Introduction to Various Aspects of Degree Theory in Banach Spaces (Mathematical Surveys and Monographs 23)

MATHEMATICAL SURVEYS AND MONOGRAPHS NUMBER 23

Introduction to VARIOClS ASPECTS OF DEGREE THEORY IN BANACH SPACES

E. H. ROTHE

American Mathematical Society Providence, Rhode Island

1980 Mathematics Subject Qassification (1985 Revision). Primary 55M25; Secondary 46-02.

Library of Congress Cataloging-in-Publication Data Rothe, Erich H. Introduction to various aspects of degree theory in Banach spaces. (Mathematical surveys and monographs, ISSN 0076-5376; no. 23) Bibliography: p. Includes index. 1. Degree, Topological. 2. Banach spaces. 1. Title. 11. Series. 514' .2 86-8038 QA612.R68 1986 ISBN 0-8218-1522-9 (alk. paper)

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Contents Preface

v

���

1

CHAPTER 1. Function-Analytic Preliminaries CHAPTER 2. The Leray-Schauder Degree for Differentiable Maps

13 26

1. The degree for linear maps

26

2. The degree if Yo is a regular value for the map f

27

3. The one-dimensional case and the case of polynomial maps

31

4. The degree for a not necessarily regular value Yo

39

5. Notes

51

CHAPTER 3. The Leray-Schauder Degree for Not Necessarily

59

Differentiable Maps

1. An extension lemma

59

2. An application of the extension lemma

62

3. The degree theory for finite layer maps

64

4. Another application of the extension lemma

65

5. Two additional properties of the Leray-Schauder degree

67

6. Generalized L.-S. maps

69

7. Notes

71

CHAPTER 4. The Poincare-Bohl Theorem and Some of Its Applications

1. The Poincare-Boh! theorem and the winding number

76 76

2. The interpretation of degree and winding number as intersection numbers

83

3. Notes

84

CHAPTER 5. The Product Theorem and Some of Its Consequences

1. The product theorem

88 88

97

2. The invariance of the domain

3. The Jordan-Leray theorem

103

4. Notes

108 iii

CHAPTER 6. The Finite-Dimensional Case 1. Some elementary prerequisites

112 112

2. The degree for mappings between finite-dimensional spaces of the

same dimension

117

3. Simplicial mappings

122

4. On subdivisions

125

5. Simplicial approximations

126

6. Notes

128

CHAPTER 7. On Spheres

138

1. Elementary properties and orientation of spheres. Degree of

mappings between spheres 2. Properties of the degree

d(f, Sf, Sr)

3. The order of the image of a sphere with respect to a point

138 142 146

4. Two approximation lemmas

155

5. Notes

157

CHAPTER 8. Some Extension and Homotopy Theorems 1. An extension and a homotopy theorem

172 172

2. Two further extension theorems

182

3. Notes

186

CHAPTER 9. The Borsuk Theorem and Some of Its Consequences

192

1. The Borsuk theorem

192

2. Some consequences of the Borsuk theorem

195

3. Notes

198

APPENDIX A. The Linear Homotopy Theorem 1. Motivation for the theorem and the method of proof

199 199

2. Background material from spectral theory in a complex Banach

space Z 3. The complexification Z of a real Banach space E

200 204

4. On the index j of linear nonsingular L.-S. maps on complex and

real Banach spaces

208

5. Proof of the linear homotopy theorem

214

6. The multiplication theorem for the indices

228

APPENDIX B. Proof of the Sard-Smale Theorem 4.4 of Chapter 2

231

References

238

Index

241

Preface As its title indicates this book is not meant to be an encyclopedic presentation of the present state of degree theory. Likewise, the list of references is not meant to be a complete bibliography. The choice of subjects treated is determined partly by the fact that the book is written from the point of view of an analyst, partly by the desire to throw light on the theory from various angles, and partly the choice is subjective. A detailed description of the subjects treated is contained in §1O of the following introduction. At this point we mention, together with relevant references, some subject matter belonging to degree theory which is not treated in this book. (a) The Leray-Schauder degree theory in Banach spaces may be extended to linear convex topological spaces as already noticed by Leray in [35]. A complete self-contained treatment may be found in Nagumo's 1951 paper [43]. (b) Already, in his 1912 paper [9], Brouwer established degree theory for certain finite-dimensional spaces which are not linear. For later developments in this direction and for bibliographies, see, e.g., the books by Alexandroff-Hopf [2], by Hurewicz-Wallmann [29], and by Milnor [39]. In their 1970 paper [20] Elworthy and Tromba established a degree theory in certain infinite-dimensional manifolds. See also the systematic exposition by Borisovich-Zvyagin-Sapronov

[6]. (c) There are various degree theories for mappings which are not of the type treated by Leray and Schauder: a theory by Browder and Nussbaum on "intertwined" maps [11] and a theory of "A-proper" maps by Browder and Petryshyn [12] in which the degree is not integer-valued but a subset of the integers. See also [10]. Moreover, the method used in the Fenske paper [21] mentioned in §8 of the following introduction applies also to mappings which are "a-contractions." Various mathematicians considered degree theory for multiple-valued maps. A survey of these generalizations and further bibliographical data may be found in the book [37] by Lloyd. (d) Some important theorems based on degree theory (e.g., those contained in §§2-3 of Chapter 5 and parts of Chapters 8 and 9 of this book) and some generalizations thereof can be derived by the use of cohomology theory in infinitev

vi

PREFACE

dimensional spaces. See the paper by Geba and Granas [22], in particular Chapter IX, and the book by Granas [25], in particular Chapter 11. (e) The axiomatic treatment of degree theory by Amann and Weiss [3]. (f) The "coincidence degree theory" of Mawhin [38]. (g) For supplementary reading and further references we mention books by the following authors: Krasnoselskir [31], 1956, J. Cronin [14], 1964, J. T. Schwartz [51], 1965, K. Deimling [15], 1974, E. Zeidler [54], 1976, F. Browder [10], 1976, G. Eisenack-C. Fenske [19], 1978, N. G. Lloyd [37], 1978. We also mention the paper [52], 1980 by H. W. Siegbert which contains an interesting exposition of the history of finite-dimensional degree theory from Gauss via Kronecker to Brouwer. (h) Ever since it was established by Leray and Schauder [36], degree theory in Banach spaces has been an important tool for the treatment of boundary value problems (including periodicity problems) for ordinary and partial differential equations, for integral equations, and for eigenvalue and bifurcation problems. These applications to analysis are not treated in the present book, but the reader will find a number of them discussed in many of the books quoted in part (g) of this preface. See, e.g., Chapters III-VI of [31], Chapters II and IV of [14], and Chapter 9 of [37]. The Cronin book [14] discusses, in particular, papers up to 1962 using the "Cesari method." For the role of degree theory in the further development of this method, now usually referred to as the "Alternative Method," the reader may consult the bibliography on pp. 232-234 in Nonlinear Phenomena in Mathematical Science, ed. V. Lakshmikantham, Academic Press, 1982. We finish this preface with a word on the organization of this book. In addition to the introduction the book consists of nine chapters. Each chapter (except Chapter 1) is divided into sections and each section into subsections. (The notation "1.2" in Chapter 3 means subsection 1.2 in that chapter; "(1.2)" refers to a formula in that chapter. However §1.2 refers to §2 in Chapter 1.) Each chapter is followed by notes containing some of the proofs and historical remarks. Finally, there are two Appendixes A and B. My thanks go to my colleagues and friends Professors Lamberto Cesari, Charles Dolph, and R. Kannan for many encouraging conversations. My thanks also go to Mrs. Wanita Rasey who, with great patience, converted an often not easily decipherable handwritten script into a readable typescript. I am obliged to the editorial board of the Mathematical Surveys and Monographs for reviewing the typescript and to the editorial staff of the American Mathematical Society, in particular to Ms. Mary C. Lane, Lenore C. Stanoch, and Holly Pappas for their cooperation and labor in transforming the manuscript into a book.

Introduction 1. It is one of the most important tasks of analysis to compute the number

N(f, fl, Yo) of solutions x E fl of the equation I(x) = Yo,

(1)

where fl is a subset of some domain space E and where Yo is a point of a range space E 1 . If fl is the open interval (a, b) on an x-axis, if 1 is a polynomial in x, and if Yo is a point on a y-axis, then the above problem is completely solved by a classical theorem of Sturm (1835) which gives a procedure to calculate the number N(f, fl, yo). However, this number may not be continuous in dependence on Yo or I. If, e.g., fl is the interval (-1, +1) on the real axis Rand 1 is the map fl ---- R given by I(x) = y = x 2 , then for any positive c < 1

N(f,o.,yo)

={

2 if Yo = c, 1 ~fyo = 0, o If Yo = -c.

Thus our number N is not continuous in Yo at Yo = o. It is just as easy to see that N(f, fl, Yo) may not change continuously if 1 is changed continuously. For instance if fl is as above and I(x) = x 2 + a with a real, then N(f,fl,O) is not continuous at 0=0. 2. The basic difference between the root count N(f, fl, Yo) and the "degree" d(f, 0., Yo) is that the latter has the above-mentioned continuity properties. There are two good reasons for insisting on these: (i) without them a small error in the computation of Yo or 1 may lead to a wrong root count; (ii) if d(f, fl, Yo) is continuous in a certain class C of mappings I, then this number, being integer-valued, is constant for all 1 E C. It may happen that C contains an 10 for which the degree is particularly easy to compute. A case often occurring in the applications is that C contains the identity map 10 = I. 3. We return to the example y = I(x) = x 2 of §1 to indicate how a root count d(f, fl, Yo) may be obtained which is continuous in Yo at Yo = O. For Yo > 0 we 1

INTRODUCTION

2

note that, with increasing x, the value of J(x) decreases as x crosses the root x- = -JYO of (1), while J(x) increases as x crosses the root x+ = +JYO of (1). This can be stated as follows: the differentials Df(x-; h) = f'(x-)h = -2JYoh and DJ(x+; h) = 2JYoh (considered as linear maps in h) both map R onto R in a one-to-one fashion. However, the first one reverses the "orientation" of R while the second one preserves it. For this reason we count x- as a negative contribution to our root count, and x+ as a positive one, i.e., we attach to xan index j(x-) = -1, set j(x+) = +1, and define d(J, 0, Yo) = j(x-) + j(x+). Thus, for 0 < Yo < 1, d(J, 0, Yo) = o. But it is natural to assign to our "degree" the value zero also if Yo < 0 since (1) has no solution for such Yo. We then only have to define d(J, 0,0) = 0 to obtain continuity at Yo· As will be seen later, this method of "root counting" can be vastly generalized and leads to a definition of the degree which has the desired continuity properties of the degree with one important exception: in general, d(J, 0, Yo) is not continuous as Yo crosses a point of the image J(oO) of the boundary 00 of 0. A simple example is the following one: let 0 again be the interval (-1, + 1) and let J(x) = 2x. It is then easily seen that

d(J 0 ) = {I if Iyol < 2, , , Yo 0 if Iyo I > 2, if one uses the same method of root counting as in the previous example. Thus ±2 are points of discontinuity. But these two points are the images of the points 1, -1 whose union is 00. It is for these considerations that in the precise definition of the degree to be given later we always suppose that

Yo

-::J

J(oO).

(2)

4. The approach to degree theory used for the simple examples of §3 can be extended to mappings in an arbitrary (real) n-dimensional Euclidean space En. In the present section we describe the basic steps of this extension. Let 0 be a bounded open set in En, and let J be a map of the closure 0 of o into En. We suppose that J E C'(O), i.e., that the differential DJ(x; h) of J exists and is continuous for all x in some open set containing O. We also assume that the condition (2) is satisfied. Step I. We consider a linear map L: En --> En which is not singular. Then det l -::J 0 and we define the index j(l) by setting if det l > 0, _ 1 if det l < 0,

'(l) = { 1 J

(3)

where det stands for determinant. Step II. Suppose Yo is a regular value for J, i.e., either (i) the equation (1) has no roots in 0, or (ii) for every root x E 0 of (1) the differential Lx (h) = D f( x; h) is nonsingular. We then set d(J, 0, Yo) = 0 (4)

INTRODUCTION

3

in Case (i). Now the fact that Yo is a regular value together with {2} implies that (I) has at most a finite number of roots. {This is a special case of Lemma 22 in Chapter I.} Let, in Case (ii), Xl, X2, ••. , Xr be these roots. Then the differentials lp{h} = Df(xp;h}, p= 1,2, ... ,r, are not singular and the indicesJ~ =j{lp} are defined. We then set T

d(f,O,yo} = LJ~ =p-q,

(5)

p=l

where p and q are the numbers of those p for which det lp > 0 and det lp < 0 resp. Step III. Suppose Yo is not a regular value for f. In this case the definition of the degree is obviously made possible by Theorems 1111 and IIh below.

111 1 . Every neighborhood N(yo} of Yo contains a point y which is a regular value for f· THEOREM

THEOREM

111 2 • A neighborhood N(yo} of Yo can be chosen in such a way

that d(f, 0, y'} = d(f, 0, y"} for any couple y', y" of regular values in N(yo}. We note that Theorem 1111 is a special case of a well-known theorem of Sard [49] which will be stated in Appendix B. It takes a fourth step to extend degree theory from maps f E G'(O} to continuous maps 0 ---> E. This extension is based on the fact that such continuous f can be uniformly approximated by maps fo E G'{O}. 5. The method sketched in the preceding section will be referred to as the differential method. In the present section we give an intuitive description of the "simplicial" approach to a degree theory in En introduced by Brouwer [8]. For each nonnegative integer r ::; n, simplices (]'T of dimension r in En are defined: (]'o is a point, (]'1 an open interval, (]'2 an open triangle, (]'3 an open tetrahedron and so on. We assume En to be oriented. This orientation induces an orientation of each n-simplex (]'n in En. This induced orientation is called the positive orientation of (]'n. Now let (],f, (]'~, ... , (]''; be a finite number of {positively) oriented disjoint nsimplices in En. Let 0 be the union of the closures of these simplices, and let o be the interior of O. Then a continuous map 0 ---> En is called simplicial if for each p = 1,2, ... , s the restriction fp of f to (]'; is affine, i.e., is of the form fp{x) = cp + lp{x} where cp is a constant and lp is linear. We set det fp = det lp and assume that lp is nonsingular. Before defining the degree for simplicial f we need the concept of "general position": a point set SeEn is said to be in general position if for r ::; n no r + 1 points of S lie in an {r - 1}-dimensional plane of En {e.g., four points of S do not lie in a two-dimensional plane of En if n ~ 3}. The definition of d(f, 0, Yo} is then given in the following steps.

INTRODUCTION

4

Step I. Let f = a be an affine map E ---+ E: a(x) = const. + l(x) where I is a nonsingular linear map E ---+ E. We then define the index j(a) by .( ) = { 1 _ 1

J a

if det I > 0, if det I < o.

n

Step II. Let f be a simplicial map ---+ En and suppose that Yo is a (simplicially) regular value for f, i.e., that Yo together with the vertices of the simplices f((T~), p = 1,2, ... , s, form a point set in general position. This implies that Yo does not lie on the boundary of any of the simplices f((T~) = fp((T~), p = 1,2, ... , s. Since the determinant of the linear part of fp is i- 0, it follows that in each (T~ the equation (1) has either none or exactly one root. Thus the number of roots in is at most finite. We set d(f, 0, Yo) = 0 if (1) has no root. If there are roots, we set

°

(7) p

where the summation is extended over those p for which (T; contains a root of (1) and where p and q denote the number of those simplices for which det fp is positive and negative resp. Step III. Now let Yo satisfy (2) but not be a regular value for f. Then the definition of d(f, 0, Yo) is made possible by Theorems III! and III2 below. THEOREM III!. Every neighborhood N(yo) of Yo contains a point y which is a (simplicially) regular value for f. THEOREM IIh. A neighborhood N(yo) of Yo can be chosen in such a way that d(f, 0, y') = d(f, 0, y") for any couple y', y" of simplicially regular values in N(yo).

We note that Theorem III! is a special case of a well-known theorem on point sets in general position (see Theorem 1.11 in Chapter 6). Thus this theorem plays for the simplicial method a role corresponding to the one that the Sard theorem plays for the differential method. The possibility of extending the degree theory to arbitrary continuous maps ---+ En is based on the fact that every such map can be uniformly approximated by simplicial maps. 6. The first one to define a degree for continuous maps in finite-dimensional spaces-now known as the Brouwer degree-was Brouwer in his 1911 paper [8]. When he introduced the simplicial method described in §5, degree theory became part of the so-called polyeder-or simplicial-topology and was treated in this way, e.g., in unpublished lecture notes by H. Hopf of 1926/27 [27] and in the topology book of 1934 by P. Alexandroff and H. Hopf [2]. The latter book treats degree theory as a special case of intersection theory and contains earlier important degree-theoretic results of Kronecker [32] and extensions thereof. 7. In their now classical 1934 paper [36] Leray and Schauder defined a degree d(f, 0, Yo), where is a bounded open set in a Banach space E and where f is

n

°

INTRODUCTION a map n -+ E of the form f(x)

=x -

5

(8)

F(x),

where F is completely continuous, i.e., where the closure of F(n) is compact, and where Yo satisfies (2). They did this by taking Brouwer's degree theory for the finite-dimensional space for granted and then using a limit process in the dimension. Calling maps of the form (8) Leray-Schauder (L.-S.) maps, their procedure may be described as consisting of the following two steps: Step I. They defined a degree for those L.-S. maps for which F(O) lies in a finite-dimensional subspace En of E. A map of this kind is called a finite layer map with respect to En. The definition of a degree for layer maps is based on the following: LEMMA (FIRST LERAY-SCHAUDER). Let the L.-S. map f be a layer map with respect to each of the finite-dimensional subspaces En and Em of E. Suppose that (2) is satisfied and that Yo E En c Em. Then the Brouwer degrees d(f, n En, Yo) and d(f, n Em, Yo) are equal (see subsections 3.2 and 3.3 in Chapter 3).

°

°

This lemma makes it possible to define d(f, 0, Yo) = d(f, On En, yo). Step II. The definition of d(f, 0, Yo) for arbitrary L.-S. maps is now made possible on account of the following two Lemmas III and 112 : LEMMA III (SECOND LERAY-SCHAUDER) (cf. Lemma 4.3 in Chapter 3). If f is an L.-S. map n -+ E and if (2) is satisfied, then, with c being a given positive number, there exists a finite layer map A: -+ E such that

n

IIf(x) - A(x)11 < c Here

II . II

for all x E

n.

(9)

denotes the norm in E.

LEMMA 112. There exists an c > 0 of the following property. If A1 and A2 are two finite layer maps such that (9) holds with A = A1 and A = A2, then

d(A1' 0, Yo) = d(A2' 0, yo). 8. It is natural to ask whether the degree theory in Banach spaces can be established intrinsically, i.e., without assuming the finite-dimensional theory and without the passage to the limit in the dimension. This does not seem possible with the simplicial method of §6 since there is no analogue in an infinitedimensional Banach space of an n-dimensional simplex in En. Since differentials in Banach space are well defined, it is different with the differential method described in §4. This method introduced in 1951 [42] in an important paper by Nagumo made degree theory in En part of differential topology. It is clear from §4 what is needed to apply the differential method in the Banach space case: looking at Step I in §4 we note that for defining an index jell for a nonsingular linear L.-S. map

6

INTRODUCTION

of a Banach space E into itself we need an analogue of the division of nonsingular linear maps l: En -> En into those of negative and positive determinant. Now, the simplest map of determinant + 1 in En is the identity map, and a simple map of determinant -1 is the map [-, which in coordinates Xl, X2, ... , Xn for En is given by x~ = -Xl, x~ = X2,"" x~ = Xn . Now the identity map is also defined in the Banach space E, and it is easy to define in E a linear nonsingular map which is the analogue of the map [- (see subsections 4.4 and 4.5 in Appendix A). If we denote this analogue again by [-, then the following linear homotopy theorem (proved in 1955 [48] and independel1tly in 1970 [20]) holds: THEOREM. The family of linear nonsingular L.-S. maps of a Banach space E into itself consists of two disjoint homotopy classes H+ and H- where [ E H+ and [- E H-.

This theorem allows us to define an index j(l) for every nonsingular linear L.-S. map E -> E by setting '(l) = { + 1 if l E H+ , J _ 1 if l E H-.

(10)

Turning to Step II of §4, we note first of all that the definition of regular value given there for En makes sense for an arbitrary Banach space E (see §17 of Chapter 1). It can be proved that also in E the equation (1) has at most a finite number of roots if Yo is a regular value for f and satisfies (2) (see §22 of Chapter 1). Moreover, according to a lemma of KrasnoselskiY [31] (see §18 of Chapter 1), the fact that f is an' L.-S. map implies that the linear map lx(h) = Df(x; h) is also of the L.-S. type. If, in particular, X is a root of (1) with a regular Yo, then lx(h) is nonsingular by definition. It follows that if Xl, X2,"" Xs are the roots of (1) with regular Yo and lp(h) = D f(xp; h), p = 1,2, ... ,s, then the indices j(lp) are well defined and the degree d(f, 0, Yo) can be defined by the first part of the equality (5) with j defined by (10). If there are no roots in 0 of equation (1), the definition is given by (4). To find a definition of the degree if Yo is not a regular value for f, we recall that the definition given in Step III of §4 in the finite-dimensional case was based on Theorems IIh and IIh of that section whose proof in turn was based on Sard's theorem. What is now needed is a theorem generalizing Sard's theorem to the infinite-dimensional case. Such a theorem was established by Smale in 1965 [53]. Using this theorem, now known as the Sard-Smale theorem, C. Fenske established in 1971 [211 the degree theory "intrinsically" for a mapping f E C"(O). (The Sard-Smale theorem had already been used in 1970 [20] to establish a degree theory for "Banach manifolds" by Elworthy and Tromba.) Both of these papers contain a proof for the generalization to Banach spaces of Theorem IIh stated in §4 above. We finally note that the method described in the last paragraph of §4 cannot be applied for arbitrary Banach spaces E: it is known from investigations of Kurzweil [33] and Bonic-Frampton [5] that not in every Banach space can an

INTRODUCTION

7

L.-S. map be uniformly approximated by differentiable maps. Thus this "intrinsic" method does not lead to the same generality as the original method of Leray and Schauder described in §7. It supposes in fact that the mappings involved are twice differentiable. (However, according to a paper [41] by Moulis one-time differentiability is sufficient for the Hilbert manifold.) 9. It is always instructive to look at a given subject from different points of view even if this sometimes entails repetitions. Therefore, this book will not give preference to any of the various approaches to degree theory sketched above but will follow a "mixed method" which will give an opportunity to make clear the relationship between these approaches. 10. In this section an outline of this book's content will be given. Chapter 1 contains preliminary material to be consulted if in the later part of the book the reader encounters definitions or theorems with which he is not familiar. Chapter 2 develops the "intrinsic" degree theory sketched in §8 of this introduction. §2.1 (i.e., §1 of Chapter 2) contains the definition of the degree for linear L.-S. maps, which is based on the linear homotopy theorem whose proof is given in Appendix A following the "spectral method" used in [48]. §2.2 contains the definition of the degree d(f, 0, Yo) for differentiable 1.-S. maps provided Yo is a regular value for f. Some of the properties of this degree are immediate consequences of the definition and are collected in Lemma 2.4; they are referred to as elementary properties in contradistinction to the continuity properties of the degree. §2.2 also contains the definition of and a later often-used lemma on layer maps, i.e., 1.-S. maps f for which the perturbation x - f(x) of the identity lies in a proper subspace of the Banach space E. Before considering the case where Yo is not a regular value for f, we treat in §2.3 the one-dimensional case because it is instructive to see the general case reduced to this elementary one. The latter section starts with the definition of the degree d(f' 0, Yo) with Yo not being a regular value for f provided f is twice differentiable. This definition is based on Theorems 4.1 and 4.2 in Chapter 2 (i.e., on the Banach space version of Theorems IIIl and IIh stated in Step III of §4 of this introduction). Theorem 4.1 follows from the Sard-Smale theorem whose proof is given in Appendix B. The proof of Theorem 4.2 uses methods contained in papers by Fenske [21J and by Elworthy-Tromba [20J in conjunction with a method employed by Nagumo in the finite-dimensional case [42, p. 491£.J by which the proof is ultimately reduced to the one-dimensional case, i.e., to Theorem 3.6 in Chapter 2. The continuity of the degree d(fO, Yo) in Yo (Theorem 4.14 in Chapter 2) is essentially a consequence of Theorem 4.2. Following again an idea used by Nagumo in the finite-dimensional case [42, pp. 492-493]' the continuity of the degree in f is proved by applying Theorem 4.14 to a Banach space "one dimension higher" than the given one. The degree theory developed in Chapter 2 assumes that the L.-8. maps are

