ′ (x, y) x + kxk2 kyk2 − ρ′ (y, x) y + + ˆ y) := R(x, 2 kxk2 kyk2 − ρ′− (x, y)2 (which is equal to R(x, y) in an i.p.s.). Then a strictly convex space X with dim X ≥ 2 is an i.p.s. if and only if kxkkykkx − yk ˆ y) = p . R(x, 4 s(s − kxk)(s − kyk)(s − kx − yk)

(4.4.1)

The proof is immediate and uses the fact that the right-hand side ˆ y) are symmetric, which implies of (4.4.1) and the numerator of R(x, ρ′− (x, y)2 = ρ′− (y, x)2 for all x, y in X and X is an i.p.s. 4.5

Circumcenters in a triangle

Let (X, k·k) be a real normed linear space with dim X ≥ 2. In the following, when speaking about points we will always refer to vectors with the initial point at the origin. Then, consider the triangle ∆xyz with sides x − y, y − z, x − z, where x, y, z belong to X, x, y are linearly independent and z is in the plane determined by x and y (see Figure 4.5.1).

→ z → y 0

△xyz → x Figure 4.5.1

In an i.p.s., the perpendicular bisectors of the three sides of a triangle all pass through the circumcenter, which is the center of the circumscribed circle (see [Coxeter (1969)], p. 12-13 and [Puig-Adam (1986)], p. 92). Then, in a natural way we have the following

118


Definition 4.5.1 The triangle ∆xyz has a circumcenter C if and only if M+ (x, y), M+ (x, z) and M+ (y, z) meet at the point C. First we consider the class of triangles ∆xy(x + y), where we can show the following result: Theorem 4.5.1 Let (X, k · k) be a strictly convex space with dim X ≥ 2. Then X is an i.p.s. if and only if there exists the circumcenter of ∆xy(x+y) for all linearly independent vectors x, y in X. Proof. If we consider ∆xy(x + y), using the linear independence of x and y, it is a straightforward computation to prove that the three straight lines M+ (x, y), M+ (x, x + y) and M+ (y, x + y) meet at a point if and only if the following equality holds (kyk2 − ρ′+ (x, y))(kxk2 − ρ′+ (x + y, x))(ky + xk2 − kyk2 − ρ′+ (y, x)) = (kyk2 − ρ′+ (x + y, y)) (ky + xk2 − kxk2 − ρ′+ (x, y))(kxk2 − ρ′+ (y, x)) (4.5.1) + 2(kxk2 − ρ′+ (x + y, x))(kxk2 − ρ′+ (y, x) − kyk2 + ρ′+ (x, y) .

If we replace x by λx (λ > 0), then by using the properties of ρ′+ , simplifying λ; dividing by 2λ the term ky+λxk2 −kyk2 −λρ′+ (y, x) in the left part of the last equality and dividing also by 2λ the term ky||2 −ρ′+ (λx+y, y) in the right part of last equality; using the equality ρ′+ (λx + y, y) = kλx + yk2 − λρ′− (λx + y, x),

and finally by taking limit when λ tends to zero, by Proposition 2.1.6, Corollary 2.1.1 and the definition of ρ′+ , we obtain ρ′+ (y, x) = ρ′− (y, x) or ρ′+ (y, x) = 0. Replacing x by −x, we have ρ′+ (y, x) = ρ′− (y, x) or ρ′− (y, x) = 0, and combining the four possible cases, we obtain ρ′+ = ρ′− . Moreover, by substituting x := u, y := v−u in (4.5.1), using the equality ρ′+ = ρ′− and the properties of ρ′+ , operating and grouping in a suitable way, we get −kuk2 − kvk2 + ku − vk2 + 2ρ′+ (v, u) = 0 ρ′+ (v − u, u)(kuk2 + kvk2 − ρ′+ (u, v) − ρ′+ (v, u)) = ku − vk2 (−kuk2 + ρ′+ (v, u)) + kuk2 kvk2 − ρ′+ (u, v)ρ′+ (v, u). By symmetry and equalities ρ′+ = ρ′− and ρ′+ (u − v, v) = −kv − uk2 − ρ′+ (v − u, u),

(4.5.2)


119

we have −kuk2 − kvk2 + ku − vk2 + 2ρ′+ (u, v) = 0, or ρ′+ (v − u, u)(kuk2 + kvk2 − ρ′+ (u, v) − ρ′+ (v, u)) = ku − vk2 (−kuk2 + ρ′+ (v, u)) − kuk2 kvk2 + ρ′+ (v, u)ρ′+ (u, v).

(4.5.3)

If (4.5.2) and (4.5.3) hold, then ρ′+ (u, v)ρ′+ (v, u) = kuk2 kvk2 , but by hypothesis, X is strictly convex and we obtain a contradiction (cf. Theorem 2.1.1 (vii)). Then, for all u, v in X kuk2 + kvk2 − ku − vk2 ′ ∈ ρ+ (u, v), ρ′+ (v, u) , 2

and changing v for −v and using ρ′+ = ρ′− ,

ku + vk2 − kuk2 − kvk2 ∈ {ρ′+ (u, v), ρ′+ (v, u)}. 2 Then for all t > 0, the function defined by f (t) :=

ku + tvk2 − kuk2 − t2 kvk2 2t

is continuous in (0, +∞) and f (t) ∈ {ρ′+ (u, v), ρ′+ (v, u)} for all t > 0, lim f (t) = ρ′+ (u, v). Then ρ′+ (u, v) = ρ′+ (v, u) for all u, v in X, and

t→0+

by Theorem 2.1.1 (x), X is an inner product space.

Note that Theorem 4.5.1 covers a general case because ∆xy(x + y) is equivalent to the triangle determined by x and y (i.e., with sides x, y and x − y) (see Figure 4.5.2).

→ y

0

→ x +

→ y

△xy(x + y)

→ x Figure 4.5.2

120


Then, with a similar proof to that of the last theorem, we have the following Corollary 4.5.1 Let (X, k · k) be a strictly convex space with dim X ≥ 2 such that there exists the circumcenter of ∆xy(αx + y) for all linearly independent vectors x, y in X and α in R fixed. Then (X, k · k) is smooth. In a real normed linear space (X, k · k) we consider the triangle ∆xyz with sides x − y, y − z and x − z, where x and y are linearly independent and z = z1 x + z2 y (z1 , z2 in R) is in the plane determined by x and y. Then by a straightforward computation, we can prove that the intersections of the three perpendicular bisectors M+ (x, y), M+ (x, z) and M+ (y, z) give us three points  A(z2 − 1) − Bz1 x+z   + (Cx + Dz) C1 =   2 2(CB + DBz1 − ADz2 )      E(1 − z1 ) + F z2 x+y + C2 = (Ax + By)  2 2(BF z1 − AE − AF z2 )       D(z1 + z2 ) + C y+z   C3 = + (Ey + F z), 2 2(−CE − CF z2 − DEz1 )

(4.5.4)

where A = kyk2 − ρ′+ (x, y), B = kxk2 − ρ′+ (y, x), C = kzk2 − ρ′+ (x, z), D = kxk2 − ρ′+ (z, x), E = kzk2 − ρ′+ (y, z), F = kyk2 − ρ′+ (z, y), and by Theorem 4.5.1, in a real normed space, the three points are, in general, different. Definition 4.5.2 The points C1 , C2 , C3 defined in (4.5.4) are called the circumcenter points of ∆xyz. We consider two classical properties concerning the circumcenter of a triangle in the Euclidean plane, and we translate these properties into a real normed linear space considering the circumcenter point C1 to obtain new characterizations of inner product spaces. Theorem 4.5.2 Let (X, k · k) be a strictly convex space with dim X ≥ 2 and let x, y, z be vectors in X with kxk = kyk = kzk. Then X is an inner product space if and only if the circumcenter point C1 of ∆xyz is the origin.


121

Proof. The direct part is a well-known result. Reciprocally, if for all x, y, z in X with kxk = kyk = kzk the circumcenter point C1 of ∆xyz is the origin, then in particular it is true for unit vectors such that x and y are linearly independent and z = z1 x + z2 y with z1 , z2 in R, z2 6= 0. In such a case, using (4.5.4) and the linear independence of x and y, we infer that the system 1 + z1 +

z2 +

A(z2 − 1) − Bz1 (C + Dz1 ) = 0, CB + DBz1 − ADz2 A(z2 − 1) − Bz1 − Dz2 = 0, CB + DBz1 − ADz2

where A, B, C and D are defined after (4.5.4) gives C = D, and therefore ρ′+ (x, z) = ρ′+ (z, x) for all x, z in SX . As a consequence, by Theorem 2.1.1, (x), is an inner product space. Theorem 4.5.3 Let (X, k · k) be a strictly convex space with dim X ≥ 2. Then X is an inner product space if and only if the circumcenter point C1 of ∆xy(λx + y) is (x + y)

kxk2 λ + ρ′+ (x, y) + ρ′+ (y, x) 2kxk2 + ρ′+ (x, y) + ρ′+ (y, x)

whenever x, y are linearly independent vectors in X with kxk = kyk and λ belongs to R. Proof. The direct part of the statement is just a verification. Reciprocally, consider ∆xy(λx + y), where x and y are unit vectors and assume C1 = kxk2 λ + ρ′+ (x, y) + ρ′+ (y, x) . Then, using expression (4.5.4) of C1 (x + y) 2kxk2 + ρ′+ (x, y) + ρ′+ (y, x) (where z1 = λ and z2 = 1) and the linear independence of x and y, we have λ + ρ′+ (x, y) + ρ′+ (y, x) Bλ 1+λ − (C + Dλ) = 2 2(CB + DBλ − AD) 2 + ρ′+ (x, y) + ρ′+ (y, x) Bλ 1 = − D. 2 2(CB + DBλ − AD) By the last equalities, with a long and tedious computation we can prove that ρ′+ (x, y) = ρ′+ (y, x) (ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ)

and

or 1 = ρ′+ (λx + y, x)).

122


Now, we claim that |ρ′+ (x, y)| = |ρ′+ (y, x)| for all x, y in SX . Let x and y be two unit linearly independent vectors in X. If λ = 1, the result is evident. Assume that λ 6= 1. If ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ) we consider the vectors x and

and

1 = ρ′+ (λx + y, x),

(4.5.5)

λx + y , then we have two cases: kλx + yk

ρ′+ (λx + y, x) = ρ′+ (x, λx + y) or ρ′+ (x, λx + y) + ρ′+ (λx + y, x) = 2(1 − λ)kλx + yk and 1 =

ρ′+

λx + y ,x . λx + kλx + yk

In the first case by Theorem 2.1.1 and (4.5.5), 1 = λ + ρ′+ (x, y) and = 1 − λ = ρ′+ (y, x).

ρ′+ (x, y)

λx + In the second case, we take the vectors x and

λx + the process.

λx+y kλx+yk

and repeat

λx+y kλx+yk

Summarizing, if we consider the sequence (bn ) defined by b1 := λx + bn−1 for all n ≥ 2, we have to bear in mind two y and bn := λx + kbn−1 k possibilities. Possibility 1. There exists n in N such that ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ), 1 = ρ′+ (b1 , x), ρ′+ (x, bk ) + 1 = 2(1 − λ)kbk k, 1 = ρ′+ (bk+1 , x) for 1 ≤ k ≤ n − 1 and ρ′+ (bn , x) = ρ′+ (x, bn ).

We will prove, by induction, that we can infer ρ′+ (x, y) = ρ′+ (y, x) = 1 − λ. For n = 1 we have proved that it is true. If it is true for n − 1, we wish to prove that it is true for n. = Using Theorem 2.1.1, we have 1 = ρ′+ (x, bn ) = ρ′+ x, λx + kbbn−1 n−1 k

ρ′+ (x,bn−1 ) ρ′ (x,b ) and kbn−1 k = + 1−λn−1 . kbn−1 k So, ρ′+ (x, bn−1 ) + 1 = 2ρ′+ (x, bn−1 ) and ρ′+ (x, bn−1 ) = 1 and by induction hypothesis, ρ′+ (x, y) = ρ′+ (y, x) = 1 − λ.

λ+

= ρ′+ (bn−1 , x),


123

Possibility 2. For all n in N ρ′+ (x, bn ) + 1 = 2(1 − λ)kbn k,

(4.5.6)

and ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ),

1 = ρ′+ (λx + y, x).

(4.5.7)

Now, if we consider ∆yx(λy + x), and we apply the hypothesis to this triangle, in a similar way to the results obtained before, we have |ρ′+ (x, y)| = |ρ′+ (y, x)| or Possibility 2’. For all n in N,

ρ′+ (y, cn ) + 1 = 2(1 − λ)kcn k,

(4.5.8)

and ρ′+ (x, y) + ρ′+ (y, x) = 2(1 − λ), where c1 = λy + x and cn = λy +

cn−1 kcn−1 k

1 = ρ′+ (λy + x, y),

for all n ≥ 2.

In this case, using (4.5.6) and (4.5.8) for n = 1, we obtain ρ′+ (x, y) = 2(1 − λ)kλx + yk − λ − 1 and ρ′+ (y, x) = 2(1 − λ)kλy + xk − λ − 1. Substituting this into (4.5.7), 2(1 − λ) = 2(1 − λ) (kλx + yk + kλy + xk) − 2λ − 2 and kλx + yk + kλy + xk =

2 . 1−λ

Since 0 ≤ kλx+yk+kλy +xk ≤ 2|λ|+2, by (4.5.7), using (i) in Theorem 2.1.1 and the fact that x, y are unit vectors, we have 2(1 − λ) ≤ 2, so λ > 0. 2 ≤ 2λ+2 and 1−λ > 0 gives 2 ≤ 2(λ+1)(1−λ), 1 ≤ 1−λ2 , Finally, 0 ≤ 1−λ λ = 0 and we obtain a contradiction. So, this case is impossible. Consequently, |ρ′+ (x, y)| = |ρ′+ (y, x)| for all x, y in SX and by ([Amir (1986)] (2.5), p. 18) X is an i.p.s.