8

INTRODUCTION

twice differentiable since this is assumed in Theorem 4.2. The object of Chapter 3 is to develop degree theory without that assumption. A first step in that direction is the extension lemma, Lemma 1.3, which-speaking vaguely-states: let C = C(T1) be the normed space of all completely continuous maps 0 ---t E (see subsection 1.1 in Chapter 3), and let C 1 be a linear subspace of C. Suppose a degree having certain specified properties has been defined for all mappings h (x) = x - F1 (x) for which F1 belongs to C 1 = C 1 (0); then the definition of the degree can be extended to all maps ](x) = x - F(x) for which F is an element of the closure 01 of C 1 in such a way that the specified properties hold for the extended degree. This lemma is applied twice. In §3.2 it is applied with (11) where C"(O) denotes the set oftwice differentiable maps 0 ---t E. It follows that the degree theory is established for all L.-S. maps in those Banach spaces for which (12) C 1 (0) = C(O), with C 1 (n) given by (11). As already pointed out in the last paragraph of §8 of this introduction, the equality (12) does not hold in all Banach spaces. It holds, however, for certain types of Banach spaces (see subsections 2.3-3.8 in Chapter 3). In particular (12) holds for finite-dimensional Banach spaces. Since in such spaces a map is L.-S. if and only if it is continuous, the degree theory is established for continuous maps of a finite-dimensional Banach space into itself. This allows us to carry out the first step of the Leray-Schauder theory sketched in §7 of this introduction and thus to establish the degree theory for finite layer maps (see §3.3). We now make a second application of the extension lemma, Lemma 1.3 (see §3.4): we apply it with C 1 (0) being the subset of those F E C(O) for which F(O) lies in some finite-dimensional subspace En = En(F) of the Banach space E. But for this C 1 (T1) the equality (12) holds on account of the second LeraySchauder Lemma III stated in §7 of this introduction (and proved in §3.4). Thus by the extension lemma the degree theory is established for all L.-S. maps in arbitrary Banach spaces. In §3.6 the degree is defined for generalized Leray-Schauder (g.L.-S.) maps, which essentially are maps of the form >.(x)f(x) where f is an L.-S. map and >. a positive function bounded away from zero. The reason for introducing this slight generalization [47] is that a central projection of an L.-S. map is in general not L.-S., while a g.L.-S. map remains g.L.-S. under central projections. It is easy to see that the assertion of the continuity theorem, Theorem 6.6 in Chapter 3, remains true if the "homotopy" is a linear convex combination of two g.L.-S. maps fo and 11, i.e., if ft(x) = (1 - t)fo(x) + tfdx), 0 ~ t ~ 1. The

INTRODUCTION

9

resulting theorem will be called the Poincare-Bohl theorem in accordance with terminology customary in the finite-dimensional case. Chapter 4, which consists mainly of an updating of papers [46] and [47], deals with some consequences of this theorem. Among these are (i) the fact that the degree d(f, 0, YO) depends only on the values of I on the boundary 80 of 0, which allows one to define the ''winding number" u(f(80), Yo) of a g.L.-S. map I defined only on 80 with respect to the point Yo (provided that Yo rt. 1(80)); (ii) Theorem 1.9 in Chapter 4, which is a generalization to Banach spaces of the classical Rouche theorem; (iii) Theorem 1.12 together with its corollaries and Theorem 1.17 (these are useful for giving existence proofs in analysis); (iv) the fixed point theorem, Theorem 1.18. §4.2 gives an interpretation of degree and winding number as "intersection"

numbers. Chapter 5 deals with the product theorem, i.e., a formula which expresses the degree d(YI, 0, zo) of a composite map gl in terms of the degrees of g and I and with consequences of that theorem. The results of this chapter are all due to Leray, who in a short 1935 Note ([34], see also [35]) established the product theorem for L.-S. maps in Banach spaces by a limit process from the finitedimensional case, used this theorem to prove the "invariance of the domain," i.e., the fact that an open set in a Banach space remains open under a one-toone L.-S. map, and deduced from the product theorem also the validity of the Jordan theorem in its generalized form in Banach spaces (see subsection 3.1 in Chapter 5). This last result will be referred to as the Jordan-Leray theorem. The proof of the product theorem will be given in §5.1. We first assume that I and g are twice differentiable and that Zo is a regular value for gf. In this case the theorem is a direct consequence of the chain rule for differentials and of Theorem 6.1 of Appendix A according to which j(hl2) = j(h)i(l2), where j denotes the index defined by equation (10) of this introduction. Thus, in this case, our theorem is established directly in Banach space without recourse to a limit process in the dimension. For the proof without the special assumptions made above, a complicated approximation procedure is needed for which we refer to the text. The invariance of the domain theorem is proved in §5.2. This proof is followed by some consequences of that theorem (subsections 2.4-2.10 in Chapter 5). §5.3 contains the proof of the Jordan-Leray theorem. §§5.1 and 5.2 are influenced by the presentation in Nagumo's paper [42] dealing with the finite-dimensional case, and for part of §5.3 we used the method employed by Lloyd [37, pp. 47-50] for the proof ofthe finite-dimensional versions of the theorem in question. Chapter 6 deals with finite-dimensional linear spaces. §6.1 contains prerequisites about subjects like point sets in general position, simplices, affine maps, and orientation.

11" 'I' l{U U U tJ' 1'1 U N

lU

Er Er,

If 0 is a bounded open subset in the space of finite dimension n, if f is a and Yo a point of then the degree d(j, 0, Yo) was continuous map n -> already defined in subsection 2.4 of Chapter 3. Since two linear spaces of the same dimension are linearly isomorphic, the degree d(j, 0, Yo) can, in an obvious way, be defined if f is a continuous map of into another linear space E-g of dimension n and if Yo is a point of E-g. It is easy to verify that this degree has the properties corresponding to those of the previously defined degree. This is done in §6.2. In §6.3 simplicial maps are introduced. Since simplicial maps are continuous, the degree theory developed in §3.2 applies to them (see in particular subsection 2.4 in Chapter 3). We note that by part (iv) of Lemma 3.7 in Chapter 6 the equality (7) of this introduction holds, an equality which has been a matter of definition in the original simplicial theory sketched in §5 of this introduction. §§6.5 and 6.6 deal with subdivision and simplicial approximations resp. Chapter 7 treats mappings between finite-dimensional spheres. Following H. Hopf's presentation in [27], the definition of the degree for such maps is reduced to the "plane" case by means of stereographic projection. §7.1 deals first with elementary geometric properties of the sphere and with its orientation. Furthermore if Sr and S2' are two spheres of dimension n, if 1 is a continuous map Sr -> S2', and if Yo is a point of S2', the degree d(j, Sr, Yo) is defined and proved to be independent of the particular choice of Yo. This allows us in §7.2 to denote this number by d(j, Sr, S2') and derive the main properties of this degree. §7.3 starts with the definition of the "order" v(j(Sf), cd of a continuous map f: Sr -> E-g - Cl where Sr is an n-sphere, where E-g is a linear space of dimension n, and where Cl is a point of E-g: let B;+l be a ball with center Cl whose closure does not intersect 1( Sr) and let S2' = a B;+ 1. If 7r is the projection of 1(Sr) on S2' from Cl and 1* = 7r1, then the degree d(j*, Sr, S2') is defined and can be proved to be independent of the particular choice of B~+l. The order v(j(Sr), cd is then defined as this degree. The main result states that the order equals the winding number u(j(Sr),ct}, or what is the same, by Definition 1.5 of the winding number in Chapter 4, that v(j(Sr),ct} = d(1,B~+l,Ct}, where 1 is an arbitrary continuous extension of 1 to the ball B~+l whose boundary is Sf· The proof presented here follows the one given by Hopf in [27]. Relations established in this proof allow us in subsection 3.13 of Chapter 7 to interpret the order and the degree as intersection numbers in more detail than in §4.2. Theorem 3.4 in Chapter 7 will be of use in Chapter 8. To describe the content of Chapter 8, we note that Definition 1.5 in Chapter 4 of the winding number u together with an elementary property of the degree implies that if f is a g.L.-S. map ao -> E - Yo and if u(j(80), Yo) ::J 0, then for every g.L.-S. extension 1 of 1 to the equation l(x) = Yo has at least one

Er

n

n

INTRODUCTION

11

solution x in O. In other words, the following assertion A holds: (A) for f to have a g.L.-S. extension 1 to n which maps n into E - Yo, the condition

u(f(ao), Yo) = 0

(13)

is necessary. We note next that Theorem 1. 7 in Chapter 4 implies (B) for the g.L.-S. maps f and 9 mapping ao into E - Yo to be homotopic in E - Yo, the equality

u(f(ao), Yo) = u(g(aO), Yo)

(14)

is a necessary condition. The object of §8.1 is the proof of Theorems 1.2 and 1.4 in Chapter 8, which state that conditions (13) and (14) are not only necessary but also sufficient. The finite-dimensional versions of these theorems are both special cases of a classical theorem by H. Hopf (see [2, p. 499ff.] and the papers by Hopf quoted on p. 621 of [2]). We therefore refer to Theorem 1.2 as the Hopf extension theorem. Theorem 1.4 will be called the Hopf-Krasnoselskil homotopy theorem since Krasnoselskir was the first to establish the theorem in the generality stated; see [31, p. 110]. (For "strictly convex" Banach spaces the theorem was established in [47], and for Hilbert spaces with 0 being a ball in [45].) The proof of the Hopf extension theorem will first be given for the finitedimensional case following the method used in [2, pp. 500-505]. It is then proved for finite layer maps in a Banach space E and finally for g.L.-S. maps. For the last two steps the "homotopy extension" theorem, Theorem 1.12 in Chapter 8, is used, whose prooffor Banach space is due to Granas (see [24, p. 37 and 23]). Krasnoselskir's proof of Theorem 1.2 assumes Hopf's finite-dimensional result and then uses a limit process in the dimension. The proof given in the present book generalizes to Banach space a method used by Hopf in the proof he gives for the finite-dimensional case in his lectures [271. In sketching this method by which the homotopy theorem is shown to be a consequence of the extension theorem, we restrict ourselves at this moment for brevity's sake to the case that the convex set VI is a ball Br+1 with center yo: we extend the map f given on Sf = aBr+ I to Br+ I, denoting the extension again by f. Due to the fact that by assumption f '# Yo on Sf, we can choose a ball B;;+1 c Br+1 with center Yo such that f( x) '# Yo for x E Br+1 - Blj. For x E S8 = a B;;+1, we set go(x) = g(x') where x' is the point on Sf which lies on the same ray from Yo as x. If = f on Sf and = go on S8, then assumption (14) together with the definition of go implies that u(f(aOo),Yo) = 0 where 0 0 = Bf+l - BO'+1. Consequently we can apply the Hopf extension theorem to obtain an extension 1 from i to with 1'# Yo. Now if Xo and Xl are points on S8 and Sf resp. which lie on the same ray from Yo, then for t E [0,1] the point xo(1- t) + tXI lies in and therefore h(x, t) = 1(xo(1 - t) + tXi) '# Yo. Since h(x,O) = go(Xo) = g(xt} and h(x, 1) = f(xt}, we obtain a homotopy of the desired property.

i

no

i

no

12

INTRODUCTION

Part of Hopf's finite-dimensional theorem states [2, p. 502ff.]: if nn+l is an open bounded set of En+l and if f is a continuous map an n+ 1 -+ Em - Yo, where m > n + 1 and where Yo is a point of Em, then f can always be extended to a continuous map nn+l -+ Em - Yo. We prove this assertion in §8.2. It will be used in Chapter 9. §9.1 contains a proof of a theorem by Borsuk [7] and its generalization to Banach space: if n is a bounded open set in the Banach space E which is symmetric with respect to the zero point () of E and if f is an L.-S. map an -+ E which is odd, i.e., satisfies f( -x) = - f(x), (15) then the winding number u(J(an), ()) is odd. The prooffor the finite-dimensional case uses the extension theorem, Theorem 2.2 in Chapter 8, and follows the proof given by J. T. Schwartz [51]. The generalization to Banach space is based on Lemma 1.9 in Chapter 9, which is due to Granas [24, p. 41]. §9.2 contains some consequences of the (generalized) Borsuk theorem. Lemma 2.2, for example, states that the conclusion of the Borsuk theorem remains true if the assumption (15) is replaced by the weaker one that for no x E an do the "vectors" f(x) and f( -x) have the same direction. Another consequence (Theorem 2.5 of Chapter 9) is a generalization to Banach spaces of a theorem conjectured by Ulam, proved by Borsuk, and now known as the Borsuk-Ulam theorem according to which a continuous map of an n-sphere sn into the space En maps at least one pair of antipodal points of sn into the same point of En (cf. [24, Chapter IV]). We finally note that topological tools (like subdivision and simplicial approximations) used in this book are defined and treated not in the degree of abstraction and generality to be found in topology books but only to the degree needed for the purpose at hand.

CHAPTER 1

Function-Analytic Preliminaries 1. The term "Banach space" will always denote a Banach space over the reals except in Appendix A. For definition and elementary properties of a Banach space the reader is referred to [18]. 2. Let E be a Banach space with norm II . II and zero element fJ. If 8 is a subset of E, then its closure and boundary will be denoted by S and 88 resp., and the empty set by 0. I denotes the identity map E onto E. By "subspace" of E we always mean "linear subspace." If Xo is a point of E and p a positive number, the set {x EEl Ilx - xoll < p} is called the ball with center Xo and radius p. It will be denoted by B(xo, pl. 3. DEFINITION. (i) Let El and E2 be closed subspaces of E and suppose every point x E E can be represented uniquely as

(1) Then we write

(2) and say: E is the direct sum of El and E 2. Each of the spaces El and E2 is said to be a direct summand of E, and El and E2 are said to be complementary to each other. We note that the uniqueness of the representation (1) is equivalent to saying that (3) where the symbol "n" denotes intersection. (ii) If El and E2 are Banach spaces, then the "product" El X E2 is defined as the space of couples (Xl,X2), Xl EEl, X2 E E2 with the linear operations (Xl, X2)

+ (Yl, Y2)

= (Xl

+ Yl, X2 + Y2), .A real.

With a proper norm, El X E2 becomes a Banach space (see [18, p. 89]), and by identifying the points (Xl, fJ) and (fJ, X2) of El X E2 with the points Xl of El 13

.I: UJ.''-.I.J...I.'-JJ.'-I"I..J.''I"I..J...J.1 .J...1V.1 .1l..1.:JJ...JJ.lYJ.J.l".l"1..1l..1J..:Jt..J

and X2 of E2 resp., the spaces El and E2 become subspaces of El x E2 and El x E2 = E l +E2. REMARK. If the direct summand El C E is given, then E2 is not uniquely determined by E. However, if E2 is of finite dimension n and E = El +E2 = El +E~, then E~ is also of dimension n. Indeed for x E E we have the direct decompositions x = Xl + X2 = x~ + x~ where X2 E E2, x~ E E2 while Xl and x~ are elements of E l . Thus X2 = Xl + X2 with Xl = x~ - Xl EEl. This direct decomposition of X2 shows that x~ is uniquely determined by X2. We thus obtain a linear map l of E2 into E~. Thus n S n' where n' denotes the dimension of E~. Since n' S n can be proved the same way, we see that n' = n. The common dimension n is called the co dimension of E l . 4. LEMMA. A finite dimensional subspace El of E is a direct summand of

E. PROOF. Let bl. b2, . .. ,bn be a base for E l . Then every X E El can be written uniquely as n

(4)

X = Lll!ibi, i=l

where the ll!i = ll!i(X) are real-valued functions with domain El which are obviously linear. But it is well known that they are also continuous (see e.g. [18, pp. 244-245]). Therefore, by the Hahn-Banach theorem [18, II.3.1O] they can be extended to continuous linear functionals Ai(X) on E. We now set n Xl

= LAi(X)bi ,

(5)

i=l

As X varies over E, then obviously Xl varies over E l , and X2 over a linear subspace E2 oLE. Now (1) is obviously true. It remains to prove (3). Let X E El n E 2. From X E E2 we see that n

X

= X-

L Ai (x)b

i

i=l

for some

X E E.

But this equality implies that

X E El

since

X

EEl.

Thus

n

X=

L Ai (x)bi

and

Ai(X) = ll!i(X).

i=l

It follows that X = (). 5. DEFINITION. A continuous linear map P: E --; E is called a projection if p 2(X) = P(x)

6. LEMMA. Let X2 are projections.

Xl,

for all X E E.

X2 be as in (5); then the maps

(6) PI:

X --;

Xl

and P2 : X --;

..'UNlJ'nUN-ANALYTIlJ

PH.~LIMINAH.I~~

15

PROOF, By (5) n

Pf(X) = P1 (xd =

L Ai(P (x»bi 1

i=1

But obviously

z =], i f;j, Therefore Pf(x) = 2::]'=1 Aj(x)bj = P1 (x). To prove (6) with P = P2, we apply P2 to the second of the equations (5) and obtain

pi(x) = P2(X2) = P2x - P2P1 (X).

(6a)

But applying P2 to the first equation (5) we obtain

(7) since P2bi = (J. (6a) together with (7) proves our assertion, since the boundedness of P1 and P2 is clear from (5), the Ai being bounded. 7. DEFINITION. A subset S of E is called compact if every sequence of points in S contains a subsequence which converges to a point of S. 8. DEFINITION. A continuous map F of a subset S of E into E is called completely continuous if for every bounded subset S1 of S the set F(Sd is compact, 9. DEFINITION. A map I: S --+ E is called a Leray-8chauder, or L.-8., map if F = I - I is completely continuous. Here as elsewhere the symbol "I: S --+ E" . means: I is a map mapping S into E. 10. LEMMA. Let 8 be a bounded subset 01 E, and let 1 be an L.-8. map E. Then (i) 1(8) is bounded; (ii) il S is closed, then I(S) is closed; (iii) il r is a compact set in E, then 1- 1 (r) is compact; (iv) let S be a bounded closed set in E, and let I: S --+ E be an L.-S. map. Suppose I maps S onto I(S) in a one-to-one lashion. Then the g = 1-1 is an L.-S. map.

8

--+

PROOF. Assertion (i) follows trivially from Definition 9 and the fact that a compact set is bounded. Prool 01 (ii). Let Yl, Y2, ... be a convergent sequence of points in 1(S) and let Yo = lim Yj·

(8)

)-+00

We have to show that there exists an Xo E S such that YO = I(xo).

(9)

16

FUNCTION-ANALYTIC PRELIMINARIES

Now since Yj E f(8) there exist Xl E 8 such that J' = 1,2, ... ,

(10)

where F is completely continuous. Consequently there exists a subsequence {Xli} of the sequence {Xj} such that the F(xjJ converge. Since by (8) the Yl i converge, it follows from (10) that the Xli converge. If Xo is the limit of that sequence, then Xo E 8 since S is closed, and assertion (9) follows from (8) and (10). Proof of (iii). If f- 1(r) is empty or finite, then the assertion is trivially true. Suppose then that f-1(r) contains an infinite sequence Xl, X2, .... We have to prove that this sequence contains a convergent subsequence for whose limit Xo

f(xo)

E

r.

(11 )

Now let the sequence Y1, Y2, ... of points in r be defined by (10). Since r is compact and F completely continuous, there exists a subsequence Ud of the integers such that the Yji converge to a point Yo E r and such that the F(xjJ converge. It then follows from (10) that the Xji converge to some point Xo and that Yo = f(xo), i.e., that Xo E f-1(r). This proves assertion (iii). Proof of (iv). Let G(y) = y - g(y). We have to prove that G(y) is completely continuous, i.e., that (a) every sequence Y1, Y2, ... of points in f CO') contains a subsequence {y ji } for which G(YjJ converges, and (b) G(y) is continuous. Now from f(x) = X - F(x), and from y = f(x), X = g(y) we see that

G(y) = y - g(y) = f(x) - X = -F(x). Now if {Yj} is a sequence in f(8), then {Xj = g(Yj)} is a sequence in 8. Since F is completely continuous and G(y) = -F(x), assertion (a) follows. The proof of assertion (b) is based on the following remark. If the sequence {Zj} in E satisfies the following two conditions:

(a) every subsequence of the {Zj} has a convergent subsequence; (;3) all convergent subsequences of the sequence {Zj} converge to the same point zo, then the sequence {Zj} converges to Zoo Indeed, if this were false, there would exist an e > 0 and a subsequence {Zji} of the {Zj} such that Ilzo - Zji I 2: e. This inequality holds for every subsequence of the {Zj,}. This leads to a contradiction since by (a) we can choose a convergent subsequence of the sequence {Zji}' and by (;3) such a subsequence converges to

ZOo This proves the remark and we return to the proof of (b). We have to show: if f (8) which converges to a point Yo, then the sequence {Zi = G(Yi)} converges to G(yo). Now by assertion (a) the sequence {Zi} satisfies assumption (a) of the remark. It remains to show that it also satisfies assumption (;3). Suppose {YnJ is a subsequence of the {Yi} for which Yl, Y2, ... is a sequence of points in

FUNCTION-ANALYTIC PRELIMINARIES

17

Zni = G(YnJ converges. Then we have to show that the limit of the zni is G(yo). Now G(YnJ = -F(xnJ with Xni = g(YnJ. Thus the F(xnJ converge. But Yni = f(xnJ = Xni - F(xnJ. Since Yni --+ Yo, it follows that the Xni converge to some point Xo. Then Yo = Xo - F(xo) = f(xo), and G(yo) = -F(xo) = -limF(xnJ = limG(YnJ as we wanted to prove. REMARK. It is easily seen that the assertions (ii) and (iii) of Lemma 10 remain valid if the assumption that 8 is bounded is replaced by the assumption that f-l(B) is bounded for any bounded set B. 11. DEFINITION. Let 1 be a linear continuous map E --+ E. Then 1 is called regular or nonsingular if it maps E onto E. 1 is called singular if it is not regular. The kernel K = K(l) of 1 is defined by K(l) = {x E E Il(x) = 9}. K(l) is obviously a subspace of E.

12. LEMMA (FREDHOLM ALTERNATIVE). Let 1 be a linear L.-8. map E --+ E, and let the range P = P( l) be defined by P{l) = {y EEl Y = l(x)

for some x E E}.

Then

(i) dimK(l) is finite; (ii) there exists a subspace K* (l) such that

E = P(l)+K*(l).

(12)

K* is of course not uniquely determined. However

(iii) dimK*(l) = dimK(l)

(13)

for every K*(l) satisfying (12). Each such K*(l) is called a co-kernel of l. (iv) I is nonsingular if and only if it is one-to-one. (v) If I is a nonsingular linear L.-8. map E --+ E then 1-1 is a linear L.-8. map E --+ E. PROOF. A prooffor assertions (i), (ii) and (iii) may be found in [4:, Chapter X, §§2, 4]. Assertion (iv) is an immediate consequence of the first three assertions. Proof of assertion (v). 1- 1 is a one-valued map E onto E which is obviously linear. By a classical theorem of Banach, it is also continuous (see [4:, Chapter III, Theorem 5, p. 41] or [18, Theorem II.2.2, p. 571. Now 1 = 1- L with completely continuous L. Let k = l(h) = h - L(h) = h - L(l-1(k)) or l-1(k) = h = k

+ L(l-1(k)).

Since L is completely continuous and l-1 is continuous, it follows that the map L(k) -L(l-1(k)) is completely continuous.

=

18

FUNCTION-ANALYTIC PRELIMINARIES

13. The differential. Let Xo be a point of E, and neE an open set containing Xo. Let f be a continuous map of n into a Banach space E. If there exists a continuous linear map lxo : n ---- E such that

1. f(xo + h) - f(xo) - lxo (h) - () h~ Ilhll - ,

(14)

then it is easily seen that lxo is uniquely determined by (14) (see e.g. [16, VIII.1]), and lxo is called the differential of f at xo, while f is called differentiable at Xo· We use the notation (15) lxo (h) = D f(xo; h). If f is differentiable at every Xo En, then it is called differentiable in n. 14. The chain rule. Let E, F, G be Banach spaces. Let Xo be a point of E and U an open neighborhood of Xo. Let f be a continuous map U ---- F. Let V be an open neighborhood of the point Yo = f(xo), and let g be a continuous map V ---- G. Let f and g be differentiable at Xo and Yo resp. Then the composite map gf: U ---- G is differentiable at xo, and

Dgf(xo; h) = Dg(yo; Df(xo; h)).

(16)

For a proof see [16, p. 1451. 15. The second differential. As is well known, a continuous linear map of E into a Banach space E is bounded, i.e., for all hE E

IIl(h)1I :-:; Mllhll

(17)

for some positive M. It is also well known that with the obvious definitions of addition and multiplication by reals the set of linear maps l: E ---- E becomes a Banach space if the norm IIlll of 1 is defined as the smallest number M which satisfies (17) (see [18, p. 61]). We denote this Banach space by A(E, E). With this notation D f(xo; .) is an element of A(E, E) if f is differentiable at xo, and if f is differentiable ir. the open set n containing xo, then D f is a map n ---- A(E, E) mapping the point x E n into the element D f(x;·) of A(E, E). If this map is continuous, we say: f is continuously differentiable, or D f is continuous. IE may be that at some point Xo E n this map of n into the Banach space E = A(E, E) has a differential. Then we denote this differential by D2 f(xo;') and call it the seco~d differential of f at Xo. D2 f(xo;') is an element of the Banach space A(E, E) = A(E, A(E, E)). If D2 f(xo;') exists for all Xo E n, we say f is twice differentiable in n. Then D2 is a map n ---- A(E, A(E, E)); if this map is continuous, we say f is twice continuously differentiable. 15a. The second differential as a bilinear map. Let E and E be Banach spaces. A bilinear map E x E ---- E is a map l2 = l2(hl. h2 ), hl. h2 E E, into E which is linear in hl and in h 2 • If l2 is continuous, then (18)

for some number M. The set of all continuous bilinear maps l2: E x E ____ E can be made a Banach space if the norm of h is defined as the smallest number

FUNCTION-ANALYTIC PRELIMINARIES

19

M satisfying (18) (see e.g. [16, V.7]). We denote this Banach space by A2 = A(E x E, E). We define a map of this space into the space A(E, A(E, E)) defined in the preceding section as follows: given 12 E A2 and hI E E we define a map Ihl : E -+ E by setting

Ihl(h 2) = 12(hl,h2).