124

4.6


Euler line in real normed space

In Chapter 3 we have defined the orthocenter points of a triangle (see Definition 3.5.2). By using those definitions and the results obtained in Section 4.5 and concerning the circumcenter points, we may now consider, in a real normed space, the existence of the Euler line (see [Coxeter (1969)]). Proposition 4.6.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for all x, y in X linearly independent and z = λx + y, λ in R+ fixed, the barycenter B1 of ∆xyz belongs to the line determined by the orthocenter point H12 and the circumcenter point C1 and B1 C1 = 12 B1 H12 (i.e., the Euler line exists) then ρ′+ = ρ′− . Proof.

According to [Tom´ as (2005)] and (4.5.4), consider C1 =

λB x + λx + y − (Cx + D(λx + y)), 2 2(CB + DBλ − AD)

where A = kyk2 −ρ′+ (x, y), B = kxk2 −ρ′+ (y, x), C = kλx+yk2 −ρ′+ (x, λx+ y) and D = kxk2 − ρ′+ (λx + y, x). H12 2C1 x + y + λx + y = + , and using the linear By hypothesis 3 3 3 independence of x and y, by a straightforward computation we obtain ρ′+ (y, x) = kxk2 or (kxk2 − ρ′+ (λx + y, x))(−k(λ − 1)x + yk2 − λkxk2 − ρ′− (y − x, x))

= (λkxk2 + ρ′− (y − x, x))(kλx + yk2 − ρ′+ (x, λx + y)). Then, for all a > 0 in R, different from ax and y, by (4.6.1) we have

ρ′+ (y,x) kxk2 ,

(4.6.1)

if we consider the points

(akxk2 − ρ′+ (λax + y, x))(−k(λ − 1)ax + yk2 − λa2 kxk2 − ρ′− (y − ax, ax)) = (λakxk2 + ρ′− (y − ax, x))(kλax + yk2 − aρ′+ (x, λax + y)).

Taking the limit when a tends to zero using (2.1.9), (2.1.10) and the properties of ρ′+ , we obtain ρ′+ (y, x) = ρ′− (y, x). Finally, if we consider in a real normed linear space (X, k · k) another height function, defined in Chapter 3, h2 (x, y) = y +

kyk2 − ρ′+ (y, x) (x − y) kx − yk2


125

and analogously to the case where we have considered h1 , we define the corresponding orthocenter point H12 , with a similar proof to Proposition 4.6.1 we obtain the following characterization of i.p.s. Proposition 4.6.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an i.p.s. if and only if for all x, y in X linearly independent and z = λx + y, λ in R+ fixed, the barycenter B1 of ∆xyz belongs to the line determined by the orthocenter point H12 and the circumcenter point C1 and B1 C1 = 12 B1 H12 (i.e., the Euler line exists). 4.7

Functional equation of the perpendicular bisector transform

In this section we discuss the preservation of the perpendicular bisector M := M+ (see Section 4.1), namely we solve the functional equation h i x+y + λ kyk2 − ρ′+ (x, y) x + kxk2 − ρ′+ (y, x) y f 2 h i f (x) + f (y) (4.7.1) + g(λ) kf (y)k2 − ρ′+ f (x), f (y) f (x) = 2 h i + kf (x)k2 − ρ′+ f (y), f (x) f (y) . We start with the following result.

Theorem 4.7.1 Let (X, k · k) and (Y, k · k) be real normed linear spaces of at least dimension two. Assume that f : X → Y is continuous at a point on rays (i.e., the function R ∋ t 7→ f (tx) ∈ Y is continuous at a point for every fixed x in X), f (0) = 0 and f satisfies for all x, y in X and real λ functional equation (4.7.1), where g : R → R is such that g 6≡ 0 and 0 ∈ g(R). Then f is linear. Proof. If f ≡ 0, then obviously f is linear. Assume that f 6≡ 0 satisfies (4.7.1). Take u ∈ X and α, β ∈ R, and substitute x := αu, y := βu in (4.7.1). Then for all λ ∈ R we have f (αu) + f (βu) α+β u = f 2 2 h i + g(λ) kf (βu)k2 − ρ′+ f (αu), f (βu) f (αu) h i + kf (αu)k2 − ρ′+ f (βu), f (αu) f (βu) .

126


Take λ ∈ g −1 ({0}). Then for all u ∈ X and α, β ∈ R f

α+β u 2

=

f (αu) + f (βu) . 2

(4.7.2)

Define function fu : R → Y by fu (α) := f (αu), α ∈ R. Then (4.7.2) becomes fu (α) + fu (β) α+β = . fu 2 2 From the regularity condition upon f , we derive that fu is continuous at a point and (see e.g. [Aczél (1966)]) there exist au , bu in Y such that fu (α) = au α + bu ,

u ∈ X,

α ∈ R.

Moreover, 0 = f (0) = fu (0) = bu , so f (u) = fu (1) = au and f (αu) = αf (u),

u ∈ X,

α ∈ R.

(4.7.3)

Putting λ := 0 into (4.7.1), we get

f

x+y 2

=

f (x) + f (y) 2h

i + g(0) kf (y)k2 − ρ′+ f (x), f (y) f (x) h i + kf (x)k2 − ρ′+ f (y), f (x) f (y) .

(4.7.4)

With substitutions x := αu, y := αv in (4.7.4), we obtain on one side f

αu + αv 2

αf (u) + αf (v) u+v = 2 2 h i 2 ′ + α g(0) kf (v)k − ρ+ f (u), f (v) f (u) h i + kf (u)k2 − ρ′+ f (v), f (u) f (v) ,

= αf

127


and on the other side f (αu) + f (αv) αu + αv = f 2 2 h i + g(0) kf (αv)k2 − ρ′+ f (αu), f (αv) f (αu) h i + kf (αu)k2 − ρ′+ f (αv), f (αu) f (αv) .

Combining the above two equalities, we get h h i i g(0) kf (v)k2 −ρ′+ f (u), f (v) f (u) + kf (u)k2 −ρ′+ f (v), f (u) f (v) = 0,

which together with (4.7.4) leads to f (x) + f (y) x+y = , f 2 2

x, y ∈ X,

so f is linear and the proof is completed.

Before stating the next theorem, we prove the following lemma. Lemma 4.7.1 Let (X, k · k) and (Y, k · k) be real normed linear spaces of at least dimension two. Assume that f : X → Y is nonzero, f (0) = 0, f is continuous at a point on rays and satisfies for all x, y in X and real λ functional equation (4.7.1), where g : R → R is such that g 6≡ 0 and 0 ∈ g(R). Then: (i) f (x) = 0 if and only if x = 0; (ii) dim f (X) ≥ 2. Proof. (i) Observe first that by Theorem 4.7.1, f is linear and we can rewrite (4.7.1) in the form h i λ kyk2 − ρ′+ (x, y) f (x) + kxk2 − ρ′+ (y, x) f (y) h i = g(λ) kf (y)k2 − ρ′+ f (x), f (y) f (x) (4.7.5) h i + kf (x)k2 − ρ′+ f (y), f (x) f (y) . Assume that f (x0 ) = 0 for some x0 ∈ X. Then by (4.7.5), for any y ∈ X, λ[kx0 k2 − ρ′+ (y, x0 )]f (y) = 0,

λ ∈ R.

(4.7.6)

128


Take y ∈ X such thatf (y) 6= 0. Since f is linear, f (ty) = 0 for any t > 0 and by (4.7.6) kx0 k2 − tρ′+ (y, x0 ) = 0, and taking limit when t tends to 0, x0 = 0. (ii) Since f is linear, f (X) is a linear subspace of Y . Assume, contrary to our claim, that dim f (X) = 1, so for each x, y in X there exists γ(x, y) ∈ R such that f (y) = γ(x, y)f (x). Take a nonzero x0 . In virtue of (4.7.5), for all y ∈ X we have kyk2 − ρ′+ (x0 , y) + kx0 k2 − ρ′+ (y, x0 ) γ(x0 , y) = 0.

Let y0 ∈ X \ {0} be such that ky0 k 6= kx0 k. If kx0 k2 − ρ′+ (y0 , x0 ) = 0, then ky0 k2 − ρ′+ (x0 , y0 ) = 0. By Theorem 2.1.1 (vi) we get ky0 k = kx0 k, which is a contradiction. So, kx0 k2 − ρ′+ (y0 , x0 ) 6= 0 and γ(x0 , y0 ) = −

ky0 k2 − ρ′+ (x0 , y0 ) . kx0 k2 − ρ′+ (y0 , x0 )

Take now instead of y0 vector αy0 with α > 0 and such that kαy0 k 6= kx0 k. We have f (αy0 ) = γ(x0 , αy0 )f (x0 ) and αf (y0 ) = f (αy0 ) = −

f (y0 ) = −

α2 ky0 k2 − αρ′+ (x0 , y0 ) f (x0 ), kx0 k2 − αρ′+ (y0 , x0 )

αky0 k2 − ρ′+ (x0 , y0 ) f (x0 ) kx0 k2 − αρ′+ (y0 , x0 )

for all α > 0 such that kαy0 k 6= kx0 k. So, we have f (y0 ) = − lim+ α→0

αky0 k2 − ρ′+ (x0 , y0 ) ρ′+ (x0 , y0 ) f (x0 ), f (x ) = 0 kx0 k2 − αρ′+ (y0 , x0 ) kx0 k2

but on the other hand f (y0 ) = − lim

α→∞

ky0 k2 − α1 ρ′+ (x0 , y0 ) ky0 k2 f (x0 ) = ′ f (x0 ), 1 ′ 2 ρ+ (y0 , x0 ) α kx0 k − ρ+ (y0 , x0 )

where ρ′+ (y0 , x0 ) 6= 0, otherwise f (y0 ) would not be finite. Hence, ρ′+ (x0 , y0 ) ky0 k2 = , ′ kx0 k2 ρ+ (y0 , x0 )


129

and inequalities |ρ′+ (x0 , y0 )| ≤ kx0 k ky0k and |ρ′+ (y0 , x0 )| ≤ kx0 k ky0 k become equalities. By Proposition 2.1.3, however, we obtain |ρ′+ (x0 , y)| = kx0 k kyk for all y ∈ X, which is a contradiction to Lemma 2.3.1. Careful reading of the proof of Lemma 4.7.1 shows that in fact we have the following. Corollary 4.7.1 Let (X, k · k) and (Y, k · k) be real normed linear spaces of at least dimension two. Assume that f : X → Y is nonzero, f (0) = 0, f is continuous at a point on rays and satisfies for all x, y in X and real λ functional equation (4.7.1), where g : R → R is such that g 6≡ 0 and 0 ∈ g(R). Then f preserves the linear independence of vectors. Now we can continue with solving (4.7.1). Theorem 4.7.2 Let (X, k · k) and (Y, k · k) be real normed linear spaces of at least dimension two. Assume that f : X → Y is nonzero, f (0) = 0, f is continuous at a point on rays and satisfies for all x, y in X and real λ functional equation (4.7.1), with g : R → R such that g 6≡ 0 and 0 ∈ g(R). Then there exists a positive constant c such that g(λ) = cλ for all λ in R and f is a linear similarity transformation and kf (x)k = √1c kxk for all x in X. Proof. By Lemma 4.7.1 for each x ∈ X \ {0} there exists a y in X such that f (x) and f (y) are linearly independent. Then, for all x, y in X such that f (x) and f (y) are linearly independent in Y , and for all λ ∈ R, by (4.7.5) we have h i h i λ kyk2 − ρ′+ (x, y) = g(λ) kf (y)k2 − ρ′+ f (x), f (y) and

h i h i λ kxk2 − ρ′+ (y, x) = g(λ) kf (x)k2 − ρ′+ f (y), f (x) .

Take x0 ∈ X \ {0}. There exists a y in X such that f (x0 ) and f (y) are linearly independent. However, if f (x0 ) and f (y) are linearly independent, so are f (x0 ) and f (ty) for all t > 0, and we have for all λ ∈ R: h i λ kx0 k2 − ρ′+ (ty, x0 ) = g(λ) kf (x0 )k2 − ρ′+ f (ty), f (x0 ) , h i λ kx0 k2 − tρ′+ (y, x0 ) = g(λ) kf (x0 )k2 − t ρ′+ f (y), f (x0 ) .

130


Taking the limit while t tends to zero, we obtain λkx0 k2 = g(λ) kf (x0 )k2 ,

λ ∈ R.

Since f (x0 ) 6= 0, g(λ) =

kx0 k2 λ, kf (x0 )k2

λ ∈ R,

whence g(λ) = cλ for all λ ∈ R and for some nonzero constant c and, consequently, kf (x)k2 = kxk2 /c for all x ∈ X (also for x = 0). Now we consider the reciprocal of the last result in order to obtain a characterization of functions satisfying (4.7.1). Theorem 4.7.3 Let (X, k · k) and (Y, k · k) be real normed linear spaces of at least dimension two. Assume that f : X → Y is nonzero, f (0) = 0, f is continuous at a point on rays and g : R → R is such that g 6≡ 0 and 0 ∈ g(R). Then f satisfies for all x, y ∈ X and real λ functional equation (4.7.1) if and only if there exists c > 0 such that g(λ) = cλ for all λ ∈ R, f is a linear similarity transformation with kf (x)k = √1c kxk for all x ∈ X and such that ρ′+ (x, y) = cρ′+ (f (x), f (y)) for all x, y in X. Proof.

It is a straightforward computation.

The equation ρ′+ (x, y) = ρ′+ (f (x), f (y)) was studied in detail together with ρ′− (x, y) = ρ′− (f (x), f (y)) in Section 2.3 under different assumptions (see also [Alsina and Tom´ as (1991)]).

Chapter 5

Bisectrices in Real Normed Spaces

Our aim in this chapter is to give various notions of bisectrices in triangles located in real normed spaces, and to show how their essential properties yield a collection of interesting characterizations of inner product spaces. 5.1

Bisectrices in real normed spaces

Let (X, k · k) be a real normed linear space with dim X ≥ 2. Given two vectors x, y in X\{0}, we may consider the vector w(x, y) = αx + (1 − α)y with α ∈ (0, 1) satisfying the property: kw(x, y) − xk kw(x, y) − yk = . kyk kxk

(5.1.1)

In the classical case of R2 endowed with the Euclidean norm, this last property states the so-called bisectrices theorem. In general, the substitution of w(x, y) in (5.1.1) by the expression αx + (1 − α)y determines α, and we have w(x, y) =

kxk kyk x+ y. kxk + kyk kxk + kyk

(5.1.2)

If y = x, in a natural way w(x, x) = x. This expression gives the reason for the following definition. Definition 5.1.1 For any triangle determined by vectors a, b, c with initial point at the origin in a real normed linear space (X, k · k) with dim X ≥ 2, the bisectrices are the three straight lines defined by a + Rw(b − a, c − a), b + Rw(a − b, c − b), 131

132


c + Rw(a − c, b − c). Thus, without loss of generality we can restrict our attention to triangles determined by vectors x and y with initial point at the origin and such that at least one of them is different from zero. Such triangles with sides x, y and x − y will have three bisectrices w(x, y), w(−y, x − y) and w(−x, y − x). A well-known theorem of Senechalle [Senechalle (1968)] states that if there exists a function φ : [0, 2] → [0, ∞) such that for all unit vectors u, v in a real normed space (X, k · k) ku + vk = φ(ku − vk), then X is an i.p.s. By means of this result we can prove the following general result involving the bisectrices. Theorem 5.1.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an i.p.s. if and only if in every triangle determined by linearly independent vectors x and y with initial point at the origin, the length of w(x, y) is determined by the lengths of the three sides. Proof. If X is an i.p.s. then it is well known that kw(x, y)k =

p 4 kxkkyks(s − kx − yk), kxk + kyk

where s = (kxk+kyk+kx−yk)/2. Thus kw(x, y)k depends only on kxk, kyk and kx − yk. Reciprocally, if there exists a function G : R+ × R+ × R+ → R such that for all linearly independent x, y in X\{0} kw(x, y)k = G(kxk, kyk, kx − yk), then by (5.1.2)

x y kxk + kyk

kxk + kyk = kxkkyk G(kxk, kyk, kx − yk) =: H(kxk, kyk, kx − yk),

where H : R+ × R+ × R+ −→ R is given by the formula H(q, s, t) =

q+s G(q, s, t) qs


133

with positive q, s, t and such that t < q + s.   H(1, 1, t) if t ∈ (0, 2) If we define φ : [0, 2] −→ [0, 2] by φ(t) := 0 if t = 2  2 if t = 0 then for all u, v in SX , we have: if u 6= ±v then

u v

= H(kuk, kvk, ku − vk)

+ ku + vk = kuk kvk = H(1, 1, ku − vk) = φ(ku − vk),

if u = v then ku + vk = 2 = φ(0) and if u = −v then ku + vk = 0 = φ(2). Applying Senechalle’s theorem, X is an inner product space.

Considering the height function h3 defined in Chapter 3, we may define the radius r(x, y) of the inscribed circumference as the norm of h3 -height of the triangle determined by one side x − y of the triangle and the incenter, i.e.,

x x y y kxkkyk ′

. (5.1.3) − ρ− , r(x, y) := 2s kxk kyk kxk kyk It is easy to check that in an i.p.s. the above expression (5.1.3) coincides with the celebrated formula for the radius of the inscribed circumference: p s(s − kxk)(s − kyk)(s − kx − yk) , (5.1.4) rc (x, y) := s kxk + kyk + kx − yk . 2 The main result in this section (see [Alsina, Guijarro and Tomás (1996)]) is the following: where s =

Theorem 5.1.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. Then X is an i.p.s. if and only if for all linearly independent vectors x, y in X r(x, y) = rc (x, y).

Proof. If we assume that r(x, y) defined by (5.1.3) is equal to rc (x, y) defined

134


by (5.1.4), then we have r(x, y)2 =

(s − kxk)(s − kyk)(s − kx − yk) , s

i.e.,

2 x y y kxk2 kyk2

x −ρ′

= (s−kxk)(s−kyk)(s−kx−yk) , , −

2 4s kxk kyk kxk kyk s

or equivalently,

2

kykx − ρ′− (y, x) y = 4s(s − kxk)(s − kyk)(s − kx − yk),

kyk

for all linearly independent vectors x and y in X. Substituting in this equality y by tz, with t > 0, dividing by t2 and taking limits when t tends to zero, we obtain

z 4s(s − kxk)(s − ktzk)(s − kx − tzk)

2

= lim+

kzkx − ρ′− (z, x) kzk t2 t→0 4 kxk + tkzk + kx − tzk tkzk + kx − tzk − kxk = lim+ 2 · 2 2 t→0 t kxk + kx − tzk − tkzk kxk + tkzk − kx − tzk × · 2 2 2 (kx − tzk + tkzk) − kxk2 kxk2 − (kx − tzk − tkzk)2 = lim+ · 2t 2t t→0 2 2 kx − tzk − kxk kzk2 = lim − t + kx − tzkkzk + 2(−t) 2 t→0+ kx − tzk2 − kxk2 kzk2 t + kx − tzkkzk × − 2(−t) 2 = −ρ′− (x, z) + kxkkzk ρ′− (x, z) + kxkkzk = kxk2 kzk2 − ρ′− (x, z)2 , and therefore, for all linearly independent x and z in X

kzk2x − ρ′− (z, x)z 2 = kzk2 kxk2 kzk2 − ρ′− (x, z)2 .

If x and z are linearly independent, then vectors z + x and z are also linearly independent and the substitution z + x in the place of x yields

2 kzk2 kz +xk2 kzk2 −ρ′− (z +x, z)2 = kzk2(z +x)−ρ′− (z, z +x)z

2 = kzk2z +kzk2x−kzk2z −ρ′− (z, x)z


135

2 = kzk2x − ρ′− (z, x)z

= kzk2 kxk2 kzk2 − ρ′− (x, z)2 ,

i.e.,

kxk2 kzk2 − ρ′− (x, z)2 = kz + xk2 kzk2 − ρ′− (z + x, z)2 . Thus for all linearly independent vectors u, v in SX , if we substitute in the last equality z := u − v, x := v, we obtain kvk2 ku − vk2 − ρ′− (v, u − v)2 = kuk2ku − vk2 − ρ′− (u, u − v)2 , and consequently, ′ ρ (v, u − v) = ρ′ (u, u − v) , − −

which in turn implies ρ′− (v, u) − kvk2 = kuk2 − ρ′+ (u, v) , i.e., 1 − ρ′ (v, u) = 1 − ρ′ (u, v) . − +

Since ρ′− (v, u) ≤ kukkvk = 1 and ρ′+ (u, v) ≤ kukkvk = 1, we deduce ρ′− (v, u) = ρ′+ (u, v), and interchanging the roles of u and v we get ρ′− (u, v) = ρ′+ (v, u). Hence, ρ′+ (v, u) = ρ′− (u, v) ≤ ρ′+ (u, v) and ρ′+ (u, v) = ρ′− (v, u) ≤ ρ′+ (v, u), i.e., ρ′+ (u, v) = ρ′+ (v, u), for all linearly independent vectors u, v from SX , so X is an i.p.s. Note. The value r(x, y) as introduced above is not the only possible definition of the radius of the inscribed circumference, since we can consider other heights like h3 (λw(x, y), γw(−x, y − x)), h3 (µw(−y, x − y), λw(x, y)), h3 (µw(−y, x − y), γw(−x, y − x)), h3 (γw(−x, y − x), λw(x, y)) or h3 (γw(−x, y − x), µw(−y, x − y)) instead of h3 (λw(x, y), µw(−y, x − y)), and moreover we can consider different height functions.

136

5.2



In a right triangle in an inner product space (X, h·, ·i) with perpendicular sides x and y, one can prove that the length of the perpendicular bisector w(x, y) is given by √ 2 kxk kyk . kw(x, y)k = kxk + kyk Comparing the above formula with (5.1.2), we get

√

kxky + kykx = 2 kxk kyk.

This provides the idea for defining in a normed space a new orthogonality relation. Definition 5.2.1 In a real normed linear space (X, k · k) with dim X ≥ 2, we consider the bisectrix orthogonality

√

x ⊥w y iff kxky + kykx = 2 kxk kyk, which for nonzero x and y means

√

x y

+

kxk kyk = 2.

In the following example, we compare various notions of orthogonality. Example 5.2.1 Consider the space (R2 , k · k∨ ). We give the form of all vectors from R2 and unit vectors which are orthogonal to (1, 0). Namely, we have (see Figure 5.2.1) (1, 0) ⊥w (a, b) if and only if (a, b) is the zero vector or √ √ √ √ {− 2 − 1, 2 − 1} × [− 2, 2] √ √ √ √ ∪ [− 2 − 1, 2 − 1] × {− 2, 2} .

(a, b) ∈

√ √ Moreover, if k(a, b)k∨ = 1, then (a, b) ∈ {( 2 − 1, −1), ( 2 − 1, 1)}.

137


y 1

−2

−1

0

1

x

Figure 5.2.1 On the other hand, for the classical orthogonality relations and those defined in Sections 2.2, 3.4 and 4.2, we have ⊥J : (1, 0) ⊥J (a, b) if and only if (see Figure 5.2.2) (a, b) ∈ {(x, y) : (y ≥ x + 1 and y ≥ −x + 1) or (y ≤ −x − 1 and y ≤ x − 1) or (x = 0 and y ∈ [−1, 1])} ; if k(a, b)k∨ = 1, then (a, b) ∈ {(0, −1), (0, 1)}. ⊥B : (1, 0) ⊥B (a, b) if and only if a = 0 and b ∈ R (see Figure 5.2.3); if k(a, b)k∨ = 1, then (a, b) ∈ {(0, −1), (0, 1)}. y

y

1

0

Figure 5.2.2

1 1

x 0

1

Figure 5.2.3

⊥P : (1, 0) ⊥P (a, b) if and only if (see Figure 5.2.4) (1 + a)2 − b2 = 1 and a ≥ 0; √ √ if k(a, b)k∨ = 1, then (a, b) ∈ {( 2 − 1, −1), ( 2 − 1, 1)}.

x

138


⊥ρ : (1, 0) ⊥ρ (a, b) if and only if a = 0 and b ∈ R (see Figure 5.2.3); if k(a, b)k∨ = 1, then (a, b) ∈ {(0, −1), (0, 1)}. y y 1 1 0

1

x

Figure 5.2.4

0

1

x

Figure 5.2.5

⊥M : (1, 0) ⊥M (a, b) if and only if (see Figure 5.2.5) (a, b) ∈ {(x, y) : (y > x + 1 and y > −x + 1) or (y < −x − 1 and y < x − 1) or (x = 0 and y ∈ [−1, 1])} ; if k(a, b)k∨ = 1, then (a, b) ∈ {(0, −1), (0, 1)}. ⊥h : (1, 0) ⊥h (a, b) if and only if (see Figure 5.2.6) (a, b) ∈ (x, y) : (y 2 = x2 − x and x ≤ 0) or (y 2 = x2 − x + 1 and x ≥ 1) or (x = 1 and |y| > 1) ; if k(a, b)k∨ = 1, then √ o n √ (a, b) ∈ (1, −1), (1, 1), 1−2 5 , −1 , 1−2 5 , 1 .


139

y

1

0

1

x

Figure 5.2.6 Proposition 5.2.1 Let (X, k · k) be a real normed linear space, dim X ≥ 2. The bisectrix orthogonality ⊥w has the following properties: (i) (ii) (iii) (iv) (v)

x ⊥w 0, 0 ⊥w y for all x, y in X; x ⊥w y if and only if y ⊥w x; If x ⊥w y, x, y 6= 0, then x, y are linearly independent; If x ⊥w y, αβ ≥ 0, then αx ⊥w βy; For each x in X and nonnegative t there exists a vector y in X with kyk = t and such that x ⊥w y; (vi) For each nonzero x, y in X there exists a positive t such that x+ty ⊥w x − ty; (vii) In an inner product space x ⊥w y if and only if hx, yi = 0. Proof. (v) For x = 0 or t = 0 the assertion is obvious. Fix a nonzero x ∈ X and t > 0, and define a function f : St → R, where St := {z ∈ X : kzk = t}, by the formula

√

x y

+ − 2 for all y ∈ St . f (y) :=

kxk t Then f is continuous and, moreover, we have √ √ xt xt = 2 − 2 > 0 and f − = − 2 < 0. f kxk kxk

So, since f is continuous there exists a y0 ∈ St such that f (y0 ) = 0, i.e., x ⊥w y0 .

140


(vi) Fix x, y ∈ X \ {0}. If x and y are linearly dependent, y = αx for some nonzero real α, then it is enough to take t := α1 . Assume then that x and y are linearly independent (so for all β ∈ R vectors x+ βy and x− βy are nonzero ones) and define a function g : (0, +∞) → R by the formula

√

x + ty x − ty

− 2, t ∈ (0, +∞).