(19)

Then Ihl E A(E, E) and the map (20) is an element of A(E, A(E, E)). Thus we assign to the element 12 of A2 the element (20) of A(E, A(E, E)). It can be shown that the map thus defined is a linear isometry onto (see [16, V.7.8]). We identify the two spaces in question by identifying elements corresponding under this linear isometry. In particular the element D 2 /(xo;·) of A(E,A(E,E)) may then be considered as an element of A2 which we denote by D2 I (xo, hI, h2). If I is twice differentiable in an open neighborhood 0 of xo, then D2 I is said to be continuous at Xo if the map D2 I: 0 -+ A2 is continuous. It can be proved that D2 is symmetric, i.e., that

D2 I(xo, hI, h2) = D2 I(xo, h2, hd (see [16, VIII.12]). Differentials Dr of order r > 2 are defined correspondingly. 16. Let 0 be an open set in E and let I be a map of 0 into the Banach space E. We say "I E C'(O)" if I is differentiable in 0 and if the differential is continuous. We say "I E C"(O)" if D 2 /(x;·) exists for all x E 0 and is continuous. If 8 is an arbitrary subset of E and I a map 8 -+ E, then I E C'(8) or IE C"(8) means: I is defined in some open subset 0 of E containing 8 and IE C'(O) or C"(O) resp. cr(8), r ~ 2, is defined correspondingly (cf. §15). 17. DEFINITION. Let n be a bounded open set in E. Let f E C'(n). Then a point Xo E n is called a regular point (with respect to f) if the linear map D f (xo j h) is not singular. The point Xo is called singular or critical if it is not regular. A point Yo in E is called a regular value for f if no solution x E of the equation I(x) = Yo (21)

n

is a singular point. Yo is called a singular or critical value if it is not a regular value. Note that in particular every Yo E E for which (21) has no solution in is a regular value for f.

n

18. KRASNOSELSKIi's LEMMA. Let F be a completely continuous map n -+ E which is differentiable in n. Then lor each x E n the linear map E -+ E

given by h -+ DF(x; h) is completely continuous. For a proof see [31, p. 136]. Obviously the lemma is equivalent to saying that the differential h - DF(xj h) of the differentiable L.-8. map I(x) = x - F(x) is, for each x E n, a linear L.-8. map.

FUNCTION-ANALYTIC PRELIMINARIES

20

19. LEMMA. Let 0 be a bounded open subset of E and f E C'(O). In addition we assume that f is L. -8. Then (i) the set of regular points in 0 is open, and (ii) the union of ao with the set 8(0) of singular points in 0 is closed. PROOF. Let Xo E 0 be a regular point for f. By Lemma 18, Lo = Df(xo;') is an L.-8. map. By assumption lo is not singular. Therefore by assertion (v) of Lemma 12, lOl exists as a linear L.-8. map and thus is bounded. But by the continuity of Df(x;·) (as a map 0 -+ A(E, E)) there exists an open neighborhood N(xo) of Xo such that for x E N(xo)

IIDf(x;·) - Df(xo; ')11 < Il l 0111-1. From this inequality and from Lemma 5.3 of Appendix A, it follows that every

x E N (xo) is a regular point for

I.

Assertion (ii) follows from (i) since (aO) the set of regular points in O.

u 8(0)

is the complement in 0 of

20. LEMMA. Under the assumptions 01 Lemma 19 the set R 01 those regular values y which satislY (22) y f/. l(aO) u open.

PROOF. The complement of R is the set

R' = f(8(0) u aO). But 8(0) u ao is obviously bounded and, by assertion (ii) of Lemma 19, also closed. Therefore by assertion (ii) of Lemma 10 the set R' is closed. Thus R is open. 21. INVERSION THEOREM. Let Xo E E, let 0 be a bounded open neighborhood olxo, and let 1 E C'(O) be a map 0 -+ E. 8uppose Df(xo;') is nonsinguLar and one-to-one. Then there exists an open neighborhood 0 1 C 0 of Xo and an open neighborhood N(yo) of Yo = f(xo) such that the map f maps 0 1 one-to-one onto N(yo) and such that the map f- 1 : N(yo) onto 0 1 is in C'(N(yo)). In addition if f E CB(O), S ~ 1, then f- 1 E CB(N(yo)) for some neighborhood N(yo) of Yo·

For a proof see, e.g., [16, X.2]. Note: if 1 is an 1.-8. map, then, by part (iv) of Lemma 12 and by Lemma 18, the assumption that Df(xo;') is one-to-one follows from the assumption that this differential is nonsingular. 22. LEMMA. Let 0 be a bounded open set in E and let f E C'(O) be an L.-8. map n -+ E. 8uppose Yo is a regular value for f, and

Yo

f/. 1(80.).

(23)

FUNCTION-ANALYTIC PRELIMINARIES

21

Then (a) the equation f(x) = Yo (24) has at most a finite number of solutions. (b) If in addition Yo = () E 0, if 0 is symmetric with respect to () (i.e., contains with x also -x), and if f is odd (i.e., f(-x) = -f(x)), then Df(x;TJ) is even in x, i.e., (25) Df(x,TJ) = Df(-x,TJ)· PROOF. (a) Suppose there exists an infinite sequence Xl, X2, ... of points in 0 which are solutions of (24). Since the set {yo} consisting of the single point Yo is compact, the set f-1(yO) is compact by part (iii) of Lemma 10. Therefore there exists a subsequence Xi l , Xi2' ... of the sequence Xl, X2, ... which converges to some point Xo Then by continuity of f, Xo is a solution of (24), and by (23) the point Xo lies in O. Now by Theorem 21 there exists a neighborhood 0 1 cO of Xo such that Xo is the only root of (24) which lies in 0 1 , This, however, contradicts the fact that the solutions Xi l , Xi2" .. converge to Xo. Proof of (b). From the assumption that f( -x) = - f(x) and from the linearity of Df(x;TJ) in TJ, we see that with TJ =-h

En.

f(x

+ h) - f(x) - Df(x;h) IIhll

f(-x

+ TJ) - f(-x) - Df(x;TJ) IITJII

By definition of the differential the left member tends to zero as h -+

().

Therefore

lim f( -x + TJ) - f( -x) - Df(x; TJ) = (). " ..... 0

IITJ II

But, by §13, this shows that D f(x; TJ) is the differential of f at -x. This proves

(25). 23. The intervals 0 ~ t ~ 1 and 0 < t < 1 will be denoted by [0,1] and (0,1) resp. If Xo and Xl are points in the Banach space E, then the segment XOX1 is defined as the set {x EEl x = (1 - t)xo + tXl, t E [0, I]}, and the ray XOXt as the 'set {x EEl x = (1 - t)xo + tXl, t ~ o}. We also say XOXt is a ray issuing from xo. A set C c E is called convex if for any couple xo, Xl of points in C the segment XOXl is contained in C. 24. Let Po, Pll" . , PN be N + 1 distinct points in E. Then the union P of the segments Si = PiPHl, i = 0, 1,2, ... , N - 1, is called a polygon with vertices Pi and edges Si· P is said to connect Po and PN. Obviously 8i n 8i+1 = PHl, i = 0,1, ... , N - 1. If these are the only intersections of two edges 8i and 8j, i '" j, then the polygon P is called simple. 25. We recall that a set SeE is called connected if it is not the union of two nonempty disjoint subsets which are open (with respect to S). With this definition the following theorem holds: if S is an open connected set in E, then any two points qo and ql in S can be connected by a simple polygon PeS. (See, e.g., [2, Theorem XX, p. 50]; the theorem is stated there only for finitedimensional spaces, but the proof holds for any (linear) metric space and thus for Our Banach space E.)

FUNCTION-ANALYTIC PRELIMINARIES

22

26. DEFINITION. Let 8 be a subset of E, let El be a Banach space {which may be equal to E}, let f: 8 -+ El be a C'-map and let Yo =I- Yl be points of E l . Then the segment u = YOYl is said to be transversal to f if for every x E 8 for which f(x) E u the differential Df(x; h) together with u spans E l .

27. THEOREM. Let the subset Y of E be precompact, i.e., have the property that its closure is compact. Then if e is a given positive number, there exist points Yl, Y2, ... ,Ys in Y such that for every point Y E Y

Ily - Yull < e

{26}

for at least one positive integer u ::; s.

PROOF. As Y varies over Y, the balls B(y,e/2) form an open covering of

Y. Consequently, there exist a finite number of these balls, say B("1h, e/2), ... , B(ys' e/2} which form an open covering of Y, Yo,.··, Ys being points of Y. Now for every Yu there exists a Yu E Y such that Ilyu - Yull < e/2. Now if Y is an arbitrary point in Y, then y E B(yu, e/2} for some a ::; s. For such a

Ily - Ytfll ::; Ily - Yull + Iyu - Ytfl < e. REMARK. This lemma states that "up to e" a precompact set lies in a finitedimensional plane, namely, the plane spanned the points Yl, Y2,··., Ys satisfying

(26). 28. DEFINITION. Let 8 be a subset of E. Then the convex hull co 8 of 8 is defined as the smallest convex set containing 8. Since the intersection of convex sets is convex, co 8 is the intersection of all convex sets containing 8. The closed convex hull co 8 is defined as the intersection of all closed convex sets containing 8. Obviously co 8 is closed. Moreover co8 = COS

(27)

(see [18, p. 415]). 29. MAZUR'S LEMMA. If 8 is compact, then co 8 is compact (see, e.g., [18, p. 416]).

30. DUGUNDJI'S EXTENSION THEOREM. (For a proof see [17 or 28, p. 57].) Let 8 0 be a closed subset of the Banach space E and let ¢o be a continuous map of So into a Banach space E l . Then there exists a continuous map ¢: E -+ El which is an extension of ¢o (i.e., whose restriction to 8 0 is ¢o) and which has the following property: ¢(E) c co¢{80 }·

(28)

31. LEMMA. Let So be a bounded closed subset of the Banach space E, and let ¢o be a completely continuous map of 8 0 into a Banach space E l . Then there exists a completely continuous extension ¢ of ¢o to E which satisfies (28).

PROOF. By Dugundji's theorem in §30, there exists a continuous extension

FUNCTION-ANALYTIC PRELIMINARIES

23

i.e., that for every bounded subset 0 0 of E the set 4>(0 0) is compact. Let then 0 0 C E be bounded. By (28) and (27)

c 4>(E) c co4>(So) c co4>(So), 4>(0 0) c 4>(E) c co4>(So) = co4>(So). 4>(0 0)

(29)

But 4>(So) is compact by assumption. Therefore, by Mazur's lemma in §29, the set co4>(So) is compact. Thus (29) shows that the closed set 4>(00) is contained in a compact set and therefore compact. (For different proofs see [31, p. 116 and 24, p. 29].) 32. Let 0 be an open rectangle in the plane E2 with coordinates Xl, X2. Let f be a C'-map E. Then to each c > 0 there is a polynomial P(XI,X2) such that

n-

for (Xl! X2) EO and i = 1,2. (See, e.g., [13, p. 68].) If f is just continuous, then the first of the above inequalities holds for some polynomial p. Corresponding statements hold for En with n > 2. 33. DEFINITION. Let H be an open subset of the Banach space E (which may be equal to E). Let K be a closed bounded subset of H. Then a component of H - K is a maximal open and connected subset of H - K. 34. We will need the following properties of components: (i) every point Xo E E - K determines uniquely a component C(xo) of E - K which contains Xo; (ii) among the components of E - K there is exactly one unbounded one; (iii) if B is a ball for which K c B, (30) then all bounded components of E - K are subsets of B; (iv) let Co be the unbounded component of E - K and let {Co.}o.EA be the set of the bounded components of E - K, where A is an index set (not containing 0). Suppose that K has a positive distance from aB. Then the components of B - K are the sets Co n B and Co., Q E A. Assertions (i) through (iii) follow easily from Definition 33 and the fact that E -11 is an open connected unbounded subset of E - K and therefore a subset of an unbounded component Co of E - K, while C(xo) = Co for every Xo E E - B. As to assertion (iv) we note that E - K = Co + Uo. Co.. If we intersect both members of this inequality by B and note that En B = B and, by (iii), Co. n B = Co., we obtain the equality B - K = Co n B + Uo.EA Co. which is easily seen to imply (iv). 35. THEOREM. Let X and Y be two arbitrary sets and let A and B be maps X ~ Y and Y _ X resp. Suppose that (a)

AB = [y,

(b)

BA = Ix,

(31)

24

FUNCTION-ANALYTIC PRELIMINARIES

where Ix and Iy denote the identity maps on X and Y resp. Then A is an isomorphism (i.e., a one-to-one and onto map 01 X onto Y) and B is an isomorphism 01 Y onto X. PROOF. Let Xl and X2 be points of X for which A(X1) = A(X2) and therefore BA(xd = BA(X2). But then by (31b) Xl = BA(xd = BA(X2) = X2· This shows that A is one-to-one. To show that A is onto, let y be an arbitrary point of B. Then by (31a) AB(y) = y. This shows that A maps the point X = B(y) onto the point y. This proves the assertion concerning A. The assertion concerning B is proved correspondingly. 36. Let S be an arbitrary set. Then a "vector space V generated by S" can be defined as follows: we consider all maps 1 of S into the reals R for which I(x) i= 0 for at most a finite number of elements xES. We define addition h + h of two such functions by setting

(/1 + h)(x) = h(x) + h(x),

(32)

and define (r f) for r E R by

(33)

(rf)(x) = rl(x). Finally for Xo E S we identify the function { 1 for x = xo, Ixo(x) = for x i= Xo

0:

with the point Xo of S. With this identification we see from the rules (32) and (33) that the function which maps the points Xl, X2, ... , Xn of S into the real numbers r1, r2, ... , rn resp. and which equals 0 for X i= Xi (i = 1,2, ... , n) may be written as the "linear combination" i=l

Obviously the space V thus defined is a vector space (over the reals). 37. Let V be the vector space defined in §36. A subset G of V is said to generate V if every element X of V may be written as n X

= 2:giTi,

gi

E G,

ri

E R,

i=l

with the integer n depending on x. A subset E of V is said to be linearly independent iffor any finite set 0"1, .•• , O"n of elements of E the relation Ti

E R,

i=l

implies that T1 = T2 = ... = Tn = O. A base of V is a subset B of V which generates V and is linearly independent. Obviously the set S is a basis for V. 38. Let B1 and B2 be two bases of V. Then B1 and B2 have the same cardinal number. For a proof see e.g. [30, IX, 1 and 2].

FUNCTION-AN ALYTIC PRELIMINARIES

25

39. THEOREM. Let Sand T be two arbitrary sets, and let V and W be the vector spaces generated by Sand T resp. We assume that there is a linear one-to-one correspondence l between the elements of V and W. Then for any bases Bv and Bw of V and W resp. c(Bv) = c(Bw), where c( ) denotes "cardinal number

(34)

0/'''

PROOF. Let Bw = l(Bv). By the properties of l there is a one-to-one correspondence between the elements of Bw and those of Bv. Therefore

c(Bv) = c(Bw).

(35)

Morever the elements of Bw are linearly independent, since a relation Bw, ri E R, implies that

I:~=1 iiVi = 0, Wi E

and therefore I: Viri = O. Since the Vi are linearly independent, we see that rl = r2 = ... = rn = O. On the other hand Bw generates W. For if y is an element of W, there is a unique point x E V such that y = lex). But since Bv is a base for V we see that n

x=Lb~ri'

i=1

b~ E Bv, ri E R.

Therefore y = I:l(b~)ri with l(b~) E Bw. Thus Bw is linearly independent and generates W, i.e., it is a base for W. Since Bw is also a base of W, we see from Lemma 38 that c(Bw) = c(Bw). This together with (35) implies the assertion (34).

CHAPTER 2

The Leray-Schauder Degree for Differentiable Maps §1. The degree for linear maps 1.1. Let I be a nonsingular linear Leray-8chauder (L.-8.) map of the Banach space E into itself. Let 0 be a bounded open subset of E and Yo a point of E. (8ee §§1, 9, 11 in Chapter 1.) We want to define the degree

d = d(l, 0, Yo).

(1.1 )

Now I maps E one-to-one on itself (assertion (iv) in §1.12). 8ince the degree is supposed to be an algebraic count of the number of coverings of Yo by 1(0), we want to set if Yo ¢ l(O), (1.2) d(l, Yo) = 1, if Yo E l(O).

0,

{O±

As pointed out in §3 of the introduction, we assume

Yo ¢ l(aO).

(1.3)

With this restriction it still remains to determine when to use the plus sign and when the minus sign in (1.2). It is natural to use the first one if I is the identity map. In the general case the decision will be based on Theorem 1.3 below. To state this theorem we need a definition. 1.2. DEFINITION. Let lo(x) = x - Lo(x) and h(x) = x - Ll(X) be two linear nonsingular L.-8. maps E ~ E. Then lo and h are called linearly L.-8. homotopic if there exists a continuous map L(x, t): Ex [0, 1] ~ E such that (a) for each t E [0,1] the map lex, t) = x - L(x, t) is a linear nonsingular L.-8. map E ~ E, and (b) l(x,O) = lo(x), lex, 1) = 11(X). A homotopy class H of nonsingular linear L.-8. maps is a family of such maps of the property: if 10 E H then all linear nonsingular L.-8. maps I which are L.-8. homotopic to lo are in H. 1.3. THE LINEAR HOMOTOPY THEOREM. The family of all nonsingular linear L.-S. maps E ~ E consists of exactly two homotopy classes H+ and H- , 26

d(J, 0, YO): YO A REGULAR VALUE

27

where H+ denotes the class containing the identity map l = I. For an example of an l E H-, see subsection 4.4 in Appendix A.

1.4. DEFINITION. I For each linear nonsingular L.-S. map E

-+

E the index

j(l) is defined by

if l E H+, if l E H-.

+ 1,

j(l) = {

-1,

(1.4)

1.5. REMARK. The reader will note: if E is finite dimensional with coordinates Xl, X2, ... , X n , then if determinant l > 0,

l E {H+,

H- , if determinant l < 0,

and the map (Xl, X2, ... , Xn) -+ (-Xl, X2,···, xn) belongs to H-. 1.6. Definition of the degree for linear L.-S. maps. Let land 0 be as in subsection 1.1, and let Yo be a point of E satisfying (1.3). We then define 0 del, 0, Yo) = { j(l)

if Yo rt leO), if Yo E l(O).

§2. The degree if Yo is a regular value for the map 2.1. Let 0 be as in Definition 1.6, and let point Yo E E satisfy

Yo

rt

J be an L.-S. map 0

J(aO).

-+

J E. Let the (2.1)

Suppose moreover that f E C'(O) and that Yo is a regular value for j (Definition 17 in Chapter 1). Then by Lemma 22 in Chapter 1 the equation I(x)

= Yo

(2.2)

has either no solution in 0 or a finite number of solutions, say Xl, X2, ••• , X r , and the differentials lp(h) = D j(xp; h), p = 1,2, ... , r, are nonsingular linear maps which by Lemma 18 of Chapter 1 are L.-S. Consequently their indices are defined (see 1.4). This makes the following definition possible. 2.2. DEFINITION. Under the assumptions and with the notations of subsection 2.1, we set d(J, 0, Yo) = 0 if the equation (2.2) has no roots in O. Otherwise we set r

d(J,O,yo)

= Lj(Dj(xp;h)).

(2.3)

p=l

2.3. The reader will notice: if E is finite dimensional, then j(Dj(x; h)) is the sign of the Jacobian of j at the point x. In the following lemma we list a number of elementary but basic properties of the degree. The proof will be given in the Notes to Chapter 2.2

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

28

2.4. LEMMA. The assumptions are the same as in Definition 2.2. In particular Yo is a regular value for f satisfying (2.1). I denotes the identity map. It is asserted:

(i) d(I, 0, Yo) =

if Yo E 0, if Yo Et

{1

0:

n.

-10, then the equation (2.2) has at least one root in O. This assertion is the base of existence proofs for solutions of (2.2). We note, however, that (2.2) may have solutions even if the degree equals zero. This is, for instance, the case if 0 is the interval (-1,+1) of the real axis, if f(x) = x 2 , and if 0 < Yo < 1 (cf. §3 of the Introduction). (iii) ("Sum Theorem"). Let 0 1 , O2 , ... ,0 8 be disjoint open subsets of 0 for which 8 8 (ii) If d(f, 0, Yo)

U Ou = n

and

Yo

Et

U f(80

u ).

u=l

u=l

Then

8

d(f, 0, Yo) =

L d(f, Ou, yo).

(2.4)

u=l

(iv) Suppose all roots x

E 0 of (2.2) lie in the open subset 0 1 of O. Then

d(f, 0, Yo) = d(f, 0 1 , yo).

(2.5)

d(f, 0, Yo) = d(fb, 0, Yo - b),

(2.6)

(v) For any point bEE where fb(x) = f(x) - b. (vi) Let Xo be a point of E, let ~ and let g(~) = f(x). Then

= X-Xo, let 0 0 =

{~E E

I x = ~+xo

(2.7)

d(f, 0, Yo) = d(g, 0 0 , yo).

(vii) With the notation of (vi) let h(~) d(f, 0, Yo

= f(x)

+ b) = d(h, 0 0 , b)

- Yo

EO},

= g(O -

for any bEE.

Yo. Then

(2.8)

Before stating a further property of the degree we need a definition. 2.5. DEFINITION. If there exists a subspace E1 -I E of E such that F(x) = x - f(x) EEl, then f is called a layer map, more precisely a layer map with respect to E 1 . Obviously f is then also a layer map with respect to any subspace E2 -I E which contains E 1. If f is a layer map with respect to a finitedimensional subspace we say: f is a finite layer map.

n

2.6. LEMMA. If the L.-8. map f(x) = x - F(x): ---+ E is a layer map with respect to a subspace E1 which is a direct summand of E (Definition 3 in Chapter 1), then for Yo E E1

(2.9) provided that f E G' , that Yo satisfies (2.1), and that Yo is a regular value for f.

d(f, n, YO): YO A REGULAR VALUE

29

PROOF. We first have to show that the right member of (2.9) is defined, i.e., that (2.1O) where 8 1 denotes the boundary of n n El (with respect to Ed. But if (2.1O) were not true, then (2.1) would not be true as follows from the inclusion

(2.11) which the reader will easily verify. We now turn to the proof of (2.9). Since Yo and F{x) both lie in Eb it follows that every root x of (2.2) lies in El and is therefore also a root of (2.12)

h{x) = Yo,

where h denotes the restriction of / to nnEI . The converse is obvious. Thus either both equations have no roots, or they have the same roots, say Xb X2, •.. , x r . In the first case (2.9) is true, both members of that equality being equal to zero by assertion (ii) of Lemma 2.4. In the second case r

d(f,n, YO) = L,j{D/{xp;h)),

hEE,

p=l

r

d{h,nnEbyo) = L,j{Dh{xp;hd), p=l

and our assertion (2.9) will be proved once it is shown that p = 1,2, ... ,T. Now, by assumption, F(x) EEl for all x E

DF(x,h)

n.

(2.13)

Therefore

E El

(2.14)

for all h E E.3 Also, by assumption, El is a direct summand in E, i.e., there exists a subspace & of E such that E is the direct sum of El and E 2 • and every h in E has the unique representation h = hI + h2, hI EEl. h2 E E 2. Thus D /(x; h) = hi

+ h2 -

DF{x; hd - DF(x; h 2 ),

(2.14a)

where, by (2.14), DF(x; ht) and DF(x; h 2 ) lie in E 1 • Therefore, (2.14a) reads in matrix form

Df(x; h)

~

(hi -D~(x;hd

-

D~;X;h,») ,

and since xp E E 1 , (2.15)

30

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

Putting a parameter t here in front of the term in the upper right corner, we see that the linear map (2.15) is L.-S. homotopic to the map

lp(h)=(DJt{~p;hd ~2)'

(2.16)

and, by the linear homotopy theorem, we conclude that P = 1,2, ... ,T.

j{D I(xp; h)) = j(lp(h)),

(2.17)

Finally it follows from (2.16) and Theorem 6.2 in Appendix A that j{lp(h)) = j{DJt{xp; hd). This equality together with (2.17) proves the assertion (2.13). We finish this section by proving some continuity statements which will be used in Chapter 4. 2.7. THEOREM. Let 0, I and Yo be as in subsection 2.1. Then there exists a positive Po with the lollowing property: ilO < P < Po then lor each y E B (Yo, p) (see §2 in Chapter 1) the lollowing assertions hold. (i) y ~ 1(£10), (ii) y is a regular value lor I, and (iii) il the equation (2.2) has solutions in 0, then the equation

I(x) = y

(2.18)

d(f, 0, y) = d(f, 0, yo).