+ g(t) :=

kx + tyk kx − tyk √ Then lim g(t) = 2 − 2 > 0 and t→0+

1 1 √

√

tx + y tx − y

+ 1 − 2 = − 2 < 0. lim g(t) = lim 1

t→+∞ t→+∞ k t x + yk k t x − yk

Since g is continuous, there exists a t0 ∈ (0, +∞) such that g(t0 ) = 0 and the proof is completed. Property (vi) from Proposition 5.2.1, together with the Amir’s characterization of i.p.s. (6.9) from [Amir (1986)], allows us to give the first characterization of i.p.s. by means of our new orthogonality ⊥w . Theorem 5.2.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for all vectors x, y in X, we have x ⊥w y

implies

kx + yk2 + kx − yk2 ∼ 2kxk2 + 2kyk2,

then the norm k · k comes from an inner product. In the following we study the relations between ⊥w and the classical orthogonalities. Relations between ⊥w and ⊥P

Directly from the definition of ⊥w , we have the following relations.

Proposition 5.2.2

Let (X, k · k) be a real normed linear space.

(i) For all nonzero vectors x, y in X the relation x ⊥w y holds if and only y x ⊥P . if kxk kyk (ii) For all u, v in SX the relation u ⊥w v holds if and only if u ⊥P v. Corollary 5.2.1 Let (X, k·k) be a real normed linear space. If for vectors x, y in X we have x ⊥P y

implies

x ⊥w y,


141

then for all nonzero x, y in X x ⊥P y

implies

y x ⊥P . kxk kyk

By Proposition 5.2.2, we can rewrite (10.16) from [Amir (1986)] in the form Theorem 5.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The norm in X derives from an inner product if and only if for all u, v in SX , if u ⊥w v, then there exists δ > 0 such that for all t ∈ [−δ, δ] we have u ⊥P tv. Relations between ⊥w and ⊥J Theorem 5.2.3 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for vectors x, y in X we have x ⊥w y

implies

x ⊥J y,

then the norm k · k comes from an inner product. Proof. Take nonzero x, y in X with kxk = kyk. There exists a t > 0 such that x + ty ⊥w x − ty. This implies x + ty ⊥J x − ty, i.e., kxk = ktyk and t = 1. So, x + y ⊥w x − y and hence, x + y ⊥w α(x − y) for all nonnegative α. That means x + y ⊥J α(x − y), which implies that also x + y ⊥J −α(x − y), so in fact, for all real α we have kx + y + α(x − y)k = kx + y − α(x − y)k, k(1 + α)x + (1 − α)yk = k(1 − α)x + (1 + α)yk, and if α 6= −1, then

x + 1 − α y = 1 − α x + y ,

1 + α 1 + α

2

x + −1 + 2

y = −1 + x + y

, 1+α 1+α

142


so, kx + βyk = kβx + yk for all real β, and from [Ficken (1944)] (see also [Amir (1986)], (2.7)) we have the desired assertion. Before giving the next characterization of i.p.s., we present the following (see [Alonso and Ben´ıtez (1989)]). Lemma 5.2.1 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for all nonzero x, y in X we have x ⊥J y

implies

y x ⊥J , kxk kyk

then the norm in X comes from an inner product. So we can state: Theorem 5.2.4 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for vectors x, y in X we have x ⊥J y

implies

x ⊥w y,

then the norm k · k derives from an inner product.

x y

+ Take nonzero x, y in X such that x ⊥J y. Then

kxk kyk =

√ √

x y

= 2. Consequently, − 2. But since also x ⊥J −y, we have

kxk kyk x y ⊥J , and on the account of Lemma 5.2.1 we have our assertion. kxk kyk

Proof.

Similarly to Theorem 5.2.2, by Proposition 5.2.2 we can rewrite (10.14) from [Amir (1986)] in the form Theorem 5.2.5 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The norm in X derives from an inner product if and only if for all u, v in SX if u ⊥w v, then there exists δ > 0 such that for all t ∈ [−δ, δ], u ⊥J tv. Relations between ⊥w and ⊥B

We start this section by quoting two characterizations of i.p.s. from [Amir (1986)] (cf. (10.4) and (10.12)).


143

Lemma 5.2.2 Let (X, k · k) be a real normed linear space with dim X ≥ 2. The following conditions are equivalent: (i) The norm k · k comes from an inner product; √ (ii) For all u, v in SX , we have u ⊥B v ⇒ ku + vk = 2 ; (iii) For all u, v in SX , we have ( u ⊥P v ⇒ u ⊥B v). Theorem 5.2.6 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for vectors u, v in SX , we have u ⊥B v

implies

u ⊥w v,

then the norm k · k derives from an inner product. √ Proof. Let u, v ∈ SX be such that u ⊥B v, then ku + vk = 2, and on account of Lemma 5.2.2 the norm in X comes from an inner product. Theorem 5.2.7 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for vectors u, v in SX , we have u ⊥w v

implies

u ⊥B v,

then the norm k · k comes from an inner product. Proof. Let u, v ∈ SX be such that u ⊥P v, then equivalently u ⊥w v, which implies u ⊥B v, and on account of Lemma 5.2.2 the norm in X comes from an inner product.

Relations between ⊥w and ⊥ρ , ⊥ρ Theorem 5.2.8 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for all vectors u, v in SX , we have u ⊥w v

implies

u ⊥ρ v,

then the norm k · k comes from an inner product. Proof. From the assumption, we have that u ⊥w v

implies

ρ′+ (u, v) = 0,

144


which in turn implies that u ⊥B v. Now by applying Theorem 5.2.7 we arrive at the desired assertion. Theorem 5.2.9 Let (X, k · k) be a real normed linear space with dim X ≥ 2. If for all vectors u, v in SX , we have u ⊥w v

implies

u ⊥ρ v,

then the norm k · k derives from an inner product. Proof.

From the assumption, we have that u ⊥w v

implies

ρ′+ (u, v) + ρ′− (u, v) = 0,

which by properties of ρ′± implies the inequalities ρ′− (u, v) ≤ ρ′+ (u, v) = −ρ′− (u, v)

and

− ρ′+ (u, v) = ρ′− (u, v) ≤ ρ′+ (u, v),

whence, ρ′− (u, v) ≤ 0 ≤ ρ′+ (u, v), i.e., u ⊥B v. By again applying Theorem 5.2.7 we complete the proof. 5.3

Functional equation of the bisectrix transform

Let (X1 , k · k1 ) and (X2 , k · k2 ) be real normed spaces such that dim X1 ≥ 2 and dim X2 ≥ 2. For any couple of nonzero vectors x and y in X1 , we consider the bisectrix segment kyk1 kxk1 y+ x : 0≤λ≤1 . λ kxk1 + kyk1 kxk1 + kyk1 Our aim in this section is to study mappings f : X1 → X2 which transform some points in the bisectrix segment generated by any couple x, y in X1 into the corresponding points of the bisectrix segment in X2 generated by f (x) and f (y), i.e., we will deal with the equation: kyk1 kxk1 y+ x f λ kxk1 + kyk1 kxk1 + kyk1 kf (x)k2 kf (y)k2 f (y) + f (x) , (5.3.1) = g(λ) kf (x)k2 + kf (y)k2 kf (x)k2 + kf (y)k2


145

where for λ0 , k0 given in (0,1) the function g : {0, λ0 , 1} → {0, k0 , 1} is such that g(0) = 0, g(λ0 ) = k0 and g(1) = 1. First we prove that the above equation implies the positive homogeneity of f . Lemma 5.3.1 Given λ0 , k0 ∈ (0, 1), let g be such that g(0) = 0, g(λ0 ) = k0 and g(1) = 1, and let f : X1 → X2 be a mapping, continuous at 0, such that f (x) = 0 if and only if x = 0. If (5.3.1) is satisfied, then necessarily f (αx) = αf (x) for all x in X1 and all positive reals α > 0, and moreover λ0 = k0 . Proof. For λ = 1, equation (5.3.1) gives the following equality kyk1 kxk1 y+ x f kxk1 + kyk1 kxk1 + kyk1 kf (x)k2 kf (y)k2 = f (y) + f (x). kf (x)k2 + kf (y)k2 kf (x)k2 + kf (y)k2

(5.3.2)

The substitution y := αx into (5.3.2), with x 6= 0 and real α, yields |α| + α kf (αx)k2 f (x) + kf (x)k2 f (αx) f x = . (5.3.3) |α| + 1 kf (x)k2 + kf (αx)k2 Since for any α < 0, |α| + α = 0 and f (0) = 0, by (5.3.3) we have for α < 0, x 6= 0, f (αx) = −

kf (αx)k2 f (x). kf (x)k2

(5.3.4)

Thus for any α > 0, by writing f (αx) = f ((−α)(−x)) and using condition (5.3.4) twice, we immediately conclude f (αx) =

kf (αx)k2 f (x). kf (x)k2

(5.3.5)

Fixing u 6= 0, for any real α, β > 0 the substitutions x := αu and y := βu into (5.3.2) give, in view of the last equality (5.3.5): 2kf (αu)k2 kf (βu)k2 2αβ 1 u = f (u), f · α+β kf (αu)k2 + kf (βu)k2 kf (u)k2 and taking norms

f 2αβ u = 2kf (αu)k2 kf (βu)k2 .

α + β 2 kf (αu)k2 + kf (βu)k2

(5.3.6)

146


Since u is fixed, let us introduce the function h : (0, ∞) → (0, ∞), h(t) := kf (tu)k2 . Then, by (5.3.6), h is a morphism with respect to the harmonic mean, i.e., 2h(α)h(β) 2αβ = . h α+β h(α) + h(β) Then the function J(t) := 1/h(1/t) satisfies the classical Jensen equation (see [Aczél (1966)]) and it is bounded below by 0, so there exist two positive constants a, b such that J(t) = at + b, i.e., h(t) = t/(a + bt). Now we let u vary again, so a and b will be functions of u and kf (tu)k2 = t/(a(u) + b(u)t). Since for t = 1 we have a(u) + b(u) = 1/kf (u)k2, we obtain kf (tu)k2 =

tkf (u)k2 , 1 + b(u)(t − 1)kf (u)k2

(5.3.7)

for all u 6= 0 and t > 0. Bearing in mind (5.3.5), we also have f (tu) =

t f (u). 1 + b(u)(t − 1)kf (u)k2

(5.3.8)

By (5.3.7) we have for all u 6= 0 and t > 0, t 6= 1, t 1 1 − . b(u) = t − 1 kf (tu)k2 kf (u)k2 Using this expression for computing b(tu) and with the help of (5.3.7), it is easy to verify that b must be homogeneous of degree zero: b(tu) = b(u),

(5.3.9)

for all t > 0 and u 6= 0. If in (5.3.1) we substitute x = y = u, we obtain for λ = λ0 f (λ0 u) = k0 f (u),

(5.3.10)

and combining (5.3.8) and (5.3.10): k0 =

λ0 , 1 + b(u)(λ0 − 1)kf (u)k2

and therefore, for any u, v 6= 0 we must have b(u)kf (u)k2 = b(v)kf (v)k2 ,

(5.3.11)


147

i.e., b(u)kf (u)k2 = C for some constant C. We claim that C must be zero. In fact, in view of (5.3.9), for all t > 0 we have b(u)kf (tu)k2 = b(tu)kf (tu)k2 = C, and using (5.3.7) tb(u)kf (u)k2 Ct = = C, 1 + C(t − 1) 1 + b(u)(t − 1)kf (u)k2 whence either C = 0 or C = 1. The case C = 1, i.e., b(u)kf (tu)k2 = 1 is inconsistent with the continuity of f at 0. Thus C = 0, and by (5.3.11) k0 = λ0 , and by (5.3.8), f (tu) = tf (u) for all t > 0 and any u in X1 . Now we can solve (5.3.1). Theorem 5.3.1 Let (X1 , k · k1 ), (X2 , k · k2 ) be real normed linear spaces with dim X1 ≥ 2 and dim X2 ≥ 2. Let, moreover, g : {0, λ0 , 1} → {0, k0 , 1} with λ0 , k0 ∈ (0, 1) be such that g(0) = 0, g(λ0 ) = k0 and g(1) = 1. Assume that f : X1 → X2 is a mapping continuous at 0 and such that f (x) = 0 if and only if x = 0. Then f satisfies (5.3.1) if and only if λ0 = k0 and f satisfies the following conditions: (i) f (tu) = tf (u) for all t > 0 and any u in X1 ; (ii) f (u + v) = f (u) + f (v) whenever kuk1 = kvk1 ; (iii) kf (u)k2 = Ckuk1 for some constant C > 0 and for any u in X1 . Proof. Sufficiency is a straightforward verification. So, let us assume that f satisfies (5.3.1). By Lemma 5.3.1 we already know that λ0 = k0 and (i) holds. Thus by (5.3.1), for all nonzero u, v such that kuk1 = kvk1 , we have 2kf (v)k2 f (u) + 2kf (u)k2 f (v) u+v = . (5.3.12) f (u + v) = 2f 2 kf (u)k2 + kf (v)k2 On the other hand, by (5.3.1) and (i) we also have kxk1 + kyk1 (kf (y)k2 f (x) + kf (x)k2 f (y)) . kf (x)k2 + kf (y)k2 (5.3.13) Thus by (5.3.12), (5.3.13) and (i), we have the following chain of equalities f (kyk1x + kxk1 y) =

kxk1 + kyk1 (kf (y)k2 f (x) + kf (x)k2 f (y)) kxk1 kyk1 (kf (x)k2 + kf (y)k2 )

148


y x 1 f (kyk1 x + kxk1 y) = f + = kxk1 kyk1 kxk1 kyk1

y y x x 2 f kxk f + f 2

f kyk1 kyk1 kxk1 1

2

2 =

y x

f kxk1 + f kyk1 2

2

2kf (x)k2 f (y) + 2kf (y)k2 f (x) = , kyk1 kf (x)k2 + kxk1 kf (y)k2

whence

2 kxk1 + kyk1 − kxk1 kyk1 (kf (x)k2 + kf (y)k2 ) kyk1 kf (x)k2 + kxk1 kf (y)k2

× {kf (y)k2 f (x) + kf (x)k2 f (y)} = 0, i.e., by (5.3.13) for all nonzero x and y in X, we obtain 2 kxk1 + kyk1 − kxk1 kyk1(kf (x)k2 + kf (y)k2 ) kyk1 kf (x)k2 + kxk1 kf (y)k2 ×f (kyk1 x + kxk1 y) = 0.