(2.19)

has solutions in O. (iv)

PROOF. £10 and therefore, by Lemma 10 in Chapter 1, l{aO) is closed. Consequently (2.1) implies that Yo has a positive distance from l{aO). This together with Lemma 20 in Chapter 1 shows that Po can be chosen in such a way that assertions (i) and (ii) are satisfied. To prove the remaining assertions, we consider first the case that (2.2) has no solutions in 0 and therefore by (2.1) has no solutions in O. Then Yo ~ 1(0), and since by Lemma 10 in Chapter 1 I{O) is closed, Yo has a positive distance PI from ICri), and (2.18) has no solutions for y E B(yo, pd. Thus (iii) is vacuously satisfied, and (iv) holds, both members of (2.19) being zero by part (ii) of Lemma 2.4.

Now suppose (2.2) has roots in O. Since Yo is a regular value, there are only a finite number, say Xl, X2, ••• , XS. Now let 81 > 0 be such that the closures of the balls B(x", 8d, (1 = 1,2, ... , s, are disjoint from each other and from £10. By Lemma 20 in Chapter 1 we can also require that the differentials D I (x; h) are nonsingular for x E B{x",8I), (1 = 1,2, ... , s. Then by the inversion theorem in §1.21 there exists a positive 82 < 81 of the following property: there is a P2 = P2 (82 ) such that for Y E B(YQ,p),

0< P < P2,

(2.20)

THE ONE-DIMENSIONAL CASE

the equation (2.18) has for each xu(y) satisfying

0'

31

= 1,2, ... , s one and only one solution x =

0< p < P2.

(2.21 )

This proves assertion (iii). It remains to prove assertion (iv). Now, since by Theorem 4.6 in Appendix A the index j(l) of a nonsingular linear 1.-S. map is invariant under a continuous change of I, it follows from the continuity of Df(x;h) in x (cf. also the proof of Lemma 20 in Chapter 1) that a positive 62 can be chosen in such a way that j(Df(xu;h)) = j(Df(x;h)) for x E B(x u ,62 ). Thus for such x s

s

u=l

u=l

L. j(Df(xu(Y); h)) = L. j(Df(xu ;h)).

(2.22)

Here, by Definition 2.2. of the degree, the right member equals the right member of the asserted equality (2.10) by (2.19). We now show also that the left member of (2.22) equals the left member of (2.19) for small enough p. Again by definition of the degree this amounts to showing that for p small enough Xl(Y), X2(Y),.··, xs(y) are the only roots in 0 of (2.18) for Y satisfying (2.20). In other words, we have to show that for such y equation (2.18) has no roots in the set 8

0' =

n-

U B(x

u ,62 ).

u=l

To see this we have only to recall that xl, X2, . .. ,Xs are the only roots of (2.2) in 0 and that XCT E B(XCT' 62 ). Consequently Yo does not lie in the closed

set f(O') and has therefore a positive distance 263 from this set, and the ball B(YQ, 63 ) does not intersect f(O'). In other words, for y E B(yo, 63 ) all roots of (2.18) lie in n - n', as we wanted to show, i.e., not in 0'. 2.8. COROLLARY TO THEOREM 2.7. Lety = !/O(t), 0 $ t $ 1, be a continuous CUM1e consisting of regular values for f satisfying (2.1). Then d(f, 0, Yo(t» is independent of t.

Indeed by assertion (iv) of Theorem 2.7 the degree is constant in ''the small," and the corollary follows by applying the Heine-Borel theorem.

§3. The one-dimensional case and the case of polynomial maps 3.1. It is instructive to consider the above special cases. In the present section El denotes the Banach space ofreal numbers x with the usual norm Ilxll = Ixl. o will be the open interval (0:, ,8), where 0: < ,8 are two fixed points in El such that an is the union of the points 0: and,8. We start by stating some auxiliary lemmas whose elementary proofs will be given in the Notes.

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

32

3.2. LEMMA. Let f be a continuous map El. It is asserted: (i) if the equation

n ---+

El and let a be a point of

f(x) = a

(3.1)

has no roots x E n, then there exists an Cl > 0 of the following property: if 9 is a continuous map ---+ El and if b is a point of El for which

n

If(x) - g(x)1 } Ib-al

(3.2)


then the equation g(x) = b

(3.3)

has no roots in n. (ii) If (3.1) has roots in [a, /3] and these roots are contained in an open subset U of [a, /3], then there exists a positive Cl such that all roots of (3.3) lie in U if (3.2) is satisfied. 4

m

3.3. LEMMA. If f is a C' -map [a, ---+ El, then the set C f of critical points in [a, /3] is the set of roots of the equation

f'(x) = 0,

x E [a, /3],

(3.4)

and the set Sf of critical values of f is the set f(Cf)· If f is a polynomial on [a, /3] (i. e., the restriction of a polynomial to [a, /3]), then the sets C f and Sf are finite. 5 3.4. LEMMA. Let f be a polynomial in [a, /3] for which a and /3 are regular points. Let V be an open set containing the set Sf of singular values for f. Then there exists an C > 0 of the following property: if 9 is a polynomial in [a, such that If(x) - g(x)1 } (3.5) If'(x) _ g'(x)1 < C for x E [a, /3],

m

then Sg C V,

(3.6)

where Sg is the set of singular values for g.6

m.

Let r be a closed bounded set in El whose points are regular values for f. Then there exists an C > 0 such that all points of r are regular values for the polynomial g in [a, if (3.5) is satisfied. 7 3.5. LEMMA. Let f be a polynomial in [a,

m

m

3.6. THEOREM. Let f(x, t) be a continuous map [a, x [0,1] ---+ El. For each t E [0,1] the map ft(x) = f(x, t) is supposed to be C'. Let ao < al be two points in El, and let

a(t) = (1- t)ao + tal,

t E [0,1].

(3.7)

THE ONE-DIMENSIONAL CASE

33

It is assumed that for t E [0,1]

n ft(i)O) = 0, and a1 = a(l) are

a(t) and that the points ao = a(O) resp. It is asserted that then

(3.8)

regular values for fo and fI

d(fo,O,ao) = d(fI,O,at}

(3.9)

(Note: for some t in the open interval (0,1) the point a(t) may be a singular value for ft. Thus our theorem is not a consequence of Corollary 2.8 even if f does not depend on t.) 3.7. We will first show that for the proof of this theorem it will be sufficient to assume that f(x, t) is a polynomial in x and t. To this end we prove the following lemma.

n

3.S. LEMMA. Let f be a C' -map = [a,,8] ~ E1. Let a be a point of E1 which is a regular value for f and satisfies the relation

art f(80).

(3.10)

Then there exists a positive c of the following property: if 9 is a C' -map satisfying the inequalities

n ~ E1

Ig(x) - f(x)1 } Ig'(x) _ f'(x)1 < c for x E [a,,8],

(3.11)

then a is a regular value for g, and d(g, 0, a) = d(f, 0, a).

(3.12)

PROOF. We consider first the case that the equation (3.1) has no roots in (a, ,8). Then d(f, 0., a) = O. On the other hand, it follows from part (i) of Lemma 3.2 with b = a that there is a e such that the equation g(x) = a also has no roots if (3.11) is satisfied. Consequently d(g, 0., a) = 0 and the assertion (3.12) follows. Suppose now that (3.1) has roots on (a, ,8). Since a is a regular value for f, there are only a finite number, and if Xl, X2, ••. ,Xr are these roots, then f'(x p ) #- 0 (cf. Lemma 3.3). Therefore we can choose a 6 such that

o< 6 < 1

(3.13)

and such that the closures of the open intervals ip = (xp - 6, xp + 6) are disjoint from each other, lie in (a,,8), and are such that If'(x)1 > 0 for X E zp, and consequently If'(x)1 ~ 2ml for x E zp (3.14) for some positive number ml. We also note that by (3.10) the point a has a positive distance 2m2 from 80, and consequently (3.15)

if

If(x) - g(x)1 <

m2

for x E [a, ,8].

(3.16)

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

34

Finally to define an 0: of the asserted property we note that by part (ii) of Lemma 3.2 with b = a there exists an 0:1 > 0 such that all roots in (0:,13) of

g(x)

=

a

(3.17)

lie in the open set (3.18) p

if 9 satisfies (3.2). We now claim that 0: = min(0:1'~18'~2)

(3.19)

has the desired property. Then let 9 satisfy the assumption (3.11) with that 0:. It is clear from the sum theorem (part iii of Lemma 2.4) that r

d(f,O,a) = ~d(f,ip,a).

(3.20)

p=1

But since all roots in (0:,13) of (3.17) lie in the set U given by (3.18), we also have by the sum theorem r

d(g,O,a) = ~d(g,ip,a).

(3.21)

p=1

Therefore to prove the asserted equality (3.12) it will be sufficient to show that p = 1,2, ... ,T.

(3.22)

Now xp E ip is by definition a root of (3.1), and by (3.14) this is the only one in ipo Consequently,

d(f, ip, a) = j(D f(x p; h)) = sign !,(x p).

(3.23)

But also (3.17) has one and only one root in i p • Indeed there is at most one since by (3.14), (3.11), and (3.19)

Ig'(x)1

~

1!,(x)l- I!'(x) - g'(x)1

~ 2~1

- 0: ~ ~1 > 0 for x E i p.

To show that (3.17) has a root in i p , we note that by (3.14) f'(x) has a constant sign in i p • It will be enough to consider the case that f'(x) > 0 in ip since the case f'(x) < 0 in ip can be treated in a similar way. Then since a = f(x p) we see from the mean value theorem that for some ~p E ip

g(xp + 8) - a = f(x p + 8) - f(x p) + g(xp + 8) - f(x p + 8) = !,(~p)8

+ g(x p + 8) - f(x p + 8).

Therefore by (3.14), (3.11) and (3.19)

g(x p + 8) - a 2: f'(f;,p) -Ig(xp + 8) - f(x p + 8)1

2:

2~18

- 0: 2: ~18 > O.

THE ONE-DIMENSIONAL CASE

35

Similarly we see that a - g(xp - 6) > O. Thus g(xp - 6) see that (3.17) has a unique root xp E ip. Therefore

< a < g(xp + 6), and we

d(g,ip,a) = j(Dg(xp; h)) = signg'(xp).

(3.24)

But by (3.14), (3.13), (3.11), and (3.19), sign g'(x p) = sign f'(x p), and since as already noticed sign/'(x) is constant for x E ip, the assertion (3.22) follows from (3.23) and (3.24). 3.9. It is now easily seen that the assertion of subsection 3.7 is true: by Lemma 3.8 there exists an c such that with It(x) and a(t) as in Theorem 3.6

d(gi' 0, a(i)) if go and gl are C' -maps

n

---->

= d(h 0, a(i)),

i

= 0, 1,

(3.25)

El for which

Igi(X) -/i(x)1 } < c, Ig:(x) - Ii(x) I

i = 0,1.

(3.26)

Now by §1.32 there exists a polynomial g(x, t) such that

Ig(x, t) - I(x, t)1 } Ig'(x, t) _ f'(x, t)1 < c for x E [Q,,6], t E [0,1]' where the prime denotes differentiation with respect to x. Then (3.26) is satisfied with go(x) = g(x, 0) and gl(X) = g(x, 1), and therefore (3.25) is satisfied for these go and gl. Now suppose Theorem 3.6 is true for polynomials. Then (3.9) is true with 10 and II replaced by go and g1:

d(go, 0, ao) = d(g!, 0, at}. But this equality together with (3.25) proves (3.9). From now on we will suppose that f(x, t) is a polynomial. We will prove Theorem 3.6 first for the special case that f does not depend on t, i.e., the following lemma. 3.10. LEMMA. Eet I( x) be a polynomial on El. Let 0, {3, n be as in subsection 3.1 and let aCt) be defined by (3.7). It is as81Jmed that lor 0 :5 t :5 1

aCt) n 1(00) and that the points Go that then

= a(O)

and a1

= a(l)

= (21 are regular values lor

d(f, 0, ao) = d(f, 0, ad.

(3.27)

I.

We assert (3.28)

PROOF. Since ao and al are by assumption regular values for I, it is clear from Borel's covering theorem that it will be sufficient to prove that to every to E [0,1] there corresponds a r = r(to) such that the interval T(to) = (to - r, to + r) has the following properties: (i) for every t E T(to) the point aCt) is a regular value for I with the possible exception of t = to;

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

36

(ii) the degree

d(f, fl, aCt))

(3.29)

is constant in T( to) if a( to) is a regular value for I, and in T( to) - {to} otherwise. Assertion (i) is obvious from Lemma 3.3. Moreover, since by Lemma 3.3 the set Sf of singular values of I is finite, the first part of assertion (ii) follows from part (iv) of Theorem 2.7. Now suppose aCto) is a singular value for f. Then the equation

(3.30)

I(x) = aCto)

has roots in (a, (3). Since I is a polynomial, there are only a finite number of them. Call them Xl, X2, ... , Xr . Consequently there exists a 8 > 0 of the following property: the closures of the intervals ip = (xp - 8, xp + 8) lie in (a, (3) and are disjoint from each other. Moreover, by Lemma 3.3 we can choose 8 such that also (3.31 ) I'(X) "# 0 for X E zp - {x p}. On the other hand, since aCt) is continuous, we see from part (ii) of Lemma 3.2 (with g(x) = I(x)) that there exists a positive 71 such that all roots X E (a, (3) of the equation

(3.32)

I(x) = aCt) lie in the union U of the ip if

t E T 1 (to) = (to - 71. to

+ 7d·

(3.33)

From this together with (3.31) we see that for t E T1 (to) - {to} the point aCt) is a regular value for f. Therefore on account of (3.27) we see that for such t the degree d(f, fl, aCt)) is defined and that by the sum theorem r

d(f,fl,a(t)) = "Ld(f,ip,a(t)) for t E T1 (to) - {to}. (3.34) p=l Now to define a subinterval T(to) of T 1 (to) containing to such that the degree (3.34) is constant for t E T(to) - {to}, we note that by (3.31) I is strictly monotone in each of the open intervals.

i-; = (xp -

8, xp),

it = (xp, Xp + 8).

(3.35)

Since I(x p) = aCto), it follows that m

= min(\/(xp + 8) - a(to)\, \/(xp - 8) - aCto)\) > O. p

Consequently there exists a 72

(3.36)

> 0 such that

\a(t) - a(to)\ <

m

(3.37)

if \t - to\ < 72· Now let r = min(71, 72) and T(to) = (to - r, to + r). To show that the degree (3.29) is constant for t E T(to) - {to} it is by (3.34) sufficient to prove that for these t p = 1,2, ... ,r,

(3.38)

THE ONE-DIMENSIONAL CASE

37

where cp is independent of t. But for fixed p the sign of f'(x) is constant in each of the intervals (3.35). We distinguish two cases: (1) signf'(x+) = -signf'(x-); (II) signf'(x+) = signf'(x-), where x+ and x- denote arbitrary points in it and if; resp. To prove (3.38) in Case I it will be enough to consider the case that (3.39) the proof for the case f'(x+) < 0, f'(x-) > 0 being quite similar. Now, by (3.39), the restriction of f to ip takes its absolute minimum at x = xp. Therefore the equation (3.32) has no roots in ip if

a(t) < f(x p ) = a(to),

(3.40)

and therefore by part (ii) of Lemma 2.4 for such t

dU, i p , a(t)) =

O.

(3.41)

We now claim that (3.41) is still true if

a(to) < a(t) < a(to)

+ m.

(3.42)

Indeed by (3.39) f decreases monotonically as x increases from Xp - h to xp. Consequently, by definition of m the equation (3.32) has exactly one root x- E if; if aCt) satisfies (3.42), and therefore if

to < t < to

+ T.

(3.43)

Similarly one sees that (3.32) has exactly one root x+ E it if t satisfies (3.43). Consequently, for such t, by the sum theorem and by (3.39) d(f,ip,a(t)) = d(f,it,a(t» +d(f,if;,a(t» = signJ'(x+) + signJ'(x-) = 0,

we wanted to show. It remains to consider the simpler Case II where sign I' (x) is constant in the whole interval ip except possibly at x = xp at which point I' may be zero. In any case I is strictly monotone in all of i p , and (3.32) has exactly one solution on x E ip if to - T < t < to + T. Consequently d(J,ip,a(t» = signl'(x), and therefore is constant in ip - {xp} by definition of Case II. 3.11. Coneluiora 01 the proof 0/ Theorem 3.6. As we saw it is sufficient to assume that /(x, t) is a polynomial. We note first of all that there exists a positive constant ml such that 88

dist(a(t), h(an» > 3ml

(3.44)

for all t E [0,1J. Indeed otherwise there would exist a sequence of points x"l E 80 and t.., E [0,1J such that for at least one of the points a., f3 of an, say 0, lim[a(t..,) - ft..,(a.)] = O. We may assume that the t"l converge to a point

38

THE L.-8. DEGREE FOR DIFFERENTIABLE MAPS

1 E [0,1]. But then a(l) = ft(a), and this equality contradicts the assumption (3.8). Next we note that if R(fto) denotes the set of all y E EI which are regular values for fto (as polynomials in [a, )3]), if J is an interval whose closure does not intersect f(80), and if (3.45) Y E R (fto) n J,

then we may define

d(fto' 0, R(fto) n J) = d(fto' 0, y),

(3.46)

since by Lemma 3.10 the right member of (3.46) has the same value for all y satisfying (3.45). In particular d(fto' 0, R(fto) n B(a(to), mil) is well defined (cf. (3.44)). Corresponding to each to E [0,1] we will now define a positive number 7(to) such that the intervals T(to) = (to - 7(to), to + 7(to)) have the following property: if hand t2 are a couple of points in [0,1] for which the intersection T(td n T(t2) is not empty, then

d(ftl' 0, R(ftl) n B(a(tIl, mil) = d(ft2' 0, R(ft2) n B(a(t2)' mil).

(3.47)

Before constructing a 7(tO) of the asserted property, we remark that (3.47) implies the assertion (3.9) of our Theorem 3.6. Indeed using Borel's theorem we conclude from (3.47) that

d(fo, n, R(fo) n B(ao, md) = d(fl, n, R(fd n B(al, md).

(3.48)

But by assumption ao = a{O) is a regular value for fo. Therefore by (3.46) the left member of (3.48) equals the left member of (3.9). In the same way one sees that the right members of (3.48) and (3.9) are equal to each other. We now turn to the definition of 7(tO). Since for fixed to E [0,1] there are at most a finite number of points in EI which are singular values for fto' there exists a positive m(to) such that

m{to) < ml

(3.49)

and such that all points of the closure of the interval

J(to, m(to))

=

B(a(to), m(to))

(3.50)

are regular values for fto with the possible exception of the point a(to). In particular the point y{to) = a(to) + m(to)/2 (3.51) is a regular value for fto. Consequently by Lemma 3.8 and the continuity properties of the polynomial ft(x) = f(x, t), there exists a positive number 7(tO) of the following property: if It - tol < 7(tO), (3.52) then y( to) is a regular value for ft and

d(ft,n,y(to)) = d(fto,n,y(to)).

(3.53)

d(f, 0, YO): YO NOT NECESSARILY REGULAR

39

We now define T{tO) as a positive number satisfying (3.54) Then

Iy{t) - y{to)1 < 2m!

(3.55)

It - tol < 2T{tO)'

(3.56)

if Indeed by (3.51), (3.7), (3.49), (3.56), and (3.54)

+ m{t)/2 - (aCto) + m{to)/2)1 la{t) - a{to)1 + (met) + m{to))/2 It - tollal - aol + ml < T{tO) + m! < 2m!.

Iy{t) - y{to)1 = la{t) ~

~

This proves (3.55). Now yet) and y{to) are regular values for It. Therefore the inequality (3.55) in conjunction with (3.44) and Lemma 3.10 shows that

d(ft,n,y{t)) = d(ft,n,y{to))

(3.57)

T{to) = B{a{to), T{tO))

(3.58)

if (3.56) is satisfied. The intervals form an open covering of [0,1] as to varies over that interval. We are now ready to prove that (3.47) holds for any couple tl, t2 of points in [0,1] for which T{ tl) n T{ t2) is not empty. Then if t2l is a point of this intersection, we see that It2 -tll :5 It2 -t211 + It21 -t11 :5 T(t1)+T(t2). We may assume that T{td ~ T{t2)' Then (3.59)

= it and t = t2. Therefore by (3.54) and (3.53) d(ft2,n,y(tl» = d(fh,O,y(tt). (3.60) But a1so (3.57) holds with t = t2 and to = tl' Therefore by (3.60)

i.e., (3.56) is satisfied with to

d(fh' n, y{t2» = d(fh' 0, y(tt». By the definition (3.46) this equality is equivalent to (3.47).

§4. The degree for a not necessarily regular value Yo The definition of the degree in this case is based on Theorems 4.1 and 4.2

below.

40

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

4.1. THEOREM. Let f E C'(o') be an L.-S. map 0' --- E, and let Yo be a point of E satisfying (2.1). Then every neighborhood of Yo contains a point a which is a regular value for f. 4.2. THEOREM. Let f E C"(o') be an L.-S. map, and let open set satisfying r n f(80) = 0. Then d(f, 0, y) is constant for all points y E

r

r

be a connected

(4.1)

which are regular values for f.

Obviously on account of these theorems the following definition is legitimate. 4.3. DEFINITION. Let f E C"(rl) be an L.-S. map 0' --- E, and let Yo be a point of E satisfying (2.1). Let U be an open connected neighborhood of Yo not intersecting f(80), and let y be a point of U which is a regular value for f. We then define the degree d(f, 0, Yo) by the equality d(f, 0, Yo) = d(f, 0, y).

(4.2)

The proofs of Theorems 4.1 and 4.2 are based on the following theorem, which we refer to as the Sard-Smale theorem and which is a special case of Smale's generalization of Sard's theorem. 4.4. THEOREM OF SARD- SMALE. Let II be a Banach space, and let E be a direct summand of II which is of finite codimension n (see the remark to §1.3). Let Z be a bounded open subset of II, and let ¢ E CT(Z), where r is a positive integer, be an L. -So map Z --- E. We assume that n
The proof of this theorem will be given in Appendix B. 4.5. Proof of Theorem 4.1. We apply Theorem 4.4 with II = E, Z = 0, and ¢ = f. Then n = 0 and the assumption (4.3) is satisfied for r = 1. Therefore our theorem is a consequence of the Sard-Smale theorem. The proof of Theorem 4.2 is considerably harder. 4.6. Start of the proof of Theorem 4.2. We have to show that under the assumptions made d(f, 0, Yo) = d(f, 0, yt} (4.4) for any couple of points Yo, Yl in r which are regular values for f. The first step for the proof of (4.4) consists in showing that it is sufficient to consider the special case that the segment YOYl (see §1.23) lies in r. Since the points Yo and Yl lie in the open connected set r, there exists a simple polygon Per connecting Yo and Yl (see §1.25). Let Po = Yo, Pi, ... , Pn = Yl be the vertices of P. Except for the vertices Po and Pn these vertices are not necessarily regular values for f. But we may as follows replace P by a simple polygon Q c r whose vertices are regular values for f: since 8r is closed and P is compact, these two sets have a positive distance. For any positive number pless

d(f, n, yo): Yo NOT NECESSARILY REGULAR

41

than this distance, the balls B(pi, p), i = 0, 1, ... , n, lie in r. Now by Theorem 4.1 we can choose for i = 1,2, ... ,n - 1 a point qi E B(pi, p) which is a regular value for f. Then the simple polygon Q whose vertices are qo = Yo, ql. ... , qn-l. qn = Yl lies in r. Indeed the edge f3i = qiqi+l of Q is given by f3i(t) = (1 - t)qi

+ tqi+l,

0 ::; t ::; 1,

while the edge PiPi+1 of P is given by

o ::; t ::; 1. Therefore

+ Ilqi+1 - Pi+ll1t < p, which obviously proves the assertion that Q c r. We thus see that (4.4) lIf3i(t) - Qi(t)11 ::; IIqi - Pill(l- t)

will be

proved once we can show that

d(f, fl, qi) = d(f, fl, qi+d for i = 0,1, ... ,n - 1. 4.7. It is now clear that for the proof of (4.4) we may assume that the segment is contained in r. The assertion (4.4) is trivial if neither the equation (2.2) nor the equation f(x) = Yl has a root in fl, since then by Lemma 2.4 both members of (4.4) equal zero. We therefore will always suppose that at least one of these equations has a root in fl, and we may then assume that (2.2) has at least one such root. Our next step consists in changing Yo and Yl into points Yo and Yl resp. which have the following property: if 0' = (1-

t)yo + tlh,

0::; t ::; 1,

(4.5)

then 0' is transversal to J (see §1.26). 4.8. We first show that if p is a positive number which is not only smaller than the distance of the segment Yotll from ar but which also has the properties asserted in Theorem 2.7 and those obtained by replacing Yo by til in that theorem, then the 8II8erted equality (4.4) will be implied by the equality

d(f, fl, Yo)

= d(f, fl, Yl)'

(4.6)

provided that

IIYO-YolI} IIYI - Yll1

<po

(4.7)

Indeed, by Theorem 2.7, every Yo E B(Yo,p) is a regular value for

(4.8)

f and satisfies d(f, fl, Yo) = d(f, fl, Yo),

(4.9)

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

42

and the equation

f{x)

=

Yo

(4.10)

has a solution. Moreover every Yl satisfying (4. 7) is a regular value for f and satisfies (4.11) d(f, 0, yd = d(f, 0, yd· (4.9) and (4.11) show that (4.6) implies (4.4). If Yo E B{yo, p) is given, we will always choose Yl = Yo + Yl - Yo· (4.12) This Yl obviously satisfies (4.7). 4.9. To find a Yo satisfying (4.7) such that the segment (j C r given by (4.5) (with Yl defined by (4.12)) is transversal to f, we introduce the Banach space II which is the direct sum of E with the real line R considered as a one-dimensional Banach space (see part (ii) of §1.3). If rl is a unit element in R, the elements of II are of the form z = x + trl with x E E, t a real number, and the norm Ilzll = Ilxll + Itl· Let Z be the open bounded subset of II defined by Z = {z = x + trl I x EO, -"I < t < 1 + "I},

(4.13)

where the positive number "I is chosen in such a way that the extension of the interval (4.5) to the interval -"I ~ t ~ 1 + "I still lies in r. Finally we define a map g: Z -+ E by g(x + trd = f(x) - t(Yl - Yo). (4.14) 4.10. LEMMA. [fyo is a regular value for g, then (j is transversal to f.