(5.3.14)

Thus for u and v linearly independent, the substitution x := u and 2kuk1 v into (5.3.14) after straightforward calculations yields y := kvk1 kf (v)k2 kf (u)k2 = . kuk1 kvk1 From this it is immediate to infer that there exists a constant C such that kf (u)k2 = Ckuk1

(5.3.15)

for all u (including the obvious case u = 0). Next, from (5.3.15), (5.3.12) and (i), it follows that whenever kuk1 = kvk1 , 2kf (v)k2 f (u) + 2kf (u)k2 f (v) kf (u)k2 + kf (v)k2 2Ckvk1 f (u) + 2Ckuk1f (v) = = f (u) + f (v) Ckuk1 + Ckvk1

f (u + v) =

and the theorem is proved.


149

In the case where both X1 and X2 are inner product spaces, i.e., k · k1 and k · k2 come from inner products, the above conditions (i), (ii) and (iii) for f are equivalent to the usual full linearity of f and the fact that f is a similitude, i.e., we have the following: Corollary 5.3.1 Under the assumptions of Theorem 5.3.1 , if moreover (X1 , k · k1 ) and (X2 , k · k2 ) are inner product spaces, then if f is a solution of (5.3.1), f is a linear similitude and λ0 = k0 . Proof. We just need to pay attention to the fact that the inner product structure and the condition (iii) make it possible to deduce the full linearity of f by means of (i) and (ii). To this end, for kxk1 = kyk1 from (ii) and (iii) we get the existence of a positive constant C such that C 2 kxk21 + C 2 kyk21 + 2C 2 hx, yi1 = C 2 kx + yk21 = kf (x + y)k22 = kf (x) + f (y)k22 = kf (x)k22 + kf (y)k22 + 2hf (x), f (y)i2 = C 2 kxk21 + C 2 kyk21 + 2hf (x), f (y)i2 , whence hf (x), f (y)i2 = C 2 hx, yi1 . Thus, even if kxk1 6= kyk1 , we have by the positive homogeneity of f : E D y x , kykf hf (x), f (y)i2 = kxkf kxk kyk 2 E D x y = kxkkykC 2 , = C 2 hx, yi1 . kxk kyk 1 Thus, C1 f is an orthogonal transformation from X1 into X2 , i.e., f is a linear similitude. 5.4

Generalized bisectrices in strictly convex real normed spaces

If (X, k · k) is a real normed linear space, we can consider the angle A+ (x, y) between two vectors x and y in X\{0} defined by A+ (x, y) = arccos

ρ′+ (x, y) . kxkkyk

150


Then given linearly independent x, y in X, one may look for a bisectrix vector b+ (x, y) = αx + (1 − α)y contained in the triangle with sides x, y and x − y and such that the angle between x and b+ (x, y) is equal to the angle between y and b+ (x, y). This yields the condition A+ (x, b+ (x, y)) = A+ (y, b+ (x, y)), i.e., ρ′ (y, b+ (x, y)) ρ′+ (x, b+ (x, y)) = + . kxkkb+ (x, y)k kykkb+(x, y)k

(5.4.1)

When (X, k·k) is strictly convex, the substitution of b+ (x, y) = αx+(1−α)y into (5.4.1) yields a formula for generalized bisectrices b+ (x, y) =

kyk kxk x+ y, kyk + kxkk(y, x) kxk + kykk(x, y)

where k(x, y) =

kxkkyk − ρ′+ (x, y) . kxkkyk − ρ′+ (y, x)

This definition is only possible if ρ′+ (x, y) and ρ′+ (y, x) are different from kxkkyk for all linearly independent x, y ∈ X, and this is guaranteed if we assume X strictly convex (see Theorem 2.1.1 (vii)). Analogously, we may define b− . Note that in an i.p.s. k(x, y) ≡ 1 for all linearly independent vectors x, y in X and in this case (cf. Section 5.1) b+ (x, y) =

kxk kyk x+ y = w(x, y). kxk + kyk kxk + kyk

The following result shows a new characterization of the i.p.s. structure [Alsina, Guijarro and Tom´ as (1997)] in terms of these generalized bisectrices b(x, y), by comparing their lengths with the lengths given in Theorem 5.1.1. Theorem 5.4.1 If (X, k · k) is a strictly convex space with dim X ≥ 2, then X is an i.p.s. if and only if for all linearly independent x, y in X kb(x, y)k = Proof.

p 2 kxkkyks(s − kx − yk). kxk + kyk

(5.4.2)

If (5.4.2) holds, then kb(x, y)k2 =

4

2 kxkkyks(s −

(kxk + kyk)

for all linearly independent x and y in X.

kx − yk)

(5.4.3)

151


Substitute y := tz with t > 0 into (5.4.3), divide by t2 and take limit when t tends to zero, then using the definition of b we obtain lim

t→0+

kb(x, tz)k2 t2

kzk kxkkzk − ρ′+ (x, z) x + kxk kxkkzk − ρ′+ (z, x) z 2 = . (5.4.4) kxk2 (kxkkzk − ρ′+ (z, x))2

On the other hand, lim

t→0+

4 1 kxk + tkzk + kx − tzk · kxktkzk · t2 (kxk + tkzk)2 2

kxk + tkzk − kx − tzk 2 kxkkzk (kxk + tkzk)2 − kx − tzk2 = lim+ t(kxk + tkzk)2 t→0 2kxkkzk tkzk2 + 2kxkkzk kx − tzk2 − kxk2 + = lim 2 −2t t→0+ (kxk + tkzk)2 ×

=2

kzk kxkkzk + ρ′− (x, z) . kxk

(5.4.5)

So, (5.3.4) and (5.3.5) yield 2

kzk kxkkzk + ρ′− (x, z) kxk

kzk(kxkkzk − ρ′+ (x, z))x + kxk(kxkkzk + ρ′+ (z, x))z 2 = , kxk2 (kxkkzk − ρ′+ (z, x))2

and therefore, after suitable computations we arrive at the equality

2

x z 1 − ρ′ ,

+ kxk kzk z x z x

. , = + 2 + 2ρ′−

z kxk kzk

1 − ρ′ , x kxk kzk +

Thus, if u :=

kzk kxk

x z , v := , then kxk kzk

2

1 − ρ′+ (u, v)

′

2 + 2ρ− (u, v) = u + v

1 − ρ′+ (v, u)

(5.4.6)

for all u, v independent vectors in SX . Note that, since X is strictly convex, |ρ′+ (u, v)| and |ρ′+ (v, u)| cannot have the value 1 = kuk · kvk and both sides of (5.4.6) are different from 0.

152


Changing the roles of u and v, we obtain

2+

2ρ′− (v, u)

2

1 − ρ′+ (v, u)

v + u = ′

1 − ρ+ (u, v)

(5.4.7)

and if we divide (5.3.6) by (5.3.7), we have 1 + ρ′− (u, v) = 1 + ρ′− (v, u)

1 − ρ′+ (u, v) 1 − ρ′+ (v, u)

2

but replacing v by −v, 1 − ρ′+ (u, v) = 1 − ρ′+ (v, u)

1 + ρ′− (u, v) 1 + ρ′− (v, u)

2

>0

and consequently,

1 − ρ′+ (u, v) 1 − ρ′+ (v, u)

2

=

s

1 − ρ′+ (u, v) . 1 − ρ′+ (v, u)

Thus, 1 − ρ′+ (u, v) = 1, 1 − ρ′+ (v, u) and ρ′+ is a symmetric function on SX × SX , and by Theorem 2.1.1(x) X is an i.p.s. The converse follows immediately. The following characterization is a generalization of a classical result (see [Coxeter (1969)], p. 18 and p. 23, Exercise 1, and see [Puig-Adam (1986)] vol. 1, p. 93) involving the bisectrices of the orthic triangle of a triangle △ (i.e., the triangle determined by the feet of the heights of △). We say that the triangle determined by x and y (i.e., the triangle with vertices 0, x, y) is obtuse-angled (in the vertex x or y) (see Figure 5.4.1) if and only if kyk2 − kx − yk2 ≥ ρ′+ (x, y) or ρ′+ (x, y) ≥ kyk2 . For the following results, we consider the height function h = h1 defined in Chapter 3.

153


x−

y

0

y

x Figure 5.4.1

Theorem 5.4.2 Let (X, k·k) be a strictly convex space such that the norm k · k is smooth and dim X ≥ 2. Then X is an i.p.s. if and only if for every obtuse-angled triangle at least one side that determines the obtuse-angle and one bisectrix of the orthic triangle are contained in the same line. Proof. In [Puig-Adam (1986)] it is shown that a norm in an i.p.s. has the above property. Conversely, consider the triangle determined by x and y, where x and y are linearly independent in X and kyk2 − kx − yk2 ≥ ρ′+ (x, y) or ρ′+ (x, y) ≥ kyk2 . By hypothesis, if b := b+ = b− , then b (h(y − x, −x) − h(x, y) + x, h(−y, x − y) − h(x, y) + y) = λ(y − x) (5.4.8) for some λ in (−1, 0). If linearly independent x, y in X satisfy ρ′+ (x, y) > 0, then for all t in ρ′ (x, y) i the interval 0, + 2 , we have ρ′+ (x, ty) > ktyk2 , and by (5.4.8) kyk b(h(ty − x, −x) − h(x, ty) + x, h(−ty, x − ty) − h(x, ty) + ty) = λ(t)(ty − x). Using the definitions of h and b, and the linear independence of x and y,

154


after simplifying we obtain the system 0 = kh(ty − x, −x) − h(x, ty) + xk kxk2 − kx − tyk2 + ktyk2 − ρ′+ (ty, x) ktyk2 − ρ′+ (x, ty) − + λ(t) × kxk2 kx − tyk2 ktyk2 − ρ′+ (x, ty) , +kh(−ty, x − ty) − h(x, ty) + tyk λ(t) − kx − tyk2 ktyk2 − ρ′+ (x, ty) 0 = kh(ty − x, −x) − h(x, ty) + xk −1 + − λ(t) kx − tyk2 +kh(−ty, x − ty) − h(x, ty) + tyk kxk2 + ρ′+ (ty − x, x) ktyk2 − ρ′+ (x, ty) −1+ − λ(t) . × ktyk2 kx − tyk2 Adding these two equalities yields kh(ty − x, −x) − h(x, ty) + xkktyk2 (kx − tyk2 + ρ′+ (ty, x) − ktyk2 ) = kh(−ty, x − ty) − h(x, ty) + tykkxk2 kxk2 − ρ′+ (x − ty, x) − ktyk2

and dividing by t2 , using the definition of h and ρ′± , their properties, grouping in a suitable way and taking limit when t tends to zero, we get

y + x (−3ρ′+ (x, y) + ρ′+ (y, x)) = kyk, provided ρ′+ (x, y) > 0.

2 kxk

But, if ρ′+ (x, y) < 0, then ρ′+ (−x, y) > 0, and applying the last property we have

y + x (−3ρ′+ (x, y) + ρ′+ (y, x)) = kyk provided ρ′+ (x, y) 6= 0.

2 kxk (5.4.9) −ρ′+ (x, y) , Moreover, for all x, y in X linearly independent and λ in R, λ 6= kxk2 ′ ρ+ (x, y + λx) 6= 0 and by (5.4.9)

y + λx + x −3ρ′+ (x, y + λx) + ρ′+ (y + λx, x) = ky + λxk.

kxk2 Taking limit when λ tends to zero, we get

y + x −3ρ′+ (x, y) + ρ′+ (y, x) = kyk

kxk2

and (5.4.9) holds for all linearly independent x, y in X.


155

Furthermore, if we consider linearly independent vectors u, v in SX , and u−v , we obtain in (5.4.9) we change y for −u and x for ku − vk

′ ′

−u + u − v 3ρ+ (u − v, u) − ρ+ (u, u − v) (5.4.10)

= 1. ku − vk2 Then, using the fact that the real strictly convex function f (t) = k − u + t(u − v)k,

t∈R

is not constantbecause u, v are linearly independent, and f (0) = f (1) = 1, 3ρ′+ (u − v, u) − ρ′+ (u, u − v) = f (0) = f (1), we conby (5.4.10) f ku − vk2 clude that for all unit vectors u, v we have 3ρ′+ (u − v, u) − 1 + ρ′+ (u, v) ∈ {0, ku − vk2 }. By changing the roles of u and v, we get 3ρ′+ (v − u, v) − 1 + ρ′+ (v, u) ∈ {0, ku − vk2 }. By adding the last two equalities and using ρ′+ (v − u, v) = kv − uk2 − ρ′+ (u − v, u), we obtain 2 − ρ′+ (u, v) − ρ′+ (v, u) = kku − vk2 , where k ∈ {1, 2, 3}, and therefore 2 − ρ′+ (u, v) − ρ′+ (v, u) ≥ ku − vk2. By changing u for −u, we have 2 + ρ′+ (u, v) + ρ′+ (v, u) ≥ ku + vk2, and ku + vk2 + ku − vk2 ≤ 4 for all linearly independent unit vectors u, v, and from Theorem 1.4.3, X is an i.p.s. Finally, we quote some additional results concerning bisectrices b+ [Alsina, Guijarro and Tom´ as (1996)], which can be proved by use of similar methods to those presented in Theorems 3.2.2 and 3.2.3. Theorem 5.4.3 Let (X, k · k) be a strictly convex space with dim X ≥ 2 and let H : R3 → R be a function of class C 2 in an open set containing the set of points X = {P = (kxk, 0, kxk) : x ∈ X}. Assume that kb(x, y)k2 = H (kxk, kyk, kx − yk) holds for all x, y in X. Then X is an i.p.s. if and only if for all P ∈ X (i) H(P ) = 0; (ii) D2 H(P ) = D3 H(P ) = 0; (iii) D23 H(P ) = −2, D23 H(P ) = 4 and D33 H(P ) = 0. Theorem 5.4.4 Let (X, k · k) be a strictly convex space with dim X ≥ 2. X is an i.p.s. if and only if kx − yk 2 for all x, y ∈ X\{0}, x 6= y, kb(x, y)k = G (kxkkyk) H kxk + kyk

156


where G, H are two functions in C 1 (R+ ) satisfying G(0) = H(1) = 0 and G′ (0)H ′ (1) = −2. 5.5

Incenters and generalized bisectrices

While the bisectrices meet at a point, the situation with the generalized bisectrices b+ is quite different in nature [Tomás (2004)]. Let ∆xyz denote a triangle with sides x − y, x − z, y − z. For every such triangle we can consider the three right-bisectrices x− Rb+(x− z, x− y),

y − Rb+ (y − x, y − z),

z − Rb+(z − y, z − x) (5.5.1)

and we can give the following: Definition 5.5.1 The triangle ∆xyz has a right-incenter I + if and only if the three right bisectrices (5.5.1) intersect at I + . Analogously, using b− we define the left-bisectrices and the left-incenter I − . In strictly convex spaces the existence of the point I + for the triangles of a certain class implies that ρ′+ = ρ′− . First, we will prove that the right-incenter I + of every triangle ∆xy(αx + y) with a fixed α 6= 0 exists only in smooth spaces. Theorem 5.5.1 Let (X, k · k) be a strictly convex space with dim X ≥ 2 and let α 6= 0 be a fixed real number. If there exists the right-incenter of ∆xy(αx + y) for all linearly independent x, y ∈ X, then the norm k · k is smooth. Proof. If we consider ∆xyz with z = αx + y, it is a straightforward computation to show that the three right-bisectrices (5.4.1) meet at a point if and only if 1 1 1 −1 −1 −1 , 1= A B C where A=

kx − yk , kx − yk + k(α − 1)x + ykk+ (x − y, (1 − α)x − y)

B=

kαxk , kαxk + ky − xkk+ (−αx, y − x)


157

and C=

k(α − 1)x + yk . k(α − 1)x + yk + kαxkk+ ((α − 1)x + y, αx)

Then, for all x, y in X 1 = k+ (−αx, y − x)k+ (x − y, (1 − α)x − y)k+ ((α − 1)x + y, αx) kx − ykk(1 − α)x − yk − ρ′+ (x − y, (1 − α)x − y) = kx − ykk(1 − α)x − yk − ρ′+ ((1 − α)x − y, x − y) kαxkky − xk − ρ′+ (−αx, y − x) × kαxkky − xk − ρ′+ (y − x, −αx) k(α − 1)x + ykkαxk − ρ′+ ((α − 1)x + y, αx) × . (5.5.2) k(α − 1)x + ykkαxk − ρ′+ (αx, (α − 1)x + y) By the substitution u =: x − y, v =: αx in (5.5.2), we have 1=

kukku − vk − ρ′+ (u, u − v) kukku − vk − ρ′+ (u − v, u) kvkkuk − ρ′+ (−v, −u) kv − ukkvk − ρ′+ (v − u, v) × · . (5.5.3) kvkkuk − ρ′+ (−u, −v) kv − ukkvk − ρ′+ (v, v − u)

If we replace v by λv (λ > 0), divide by λ and take the limit when λ tends to zero, using Proposition 2.1.6 and the properties of ρ′± , we get kvkkuk − ρ′+ (v, u) kukku − λvk − ρ′+ (u − λv, u) = λ kvkkuk − ρ′+ (u, v) λ→0+ kukkvk + ρ′− (u, v) kukku − λvk − ρ′+ (u, u − λv) . (5.5.4) × · lim ′ + kukkvk + ρ− (v, u) λ→0 λ lim

However, using the definition of ρ′± and Theorem 2.1.1 kukku − λvk − ρ′+ (u − λv, u) λ λ→0 ku − λvk(kuk − ku − λvk) − ρ′+ (u − λv, λv) = lim+ λ λ→0 ku − λvk kuk2 − ku − λvk2 = lim − ρ′+ (u − λv, v) kuk + ku − λvk λ λ→0+ = ρ′− (u, v) − ρ′+ (u, v), lim+

and, furthermore, lim+

λ→0

kukku − λvk − ρ′+ (u, u − λv) = 0. λ

158


Then, on substituting the above result in (5.5.4), we obtain ρ′+ (u, v) = ρ′− (u, v) for all u, v in X. Observe that the above theorem implies the corresponding theorem for triangles ∆xyz, where z is in the plane determined by x and y. For α = 1, the triangle ∆xy(x + y) is congruent (in the sense that it has the same angles and the sides have the same length) to the triangle determined by the vectors x and y. Thus, in a strictly convex space, the intersection of every pair of the three right-bisectrices of ∆xyz gives us (if they exist) the three points A (B(y − x) + (1 − B)(y − z)), 1 − B + AB C =x− (A(x − z) + (1 − A)(x − y)), 1 − A + AC B (C(z − y) + (1 − C)(z − x)), =z− 1 − C + CB

I1+ = y − I2+ I3+ where

kx − yk , kx − yk + kx − zkk+ (x − y, x − z) ky − zk , B= ky − zk + ky − xkk+ (y − z, y − x) kz − xk C= . kz − xk + kz − ykk+ (z − x, z − y) A=

(5.5.5)

Analogously, if we consider the left-bisectrices we obtain the points I1− , I2− and I3− . Definition 5.5.2 The points I1+ , I2+ , I3+ (resp. I1− , I2− , I3− ) are called the right-incenter points (resp. left-incenter points) of ∆xyz, and ∆I1+ I2+ I3+ (resp. ∆I1− I2− I3− ) is called the right-incentric triangle (resp. left-incentric triangle) of ∆xyz. Observe that I1+ , I2+ , I3+ are affinely independent or I1+ = I2+ = I3+ , because b± (x, y) = (1 − λ) x + λy with 0 < λ < 1. We now prove that the equality between right- and left-incenter points implies the smoothness of the norm (i.e., ρ′+ = ρ′− ). Proposition 5.5.1 Let (X, k · k) be a strictly convex space with dim X ≥ 2. Then the norm k · k is smooth if and only if I1+ = I1− .


159

Proof. One implication is obvious. Conversely, if I1+ = I1− , it is a straightforward computation to prove that this is equivalent to k+ (u, v) = k− (u, v) for all u, v in X linearly independent, i.e., kukkvk ρ′− (v, u) + ρ′+ (u, v) − ρ′− (u, v) − ρ′+ (v, u) = ρ′+ (u, v)ρ′− (v, u) − ρ′+ (v, u)ρ′− (u, v), and replacing u by −u, kukkvk −ρ′+ (v, u) − ρ′− (u, v) + ρ′+ (u, v) + ρ′− (v, u) = ρ′− (u, v)ρ′+ (v, u) − ρ′− (v, u)ρ′+ (u, v). On combining these equations, we obtain ρ′− (u, v)ρ′+ (v, u) = ρ′− (v, u)ρ′+ (u, v), and ρ′+ (u, v) − ρ′− (u, v) = ρ′+ (v, u) − ρ′− (v, u). From this it is very easy to prove that for all u, v in X, ρ′− (v, u) = ρ′− (u, v)

or

ρ′− (v, u) = ρ′+ (v, u).

(5.5.6)

For all λ > 0 if we change u by v + λu in (5.5.6), we obtain kvk2 + λρ′− (v, u) = ρ′− (v + λu, v)

or

ρ′− (v, u) = ρ′+ (v, u).

If we had ρ′− (v, u) 6= ρ′+ (v, u), then for all λ > 0 kvk2 + λρ′− (v, u) = kv + λuk2 − ρ′+ (v + λu, λu), kv + λuk2 − kvk2 = ρ′− (v, u) + ρ′+ (v + λu, u) and taking limit when λ λ tends to zero, (see Proposition 2.1.6) we would obtain ρ′+ (v, u) = ρ′− (v, u), which is a contradiction. So ρ′− = ρ′+ and the norm is smooth. i.e.,

Next, we consider some classical properties concerning the incenter and the incentric triangle of a given triangle and use these properties to obtain new characterizations of inner product spaces. Theorem 5.5.2 Let (X, k · k) be a strictly convex space with dim X ≥ 2. Then X is an i.p.s. if and only if for all linearly independent x, y in X and

160


for all z in the plane determined by x and y, the right incenter point I1+ of the triangle ∆xyz is the point ky − zkx + kx − zky + kx − ykz . kx − zk + kx − yk + ky − zk Proof. In an i.p.s. it is well known that the incenter of ∆xyz is given by ky − zkx + kx − zky + kx − ykz , whenever z is in the plane determined by kx − zk + kx − yk + ky − zk x and y. Conversely, we have for all x, y linearly independent and z ∈ lin {x, y} such that x, y, z are not collinear A (B(y − x) + (1 − B)(y − z)) 1 − B + AB ky − zkx + kx − zky + kx − ykz , = kx − zk + kx − yk + ky − zk

I1+ = y −

where A and B are defined in (5.5.5). By the linear independence of x, y, if z = z1 x + z2 y, where z1 and z2 are in R, we obtain the system ky − zk + kx − ykz1 A (B + (1 − B)z1 ) = 1 − B + AB kx − yk + ky − zk + kx − zk A kx − zk + kx − ykz2 1− . (B + (1 − B)(1 − z2 )) = 1 − B + AB kx − yk + ky − zk + kx − zk Now, it is a straightforward computation to obtain the following equation: B(kx − yk + ky − zk) = ky − zk

if

z1 + z2 6= 1

and k+ (y − z, y − x) = 1, i.e., ρ′+ (y − z, y − x) = ρ′+ (y − x, y − z). Or, if u := z − x and v := y − z we have ρ′+ (v, u+v) = ρ′+ (u+v, v)

for all linearly independent u, v in X. (5.5.7)

Actually, it is easy to see that (5.5.7) is valid for all u, v in X. Then ρ′+ (v, u) + kvk2 = ku + vk2 − ρ′− (u + v, u) and replacing u by λu (λ > 0) and grouping in a suitable way, kv + λuk2 − kvk2 = ρ′+ (v, λu) + ρ′− (λu + v, λu).

(5.5.8)

161


Dividing by 2λ and taking the limit when λ tends to zero using Corollary 2.1.1, we obtain ρ′− (v, u) = ρ′+ (v, u) for all u, v in X. By replacing u, v by v, u in (5.5.8), using ρ′− = ρ′+ , (5.5.7) and properties of ρ′+ , we get kuk2 + ρ′+ (u, v) = ku + vk2 − kvk2 − ρ′+ (v, u). Then ku + vk2 = kuk2 + kvk2 + ρ′+ (u, v) + ρ′+ (v, u) and replacing u by −u 2 ′ ′ 2 yields ku − vk2 = kuk2 + kvk − ρ− (u, v) − ρ− (v, u). This leads to ku + vk + 2 2 2 ku − vk = 2 kuk + kvk for all u, v in X, so X is an i.p.s.

Theorem 5.5.3 Let (X, k · k) be a strictly convex space with dim X ≥ 2. Then X is an i.p.s. if and only if, for all x, y in X linearly independent and for all z in the plane determined by x and y and such that kz −xk = ky −zk, the incenter points I1+ and I1− of the isosceles triangle ∆xyz coincide and are equal to ky − zkx + kx − zky + kx − ykz . kx − zk + kx − yk + ky − zk Proof. Analogously to the last theorem, if we assume for all x, y in X linearly independent, z in the plane determined by x and y with kz − xk = ky − zkx + kx − zky + kx − ykz , we obtain ky − zk that I1+ = kx − zk + kx − yk + ky − zk ρ′+ (v, u + v) = ρ′+ (u + v, v) for all u, v in X such that kuk = kvk, (5.5.9) and considering I1− instead of I1+ , we have ρ′− (v, u + v) = ρ′− (u + v, v) for all u, v in X such that kuk = kvk. (5.5.10) Then, if u, v are unit vectors, by (5.5.9) and (5.5.10) 1 + ρ′+ (v, u) = ku + vk2 − ρ′− (u + v, u) = ku + vk2 − ρ′− (u, u + v) = ku + vk2 − 1 − ρ′− (u, v), and ku + vk2 = 2 + ρ′+ (v, u) + ρ′− (u, v). By changing v for −v we obtain ku − vk2 = 2 − ρ′− (v, u) − ρ′+ (u, v)

(5.5.11)

162


and adding the above two equalities side by side, we get ku + vk2 + ku − vk2 = 4 + ρ′+ (v, u) + ρ′− (u, v) − ρ′− (v, u) − ρ′+ (u, v). Nevertheless, by symmetry in (5.5.11), ρ′+ (v, u) + ρ′− (u, v) = ρ′+ (u, v) + so ku + vk2 + ku − vk2 = 4 for all u, v in SX and (X, k · k) is an i.p.s. ρ′− (v, u),

Chapter 6

Areas of Triangles in Normed Spaces

In inner product spaces one finds various equivalent expressions for computing the area of a triangle. These expressions involve sides, heights, angles, etc., and allow us to formulate their natural generalizations in normed linear spaces, most of them using norm derivatives. In doing this, many new characterizations of i.p.s. are obtained.