PROOF. By (4.5) and (4.14) the equation

g(x + trd = Yo

(4.15)

+ t(Yl -

(4.16)

is equivalent to

f(x) = Yo

Yo) E (j.

But if Yo is a regular value for g, then for x, t satisfying (4.16) and therefore (4.15) the differential

Dg{x + trl; h + rrd

=

D f(x; h) - r(Yl - Yo),

(4.17)

where h E E and T, a real number, is a map II onto E. This proves the transversality of u.

4.11. LEMMA. There exists a Yo E B(yo,p) which is a regular value for g. PROOF. The subspace E of II is of codimension n = 1. By assumption C"{O). Therefore, by (4.14), g E C"(Z). Thus the assumption (4.3) of the Sard-Smale theorem 4.4 is satisfied, and our lemma follows from that theorem. 4.12. To finish the proof of Theorem 4.2 we have to show: if Yo and Yl are chosen as above, then (4.6) holds. We will follow the lines of the proof given by Elworthy and Tromba [20, pp. 74 ff.] in the case of Banach manifolds

f

E

d(f, 0, YO): Yo NOT NECESSARILY REGULAR

43

with considerable simplifications made possible by the facts that we deal with a "plane" space and that the "transverse" set a is a straight line segment. We recall that the sets f-l(yO) and f-1(Yl) are finite and that (see subsection 4.7) we may assume that f-l(yO) is not empty, while f-l(yd mayor may not be empty. Denoting the elements of f-l(yO) by Xl. X2, ... , Xr and those of f-l(yd, if this set is not empty, by Wl,W2, ..• ,Ws , we see from Definition 2.2 that our assertion (4.6) is equivalent to

~ ·(Df( .. » = { 'L.j=l j(Df(wa ;·)) if f-1(Yl) i= 0, xp, 0 if f-l(yd = 0.

~J

(4.18)

For the proof of (4.18) we will need the following properties (a)-(d) of the set

f-l(a): (a) f-l(a) is the union of maximal connected disjoint subsets called path components each of which is determined by any of its points; (b) if Xo is one of the points (4.19) then the path component Co determined by Xo may be written in the form X

o ~ s ~ 1,

= xo(s),

xo(O) = Xo

(4.20)

with xo(s) being continuously differentiable. Xo will be called the starting point of Co; (c) with Co as in (b) the point xo(1) is again one of the points (4.19); (d) with the above notations

xo(1)

i= xo(O) =

Xo.

(4.21)

A sketch for a proof of these properties will be given after assertion (4.18) has been established.

For the proof of (4.18) we establish certain relations between the indices j occurring in the sums of that equality. We first consider the case that f- 1 (Yl) is not empty. Let Cp : X = xp(s) be the path component with xp(O) = xp for some integer p ~ r. By properties (c) and (d) there are two possible cases: { Case I. Case II.

xp(1) xp (1)

= X a, = Wa ,

1 ~ u ~ r, u '" p; 1 ~ U ~ s.

(4.22)

We will prove that

·(Df(

J

.» -- { j(Df(wai.» - j(Df(xITi·»

xp,·

in Case I, in Case II.

(4.23)

It is easy to see that (4.23) implies our assertion (4.18): iffor some p ~ r Case I takes place, then by (4.23) the pth and uth terms in the sum of the left member of (4.18) cancel each other. Thus the "reduced sum" obtained by keeping only those terms for which Case II holds has the same value as the original sum. Let

44

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

r' be the number of terms in the reduced sum, and let s' be the number of the correspondingly reduced sum of the right member of (4.18). Then r' ::; s'.

(4.24)

This is obvious if r' equals 0 or 1. But if r' 2:: 2 and if P1 and P2 are two different p-values ::; r', then the path component xP1 (s) and xP2 (s) starting at xP1 and xP2 resp. are disjoint, and therefore xP1 (1) f:. xP2 (1); and since we are in Case II, W U1 = x P1 (1) and W U2 = x P2 are different members of the reduced sum of the right member of (4.18). This implies (4.24). But since s' ::; r' is proved the same way, we see that r' = s', and the equality of the reduced sums (and therefore the original sums) in (4.18) follows from the part of (4.23) referring to Case II. This proves (4.18) in case f- 1 (fh) is not empty. But if this set is empty, there are no roots Wu of f(x) = 'iit, and therefore Case I in (4.22) holds for all terms in the sum of the left of (4.18), and by the previous argument this shows that this sum is zero as was to be proved. It remains to prove (4.23). It is clear that the indices appearing in that equation are defined, i.e., that the differentials at xp = xp(O) and Xu ( = xp(1) in Case I) and Wu ( = xp(1) in Case II) are nonsingular since the points Yo and Y1 are regular values for f and f{xp) = f{xu) = Yo and f(w u ) = Y1' However the differentials of f at X = xp{s) are not necessarily nonsingular for all s E [0,1] since f{xp{s)) E (1, i.e.,

f{xp{s)) = Yo

+ t{Y1 -

Yo),

0::; t = t{s) ::; 1,

(4.25)

and since not all points y E (1 are necessarily regular values for f. We therefore will define for each s E [0,1] an auxiliary L.-S. map 4>8 (x) which has the following properties: (A) the differential of 4>8 at xp{s) is nonsingular; (B) the index of the differential referred to in (A) is independent of S; (C) for those s E [0,1] for which the differential of f at x = xp(s) is nonsingular where t = t(s) is given by (4.25)

j(Df{xp{s); h)) = - sign :!J{D4>(xp{s); h)).

(4.26)

The definition of 4>8 is as follows: for a fixed s = So E [0,1] we denote by Xl the (one-dimensional) tangent space to xp{s) at s = So, and by X 2 a complementary space to Xl. If moreover 6 is the unit vector in Xl directed in the direction of increasing s, then every x E E has the unique representation (4.27) On the other hand we denote by Y1 the one-dimensional subspace of E generated by Y1 - Yo, and by Y2 a complementary subspace to Y 1 (independent of so). If el = (Yl - Yo)IIYl - YolI- l , then every y E E has the unique representation y = (Jel

+ Y2,

(J real, Y2 E Y2 .

(4.28)

d(f, n, yo): YO NOT NECESSARILY REGULAR

45

The projections y -- (jel and y -- Y2 will be denoted by 71"1 and 71"2 resp. Finally if ao is the unique real number in the representation (4.27), for x = x(so) we define 4>80 (X) = f(xp(so)) + (a - ao)el + 71"2 (/(X) - f(xp(so)). (4.29) To prove property (A) we note first of all: if

h = 6hl

+ h2,

hi real, h2 E X2,

(4.30)

then Df(xp(so)j hd maps the tangent space Xl to x(s) at s = So into the tangent space to f(xp(s)) at s = So. But f(xp(s)) E (j C Yl for aIls. Thus this tangent space is Yl , and we see that (4.31) and therefore

7r2Df(xp(so)j 6hd = O. Since 7r2el

(4.32)

= 0 we see from (4.29) that also (4.33)

7r2D4> 80(xp(so)j 6hd = O. Taking this into account we see from (4.29) and (4.32) that

D4>80(xP(so)j h) =

(h~el 7r2Df(x~(so)j h2) ) .

(4.34)

To prove that this map is not singular, it will be sufficient to show that

7r2Df(xp(so)j h2) maps X 2 onto Y2. Then let Y2 be an arbitrary point of Y 2. Since (f is transversal to f, the differential D f (xp (so) j h) together with Y1 spans E. Thus there is an hE E and a real a such that Y2 = Df(xp(so)j h) + ael' But Y2 = 7r2Y2 since Y2 E Y2. Thus we see that Y2 = 7I"2Df(xp(so)j h). Now if h = hlel +~, hi real, ~ E E2, we see that Y2 = Df(xp(so)j 6hd + 1I"2Df(zp(so); ~). But here, by (4.32), the first term at the right equals zero. ThUB Y2 is the image of the point ~ E X2 as we wanted to show. To prove property (B) of 4>. we note that the map (4.34) depends continuously on SO 18 follows from the continuous differentiability of f and xp(s) since 11"2 is by de8Dition independent of So. The constancy of the index now follows from Theoniaa 5.2 and 4.12 of Appendix A. To prove property (C) we consider an So for which the differential of f at xp(so) is not singular. From (4.31) and (4.30) we see that

D/(zp(tlo);h)

= (Df(Xp(So)jh1ed o

1I"1Df(ZP(SO)jh2)). 7r2Df(xp(so)jh2)

(4.35)

Denoting for a moment the right member of (4.25) by lo(t) we see from (4.25) and definition of el that D/(z,(SO)jelh)

= ~o t

(ddt) S

hi 8=80

.

= elhlllJh -Yoll (~!) 3=30

(4.36)

46

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

Substituting this in (4.35) one sees easily that the map (4.35) is linearly L.-S. homotopic to the map

mo(h) = (e1h1(dt/dS)s=so

o

0

)

'Tr2Df(xp(so); h2)

and that therefore the index of the differential (4.35) equals j(mo). But using the linear homotopy theorem, the multiplication theorem 6.1 of Appendix A for indices, and the fact that the index of an 1- map equals -1 (see Definition 4.4 and Lemma 4.5 of that Appendix) one sees easily that j(mo) equals sign(dt/ds)s=so times the index of the map D¢so given by (4.34). (Note that dt/ds #- 0 for s = So since otherwise by (4.35) and (4.36) the differential (4.35) would be singular.) The assertion (4.23) follows now easily from the properties (A), (B), and (C) of the auxiliary map ¢s' Consider Case 1. xp = xp(O) and f(x p) = Yo' Thus as s moves increasingly from s = 0 to positive s the point f(xp(s)) moves from Yo to interior points of u, i.e., from t = 0 to positive t (cf. 4.5). Therefore dt/ds > 0 and by property (C),

j(D f(x p; h)) = j(D¢o(xp(O); h)).

(4.37)

But as s tends to 1 increasingly the point f(x p (l)) = Yo is approached through interior points of u, i.e., t approaches t = 0 through positive t. Thus dt/ds < 0 and, by property (C)

-j(Df(xu ; h)) = j(D¢l(X(l); h)).

(4.37a)

But by property (B) the right members of (4.37) and (4.37a) are equal. This proves (4.23) in Case 1. Case II is treated correspondingly. This finishes the proof of (4.18), and we start outlining a proof of properties (a)-(d) of the set f-l(U) (stated shortly after equation (4.18)). We note first of all that by §1.1O this set is compact since f is an 1.-8. map and u is compact. Now let Yo be a point of u and let

Yo = f(xo).

(4.38)

We then assert the following statements (0:) and (13): (0:) If Yo is an interior point of u, then there is an open interval i which is diffeomorphic to a connected open piece of f-l(U) containing Xo. (Two sets are called diffeomorphic if there is a one-to-one relation between them which is onto and continuously differentiable both ways.) (13) If Yo = Yo, then there exists an € > 0 such that the half-open interval 0 :::; r < € is diffeomorphic to a connected piece of f-l(U) with T = 0 corresponding to Xo under the diffeomorphism. The corresponding statement holds if Yo = Y1 . Now assertion (0:) is obvious if Yo is a regular value for f. For then there exists a ball B(yo, r) which is diffeomorphic to an open neighborhood of Xo and whose radius r satisfies the inequality r < min(llyo - Yoll ,IIYl - yol!)·

(4.39)

d(f, fl, yo): YO NOT NECESSARILY REGULAR

47

Then obviously the restriction of this diffeomorphism to the interval B(yo, r) n (j has the required properties. The corresponding argument proves assertion ({3) since Yo and Yl are regular values for f. It remains to prove (0:) for the case that Yo is not a regular value of f. Without loss of generality we may assume Xo

= Yo = f(xo) = (}

( 4.39)

(as one sees by using a translation ~ = x - Xo, TJ = y - yo). Now by the transversality of (f with respect to f the range R of D f (xo; h) together with the I-space generated by Yl - Yo spans E, i.e., the latter subspace is a co-kernel K* of the differentiallo(h) - Df(xo; h): E=R+K*

(4.40)

and by (4.39) Yo = (} C

(j

C K*.

(4.41 )

K* is one dimensional and since, by the KrasnoselskiY Lemma 18 of Chapter 1, the differentiallo is L.-8., it follows from Lemma 12 of that chapter that its kernel K is also one dimensional. Then there is a subspace X 2 of E complementary to K such that (4.42)

It follows from (4.40) and (4.42) that the assumptions of Lemma 5 of Appendix B are satisfied with

{

= E, p = 0, Z = fl, Zo = Xo = 0, III = X 2 ,

= x, = f.

z

II

¢

(4.43)

This lemma states: there are "coordinate transformations" y = h, c;- = h1(x) given by equations (24) and (25) of Appendix B resp. which have the following property: there are open neighborhoods U and V of the zero points for x and c;resp. such that hi is a diffeomorphism of U on V under which the zero points correspond to each other. Moreover if.,p is the map f in the "new coordinates," i.e., if \be,) = hfhl1(~), then, for ~ E V, .,p(~) is of the form given in equation (5) of Appendix B, where ~l and ~2 are determined by

(4.44)

(d. 4.42), and where x(~) C K*.

(4.45)

Since V is an open neighborhood of 0, we may choose e > 0 such that the interval ie defined by ~l = 0, k21 < e lies in V. By (5) in Appendix Band (4.45), we see that .,p(ie) C K*. But from (4.41) we see that for small enough e

t/J(ie) C i.e., ie E t/J-l(u).

(f,

(4.46)

48

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

"Transforming back" it is now easy to see from the properties of hand h1 that

hl1(ieJ

C

f-1(U)

and that this map of i~ into f-1(U) has the desired properties. From properties (Q) and (f3) just proved we conclude that a path component Co of f-1(a) is a one-dimensional C1-manifold (see, e.g., [1]). Moreover Co is closed, and as a subset of the compact set it is compact. Now it is known that a compact one-dimensional C1-manifold is diffeomorphic either to a circle or a line segment. We finally show that the first cannot happen if Xo is one of the points (4.19). Then f(xo) equals either Yo or Y1; each of these points is a regular value of f. It will be sufficient to consider the case f(xo) = Yo. Then there exists an r > 0 such that the ball B(xo; r) and the open neighborhood Vo = f-1(B(yo, r)) of Xo are diffeomorphic. Now if the path component Co containing Xo is diffeomorphic to a circle, it may be represented by x = Xo (8) where {

0~

8

~ 1,

xo(O)

=

xo(l)

=

Xo,

XO(8t} :f. X(82)

(4.47)

if at least one of the disjoint points 81. 82 is different from zero or 1. Moreover f(XO(8)) E U, i.e., by (4.5)

f(XO(8)) = Yo + t(Y1 - Yo) = Yt

(4.48)

for some t E [0,1]. By (4.47) there exist points 81,82 in the open interval (O,~) such that f(XO(8)) C U for 0 ~ 8 ~ 81 < ~, 8)) C U for 0 ~ 8 ~ 82 and, by (4.48), points tl, t2 in [0,1] such that

f (xo (1 -

f(XO(8t}) =

Ytll

<

~,

f(xo(1- 82)) = Yt2·

Now let to = min(t1, t2). Then there exist 8',8" such that 1 O < 8' < _ 81 < "2'

0 < 8" < 1 _ 82 < "2'

(4 .49)

and such that f(XO(8')) = f(xo(1 - 8")) = y(to). But by the diffeomorphism between B(Yo,r) and Vo, this implies that XO{8') = xo(1- 8"). This equality in turn implies that 8' = (1 - 8") as seen from (4.47). Thus 8' + 8" = 1. This however leads to a contradiction since 8' + 8" < 1 by (4.49).

4.13. LEMMA. The a88ertion8 of Lemma8 2.4 and 2.6 are true without the a8sumption that Yo is a regular value for f provided that f E C". The proof will be given in the Notes. 8 ,g We finish this section by proving the important continuity properties of the degree: under proper assumptions d(f, 0., y) is continuous in Y (Theorem 4.14) as well as in f (Theorems 4.15 and 4.16). Note that continuity is equivalent to constancy since the degree is integer-valued.

d(f,n, YO): YO NOT NECESSARILY REGULAR

49

4.14. THEOREM. Let f be an L.-8. map 0 ~ E which lies in C"(O). Let y = y(t), 0 ~ t ~ 1, be a continuous curve in E satisfying

yet)

~

f(80)

for 0

~

t

~ 1.

(4.50)

Then d(f,O,y(t)) is independent oft (cf. Definition 4.3). PROOF. Assumption (4.50) says that the curve yet) lies in one component of E - f(80). The assertion follows therefore immediately from Theorem 4.2. 4.15. THEOREM. Let

ft(x) = f(x, t) = x - F(x, t), where F(x, t) is a completely continuous twice-differentiable map 0 Let Yo E E satisfy Yo i- f(x, t)

for (x, t) E 80 x [0,1].

(4.51 ) X

[0, 1]

~ E.

(4.52)

Then (4.53)

is independent of t. PROOF. As already mentioned in the introduction, the proof follows an idea used by Nagumo [42, pp. 492, 493] in the finite-dimensional case: namely, to apply Theorem 4.14 to a certain mapping defined on a subset of the Banach space II (cf. subsection 4.9) which is the direct sum of E with the axis of the reals. Since dt is an integer-valued function of t, it will be sufficient to prove that dt is continuous at every point to E [0,1]' i.e., to show that there exists a positive 6 = 6(to) such that d(ftl' n, Yo) = d(fto' n, Yo) (4.54) if

Ih -tol < 6, Now for arbitrary 'o,tl in [0, 11 we set

h, to E [0,1].

rex) = (1 - s)fto(x) + sft, (x),

0 ~ s ~ 1.

(4.55) (4.56)

It is clear from (4.51) that rex) is an L.-S. mapping. In order to see that the degree d(/., 0, flo) exists we will prove: there exist positive numbers e and 6 such that IIr(x) - Yo II ~ e (4.57) for all x E and to, tl E [0,1] satisfying (4.55). If there were no such couple £,6, there would exist sequences of points t~,tr,sll, v = 1,2, ... , in [0,11, a sequence of positive numbers ell, and a sequence of points XII E 8n such that

ao

It1- tol

~ 0,

(4.58)

and

YII = (1 - sll)f(xll , to) + Sll f(x ll , tl) ~ Yo

(4.59)

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

50

as v ->

(see 4.56) and (4.57)). Obviously we may assume that the sequences are convergent:

00

to, t'l, SV

lim

£1--00

to

=

lim

v--co

t'l

lim

= II,

SV

=

v--+oo

s.

(4.60)

But by (4.59) and by (4.51)

Yv = Xv - [(1 - sV)F(xV, to)

+ SV F(xv, tJ.)j-> Yo,

(4.61)

and by the complete continuity of F the terms in the bracket are convergent for some subsequence Vi, i = 1,2, ... , of the integers v. It now follows from (4.61) and (4.60) that the XVi converge to a point x, and we see that

Yo = (1- s)/(x,lt}

+ s/(x, It}

=

/(x, It}.

(4.62)

But x E ao since XVi E ao and ao is closed. Thus (4.62) contradicts the assumption (4.52). Then let to be a given point in [O,lj, and let €,b be chosen such that (4.57) holds for tl satisfying (4.55). With such a fixed tl, our assertion (4.54) is, by (4.56), equivalent to the equality d(f°, 0, Yo) = d(fl, 0, Yo) and will therefore be implied by the assertion

d(r, 0, YO) = const.

for 0 ::; s ::; 1.

(4.63)

To prove this assertion we introduce the Banach space TI defined in subsection 4.9 and use the notation employed there. We define the cylinder Z C TI by

Z = {z =

X

+ tTl

E TI

I X EO,

-'Y/

< t < 1 + 'Y/},

(4.64)

where (4.65)

and define a map g: Z -> E by

g(z) = STI

+ r(x),

z

=

STI

+ X.

(4.66)

Then by (4.56) and (4.51)

g(z) = STI + X + (1 - s)F(x, to) + sF(x, tt} = z + (1 - s)F(x, to) + sF(x, tt}.

(4.67)

This shows that g is an 1.-8. map Z -> TI. If we set Zt

=:

tTl

+ Yo,

o ::; t ::; 1,

(4.68)

then as will be verified later Zt

¢. g(aZ).

(4.69)

But by Theorem 4.14 with E, 0, / replaced by TI, Z, g resp., it follows from (4.69) that dt = d(g, z, zd = const. for t E [0,1). (4.70) This shows that our assertion (4.63) will be proved once it is established that, for 0 ::; t ::; 1, d(ft, 0, YO) = d(g, Z, Zt). (4.71)

d(f,n,yo): YO NOT NECESSARILY REGULAR

51

The proof of this equality will be based on Lemma 4.13, on part (vii) of Lemma 2.4, and on Lemma 2.6. We set, for z defined in (4.66), ~

= z - rlt = rt(s - t) + x, h(~) =

(4.72)

g(z) - rlt.

(4.73)

It then follows from the lemmas quoted above and from (4.68) that d(g, Z, Zt)

= d(g, Z, trl + Yo) = d(h, Z -

trl, yo).

(4.74)

Now by (4.73), (4.67), and (4.72) h(~)

= ~ + (1 - s)F(x, to) + sF(x, td.

(4.75)

This shows that h is a layer map with respect to E. Therefore by the lemmas quoted above (4.76) d(h, Z - trl, Yo) = d(hE' 0, Yo),

°

where hE denotes the restriction of h to = (Z - trd n E. But t = s and ~ = x on E as is seen from (4.72). Therefore by (4.75), (4.56), and (4.51) hE(X)

= x + (1- t)F(x, to) + tF(x, td = ft(x)

(4.77)

The assertion (4.71) follows now from (4.77), (4.76), and (4.74). It remains to verify (4.69). Now the boundary az of the cylinder Z (see (4.64)) is the union of three parts: the "lateral surface" (a Zh, the "bottom" (oZh, and the "top" (oZh. More precisely,

I x E ao, -TJ ~ s ~ 1 + TJ}, (OZ)2 = {z E II I x E n, 8 = -1J}, (aZ)3 = {Z E II I x E n, 8 = 1 + 1J},

(oZh = {z

E II

where z = "18+2: and where" satisfies (4.65). Now if (4.69) were not true, then, by de6Diticm (4.68) of Zt, at least one of the following three relations would hold:

+ Yo = + Yo = tri + Yo =

tr1 tri

g(z)

for some z E (oZh and some t E [O,IJj

g(z)

for some z E (oZh and some t E [O,IJj

g(z)

for some z E (oZla and some t E [O,IJ.

Now by (4.66) the first of these relations implies that

trl + Yo = 8 r l + rex)

(4.78)

for some 2: E 00. Since rex) E E, it follows that 8 = t, Yo = r(x), x E ao. But this contradicts (4.57). The second relation implies (4.78) with t E [O,IJ and some x E n. It follows that 8 t E [0,1), and this contradicts the fact that 8 = -1J. The impossibility of the third relation is proved correspondingly.

=

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

52

4.16. THEOREM. The conclusion of Theorem 4.15 remains valid if the assumption that the map F(x, t): Ox [0,1] ---- E is completely continuous is replaced by the following two: (a) F(x, t) is, as a function oft, continuous and uniformly so as x varies over 0; (b) for fixed t E [0,1] the map Ft(x) = F(x, t) is completely continuous in x. PROOF. As will be verified in the Notes,lO the above assumptions (a), (b) together imply the assumption of Theorem 4.15. 4.17. The assumption of Theorem 4.15, i.e., the complete continuity of the map F: 0 x [0,1] ---- E does not imply assumption (a) of Theorem 4.16. An example will be given in the Notes l l .

§5. Notes to Chapter 2

Notes to §1. 1. Definition 1.4 of the index is equivalent to the definition given in subsection 4.3 of Appendix A. This follows from Lemma 4.5 and Theorem 4.6, both in Appendix A.

Notes to §2. 2. Proof of Lemma 2.4. (i) If Yo E 0, then x = Yo is the only solution of equation (2.2) with f = I. Since the differential of I is the identity map, the

right member of (2.3) equals i(I) = 1. If Yo i 0, then equation (2.2) with f = I has no solution in 0, and the degree equals zero by Definition 2.2. (ii) Suppose (2.2) has no roots. Then the degree is zero by Definition 2.2. (iii) This assertion follows directly from the definitions involved since every root x E 0 of (2.2) lies in exactly one of the 0<7' (iv) Let O2 = 0 - 0 1 . Then O2 contains no roots of (2.2). Therefore d(J, O2 , Yo) = 0 and the assertion (2.5) follows from (2.4) with s = 2. (v) A point x E 0 is a root of (2.2) if and only if x is a root of the equation (1)

Therefore if (2.2) has no roots in 0, then (1) has no roots in 0 and both members of (2.6) are zero. Suppose now that xl, X2,' .. ,Xr are the roots of (2.2). Then (2.3) holds. But D fb(x p ; h) = D f(x p ; h) since f and fb differ by a constant. Consequently the right member of (2.3) equals r

Li(Dfb(x p ; h)). p=1

But this sum equals d(Jb, 0, Yo - b) since the xp are the roots of (1). (vi) x E 0 is a root of (2.2) if and only if ~ = x - Xo is a root of g(~) = yO

(2)

NOTES

53

in 00. Therefore if (2.2) has no roots in 0, then (2) has no roots in 00 and both members of (2.7) are zero. Now suppose Xl, X2, ... , xr are the roots of (2.2). Then el = Xl - XO, 6 = X2 - Xo,···, er = Xr - Xo are the roots of (2) in 0 0 . Since D I(xp; h) = Dg( p; h), the assertion (2.7) follows directly from Definition 2.2. (vii) By (v) and (vi)

e

d(f, 0, Yo + b) = d(f - Yo, 0, b) = d(g - Yo, 00, b) = d(h, 0 0 , b). 3. We have to show: if F(x) E El for all X EO, then (2.14) holds. Now it follows easily from the definition of the differential that for real t

DF(x; h) =

l~ [F(X + th; - F(X)] .

By assumption the term in the bracket is an element of E 1 • Since El is closed, our assertion follows.

Notea to §3. 4. Prool 01 Lemma 3.2. If (3.1) has no roots in [a,,6], then for some positive

for all

X

I/(x) - al ~ 3cl E [a, ,61, and (3.2) implies that for all these Ig(x) -

bl

~

(3) X

I/(x) - al-l/(x) - g(x)I-lb -

al

~ Cl

> O.

(4)

This proves assertion (i) of our lemma. If all roots of (3.1) lie in the open subset U of fl, then no root of that equation lies in the closed set U' = [a,,B] - U. Consequently for some positive C1 the inequality (3) holds for all x E U'. But then (3.2) implies that the inequality (4) holds for all x E U', i.e., that every root of (3.3) lies in U. 5. Lemma 3.3 follows immediately from the relevant definitions since Df(x; h) = r(s)la.1Dd since equation (3.4) has at most a finite number of solutions if f iaa~.

6. PrIIII/ 0/ Lemma 3.4. By Lemma 3.3 and by assumption, I( C,) = 8, c V. If 0, is not empty, it consists, by Lemma 3.3, of a finite number of points, say Zl,S2, ••• ,s,., and 8, consists of the points l(xt)'/(x2), ... ,/(xr ) which lie in V. There£ore there exists a 8 > 0 such that for p = 1,2, ... , r

B(f(x p ), 28)

C

(5)

V.

By the continuity of I there exists a positive) such that f(B(x p ,))) C B(f(zp),6). Therefore if U = Up B(x p, )), then I(U) C

UB(f(x p

p ),

8).

(6)

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

54

Now by part (ii) of Lemma 3.2 with f and 9 replaced by f' and g' resp. and with a = b = 0, there exists a positive £1 of the following property: if 9 is a polynomial in [a, i3l for which

Ig'(x) - !,(x)1 < £1

for x E [a,.Bj,

then the set Cg of roots in [a,.Bj of the equation g'(x) U of Cf: C g C U. Now suppose assumption (3.5) is satisfied with £ therefore (8), is true. Consequently by (6)

f(C g ) c f(U) c

(7)

= 0 lies in the neighborhood (8)

= min(c1,8). Then (7), and

UB(f(xp), 8).

(9)

p

But by (3.5), If(x) - g(x)1 < £ S 8. Therefore, by (9) and (5), Sg = g(Cg) C Up B(f(xp), 28) C V. This proves our lemma if Cf is not empty. But if Cf is empty, then by the first part of Lemma 3.3 the equation f'(x) = 0 has no roots in [a, .Bj. Consequently by Lemma 3.2 (with f and 9 replaced by f' and g' resp., and with a = b = 0) there exists a positive £ such that the equation g'(x) = 0 has no roots in [a,.Bj if If'(x) - g'(x)1 < £ for x E [a, .Bj. Thus for such 9 the set Cg , and therefore the set Sg, is empty. 7. Proof of Lemma 3.5. By assumption Sf n r = 0. Therefore the set Sf containing at most a finite number of points has a positive distance from the closed set r. Thus there exists an open set V such that Sf C V and

Vnr=0.

(10)

But by Lemma 3.4 there exists a positive £ such that (3.5) implies (3.6). Therefore for such g, Sg nrc V n r and, by (10), Sg n r = 0. Notes to §4. 8. Proof of the part of Lemma 4.13 relating to Lemma 2.4. Assertion (i) is vacuously true since every point Yo E E is a regular value for the identity map I. For the proof of the remaining assertions of Lemma 2.4, we recall that the assumption (2.1) implies the existence of a Po such that, for 0 < p < Po,

Y ¢. f(80)

for y E B(yo, p),

(11)

and that, by Theorem 4.2, Po may be chosen in such a way that

d(f, 0, y)

=

d(f, 0, Yo)

(12)

for those points y E B(yo, p) which are regular values for f. Proof of assertion (ii). It follows from Theorem 4.1 that there exists a sequence {Yi} of regular values for f which satisfy lim Yi = Yo. Yi E B(yo,p), (13) i----+oo

NOTES

55

By assumption, d(f, 0" Yo) ~ O. It therefore follows from (11) and (12) that d(f, 0" Vi) ~ 0 for i = 1,2, ... , and from Lemma 2.4 that the equation f(x) = Yi has a root Xi E n. Thus i = 1,2, ....

(14)

It now follows from (13), (14), and the complete continuity of F that some subsequence {Xiv} of the {Xi} converges to some point Xo E fl, and that f (xo) = Yo. But Xo E 0, by (2.1). Proof of assertion (iii). Since the sum in (2.4) is finite, we can choose Po in such a way that not only does (12) hold for 0 < P < Po but also for a = 1,2, ... ,s and Y E B(yo, p) (15)

if y is a regular value for

f

and for such y (16)

Then by Lemma 2.4, the equation (2.4) holds with Yo replaced by y. But by (12) and (15) it then holds also for Yo. Proof of assertion (iv). This assertion follows from assertion (iii), as was shown in the proof of Lemma 2.4. Proof of assertion (v). Let y be a regular value for f satisfying (11) and (12). It is then easily verified that y - b is a regular value for fa = f - b and that

d(fo, 0" y - b) - d(fo, 0" Yo - b). But since y is a regular value for

(17)

f, we see from Lemma 2.4 that

dC/, n, y) = dC/o, n, y - b),

(18)

and we see from (12) and (17) that (18) remains true if y is replaced by Yo. ProoJ oj aB8ertion (vi). Let 11 be a regular value for f satisfying (11) and (12). It is easily verified that 11 is a regular value for g and that

d(g, 0 0,11) = d(g, 00, 1/0). Since 11 is a regular value for

(19)

I, we see from Lemma 2.4 that

d(f, 0, y) = d(g, 00, V).

(20)

But by (q) aDd (19) the relation (20) remains true if y is replaced by Yo. Finally 88 shown in the proof of Lemma 2.4 the assertion (vii) follows from assertions (v) and (vi).

9. Prool oj the part 01 Lemma 4.13 relerring to Lemma 2.6. We have to show that (2.9) holds even if Yo is not a regular value for I. Now by (2.1) there exists a neighborhood N of Yo whose closure does not intersect 1(00,). Consequently NF~'N n El is a neighborhood (with respect to Ed of Yo such that y rt. 1(00)

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

56

for y E N 1. By the Sard-Smale theorem there exists a Yo value for the restriction of f to 0 n E 1 • Then

E N1

which is a regular

d(f, 0, Yo) = d(f, 0, Yo)·

(21)

But since F(O) and Yo lie in E 1 , all roots of f(x) = x - F(x) = Yo lie in E 1 . Moreover for each such root x we see: if () = D f(x; h) = h - DF(x; h), then hE Et, and therefore h = () since D f(x; h) is not singular for the restriction of f to E1 nO. This shows that Yo is a regular value also for f. Therefore by Lemma 4.6 the right members of (21) are equal. 10. Note to the proof of Theorem 4.16. We have to show that assumptions (a)

and (b) of our theorem together imply the complete continuity of F. We first prove the continuity. For x and Xo in 0 and for t and to in [0,1]'

IIF(x, t) - F(xo, to)11

~

IIF(x, t) - F(x, to)ll

+ IIF(x, to) - F(xo, to)ll.

(22)

By assumption (a) there corresponds to a given c > 0 a b > 0 independent of x such that the first term in the right member of (22) is less than c / 2 if It - to I < b. The continuity of F(x, t) at every point (xo, to) E 0 x [0,1] now follows from (22). For the proof of the complete continuity of F it remains to show: every sequence {xv, tv} E n x [0,1] contains a subsequence {XVi' tv.} for which the sequence F(XVil tv,) converges. We assert first that it will be sufficient to show that to every TI > 0 there corresponds a subsequence {XVi' t vi } of the sequence {xv, tv} such that (23) for all i, j = 1,2, .... Indeed, let TIl, Tl2, ... be a sequence of positive numbers converging to zero. Assuming (23), we can select a subsequence {x~, t~} of the sequence {xv, tv} such that

IIF(xLtt) - F(x},t})11 < TIl for all i, j. Going on in this way we obtain for every positive integer v a sequence {xi, which is a subsequence of the sequence {xf, for f..L = 1,2, ... , v-I, and for which IIF(xi,ti) - F(xj,tj)1I < TIl' (24)

tn

tn

for i,j = 1,2, .... We claim that the "diagonal" sequence {x!,tH is a Cauchy sequence, i.e., to every c > 0 there corresponds an integer Vo such that

(25) for i,j ~ Vo. To see this we choose Vo in such a way that TIl' < c for v Then by (24) with v = Vo

~

Vo. (26)

for all i, j. But for i 2': Vo the sequence {x!,

tn is a subsequence of the sequence

NOTES

57

{XrO,trO}. Thus (26) implies that (25) holds, i.e., that the sequence F(x~,t~) converges. It remains to show that (23) holds for some subsequence {XVi' t vi } of the given sequence {Xi, td. Obviously we may assume that the ti converge and therefore form a Cauchy sequence. Consequently, by assumption (a) there exists a positive integer t such that (27)

n.

But by assumption (b), F(x,tt) is completely for i > t and for all X E continuous. Therefore there exists a subsequence {Xi,,} of the sequence {Xi} such that the sequence {F(Xi" , tt)} converges. Consequently there exists an integer t such that (28) for i o , if3 ~ £.: We now omit from the sequence {Xi,,} all those elements for which io ::; max(t, z). Then (28) holds for all elements of the sequence thus obtained. Now for this sequence

IIF(Xi" , tiJ - F(Xi/3' ti/3)11 ::; IIF(Xi",tiJ - F(Xia,tt)1I

+ IIF(Xi",tt) - F(Xi!3,tt)1I

+ IIF(Xi!3' tt) - F(Xi!3' ti/3)II· The first and the last of the three summands in the right member of this inequality are less than 17/3 by (27) and the middle one by (28). This proves (23). 11. Note to 4.17. Let E be the Hilbert space of points X = (Xl, X2, . .. ) where the Xi are real numbers with L: x~ < 00, with the usual definitions of linear operations, and with the scalar product (x,y) and norm IIxll = (x,x)I/2. Let (1 be the unit ball in E. We define a map F: x [0, IIi -+ E as follows:

n

JI = F{x, t) = (JlI (x, t), Jl2(X, t), ... ),

where

00

Jll(X, t)

= Xl + E x~tV,

J/2(x, t)

= Jl3(X, t) = ... = O.

(29)

1.1=2

We wiD sbow: (i) F is completely continuous; (ll) condition (a) of Theorem 4.16 is not sMia6ed. Proof·of (i). x [0,1]) is a bounded set in a one-dimensional subspace of~. ~e it will be sufficient to show that F(x, t) is continuous at every point (o,to) of X [O,lJ, Le., that to every e > 0 there corresponds a positive .5 = 6(x, to) such that the norm of

Fen

n

~o =

F(a + h, t) - F(a, to)

c

(30)

is -than e if Ilhll and It - tol are less than (provided that a + hEn and t,e [0,1)). Without loss of generality we assume e < 1. IT a = (at, a2, .. .), and

THE L.-S. DEGREE FOR DIFFERENTIABLE MAPS

58

h = (hI, h 2 , •. . ), and Llo = (Ll?, Llg, ... ), then by (29) and (30)

Ll~ = hI + ~[(av + hv )2tv - a~toJ, }

(31)

L\.g = L\.g = ... = O. We now choose Moreover, since

L

11811 < e/8. a~ converges, we can choose an integer N such that

(32)

00

~ a~ < e2 /8.

(33)

N+l

Then using the Schwarz inequality and taking into account the fact that the nonnegative numbers t, to, and e are all ::; 1, we see from (32) and (33) that

f:

II N+1 [(av + hv )2t

v-

a~tolll ::; f: [2a~ + 21 hvavl + h~l N+l

00

::; 2 ~ a~

( 00

N+l

::; 2

00) 1/2

+ 2 ~ h~. ~ a~ N+l

N+l

f: a~ + (f: a~) 211hll

N+l

1/2

00

+ ~ h~ N+l

+ IIhl1 2 <

~.

N+l

But since Ihvl ::; Ilhll it is obvious that there exists a 81 > 0 such that the finite sum

Ihl

+ ~[av + h~)tV -

a~toll < ~

if Ihl and It - tvl are less than 81 . It is now clear from (31) that IL\.?I < e if Ihl and (t - to) are less than 8 = min(e/8, 8d. This finishes the proof of assertion (i), and we turn to the proof of assertion (ii). Suppose that property (a) of Theorem 4.16 is satisfied. Then there would exist a 8 > 0, independent of x, such that IIF(x, t) - F(x, 1)11 < 1/2

for all x

E

(34)

0 and all t satisfying 0< 1- t < 8.

(35)

However, if x j is the point of 0 whose jth coordinate is 1 and whose other coordinates are zero, then by (29) for j ~ 2, F(x j , t) -F(xj , 1) = (t J" -1, 0, 0, ... ), and therefore for each t E [O,lJ IIF(x j , t) - F(x j , 1)11 = 1- tj.

(36)

Now choose a fixed t satisfying (35) and then a fixedj so great that 1- t > 1/2. For such t and y we conclude from (36) that IIF(x~ t) - F(x j , 1)11 > 1/2. But this contradicts the fact that (34) holds for all x E n and t satisfying (35). j

CHAPTER 3

The Leray-Schauder Degree for Not Necessarily Differentiable Maps §1. An extension lemma 1.1. As in Chapter 2, {1 denotes a bounded open set in the Banach space E. The family C(l1) of completely continuous maps F: 11 -+ E is, with the obvious definitions of addition and multiplication by (real) scalars, a linear space. Moreover, if we define

IIFII = sup IIF(x)ll,

(1.1)

zEn

then it is easily seen that C(l1) is a closed linear subspace of the space of bounded maps 11 -+ E with norm (1.1) (see e.g. [31, pp. 15-16]). In what follows C(l1) will always denote this subspace. If {1 is fixed we will often write C for C(l1). 1.2. In Chapter 2 the Leray-8chauder degree was defined for maps f: 11 -+ E for which the elements F = 1- f of C(11) belong to C"(l1). We want to extend the definition of the degree d(f, n, Yo) to L.-8. maps for which F = I - f lies in the closure of this subspace of C(ll), always assuming that I/O

rt J(aO).

(1.2)

The "extension lemma" in subsection 1.3 below serves this purpose. Before stating it we make the following REMARK. If C l (Il) is the set of twice differentiable completely continuous maps on (cf. §1.16), i.e., if Cl(n) = C(n)nc"(n) and ifCl(U) is defined correspondingly for every bounded open set U c E, then C l (Il) obviously satisfies the following conditions (i)-(iii): (i) Cl(n) is linear; (ii) if F E Cl(n), then for every element bEE the map Fb: n -+ E given by Fb(z) = F(z) - b belongs to C l (11); (iii) if 0_ denotes the translate of {1 by the element -zo E E, i.e., if nzo = {e EEl z + Zo En}, then the map G: nzo -+ E belongs to Cl(nzo ) if and only if the map F:n -+ E given by F(x) = xo + G(z - zo) belongs to Cl(Il).

n

=e

59

60

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

In addition we know from our previous results for Gil maps that the following statements (do)-(d2 ) are true (under the assumption made and with the notations used in this remark): (do) if F E G1(0), G E G1 (0",0)' f(x) = x - F(x), g(x) = x - G(x), 0 is an arbitrary open subset of 0, Yo tt f(80) and Yo tt f(80), then degrees d(f, 0, Yo), d(g, 0"'0' Yo), and d(f, 0, Yo) exist and satisfy the following conditions (dt} and (d 2 ). (d 1 ) The part of Lemma 4.13 in Chapter 2 referring to Lemma 2.4 in Chapter 2 holds. (d 2 ) If for each t E [0,1], the map Ft given by Ft(x) = F(x, t) satisfies conditions (a) and (b) of Theorem 4.16 in Chapter 2 and if (1.2) holds with f replaced by It = I - Ft , then d(ft, 0, Yo) is independent of t. 1.3. EXTENSION LEMMA. Let 0, xo, and 0"'0 be as in the Remark in subsection 1.2, let C(Oxo) be the space of completely continuous maps Oxo ....... E (such that C(11o) = C(11)). For any subfamily C = CeO) of C(O) we denote by A(C) the family of L.-8. maps f(x) = x - F(x) where FE C. Now let C1 (11) be an arbitrary subset of C(O) which satisfies the conditions (i)-(iii) and (do)-(d2 ) stated in the Remark in subsection 1.2. It is then asserted that the definition d(f,O,yo) for f E A(Ct} can be extended to all f E A(G\) in such a way that the properties (dt} and (d 2 ) are preserved and that, in addition, for these f the following assertion (d3 ) holds: let y = y(t) be a continuous curve in E, 0 ~ t ~ 1, and suppose that for each such t the condition (1.2) holds with Yo replaced by y(t). Then d(f, 0, y(t)) is independent of t. PROOF. Let f be an element of A(G\) for which (1.2) holds. To define a degree for f we note first that, on account of (1.2), Yo has a positive distance p from f(80). Therefore

Yo Now F = I that

f

tt B(f(x), p)

for all x E 80.

(1.3)

E V 1 by assumption. Consequently there exists an F1 E G1 such

11F1 - FII < p.

(1.4)

If h(x) = x - F1(X) then by (1.1)

Ilh(x) - f(x) II = 11F1(x) - F(x)11

~

11F1 - FII < p

for all x E rI. In particular,

h(x) E B(f(x),p)

for x E 80.

(1.5)

From this together with (1.3) we conclude that (1.2) holds with f replaced by h. Since It E A(Ct} we thus see that the degree d(h, 0, Yo) is defined. We show next that this degree is independent of the particular choice of It E A( C 1) provided that (1.4) holds: let, for i = 0,1, fi(x) = x - Fi(x) be maps in A(Gd for which (1.4) holds with F replaced by Fi· To show that indeed d(fP,O,yo) = d(ft,O,yo)'

(1.6)

AN EXTENSION LEMMA

61

we recall that (1.4) implies (1.5). Therefore fUx) E B(f(x),p) for i = 0,1 and x E ao. Since a ball is convex, it follows that for t E [0, 1]

ff(x) = (1 - t)fP(x)

+ til (x)

E

B(f(x), p)

for x E ao. Therefore Yo ~ n(x) for t E [0,1] and x E ao by (1.3). Since n E A(C l ) the asserted equality (1.6) follows from assumption (d 2 ) of our lemma. It is now clear that the following definition is legitimate. 1.4. DEFINITION. Let I, F, and p be as above. Let Fl E C l satisfy (1.4) and let Jt(x) = x - Fl(X). Then the degree d(f, 0, Yo) is defined by the equality (1.7)

d(f,O,yo) = d(Jt,O,yo).

1.5. It remains to show that the degree thus defined has the properties asserted in the extension lemma in subsection 1.3. The verification of property (dt} will be given in the Notes, 1 and we turn to the proof of (d 2 ). By the Borel covering theorem it will be sufficient to show that under the assumption made for (dd there corresponds to every to E [0,1] a positive TO = To(to) such that

d(ft, 0, Yo) = d(fto, 0, Yo)

(1.8)

It -

(1.9)

if

tol < TO,

t E [0,1].

Now Yo has a positive distance 2po = 2po(to) from fto(aO). Therefore

Yo

~

B(fto(x),po)

for all x E 190.

( 1.10)

But by assumption (a) we can choose a TO such that 1I!t{x) - Ito{x)1I < Po/2

for all x

En

(1.11)

and all t satisfying (1.9). Now let tl be such a fixed t-value. Then (1.11) holds with t = t1, and since /&0 and /&1 are by assumption elements of A{?:\) there exist map8 If and /f in A(Cl) such that ,

'.\

IIn(x) - 1&, (x)1I < Po/2

(1.12)

b.'.~O.l and all x E n. Then

!f(x) E B(fto(x),Po/2) C B(fto(x),Po).

But illDcelil(x) - Ito (x) II ~ II/Hx) (1.~~) _(1 •.11) (with t

,

i\um. (1.!3),

Ih(x)1I

= it) that also

+ II/tl(x) -

!lex) E B(fto(x), Po)·

(1.13)

fto(x)ll, we see from (1.14)

(1.14), and the convexity of a ball, we conclude that If{x) =

(1~1JA~)+6IHx) E B(fto(x), po) for 0 E [0,1) and therefore, by (1.10), that ., :J·Ii(~} for these 0 and for x E ao. Since ff E A(C!), it follows from our ~onClthat

d(ff,O,yo) = d(iP,O,yo).

(1.15)

62

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

Now assertion (d2) will be proved if we prove (1.8) for t = tt since tt is an arbitrary t-value satisfying (1.9). For the proof of (1.8) (with t = tt) it will, because of (1.15), be sufficient to verify that, for i = 0, 1,

d(Jt,O,yo) = d(!t;,O,yo). But this equality follows from Definition 1.4 on account of (1.12) and the following inequalities still to be proved: dist(yo, fdaO)) 2:: Po/2,

i = 0, 1.

For i = 0 this inequality is a consequence of (1.10). For i (1.16) follows from the inequality

(1.16)

=

1 the inequality

Ilfdx) - yoll 2:: Ilfto(x) - yoll-Ilfto(x) - fdx)11 together with (1.10) and (1.11) (with t = tt). Proof of (d3 ). Let ft(x) = f(x) - y(t). Then d(J,O,y(t)) = d(/t,O,O) by (dt). But the right member of this equality is independent of t by (d 2 ). §2. An application of the extension lemma

2.1. Let G1 = G1 (T1) be the subspace of those elements of G = G(O) which belong to G"(T1) (see §1.16). Then, by the results of Chapter 2, G1 satisfies the assumptions of the extension lemma in subsection 1.3, and consequently a degree theory is established for those L.-8. maps f = I - F for which F lies in the closure G't of Gb Le., those F which belong to A(Gt}. 2.2. Especially if the Banach space E has the property that

G\ =G

(2.1)

where G1 is as above, then it follows that our degree theory is established for all L.-8. maps f: 0 -+ E satisfying (1.2). However (2.1) does not hold in every Banach space. Indeed the relation (2.1) is equivalent to the following statement: to every completely continuous F: -+ E and every E: > 0 there corresponds a completely continuous F2 E G" (rl) such that

n

(2.2)

for all x E T1. An example of a Banach space in which such approximation is not always possible will be given in the Notes. 2 2.3. In the remainder of this chapter we discuss some important classes of Banach spaces in which the approximation (2.2) is always possible and in which therefore a degree theory is established for arbitrary L.-8. maps. 2.4. Let E be a finite-dimensional Banach space. Then every closed bounded set in E is compact. This implies that every continuous map of a subset of E into E is completely continuous and also an L.-S. map, since f(x) = x - F(x) is continuous if and only if F(x) is continuous. Thus C = C(n) = A(C) is the set of continuous maps n -+ E. C t = Ct (n) defined in subsection 2.1 is identical

AN APPLICATION OF THE EXTENSION LEMMA

63

with the set G" (0). But it is well known that every continuous map F: n - t E can be uniformly approximated by map F2 in G" (see, e.g., [13, p. 68]), i.e., that the approximation (2.2) is always possible. Thus (2.1) holds, and by our extension lemma in subsection 1.3, a degree theory with the properties (dd-(d3 ) stated in that lemma is established for continuous maps. We finally note that conditions (a) and (b) of property (d 2 ) (see subsection 1.3) may be replaced by the condition that the map f: n x [0, 1] - t E is continuous since a continuous map of a closed bounded set in a finite-dimensional space is uniformly continuous. 2.5. In order to obtain a class of infinite-dimensional Banach spaces in which the approximation (2.2) is always possible, we start with the following remark. Let E be a Banach space of the following property: to every open bounded set {1 in E, to every continuous map G of n into a Banach space E l , and to every TJ > 0, there exists a G" map G2: n - t El such that

IIG(x) - G2 (x)11 < TJ for all x E 11.