6.1

Definition of four areas of triangles

If (X, k · k) is a real normed linear space with dim X ≥ 2 and x, y in X are linearly independent vectors, we consider the following expressions for the area of the triangle determined by x and y 1q kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x), 2 1 Ah (x, y) := kh(x, y)kkx − yk, 2 p AH (x, y) := s(s − kxk)(s − kyk)(s − kx − yk), A(x, y) :=

Ah′ (x, y) :=

p 1 kx − yk kxk2 − kx − h(x, y)k2 , 2

where h is the height vector (see Section 3.1) given by

h(x, y) := h1 (x, y) = y +

kyk2 − ρ′+ (x, y) (x − y) kx − yk2

163

164


for x 6= y (because x, y are independent) and s is the semiperimeter of the triangle with sides x, y, x − y s=

kxk + kyk + kx − yk . 2

If x and y in X are dependent vectors, then we define A(x, y) = Ah (x, y) = AH (x, y) = Ah′ (x, y) = 0. Example 6.1.1 If we consider R2 with the norm k(x, y)k+ = |x| + |y|, then, for x = (1, 0) and y = (a, b), a, b > 0, if a ≥ 1 we have A(x, y) = AH (x, y) = Ah′ (x, y) = 0 and Ah (x, y) = b, and if a < 1 we haveA(x, y) = b | − 4a2 + 5a + b − 1| + |1 − 4ab − a + 3b| , 0, Ah (x, y) = 2(1−a+b) p AH (x, y) = ab(1 + b)(1 − a), 1p (1 − a + b)2 − ((1 − a + b)2 − (a + b)(a + b − 1))2 . Ah′ (x, y) = 2

These expressions show how we can obtain different results for areas of triangles when dealing with norms which are not derivable from inner products. 6.2

Classical properties of the areas and characterizations of inner product spaces

We will provide new characterizations of inner product spaces, based on properties of the areas defined above. Theorem 6.2.1 Let (X, k · k) be a real normed linear space dim X ≥ 2. The following conditions are equivalent: (i) (ii) (iii) (iv)

X is an i.p.s.; For all x, y in X, A(x, y) = A(x, x + y); For all x, y in X, AH (x, y) = AH (x, x + y); For all x, y in X, Ah′ (x, y) = Ah′ (x, x + y).

Proof. To prove this theorem, it is only necessary to check that properties (ii), (iii) and (iv) imply (i). (ii) ⇒ (i): By hypothesis kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x)

= kxk2 kx + yk2 − ρ′+ (x + y, x)(kxk2 + ρ′+ (x, y))

(6.2.1)


165

for all x, y in X. By suitable grouping, changing y for ty, t > 0 and using the equality ρ′+ (x + ty, x) = ρ′+ (x + ty, x + ty − ty) = kx + tyk2 − tρ′− (x + ty, y), we get kxk2 (ρ′− (x + ty, y) − tkyk2 ) = ρ′+ (x, y)(ρ′+ (x + ty, x) − tρ′+ (y, x)), and taking limit when t (t > 0) tends to zero, using Proposition 2.1.6 and Corollary 2.1.1, we obtain ρ′− (x, y) = ρ′+ (x, y) for all x, y in X. Take two unit vectors u, v in X and let us distinguish three cases: Case 1. |ρ′+ (u, v)| 6= 1. Since ρ′+ = ρ′− , by the properties of ρ′+ and (6.2.1) with x := u, y := v, we have ρ′+ (u + v, v) = ρ′+ (u + v, u + v − u) = ku + vk2 − ρ′− (u + v, u) = ku + vk2 − ρ′+ (u + v, u) ku + vk2 − 1 + ρ′+ (u, v)ρ′+ (v, u) = ku + vk2 − 1 + ρ′+ (u, v) 1 ku + vk2 ρ′+ (u, v) + 1 − ρ′+ (u, v)ρ′+ (v, u) . = ′ 1 + ρ+ (u, v) Substitution of this expression into (6.2.1) with x := v, y := u yields 1 + ρ′+ (v, u) ku + vk2 ρ′+ (u, v) + 1 − ρ′+ (u, v)ρ′+ (v, u) ′ 1 + ρ+ (u, v) = 1 − ρ′+ (u, v)ρ′+ (v, u).

ku + vk2 −

A straightforward computation gives 1 − ρ′+ (u, v)ρ′+ (v, u) (2 + ρ′+ (u, v) + ρ′+ (v, u) − ku + vk2 ) = 0.

Since ρ′+ = ρ′− , then |ρ′+ (u, −v)| 6= 1, and changing v for −v in the above expression, we get (1 − ρ′+ (u, v)ρ′+ (v, u))(2 − ρ′+ (u, v) − ρ′+ (v, u) − ku − vk2 ) = 0. Conditions |ρ′+ (u, v)| 6= 1 and |ρ′+ (v, u)| ≤ 1 imply |ρ′+ (u, v)ρ′+ (v, u)k 6= 1, so the first factor cannot vanish, necessarily the sum of the last two equalities yields ku + vk2 + ku − vk2 = 4.

166


Case 2. |ρ′+ (v, u)| 6= 1. Starting with ρ′+ (u + v, u) and applying the same argument to that used in Case 1, we get ku + vk2 + ku − vk2 = 4. Case 3. |ρ′+ (u, v)| = |ρ′+ (v, u)| = 1. If ρ′+ (u, v) = 1 and ρ′+ (v, u) = −1, then (6.2.1) with x := v, y := u yields 2 = ku + vk2 − (1 + ρ′+ (v, u))ρ′+ (u + v, v) = ku + vk2 , and (6.2.1) with x := u, y := −v gives 2 = ku − vk2 . Whence, ku + vk2 + ku − vk2 = 4. The case ρ′+ (v, u) = 1 and ρ′+ (u, v) = −1 yields the same result. If ρ′+ (u, v) = ρ′+ (v, u) = 1, then (6.2.1) with x := u and y := −v yields ku−vk2 = (1−ρ′+ (u, v))ρ′+ (u−v, u) = 0, so u = v and ku+vk2 +ku−vk2 = 4. If ρ′+ (u, v) = ρ′+ (v, u) = −1, then (6.2.1) with x := u and y := v implies u = −v and ku + vk2 + ku − vk2 = 4. Whence, in any case ku + vk2 + ku − vk2 = 4 for all u, v in SX and, consequently, on account of Theorem 1.4.3, X is an i.p.s. (iii) ⇒ (i): Using the hypothesis, by a straightforward computation, we have for all x, y in X kxk2 + kyk2 − kx − yk2 and therefore

2

= kxk2 + kyk2 − kx + yk2

2

kx + yk = kx − yk or kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2 .

,

(6.2.2)

Let x and y be two vectors in X such that x ⊥P y, then kx + yk2 = kxk2 + kyk2 . By (6.2.2), we have two cases: In the first case, kx − yk = kx + yk and kx − yk2 = kxk2 + kyk2 , i.e., x ⊥P −y. In the second case, kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2 and kx − yk2 = kxk2 + kyk2 , i.e., x ⊥P −y. Then, if x ⊥P y, x ⊥P −y and by Theorem 1.5.3, X is an i.p.s.

167


(iv) ⇒ (i): By hypothesis p p kx − yk kxk2 − kx − h(x, y)k2 = kyk kxk2 − kx − h(x, x + y)k2 (6.2.3)

for all x, y in X with x − y 6= 0 6= y. Using (6.2.3) for unit vectors u, v with u−v 6= 0 6= v by a straightforward computation, we have 2ρ′+ (u, v) 3 − ku − vk2 − ku + vk2 = −2 − ku + vk4 + 4ku + vk2 + ku − vk4 − 3ku − vk2 ,

(6.2.4)

and changing the roles of u and v, we get 2ρ′+ (v, u) 3 − ku − vk2 − ku + vk2

= −2 − ku + vk4 + 4ku + vk2 + ku − vk4 − 3ku − vk2 .

(6.2.5)

If ku − vk2 + ku + vk2 6= 3, by (6.2.4) and (6.2.5), ρ′+ (u, v) = ρ′+ (v, u). If ku − vk2 + ku + vk2 = 3, by (6.2.4) −2 − ku + vk4 + 4ku + vk2 + ku − vk4 − 3ku − vk2 = 0, so the last two equalities yield ku − vk2 = 1 and ku + vk2 = 2.

(6.2.6)

Changing u for −u in (6.2.4), we also have ku − vk2 + ku + vk2 = 3, so −2 − ku − vk4 + 4ku − vk2 + ku + vk4 − 3ku + vk2 = 0 and by (6.2.6) we have a contradiction. Consequently, ρ′+ (u, v) = ρ′+ (v, u) for all u, v in SX and X is an i.p.s. Theorem 6.2.2 Let (X, k · k) be a real normed linear space, dim X ≥ 2. The following conditions are equivalent: (i) X is an i.p.s.; (ii) For all x, y, z in X, A(z, x + y)2 + A(z, x − y)2 = 2A(z, x)2 + 2A(z, y)2 ; (iii) For all x, y, z in X, AH (z, x + y)2 + AH (z, x − y)2 = 2AH (z, x)2 + 2AH (z, y)2 ; (iv) For all x, y, z in X, Ah′ (z, x + y)2 + Ah′ (z, x − y)2 = 2Ah′ (z, x)2 + 2Ah′ (z, y)2 .

168


Proof. It is obvious that (i) implies all other conditions. We will show that (ii) implies (i). Assume (ii). Let us fix z in X and let fz : X → R+ be the function defined by fz (x) = A(z, x). Then fz satisfies the equation fz2 (x + y) + fz2 (x − y) = 2fz2 (x) + 2fz2 (y).

(6.2.7)

Defining a function F : X 2 → R by F (x, y) :=

1 2 fz (x + y) − fz2 (x) − fz2 (y) , 2

we can prove using (6.2.7), that F is additive in both variables. Further, from the definition of F and from (6.2.7) we have F (x, x) = fz2 (x), and from the biadditivity fz2 (λx) = λ2 fz2 (x) for all rational λ. Now, for fixed x, y ∈ X define a function p : R → R by p(λ) := fz2 (λx + y) = F (λx + y, λx + y) = fz2 (λx) + 2F (λx, y) + fz2 (y). We have p(λ) ≥ 0, λ ∈ R and 0 ≤ p(λ) = λ2 fz2 (x) + 2λF (x, y) + fz2 (y) for λ ∈ Q. Consider a function q : R → R given by q(λ) := λ2 fz2 (x) + 2λF (x, y) + fz2 (y),

λ ∈ R.

Since both p and q are continuous and p|Q = q|Q , so p = q. Moreover, p(λ) ≥ 0 for all λ ∈ R, so 4F (x, y)2 − 4fz2 (x)fz2 (y) ≤ 0 and |F (x, y)| ≤ fz (x)fz (y). Finally, for fixed x, y in X, we have p p fz (x + y) = F (x + y, x + y) = fz2 (x) + 2F (x, y) + fz2 (y) ≤

p fz2 (x) + 2fz (x)fz (y) + fz2 (y) = fz (x) + fz (y).

So, we have the subadditivity of fz . If x := z, from the fact that fz (z) = 0, we have fz (z + y) ≤ fz (y). Moreover, fz (y) ≤ fz (z + y) + fz (−z) = fz (z + y). Therefore, fz (y) = fz (z + y), i.e., A(z, y) = A(z, y + z) for all y, z in X, and by Theorem 6.2.1 X is an i.p.s. The other equivalences between (iii), (iv) and (i) are easily established following the basic ideas used in the case (ii).


6.3

169

Equalities between different area functions

Let us now consider the consequences of equality between the differently defined areas, obtaining, as we will show, new characterizations of inner product spaces. Theorem 6.3.1 Let (X, k · k) be a real normed linear space dim X ≥ 2. The following conditions are equivalent: (i) (ii) (iii) (iv) (v) (vi)

X is an i.p.s.; For all x, y in X, For all x, y in X, For all x, y in X, For all x, y in X, For all x, y in X,

A(x, y) = Ah (x, y); A(x, y) = AH (x, y); Ah (x, y) = AH (x, y); Ah′ (x, y) = A(x, y); Ah′ (x, y) = AH (x, y).

Proof. It is obvious that (i) implies all the other conditions. So, in order to prove this theorem, it is necessary to check that any of the remaining properties implies (i). (ii) ⇒ (i): Let x, y in X be linearly independent. If we replace y by ty with t > 0 in A(x, y) = Ah (x, y), we obtain 2

2

kxk kyk −

ρ′+ (x, y)ρ′+ (y, x)

2

tkyk2 − ρ′+ (x, y) 2

(x − ty) = y +

kx − tyk , 2 kx − tyk

and if we take limit when t tends to zero, we get

2

ρ′+ (x, y)

kxk2 . x kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x) = y −

kxk2

(6.3.1)

Now, if we change y for x + y in (6.3.1), the properties of ρ′+ and (6.3.1) yield kxk2 kx + yk2 − ρ′+ (x, x + y)ρ′+ (x + y, x) = kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x), i.e., A(x, y) = A(x, x + y) for all x, y in X (also linearly dependent) and by Theorem 6.2.1, X is an i.p.s.