(2.3)

We show that in such space E the approximation (2.2) is always possible. (Note that in (2.2), F and F2 are both completely continuous.) The proof is based on the "second Leray-Schauder lemma" stated and proved in subsection 4.3. By this lemma there exists to every completely continuous F: n - t E and c > a continuous map G of into a finite-dimensional subspace El of E such that

°

n

IIF(x) - G(x)11 < c/2 for all x

En.

(2.4)

But by our assumption on E there exists a G"-map G2 : n - t E such that (2.3) is satisfied with TJ = c/2. Then by (2.4)

IIF(x) - G2 (x)11 < c for all x

En.

(2.5)

Now G2 is completely continuous. Indeed F is completely continuous and therefore F(n) is bounded. It follows from (2.4) and (2.5) that the subset G2 (n) of the finite-dimensional space El is bounded, and this implies the complete continuity of G2 since G2 is continuous. Thus (2.5) shows that (2.2) is satisfied with F2 = G2 . 2.6. DEFINITION. A Banach space E is called 2-smooth if there exists a bounded set U E E and a G" map of E into the reals which is zero on E - U but not identically zero.

2.7. LEMMA. Let E be a separable 2-smooth Banach space. Then the approximation (2.3) is always possible. This is a special case of a more general theorem by Bonic and Frampton (see [5, Theorem 2]). We refer the reader to their proof. 2.8. It follows from subsections 2.7, 2.5 and 2.2 that a degree theory with the properties (dt}-(d 3 ) has been established for all 2-smooth separable Banach spaces. If p, is the Lebesgue measure in a finite-dimensional Euclidean space Rk, if S is a Lebesgue measurable set in Rk, if for any real-valued measurable g defined on S, [g] is the family of those g for which g(x) = g(x) except for a set

64

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

of measure zero, then it is well known that with proper definitions those [g] for IglP dJ.L < 00 form a Banach space Lp for p ~ 1. It is also well known that such Lp is separable (see, e.g., [18, p. 125]). But for p ~ 2 these spaces are also 2-smooth (see [5, pp. 881, 882]).

Is

§3. The degree theory for finite layer maps

3.1. As pointed out in subsection 2.2 the application of the extension lemma of subsection 1.3 to L.-S. maps which are in Gil does not yield a degree theory for all L.-S. maps in all Banach spaces. But in subsection 3.4 of Chapter 2 a degree theory was established for arbitrary continuous maps in finite-dimensional spaces. Based on this fact and on Leray-Schauder's "first lemma" in subsection 3.2, we establish in the present section degree theory for L.-S. maps which are finite layer maps (see Definition 2.5 in Chapter 2). Using Leray-Schauder's "second lemma" we will show in the next section that application of the extension lemma to finite layer maps yields a degree theory for arbitrary L.-S. maps in all Banach spaces (cf. §7 of the Introduction). 3.2. FIRST LERAY-SCHAUDER LEMMA. Let E2 be a Banach space of finite dimension n2, and let El be a subspace of E2 of dimension nl < n2. Let 0 be a bounded open set in E2 which intersects E 1 . Let f(x) = x-F(x) be a continuous map n - E with the property that F(I1) EEl. Finally let Yo be a point in El for which (3.1) Yo tt f(80).

Then (3.2) PROOF.

The assumption (3.1) implies the existence of p such that

0< p < IIyo - f(x) II

n

for x E 80.

(3.3)

Since F maps the bounded subset of the finite-dimensional space E2 into the subspace El of E 2, there exists a Gil map F1 : El such that

n-

11F1(x) - F(x) II < p for all x E

n.

(3.4)

Now let h(x) = x - Fl(X) and

ft(x)

= (1- t)f(x) + th(x) = x -

[(1 - t)F(x)

+ tFdx)]

(3.5)

for 0 ~ t ~ 1. Now itfollows from (3.4) that IIf(x)- h(x)II = IIF(x)-Fl(X)II < p, Le., h(x) E B(J(x),p). Since obviously f(x) E B(J(x),p) we see that ft(x) E B(J(x),p) for 0 ~ t ~ 1 and all x En. But for x E 80 we see from (3.3) that Yo tt B(J(x),p). Thus ft(X) =f. Yo for these x and all t E [0,1], and it follows from subsection 2.4 that d(J, 0, Yo) = d(h, 0, Yo), d(J, OnEl,Yo) = d(h, OnE1,yo). But by Lemma 2.6 and (11), both in Chapter 2, the right members of these two equalities are equal. This proves the assertion (3.2).

ANOTHER APPLICATION OF THE EXTENSION LEMMA

n

a bounded open set in E, and let f: -+ E be a finite L.-8. layer map. Let Eo and El be finite-dimensional subspaces of E with respect to which f is a layer map. Let Yo be a point of E which satisfies (3.1) and which lies in Eo n E l . Then 3.3.

n

LEMMA. Let E be a Banach space,

65

d(f,O n E l , Yo) = d(f, 0 n Eo, yo).

(3.6)

PROOF. Let E2 the subspace of E spanned by Eo and E l . Then by Lemma 3.2 both members of (3.6) equal d(f,O n E 2, yo). Lemma 3.3 allows us to make the following definition. 3.4. DEFINITION. Let f: 0 -+ E be a finite 1.-8. layer map. Let Yo satisfy (3.1) and let El be a finite-dimensional subspace of E with respect to which f is a layer map and which contains Yo. We then define d(f, 0, Yo) by the equality d(f, 0, Yo) = d(f, 0 n E l , yo). 3.5. LEMMA. The assertions of Lemma 4.13 in Chapter 2 remain true if the

assumption that f E C" is replaced by the assumption that f is a finite L.-8. layer map. The proof will be given in the Notes. 3 3.6. LEMMA. Let ft(x) = f(x, t) be a continuous map 0 X [0, 1] into E which is an L.-8. map for each t E [0,1]. Moreover, we assume that ft is a "uniform finite layer map," i.e., that there exists a finite-dimensional subspace El of E (independent of t) such that

F(x, t) = x - f(x, t) EEl

for all (x, t) EO X [0,1].

(3.7)

Finally it is assumed that Yo

~

ft(aO)

for t E [0,1].

(3.8)

Then d(ft, 0, Yo) is independent of t. PROOF. By our assumptions ft is a finite L.-8. layer map with respect to E 1 . Without restriction of generality we may assume that 0 n El is not empty. We also may assume that Yo EEl. Then by Definition 3.4

(3.9) if Yo EEl, and here the right member is independent of t by subsection 2.4.

§4. Another application of the extension lemma 4.1. As in subsection 1.1 let C = C(O) denote the space of completely continuous maps F: n -+ E. In the present chapter Cl will denote the subspace of those F E C which map into a finite-dimensional subspace El = El (F) of E.

n

4.2. LEMMA. Cl satisfies the assumptions of the extension lemma z"n subsection 1.3 and condition (d3 ). PROOF. That assumptions (i)-(iii) of that lemma are satisfied is obvious. As to conditions (dt}-(d 3 ) we note that (dd follows from Lemma 3.5 and that, as

66

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

shown at the end of the proof of the extension lemma, (d 3 ) is a consequence of (d2). It thus remains to prove (d 2 ). Then let ft(x) = f(x, t) = x - F(x, t), where Ft(x) = F(x, t) E C1 for every t E [0,1]. Assume that Yo

rt.

ft(iJO)

for t E [0,1]

(4.1)

and that F(x, t) is continuous in t uniformly for x E O. We have to show that d(ft, 0, Yo) is independent of t. The proof is quite similar to the argument given in subsection 1.5: it will be sufficient to show that to each to E [0,1] there corresponds a TO = To(to) such that (1.9) implies (1.8). Now by assumption (4.1) there exists a Po > 0 such that (1.10) holds, and by our uniformity assumption we can choose a TO such that (1.9) implies (1.11). Again let it be a fixed t-value satisfying (1.11). Then for x E 00 the point fd x) lies in the ball B (fto (x), Po /2), and since fto (x) lies in that ball,

for x E 00 and 0 ::; (J ::; 1. Now, by assumption, Fto(x) and Ftl (x) lie in respective finite-dimensional subspaces Eo and E1 of E, and therefore both lie in the subspace E2 spanned by Eo and E 1. Thus fO is for each (J E [0,1] an L.-S. map which is a layer map with respect to the finite-dimensional space E 2 , and the asserted equality d(fto, 0, Yo) = d(ftl' 0, YO) will follow from Lemma 3.6 once it is shown that Yo rt. fO(aO) for all (J E [0,1). To prove this we have only to note that by (4.2), (1.10), and (1.11) with t = t1 for x E 00 and t E [0,1]'

Ilfo(x) - yoll = Ilfto (x) - Yo ~

+ (JUtl (x)

- fto (x)) II

Ilfto(x) - yoll-llfh(X) - fto(x)11

~ Po/2

> O.

This finishes the proof of Lemma 4.2, and by the extension lemma a degree theory having properties (dt}-(d 3 ) is established for all 1.-8. maps f(x) = x F(x) where F E G\. We will now prove that (2.1) holds, i.e., establish the theory for all L.-S. maps. Now by our definition the equality (2.1) is equivalent to the following lemma.

4.3 SECOND LERAY-SCHAUDER LEMMA. To every completely continuous map F: -+ E and every € there corresponds a finite-dimensional subspace E1 of E and a continuous map F 1: -+ E1 such that

n

n

IIF(x) - Fdx)11 < € for all x E o.

(4.3)

PROOF. Let Y = F(O). Then the set Y is compact. Consequently by §1.27 there correspond to given c > 0 points Yl, Y2, ... , Y8 such that Yu E F(O),

(1

= 1,2, ... ,8,

(4.4)

and such that for every y E Y (4.5)

TWO ADDITIONAL PROPERTIES

67

for at least one positive integer aSs. We now define s real-valued continuous functions f.L(1 on Y by setting, for a = 1,2, ... , s, ( ) _{c-IIY-Y(111 0

I1-(1Y -

ifIIY-Y(1II
(4.6)

All 11-(1 are nonnegative and for fixed Y E Y at least one of the 11-(1 (y) is positive by (4.5). Therefore

(4.7) is a continuous map of Y into the finite-dimensional space EI spanned by the points YI, Y2, ... , Ys' Then (4.8) Now it follows from (4.6) that the norms of those terms in the numerator of the right member of (4.8) for which Ily - Y(1 II ~ c are zero while the norms of the remaining ones are not greater than cl1-(1(Y)' Thus we see from (4.8) that Ila(y) - yll


(4.9)

for all y E Y. Thus

Ila(F(x)) - F(x) II < c

n.

for all x E Since a is a continuous map Y continuous, it follows that FI(X) = a(F(x))

(4.10) -+

EI and F is completely

(4.11)

is a completely continuous map n -+ E l . Since Fl satisfies (4.3) by (4.10) and (4.11) our lemma is proved. 4.4. The degree thus defined for arbitrary L.-S. maps n -+ E is known as the Leray-Schauder degree. It has properties (d l )-(d3) asserted in subsection 1.3.

§5. Two additional properties of the Leray-Schauder degree

5.1.! Dealing with twice-differentiable L.-S. maps we proved the continuity theorems in subsections 4.15 and 4.16 of Chapter 2 and showed that the former impliea the latter but that the converse implication is not true. Now property (d2) proved for the Leray-Schauder degree corresponds to Theorem 4.16 in Chapter 2. ~,the property corresponding to Theorem 4.15 in Chapter 2 holds also for the LeraY-Schauder degree as will be shown in Theorem 5.3 below. We first state a lemma needed for the proof of that theorem . . ,5r~·c LEMMA. Let F(x, t): n x [0, I] -+ E be completely continuous. Then e > 0 there corresponds a finite-dimensional subspace El of E and a ~te'" continuous map n x [0, I] -+ El such that for all (x, t) in n x [0, I]

to.

"*"

IIF(x,t) - F1(x,t)1I < c

(5.1)

68

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

PROOF. Let Y = F(fl x [0, 1]). Then Y is compact on account of the complete continuity of F. Therefore by the argument used in the proof of Lemma 4.3, the space El and the map a(y): Y -+ El can be defined in such a way that (4.9) holds for all y E Y. Then F1(x,t) = a(F(x,t)) satisfies (5.1). 5.3. THEOREM. Let ft(x) = x-F(x, t) where F: continuous. Suppose that

Yo

~

ft(an)

nx [0, 1]

-+

for t E [0,1].

E is completely (5.2)

Then dUt, n, Yo) is independent of t. PROOF. It follows from (5.2) that there exists an c > 0 which is less than the distance of Yo from ft(an). Then let F1(x,t) satisfy the assertions in Lemma 5.2 with that c, and let fl(x) = x - F1(x,t). Then by Definition 1.4

dUt, n, Yo) = dUl, n, yo).

(5.3)

But fl is a layer map with respect to the finite-dimensional space El (which is independent of t). Therefore by Lemma 3.6 the right member of (5.3) is independent of t. 5.4. THEOREM. Let ft be as in Theorem 5.3. Let Yo = yo(t) be a continuous

curve in E satisfying yo(t)

~

h(an)

for t E [0,1].

(5.4)

Then d(h, n, yo(t)) is independent of t. PROOF. We note first that there exists a positive c such that dist(yo(t), h(an)) > c for all t E [0,1].

(5.5)

Indeed otherwise there would exist a convergent sequence h, t2, ... of numbers in [0,1] and a sequence of points Xl, X2, ... in an such that (5.6) Then with to = lim ti, the y( ti) converge to the point Yo (to) of our continuous curve Yo = yo(t), and by (5.6) lim ft;{Xi) = lim(xi - F(Xi, ti)) = yo(to). By the complete continuity of F(x, t) we now see that some subsequence of the Xi converges to a point Xo and that (5.7) Now an x [0,1] is a closed set, and since !(x, t) is an L.-S. map f(an x [0,1]) is also a closed set. Since all Xi E an it follows that Xo E an. Thus (5.7) contradicts our assumption (5.4). This contradiction proves the existence of an c > 0 satisfying (5.5).

GENERALIZED L.-S. MAPS

To prove our theorem it is sufficient to show that d(ft, n, yo{t)) is constant in a small enough neigl.lborhood of every to E [0,1]. With 6 as above let 8 > 0 be such that IIYo{t') - Yo (til) II < 6/2 (5.8) for any couple

t', til of points of [0, 1] satisfying It' - till

< 8.

(5.9)

x an,

Now let E and let tl be a fixed t-value satisfying Itl - tol < 8. Then we see from (5.5) and (5.8) that, for any t in the interval bounded by to and tl,

11!t{x) -

yo{tt}II ~

11!t{x) -

yo{t)II-llyo{t) - yo{tdll ~ 6 - 6/2> O.

(5.1O)

Therefore, by Theorem 5.3, d(ftl,n,y{td) = d(fto,n,y{td), which implies that

d(ftp n, y{tt}) - d(fto' n, y{to)) = d(fto' n, y{tt)) - d(fto' n, y{to)). Here the right member equals zero by assertion (d 3 ) in subsection 1.3, since for E and t in the interval with end points to and tl, the inequality II Ito (x) - yo(t) II > 0 holds, as is seen by a proof quite similar to that of (5.10). 5.5. The computation of the Leray-Schauder degree for an arbitrary L.-S. map can be reduced to that of a map taking place in a finite-dimensional space. More precisely we assert: let I be an L.-8. map ~ E and let Yo be a point 01 E satisfying Yo fI. f(an). Then there exists a subspace En of E of finite dimension n containing Yo and having a nonempty intersection nn with n, and a continuous map If: On ~ En such that

x an

n

(5.10) PROOF. With 0 1 as in subsection 4.1 the left member of (5.4) is given by Definition 1.4 where Ft(Il) = (1 - ft)(n) satisfies (1.4) and lies in some finitedimensional space El c E. But then there exists also a finite-dimensional En such that IHIl) C El c E" c E and such that E" contains Yo and points of O. Thue.:li is a flnitelayer map with respect to E". Therefore by Definition 3.4 of the flnite layer maps the right member of (1.7) equals d(fr, on, Yo) if Jr ~ the restriction of It to This proves (5.10). 1.8.. 8.Du.R.K. The proof of (5.10) shows that the following conditions on E" 81'8 IIIfBeient for (5.10) to hold: (i) if P = dist(l/O, an), then there exists a contiDuaaa map FI:n ~ En such that I!FI(X) - F(x)1I < Pi (ii) Yo E En.

.,.lor

nn.

§6. Generalized L.-S. mapa . I' U. In this section a slight generalization of L.-S. maps will be introduced. The mothation for this was given in §10 of the introduction. ..~ DEFINITION. Let 8 be a closed subset of the Banach space E. A map I: 8 - E is called a generalized L.-S. (g.L.-S.) map if it is of the form

I(x)

= A(X)X -

F(x),

XES,

(6.1)

70

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

where F is completely continuous and where>. is a real-valued continuous function satisfying xE S, (6.1a) 0< m ~ >.(x) ~ M, for some constants m and M. 6.3. Before defining a degree for g.L.-S. mappings we note that if >. are as in (6.1) and (6.1a), if

f, F, and

Fo(x) = yo(1/>.(x) - 1) + F(x)/>.(x)

(6.2)

fo(x) = x - Fo(x),

(6.3)

and then direct calculation shows that x is a root of

fo(x) = Yo

(6.4)

f(x) = Yo·

(6.5)

if and only if x is a root of

6.4. DEFINITION. Let

f

be a g.L.-S. map n ---- E, and let the point Yo satisfy

rt /(80.).

Yo

(6.6)

We then set

d(f, 0., Yo) = d(fo, 0., yo).

(6.7)

Note that by subsection 6.3 the assumption (6.6) implies that Yo rt fo(80.) such that the right member of (6.7) is defined, since fa is obviously an L.-S. map.

6.5. LEMMA. Let f and Yo be as in subsection 6.4. Then f has the elementary properties (i)-(vii) stated in Lemma 2.4 of Chapter 2. Since d(fo, 0., Yo) has these properties, the lemma follows easily from (6.7) and (6.3).

6.6. THEOREM. Let the map F(x, t): n x [0,1J ---- E be completely continLet >'(x, t) be a continuous map of x [0,1J into the reals. Suppose, moreover, the existence of two constants m, M such that for (x, t) En x [0,1]

n

uous.

o < m ~ >'(x, t)

~

M.

(6.8)

Let h(x) = f(x, t) = >'(x, t)x - F(x, t),

(6.9)

and let Yo be a point of E satisfying, for all t E [0,1]' Yo

rt ft(80).

(6.10)

Then d(ft, 0., Yo) is independent of t. PROOF.

Let

Fo{x, t) = Yo {1/>'{x, t) - 1) + F(x, t)/>'(x, t)

(6.11)

NOTES

71

and

fJ{x} = fo(x, t} = x - Fo(x, t}. Then by (6.1O) and subsection 6.3 for 0 :::; t :::; 1 Yo

rf- fJ{ao.}

(6.12)

and by Definition 6.4

d(ft, 0., Yo}

= d(fJ, 0., yo}.

(6.13) Now from the assumed complete continuity of F(x, t} together with (6.8) and (6.11), we see that the map Fo(x, t}: n x [0,1] ---+ E is completely continuous. Taking into account also (6.12) we see from Theorem 5.3 that the right member of (6.13) is independent of t.

6.1. COROLLARY 1 TO THEOREM 6.6. The conclusion of Theorem 6.6 remains true if the assumption that F(x, t} is completely continuous is replaced by the assumptions (a) and (b) of Theorem 4.16 in Chapter 2. This is obvious since the latter assumptions imply the complete continuity of

F(x, t} as proved in Note 10 to Chapter 2. 6.S. COROLLARY 2 TO THEOREM 6.6. Let f be a g.L.-S. map n ---+ E. Let y = yet}, 0 :::; t :::; 1, be a continuous curve in E satisfying y(t} rf- f(ao.}. Then d(f, 0., y(t)) is independent oft. This follows by the argument used for the proof of (d 3 ) in subsection 1.5.

§1. Notes to Chapter 3

Note, to §1. 1. Proof of assertion (dd in subsection 1.5, i.e., of the assertions (i}-(vii) of Lemma 2.4 in Chapter 2. Let At be the family of L.-S. maps f(x) = x - F(x}:

n-

E for which F

eO., and A. the family of those L.-S. maps for which

Fel'l. AIaertion (i) is trivial since the identity map I belongs to At. PrrIItJI/ .0/ ONemon (ii). Let lEAl. and let 110 and p be 88 in Definition 1.4. 'ften there exists a sequence In(x) = x - Fn(x) of maps in At such that III,. -/I = !lFn - FII - 0 88 n - 00, i.e., that nl~ In (x)

1IJ1iformb'~for

%E

= I(x)

(1)

n, and we may suppose that II/n(x) -/(%)11

= IlFn(%) -

F(%)II < p

(2)

for % e~ n = 1,2, .... Then by Definition 1.4 d(fn, 0., Yo} = d(f, 0., yo), n = 1,2,. . . .

~here, by assumption, the right member is different from zero.

It WIowa that there exists an ~aif),

.~ ,

.

Xn

(3) Since

In E Al

E 0. such that

. fn(x n} = Yo·

(4)

72

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

But and (6)

Since F is completely continuous, there exists a subsequence {x ni } of the sequence {x n } for which the F(xnJ converge. It then follows from (5) and (6) that the X ni converge to some point x E r1 and that x satisfies the equation Yo = x - F(x) = f(x). But x ~ afl by (1.2). Thus the solution x lies in fl. Proof of assertion (iii). Let flo = fl and for u = 1,2, ... , s let flu be as in part (iii) of Lemma 2.4 in Chapter 2. Let f E A1 . We assume

(7)

u = 0,1, ... , s.

We want to prove 8

d(f, fl, Yo) =

L d(f, flu, yo). 0'=1

Now from (7) we conclude the existence of a p > Ilyo - f(x)11 > p for x E afl u , But there exists an h(x)

=x -

U

(8)

°

such that

(9)

= 0, 1, ... , s.

F1(x) in A1 such that

Ilf(x) - h(x)11 = IIF(x) - F1(x)11 < p for all x E U

= 0, 1, ... , s,

n.

(10) (11)

for such h, by definition. Now by assumption the equality obtained from (8) by replacing f by h holds. Therefore, by (11), the assertion (8) holds. Assertion (iv) is a consequence of assertions (ii) and (iii). Proof of assertion (v). Let f E Ab let p be as in (1.3), and let h(x) = X-F1(X) be an element of A1 for which (1.4) holds. Then by Definition 1.4

d(f, fl, Yo) = d(f1, 0, yo).

(12)

On the other hand, ifg(x) = f(x)-b and gl(X) = fdx)-b, then Ilg(x)-gdx)11 = Ilf(x) - h(x)11 < p, and therefore

(13) Now our assertion is

d(f, fl, Yo) = d(g, 0, Yo - b).

(14)

But by assumption the equality obtained from (14) by replacing f and g by It and gl resp. holds. Therefore by (12) and (13) the assertion (14) is true. Proof of assertion (vi). Let f(x) = x - F(x), where

FE C 1 (r1)·

(15)

YO ~ f(80),

(16)

Suppose

NOTES

73

and let g(~) =

f(x)

(17)

= x - xo.

(18)

where ~

We have to prove that

d(g,Oro,YO) = d(j,O,yo).

(19)

Now by (16) there exists a p > 0 such that

IIYo - f(x)11 > p for x E

ao.

(20)

But by (15) there exists an F1 E C 1 (0) such that 11F1(X) - F(x)11 < p, and therefore (21) Ilh(x) - f(x)11 < p where h(x) = x - F1(X). Therefore by Definition 1.4

d(j, 0, Yo) = d(h, 0, yo).

(22)

Now if g1(~) = h(x) with ~ given by (18), then g1 E C 1 (Oro)' Moreover Ilg(~) - g1(~)11 = Ilf(x) - f1(X)11 < p by (21). But since x E ao is equivalent to ~ E ao ro , we see from (20) and (17) that Ilg(~) - yoll > p for ~ E ao ro , and it follows from our definition that (23) But since (dt) holds for F1 E C 1 (0), the right members of (23) and (22) are equal. This proves the assertion (19). Assertion {vii} is a consequence of assertions (v) and (vi).

2. Aft. emmple lor the 48SemOft at the efld of subsection 2.2. Let it be the space ~ elements are the points x = (Xl, X2, •• •), where the Xi are real n~ subject to the condition that

(24) i=l

It is wellimown that with the natural definition of addition and multiplication by -Jaiaabers and with the sum (24) taken as norm IIxll this space is a Banach ....,.(eie, e.g., [18, Chapter IV.8]). Now let n be tne unit ball B(O, 1) in h, ... ~. unit element in h (i.e., an element of norm 1), and let E1 be the an: .",~ subspace of II generated by e1· Let F: n --. E1 be defined by

F(x) =

.r:. olfriously completely continuous.

e111xll·

(25)

74

L.-S. DEGREE FOR NOT NECESSARILY DIFFERENTIABLE MAPS

Assertion A. The map F furnishes the desired example, i.e., there is an e > 0 such that no map G: rl ---> E has the following three properties: G is completely continuous, G E C' (rl), and

IIF(x) - G(x)11 < e

(26)

for all x EO. The proof is based on assertions Band C below. Assertion B. Let ¢ E C' be a completely continuous map of 0 into the reals. Then ¢(rl) c ¢(80). (27) This assertion is a special case of a more general theorem by Bonic and Frampton for which we refer the reader to [5, p. 896, Corollary 2]. Assertion C. There exists an e > 0 such that for no real-valued map ¢ E C'(O) do we have (28) Illxll - ¢(x)1 < e for all x E O. Proof of assertion C. Suppose C not to be true. Then there exists a C' map ¢: 0 into the reals such that (28) is true with e = 1/3. Now

I¢(x) - ¢(O)I ~ Ilxll-I¢(x) - Ilxlll-I¢(O)I· Since Ilxll = 1 for x E 80 and since I¢(O)I = I¢(O) - 01 < 1/3 by (28) (with e = 1/3), we see from the above inequality that

I¢(x) - ¢(O)I

~

1/3 for all x E 80.