170


(iii) ⇒ (i): By substituting y by ty with t > 0 in A(x, y) = AH (x, y) we obtain, by a straightforward computation: kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x) kx − tyk2 − kxk2 tkyk2 + + kykkx − tyk = 2 2t tkyk2 2 kxk − kx − tyk2 + + kykkx − tyk . × − 2 2t

So taking limit when t tends to zero, we have, for all x, y in X, ρ′+ (x, y)ρ′+ (y, x) = ρ′− (x, y)2 , and, by symmetry, we can deduce that |ρ′− (x, y)| = |ρ′− (y, x)|. It can be easily obtained from (2.5) in [Amir (1986)], that X is an i.p.s. (iv) ⇒ (i): See Theorem 3.2.1. (v) ⇒ (i): By hypothesis, for all x, y in X, we have the following property kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x) = kx − yk2 kxk2 − kx − h(x, y)k2 , i.e.,

kxk2 kyk2 − ρ′+ (x, y)ρ′+ (y, x) = kxk2 kx − yk2 − (kx − yk2 − kyk2 + ρ′+ (x, y))2 . (6.3.2) Substituting tx with t > 0 in the place of x in (6.3.2), dividing by t2 and taking limit when t tends to zero, we obtain for all x, y in X the equality ρ′+ (x, y)ρ′+ (y, x) = (ρ′+ (x, y) − 2ρ′− (y, x))2 .

(6.3.3)

Exchanging the roles for x and y in the above equality we get ρ′+ (x, y)ρ′+ (y, x) = (ρ′+ (y, x) − 2ρ′− (x, y))2 ,

(6.3.4)

|ρ′+ (x, y) − 2ρ′− (y, x)| = |ρ′+ (y, x) − 2ρ′− (x, y)|.

(6.3.5)

so

Substituting −x in the place of x in (6.3.4) we obtain ρ′− (x, y)ρ′− (y, x) = (ρ′− (y, x) − 2ρ′+ (x, y))2 ,

(6.3.6)

and, by symmetry ρ′− (y, x)ρ′− (x, y) = (ρ′− (x, y) − 2ρ′+ (y, x))2 ,

(6.3.7)


171

so |ρ′− (y, x) − 2ρ′+ (x, y)| = |ρ′− (x, y) − 2ρ′+ (y, x)|.

(6.3.8)

For fixed x, y in X denote A := ρ′+ (x, y), B := ρ′+ (y, x), C := ρ′− (x, y), D := ρ′− (y, x) From (6.3.3)-(6.3.8) we have AB = (A − 2D)2 = (B − 2C)2

(6.3.9)

CD = (C − 2B)2 = (D − 2A)2

(6.3.10)

|A − 2D| = |B − 2C|

|C − 2B| = |D − 2A|. This leads to four cases: A − 2D = B − 2C (a) D − 2A = C − 2B this immediately yields A = B. (b)

A − 2D = B − 2C D − 2A = −C + 2B and from (6.3.9), we get 2C = 3A+5B 2 2 3A + 5B , AB = B − 2 whence 9A2 + 14AB + 9B 2 = 0, which has no real solutions, so this case cannot appear.

(c)

A − 2D = −B + 2C D − 2A = C − 2B , and by (6.3.10) from the system we have 2A = 3C+5D 2 2 3C + 5D CD = D − 2 9C 2 + 14CD + 9D2 = 0. Again we derive that such a case cannot appear.

172

(d)


A − 2D = −B + 2C D − 2A = −C + 2B this leads to A = −B, and because of (6.3.9) we have A = B = 0. We conclude that A = B, so for arbitrary x, y in X we have ρ′+ (x, y) = so the norm in X comes from an inner product.

ρ′+ (y, x),

(vi) ⇒ (i): If we substitute x for tx with t > 0 in A2h′ (x, y) = A2H (x, y), we obtain

t2 kxk2 ktx − yk2 − (ktx − yk2 − kyk2 + tρ′+ (x, y))2 1 (kyk + ktx − yk)2 − t2 kxk2 t2 kxk2 − (ktx − yk − kyk)2 . = 4 Dividing by t2 , and taking limit when t tends to zero, we have

2

2

kxk kyk −

(ρ′+ (x, y)

−

2ρ′− (y, x))2

ρ′− (y, x)2 2 , = kyk kxk − kyk2 2

i.e., ρ′+ (x, y) = aρ′− (y, x), where a ∈ {1, 3}. But if we replace x by y and y by −x, we get ρ′− (y, x) = bρ′+ (x, y), where b ∈ {1, 3} and, by the last two equations ρ′+ (x, y) = abρ′+ (x, y). Therefore ρ′+ (x, y) = 0 (and in this case ρ′− (y, x) = 0) or ab = 1, and a = b = 1, so ρ′+ (x, y) = ρ′− (y, x). Finally, in all cases ρ′+ (x, y) = ρ′− (y, x) ≤ ρ′+ (y, x) for all x, y in X and, by symmetry, ρ′+ (x, y) = ρ′+ (y, x) for all x, y in X, so X is an i.p.s.

6.4

The area orthogonality

In an inner product space, two vectors x and y are orthogonal if and only if the area of the triangle determined by x + λy and y is 12 kxkkyk for all λ in R (see Figure 6.4.1).


x

173

λy

+

y

x Figure 6.4.1 Based on this property we can define a new orthogonality relation. Definition 6.4.1 In a real normed linear space (X, k · k) with dim X ≥ 2 we consider the area orthogonality x ⊥a y if A(x + λy, y) = Example 6.4.1 if a = b = 0.

1 kxkkyk for all λ in R. 2

In R2 with the taxicab norm (1, 0) ⊥a (a, b) if and only

The following propositions contain some basic properties of this orthogonality relation. Proposition 6.4.1 Let (X, k · k) be a real normed space with dim X ≥ 2. The orthogonality ⊥a has the following properties for all x, y in X, and for all t in R: (i) (ii) (iii) (iv) (v) (vi)

x ⊥a 0 and 0 ⊥a x; x ⊥a x if and only if x = 0; x ⊥a tx if and only if tx = 0; If x ⊥a y, then ax ⊥a by, a, b in R such that ab ≥ 0; If x ⊥a y and x, y 6= 0, then x and y are linearly independent; If the norm derives from an inner product h·, ·i, the relation x ⊥a y is equivalent to the usual orthogonality hx, yi = 0.

Proposition 6.4.2 If (X, k · k) is a real normed linear space with dim X ≥ 2 and x, y are two vectors in X such that x ⊥a y, then

174


ρ′+ (x, y) = 0; A(x, y) = A(x + λy, y) for all λ in R; x ⊥B y; ρ′+ (y, x) = ρ′− (y, x).

(i) (ii) (iii) (iv)

Proof. (i) By the hypothesis kx + λyk2 kyk2 − ρ′+ (x + λy, y)ρ′+ (y, x + λy) = kxk2 kyk2

(6.4.1)

for all λ in R, in particular for λ = 0, we have ρ′+ (x, y)ρ′+ (y, x) = 0. If ρ′+ (y, x) 6= 0 then we obtain (i). But if ρ′+ (y, x) = 0, then by (6.4.1) for all λ > 0 λkyk2 ρ′+ (x + λy, y) kx + λyk2 − kxk2 kyk2 = , 2λ 2λ and taking limit when λ tends to zero, using Proposition 2.1.6 we get ρ′+ (x, y)kyk2 =

1 kyk2ρ′+ (x, y), 2

which means ρ′+ (x, y) = 0. (ii) It is immediate after putting λ := 0 in the definition of the orthogonality ⊥a . (iii) By (i) we have ρ′+ (x, y) = 0 ≥ ρ′− (x, y) and x ⊥B y (see Proposition 2.1.7). (iv) Changing λ for λ1 with λ > 0 in (6.4.1), dividing by λ2 and using the properties of ρ′± , we obtain ρ′− (λx + y, x)kyk2 − kλx + yk2 ρ′+ (y, x) + λρ′− (λx + y, x)ρ′+ (y, x) = λkxk2 kyk2 . By Proposition 2.1.6 and Corollary 2.1.1, tending with λ to zero, we get ρ′− (y, x)kyk2 − kyk2 ρ′+ (y, x) = 0, i.e., ρ′+ (y, x) = ρ′− (y, x).


175

As an immediate consequence of Proposition 6.4.2 and properties of the Birkhoff orthogonality, we have the following. Proposition 6.4.3 If (X, k · k) is a real normed linear space with dim X ≥ 2, then the following conditions are equivalent: (i) X is an inner product space; (ii) u − v ⊥a u+ v, for all u, v in SX ; kxk (iii) x − kxk kyk y ⊥a x + kyk y , for all x, y in X;

(iv) x ⊥J y implies x ⊥a y, for all x, y in X; (v) x ⊥P y implies x ⊥a y, for all x, y in X; (vi) If dim X ≥ 3, then x ⊥B y implies y ⊥a x, for all x, y in X.

Appendix A

Open Problems

In this monograph we have selected the main characterizations of i.p.s. obtained so far using norm derivatives. However, this topic is open to the consideration of further geometrical problems. Since we are convinced that functional equations and norm derivatives merit more attention in the future, we present here a short list of open problems. In this appendix, (X, k · k) always denotes a real normed space with dim X ≥ 2. 1. Characterize all functions f : X → X which are ρ′+ -orthogonal transformations, i.e., ρ′+ (f (x), f (y)) = ρ′+ (x, y), for all x, y in X. Analogously, characterize ρ′− -orthogonal transformations. 2. Find all functions f : X → X such that ρ′+ (x, f (y)) = ρ′+ (f (x), y), for all x, y in X. If this functional equation holds for all linear transformations f whose matrices are symmetric, does k · k derive from an inner product? 3. An integer n ≥ 1 being fixed, characterize all functions f : X → X such that for all x, y in X the following functional equation holds: f

kykn x + kxkn y kxkn + kykn

=

kf (y)kn f (x) + kf (x)kn f (y) . kf (x)kn + kf (y)kn

4. Characterize all functions f : X → X such that: f

ρ′+ (x, y) y kxk2

=

ρ′+ (f (x), f (y)) f (y), kf (x)k2

for all x, y in X. 177

178


5. Characterize all functions f : X → X which satisfy the ⊥-orthogonal additivity condition: f (x + y) = f (x) + f (y) for all x, y in X with x ⊥ y, where ⊥ denotes one of the orthogonal relations considered in this book, i.e., ⊥∈ {⊥h , ⊥M , ⊥w , ⊥a }. 6. Solve the functional equation ρ′+ (f (x), f (y)) ρ′+ (x, y) x = f (y) − f (x) f y− kxk2 kf (x)k2 where f : X → X is injective and f (x) 6= 0 whenever x 6= 0 (cf. Lemma 3.7.1 and (3.7.15)). 7. Characterize all functions f : X → X such that A(f (x), f (y)) = A(x, y), for all linearly independent vectors x, y, where A is one of the area functions defined in Section 6.1. 8. Is it possible to find a normed linear space X such that dim X = 2 whose norm does not derive from an inner product, but x ⊥B y imply y ⊥a x for all x, y in X? 9. Find characterizations of i.p.s. by formulating in terms of the functionals ρ′± , geometrical properties in real normed spaces which generalize classical properties of the Euclidean plane, such as Fagnano’s problem in a triangle, properties of cyclic quadrilaterals, Feuerbach’s theorem, Fermat point, etc.

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Index

ρ-orthogonal additivity, 51 ρ-orthogonal set, 40 ⊥ρ , 28 ⊥ρ , 26 ⊥a , 173 ⊥B , 11 ⊥h , 81 ⊥J , 11 ⊥M , 107 ⊥P , 12 ⊥w , 136 ρ′± (x, y), 15-26 ρ′± -orthogonal transformations, 30-35

equivalence of two norm derivatives, 35-38 equivalent functionals, 37 Euler line, 124, 125 examples of norm derivatives, 17, 18, 19 examples of normed spaces, 5, 6 extensions of the norm derivatives, 45-51 Fourier coefficients, 10 functional equation of the bisectrix transform, 144-149 functional equation of the height transform, 94-102 functional equation of the perpendicular bisector transform, 125-130

angle in a normed space, 149 area orthogonality, 173 areas of triangles in normed linear spaces, 163-175

Gateaux (weak) derivative, 24 generalized bisectrices, 149 Gram-Schmidt orthogonalization, 10

Banach space, 4 Birkhoff inequalities, 50 Birkhoff orthogonality, 11 Birkhoff’s orthogonal set, 39 bisectrices in normed spaces, 131 bisectrix orthogonality, 136

height functions and classical orthogonalities, 74-80 heights in normed linear spaces, 57-91 height orthogonality, 81 Hilbert space, 7

circumcenter points, 120 circumcenters, 117 circumscribed circumferences, 115 classical orthogonality relations 11-13

incenter points, 158 incenters, 156 incentric triangle, 158 187

188


inner product, 7 inner product space, 7 isometry, 4 isosceles trapezoid property, 91-94 James identity, 49 James orthogonality, 11 Jordan-von Neumann law, 8 Lagrange’s identity, 41 left bisectrices, 156 left incenter, 156 left incenter points, 158 left incentric triangle, 158 left side derivatives, 15 norm, 3 norm derivatives, 3, 15-26 norm derivatives and Lagrange’s identity in normed linear spaces, 41 normed linear spaces, 3 open problems, 177, 178 orthic triangle, 152 orthocenter points, 87-91 orthocenters, 85-89 orthogonal complement, 10 orthogonalities based on norm derivatives, 26-30 orthonormal basis, 10 parallelogram law, 8 Parseval’s relation, 10

perpendicular bisectors, 103-123 perpendicular bisector orthogonality, 107 perpendicular bisectors and classical orthogonalities, 111-115 polarization identity, 9 projection theorem, 10 projections, 38-41 properties of ρ′+ and ρ′− , 19 Pythagorean identity, 10, 47 Pythagorean orthogonality, 12, 13 radius of the circumscribed circumference, 115 radius of the inscribed circumference, 133 Riesz lemma, 10 right bisectrices, 156 right incenter, 156 right incenter points, 158 right incentric triangle, 158 right side derivatives, 15 smooth real normed linear space, 24 space C(K), 6 space c, 5 space c0 , 5 space l1 , 5 space lp , 6 space Lp , 6 space l∞ , 5 space Rn , 5 strictly convex normed linear spaces, 7

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