(29)

But ¢(O) E ¢(80) by (27). Therefore there exists an x E 80 such that I¢(x) - ¢(O)I < 1/3 in contradiction to (29). Proof of assertion A. Suppose A not to be true. Then there exists a sequence el, e2, ... of positive numbers converging to 0 and a sequence G I , G 2 , •.• of completely continuous C' maps rl ---> E satisfying

IIF(x) - Gn(x)11 < en

(30)

for all x E rl. Now let E2 be a subspace of II complementary to E 1 , and let PI and P2 be the corresponding projections (see §1.6). Since P1 is bounded, we see from (30) that IIPI(F(X) - Gn(x))11 < Men for some positive constant M. Now, by (25), P1 F(x) = F(x) = e111xll and PIGn(x) = el¢n(X), where rPn is a real-valued C' map. We thus see that Illxll - ¢n(x)1 ::; Men. Since en ---> 0 this inequality contradicts assertion C. Notes to §3. 3. Proof of Lemma 3.5. First we show that assertions (i)-(vii) of Lemma 2.4 in Chapter 2 hold. Assertion (i) is again obvious. For the proof of assertion (ii)-(vii) we suppose without loss of generality that f is a layer map with respect to a finite-dimensional subspace El of E which contains Yo· We note moreover that by subsections 2.1-2.4 and by subsectio.n 1.5 the assertions (ii)-(vii) are true for continuous maps rll ---> EI where 0 1 1S a bounded open set in E 1·

NOTES

75

°

Proof of (ii). If d(f, 0, Yo) # 0, then by definition d(f, n El. Yo) # 0, and the equation f(x) = Yo has a solution x E 0 1 = On El co. Proof of (iii). By definition d(f, 0(7, Yo) = d(f,O(7 n E 1 , Yo), u = 0, 1, ... ,s, (31) where 0 0 = 0. By assumption 0(7 n OT = 0 for u, T = 1,2, ... , sand U # T, and U~ 0(7 = = O. Therefore, for these u, T, (0(7 nEd n (OT n Ed = (2) and U~ fl(7

no

n El = fl n E 1 .

Consequently, 8

d(f, On E 1 , Yo) = L d(f, 0 0 n El. yo). (7=1 From this equality together with (31) the assertion 8

d(f,O,yo)

= Ld(f,O(7 nE1 ,yo) (7=1

follows. Assertion (iv) follows from assertion (iii). Proof of assertion (v). Without restriction of generality we may assume bE E I • Then fb = f - b is also a layer map with respect to E 1 • Thus by definition

d(fb, 0, Yo - b)

°

= d(fb, n E I , Yo -

b),

d(f, 0, Yo)

°

= d(f, n E I , yo).

(32)

But the right members of these equalities are equal and the asserted equality of the left members follows. Proof of assertion (vi). Without restriction of generality we may assume that Xo EEl. Then if ~ = x - Xo and g(~) = f(x) = x - F(x) = ~ - [F(~ + xo) - xo], we see that, with f also, g is a layer map with respect to E 1 • Therefore by definition not only (32) but also the equality

d(g,OZD'Yo) = d(g,OZD nEl.Yo) (33) holds. But the right members of (32) and (33) are equal and the asserted equality of the left members of these equalities follows. Aaaertion (vii) is a consequence of asaertioDS (v) and (vi). Finally we have to verify that part of Lemma 4.13 which refers to Lemma 2.6 in Chapter 2. Again let f be an L.-S. layer map with respect to the finitedimeneionalsubspace EI of E which we suppose to contain 1/0. Let E be a not necesaarily finite-dimensional subspace of E with respect to which f is a layer map. We have to prove (34) Now by de8nition

= d(f,OnEI ,1/O). (35) But I _ia also a layer map with respect to E n E I • Therefore we may suppose d(f,O,yo)

EI C E. But then by definition

d(f, OnE, Yo)

= d(f,O nEili/o),

and the assertion (34) follows from (35) and (36).

(36)

CHAPTER 4

The Poincare-Bohl Theorem and Some of Its Applications §1. Poincare-Bohl theorem and the winding number 1.1. In a Banach space E of finite dimension, the special case of Theorem 6.6 in Chapter 3 that the map ft (defined by (6.9) of Chapter 3) is a linear convex combination of two continuous maps fo and h is known as the PoincareBohl theorem. We will use this terminology also for arbitrary Banach spaces. As we will see, this special case of Theorem 6.6 has a number of important consequences. 1.2. THE POINCARE-BoHL THEOREM. As always let E be a Banach space and 0 a bounded open subset of E. For i = 0, 1, let

Ji(X) = Ai(X)X - Fi(x)

(1.1 )

be a g.L.-B. map n --+ E (see §3.6). Let

hex) = (1 - t)fo(x)

+ th(x),

tE[O,I],

(1.2)

and suppose Yo

tj: ft(iJO)

(1.3)

for all t E [0,1]. Then dUt, 0, Yo) is independent of t. PROOF. By (Ll) and (1.2)

ft(x)

=

At(X)X - Ft(x),

( 1.4)

where

At(X) = (1- t)AO(X) + tAl (x), Ft(x) = (1 - t)Fo(x) + tFl(X).

(1.5) (1.6)

Now by Definition 6.2 in Chapter 3 AO(x) and Al(X) both lie in some interval [m, M] with 0 < m < M for all x E n. By (1.5) we therefore see that the same is true for At(X). Thus the assumption (6.8) of Theorem 6.6 in Chapter 3 is satisfied with A(x, t) = At(x). Moreover, since Fo and Ft are completely continuous, it 76

THE WINDING NUMBER

77

follows from (1.6) that the map F(x, t) = Ft(x), as a map n x [0,1] ---> E is completely continuous. Thus all assumptions of Theorem 6.6 are satisfied, and the Poincare-Bohl theorem is shown to be a special case of Theorem 6.6. 1.3. THEOREM. Let 10 and that

h

be two g.L.-S. maps

n --->

lo(x) = hex) lor x E 80.

E, and suppose (1.7)

If, moreover,

Yo

rt. 10(80),

(1.8)

then

d(h, 0, Yo) = d(fo, 0, yo).

(1.9)

(In other words: the degree 01 a map depends only on the boundary values 01 that map.)

PROOF. Let It be defined as in (1.2). We then see from (1.7) that hex) = lo(x) for x E 80 and all t E [0,1]. Therefore the assumption (1.8) of the present theorem implies the assumption (1.3) of the Poincare-Bohl theorem. Consequently assertion (1.9) follows from the Poincare-Bohl theorem. 1.4. THEOREM. Let 10 be a g.L.-S. map 80 g.L.-S. map ---> E which is an extension 0110'

n

--->

E. Then there exists a

PROOF. Let

lo(x)

= A(x)x -

Fo(x),

x E 80.

(1.10)

Then by definition Fo is completely continuous, and Ao satisfies 0< m

~

AO(x) ~ M,

x E 80,

(1.11)

for some constants m and M. Applying Lemma 31 in Chapter 1 with 8 0 = 80 and t/Jo = Fo, we see that there exists a completely continuous extension F of Fo to Applying the same lemma with So = an and tPo = AO, we see that there exists a continuous extension ~ of ~o to n satisfying ~(n) C cO(Ao(an)). Since the interval 1m, M] is convex, it follows that (1.11) remains valid if Ao is replKed by ~. It is thus clear that I(x) = ~(x)z - F(z) is a g.L.-S. extension of

n.

10 to n. Theorems 1.3 and 1.4 together make the following definition possible. 1.5•. ,QEFINm:ON. Let 10 be a g.L.-S. map an - E. Suppose (1.8) holds. Then the winding number u(fo(80), Yo) of 10(aO) with respect to Yo is defined by theequa1ity u(fo(ao), Yo) = d(f, 0, Yo), where 1 is a g.L.-S. extension of 10 ton. . 1~8. ExAMPLE. Let E = E2, the two-dimensional Euclidean plane. Let n be the UDit ball B(D,l), and 8 1 = 8B(0, 1) the one-dimensional unit sphere (unit ~). Let 10 be a continuous map 8 1 _ E2 satisfying (1.12)

78

THE POINCARE-BOHL THEOREM AND APPLICATIONS

It is then possible to define an integer n = n(fo) which indicates the number of times "f(S1) winds around 0" and to prove that

n = n(fo) = u(fO(Sl), 0).

(1.13)

This motivates the use of the term ''winding number" in Definition 1.5. For the precise definition of the number n = n(fo) and the proof of (1.13), see the Notes. l The following theorem shows that the winding number u(f (aO), YO) is "continuous in f." 1.7. THEOREM. Let

ft(x) = A(X, t)x - F(x, t)

(1.14)

be a map aox [0,1] - E. We assume that (i) A(X, t) is real valued and continuous and satisfies (1.15) o < m ~ A(X, t) ~ M for (x, t) E ao x [0,1] for some constants m, M; (ii) the map F(x, t): aOx[O, 1] is completely continuous; (iii) (1.16) Yo ~ ft(aO). Then u(ft(aO), Yo) is independent of t. PROOF. Let II be the product space E x [0,1]. Then So = ao x [0,1] is a closed subset of II. If we apply Lemma 31 of Chapter 1 with E replaced by II, we see by the argument used in the proof of Theorem 1.4 that there exists a completely continuous extension Fl of F mapping n x [0,1] into E, and a real-valued continuous extension Al of A to n x [0,1] such that (1.15) remains true if A is replaced by A1. Then

fl(x) = Al(X, t)x - F1 (x, t) is a g.L.-S. extension of ft to n x [0,1] for which (1.16) remains true if f is replaced by fl. Therefore by Definition 1.5

u(ft(aO)) = u(fl(l)O)) = d(fl,O,yo)' and our assertion follows from Theorem 6.6 of Chapter 3. We state explicitly the following important special case of the last theorem. 1.8. POINCARE-BoHL THEOREM FOR THE WINDING NUMBER. For i = 0,1, let fi(X) be a g.L.-S. map ao - E, and for t E [0,1] let ft(x) = (1 - t)fo(x) + tft(x). Suppose that Yo ~ ft(aO) for all t E [0,1]. Then u(/t(aO), Yo) is independent of t. The next theorem may be considered as a generalization of the classical Rouche theorem in the theory of functions of a complex variable.

THE WINDING NUMBER

79

1.9. ROUCHE THEOREM. Let fo and !I be two g.L.-8. maps suppose that Yo ~ fo(aO)

ao -+ E.

We

( 1.17)

and that moreover

11!I(x) - fo(x)11 < Ilfo(x) - yoll for x E a~.

(1.18)

Then u(f1(aO), YO) exists and equals u(fo(aO), Yo). PROOF. For t E [0, 1] let hex) = (1 - t)fo(x) + t(!I(x) - fo(x)). Therefore

+ t!I(x).

Then hex) - Yo

=

fo(x) - Yo

Ilh(x) - Yo II ;::: Ilfo(x) - yoll-II!I(x) - fo(x)ll· This holds for all x E fl. But for x E ao the right member is positive by assumption (1.18). Thus the assumption of Theorem 1.8 that Yo ~ ft(aO) for all t E [0,1] is satisfied, and that theorem implies the assertions of the present one.

1.10. COROLLARY TO THEOREM 1.9. In addition to the assumptions of that theorem we suppose that fo is the identity map I on ao and that Yo E O. Then u(!I(ao), YO) = 1, (1.19)

and for every g.L.-8. extension 11 of !I the equation 11(X)=YO

(1.20)

u(h(aO), Yo) = u(fo(ao), Yo).

(1.21)

has at least one solution x EO. PROOF. By Theorem 1.9 But since the identity I on n is an extension of fo, the right member of (1.21) equals, by definition, the degree d(I, 0, Yo), and thus equals one since Yo E O. This proves (1.19). But for any extension 11 to n of ft, by definition d01 ,O,J/O) = u(!I(aO),Yo). Thus by (1.19), d01 ,n,Yo) = 1 which implies our second 888ertion. 1.11. In the next theorem we will use the notion of a "star domain." An open bounded set 0 C E is called a star domain if it contains a point Yo such that issuing from Yo (see §1.23) intersects in exactly one point. is then called a star domain with respect to Yo. A bounded open convex domain is a star domain with respect to any of its points. For a proof of this assertion see the Notes. 2

every ray

an

n

1.12. THEOREM. Let the open bounded subset 0 of E be a star domain with rellJ*t to the point Yo E O. Let fo and !I be g.L.-8. maps an -+ E satisfying Yo ~

/i(an),

i

= 0,1.

(1.22)

THE POINCARE-BOHL THEOREM AND APPLICATIONS

80

Suppose that for no x E an do the vectors fo (x) - Yo and h (x) - Yo have opposite direction, i.e., for no positive>. and x E an do we have h(x) - Yo = ->.(fo(x) - yo).

(1.23)

u(h(aO),yo) = u(fo(aO),yo).

(1.24)

Then

PROOF. For t E [0, 1] let

ft(x) = (1 - t)fo(x)

+ th(x).

(1.25)

Then our assertion (1.24) will follow from Theorem 1.8 provided we can show that

Yo tI- ft(aO)

(1.26)

for t E [0,1].

Now we see from (1.25) and assumption (1.22) that (1.26) is true for t = 0 and t = 1. Now suppose (1.26) were not true for some t in the open interval (0,1). Then for such t and some x E a~, ft(X) = Yo, i.e., by (1.25) h(x) - Yo = -(1 - t)t- 1 (fo(x) - yo). But this contradicts the assumption (1.23) since (1 - t)/t > O.

1.13. COROLLARY 1 TO THEOREM 1.12. Under the assumption of that theorem, (1.27) d(]l,n,yO) = d(]o,n,yo)

for every g.L.-S. extension 11 and 10 to n of II and fo resp. By definition of the winding number this is an obvious consequence of Theorem 1.12.

1.14. COROLLARY 2 TO THEOREM 1.12. Let h be a g.L.-S. map an ..... E, where 0 is a star domain with respect to the point Yo EO. Suppose that for all x E ao and all positive >.

h(x) - Yo

# ->.(x - yo)·

Then for every g.L.-S. extension 11 of h to

(1.28)

n the equation (1.29)

has at least one solution x E O.

n,

PROOF. If fo is the identity map on ao and 10 the identity map on we see from (1.28) that fo and h satisfy the assumptions of Theorem 1.12. It follows, by Corollary 1, that (1.27) holds. But the rig~t member of this equality equals unity since 10 is the identity map. Thus d(f l' n, Yo) = 1 which implies our assertion.

THE WINDING NUMBER

81

1.15. COROLLARY 3 TO THEOREM 1.12. Let 0 and Yo be as in that theorem. Let the g.L.-S. maps fo and It: ao ----> E satisfy (1.22). Finally suppose that u(fo(ao),yo) i- u(lt(aO),yo). (1.30) Then there exists an Xo E

ao

and a positive Ao such that

It (x)

- Yo = -Ao(fo(x) - Yo).

(1.31 )

Indeed if this assertion were not true, then (1.23) would be true for all x E ao and all A > o. Consequently by Theorem 1.12 the equality (1.24) would hold in contradiction to our assumption (1.30). 1.16. COROLLARY 4 TO THEOREM 1.12. Let 0 be a bounded open set in a Hilbert space E with scalar product (.,.). Let fo and It be g.L. -So maps 11 ----> E. We assume moreover that 0 is a star domain with respect to the point Yo E 0 and that (/o(x) - Yo,

It (x)

- Yo)

> 0 for x

E ao.

( 1.32)

Then u(fo(ao),yo)

= u(lt(aO),yo)

( 1.33)

= d(] 1,0, Yo)

(1.34)

and d(] 0,0, Yo)

for any g.L.-S. extensions 10 and 11 to

n of fo

and

It

resp.

PROOF. By definition of the winding number, assertion (1.33) implies assertion (1.34). It thus is sufficient to prove the first of these equalities. Suppose then that (1.33) were wrong. Then by Corollary 3 there would exist a AO > 0 and an E ao such that (1.31) was true with x = xo. Scalar multiplication of the latter equality by fo(x) - Yo yields:

Xo

(Jo(xo) - Yo, h(xo) - Yo} = -,xollfo(x) -

yol12 < 0

in contradiction to assumption (1.32).

1.1'1. THEOREM. Let f be a g.L.-S. map of the Hilbert space E into itself and let Yo be a point of E. We suppose that uniformly

.

(

X-yo) = +00,

hmsup f(x) - f(y), II II 112:11 . . . . 00 x - Yo

i.e., that there exists an increasing sequence of positive real numbers Tl, T2, ..• which tendB to 00 and in addition has the following property: to every positive c there cOfTesponds an integer n(c) such that, for n ~ n(c),

( f(x) ifllx -

Yol/ = rn.

f(yo),

X-yo)

IIx _ yoll > c

(1.35)

It is asserted that the equation f(x) = Yo

(1.36)

THE POINCARE-BOHL THEOREM AND APPLICATIONS

82

has a solution in the ball B(Yo,r no ) with no = n(co) where

Co =

Ilf(yo) -

yoll·

(1.37)

PROOF. The assertion is trivial if Co = 0 since then x = Yo is a solution of (1.36). But for Co > 0 it will be sufficient to show that ( 1.38) Now

(f(x) - Yo, x - Yo) = (f(x) - f(yo), x - Yo) - (YO - f(yo), x - Yo),

Ilx - yoll collx - yoll

and therefore by Schwar~'s inequality, by (1.37) and (1.35), for

(f(x) - Yo, x - Yo)

~

(f(x) - f(yo), x - Yo) -

> co[llx - Yoll-Ilx - yoll]

= rno'

= O.

We thus see that the assumption (1.32) of Corollary 4 to Theorem 1.12 is satisfied with 0 = B(yo,rno ) and with 70 = f, 71(X) = x. Consequently, by that corollary, (1.34) holds with the right member of this equality being unity since 71 is the identity map. This proves the assertion (1.38). So far the applications of the Poincare-Bohl theorem 1.8 concerned existence proofs. We conclude this section with an application to a fixed point theorem. 1.18. THEOREM. Let 0 be an open bounded star domain in the Banach space E, and let F: 11 ---+ E be completely continuous. We assume that F(aO)

en

(1.39)

and assert that 11 contains a fixed point for F, i.e., a point x satisfying

F(x) = x.

(1.40)

PROOF. If ao contains a fixed point, then our assertion is true. Suppose now

F(x)

i- x

for x E a~.

(1.41)

We will then prove that there is a fixed point in O. To this end we consider the L.-S. map f(x) = x - F(x) + Yo, (1.42) where yo is a point in 0 with respect to which 0 is a star domain. Then the assertion (1.40) is equivalent to

f(x) = yo,

(1.43)

and the assumption (1.41) is equivalent to

f(x)

i- Yo

for x E ao.

(1.44)

Now to prove that (1.43) (and therefore (1.40)) has a solution in 0, it will, by Corollary 2 to Theorem 1.12, be sufficient to show that the assumption (1.28) of that corollary is satisfied with II = f. Suppose (1.28) were not true, i.e., f(x) - Yo = --X(x - YO)

(1.45)

INTERSECTION NUMBERS for some x E an and some

.x > O.

83

Then for such x and

.x

F(x) - Yo = (x - yo)(1 +.x)

(1.46)

as seen from (1.42) by elementary calculation. But since x E an, since 0 is a star domain with respect to Yo, and since 1 +.x > 1, we see from (1.46) that F(x) lies in the exterior of n in contradiction to our assumption (1.39). Thus (1.45) is shown to be wrong. §2. The interpretation of degree and winding number as intersection numbers 2.1. As in §4.llet 0 be a bounded open set in the Banach space E and let

f: n ---+ E be a g.L.-S. map. Finally let Yo be a point of E satisfying Yo ~ l(aO).

(2.1)

We know that if the equation I(x) = Yo has no solution in 0, i.e., if Yo ~ 1(0), then d(f, 0, Yo) = O. On the other hand, if this degree is different from zero, then the above equation has at least one root x E 0, i.e., Yo E 1(0). We therefore consider the degree as a measure for the intersection of Yo with 1(0) and define the intersection number i(f(O), Yo) by setting

i(f(O), Yo) = d(f, n, Yo).

(2.2)

To obtain an interpretation of the winding number u(f (aO), Yo) as intersection number we need the following theorem. 2.2. THEOREM. Let assume that

I

be a g.L.-S. map

an

---+

E. In addition to (2.1) we

u(f(ao), Yo) -I- O. Then every ray issuing from Yo (see §1.23) intersects PROOF. Let (2.3)

1 be a g.L.-S. extension of /

to

n.

(2.3)

/(an).

Then by definition and by

d(J,n,yo) = u(f(aO),yo) -I- O. (2.4) On the other hand, len) is a bounded set. Consequently, there exists an R > 0 such that the equation J(x) = y has no solution for Iyl > R, and therefore d(J, 0, y)

=0

for

Ilyll > R.

(2.5)

Suppose now there were a ray issuing from Yo and not intersecting l(aO). If we choose a point Yl on the ray with IIYlll > R, we arrive at a contradiction. For on the one hand, the segment on this ray given by the points Yt = (1 - t )yO + tYI for 0 :S t :S 1 does not intersect l(aO). Therefore by subsection 6.8 in Chapter 3 d(f, 0, Yo) = d(f, 0, yd· (2.6) On the other hand, (2.5) holds for y = Yb and the two members of (2.6) are not equal as seen from (2.4) and from (2.5) (with y = yd·

84

THE POINCARE-BOHL THEOREM AND APPLICATIONS

2.3. An obvious corollary to Theorem 2.2 is the assertion: if there exists a ray issuing from Yo which does not intersect 1(80.), then u(f(8n), Yo) = 0. 2.4. Theorem 2.2 and its corollary motivate the following definition. If 1 is a g.L.-S. map: 80. -+ E, if Yo satisfies (2.1), and if ho is a ray issuing from Yo, then the intersection number i(f(8n), ho) of 1(80.) with ho is given by

i(f(8n), ho ) = u(f(8n), yo)·

(2.7)

A more detailed motivation for this definition can be given in the finitedimensional case (see subsection 3.12 in Chapter 7). 2.5. If 1 is as in subsection 2.4 and if 1 is a g.L.-S. extension of 1 to n, then (2.2) holds with 1 replaced by But with this substitution the right member of (2.2) is-by definition of the winding number-equal to the right member of (2.7). We therefore see that

r

i(f(8n), ho) = i(1(n), yo).

(2.8)

2.6. We will use formula (2.8) to derive the "duality formula" (2.12) below. To do this we need some natural definitions. If Yl is a point of ho different from Yo satisfying Yl ~ 1(80.), we set 8(yo, yd = Yl - Yo and set

i(f(n), 8(yo, yt}) = i(f(n), yt} - i(f(n), Yo).

(2.9)

We now subject Yl to the additional condition iiYlii > R where R is the number defined in the proof of Theorem 2.2. Then (2.5) holds with y = Yl. Therefore, by (2.2), i(1(n), yt} = 0, and we see from (2.9) that for such Yl

i(1(n), 8(yo, yd) = -i(1(n), yo).

(2.10)

We note next that by our choice of Yl all intersection points of 1(80.) with ho lie in the open segment (Yo, yd of ho. Therefore we define

i(f(8n), (Yo, yd) = i(f(8n), ho).

(2.11)

But (2.11), (2.8), and (2.10) together yield the formula

i(f(8n), (Yo, yd) = -i(1(n), 8(yo, yd)·

(2.12)

§3. Notes to Chapter 4 Notes to §1. 1. We use the notation of subsection 1.6 of this chapter. Let (r, ¢) be polar coordinates of the point x E E2 with () as pole. For r :f:. 0, (r, ¢) = (r, ¢d if and only if ¢1 - ¢ is an integer multiple of 211". If s = ¢/211" we define a one-valued continuous map of the real s-axis onto 8 1 by setting x(s) = (1, s). Then

x(sd =

X(S2)

Given a continuous map

if and only if Sl

-

S2

is an integer.

(1)

f: 8 1 -+ E2 - (), we set g(s)

= I(x(s)).

(2)

NOTES

85

From (1) we see that if S1 - S2 is an integer, then (3)

g(st} = g(S2)' Now let

(p(s),~(s))

be polar coordinates for g(s):

g(s) = (p(s), ~(s)),

(4)

where by (1), p(s + 1) = p(s) > 0 and where again the polar angle ~(s) is determined up to integer multiples of 27r. Thus ~(s) is not necessarily continuous. However, it can be proved that there exists a real-valued function A(s) = Af(s), called an angle function for f, which has the following two properties: (a) A(s) is one-valued and continuous for all real s, and (b) for each fixed s, A(s) equals one of the values of the polar angle ~(s). For the verification of the existence of such a function, we refer the reader to [2, p. 463J. It is clear that such an A is not unique. For if A( s) is an angle function, then for each integer k, Ak(S) = A(s) + 27fk is also an angle function. From property (b) it is also clear that to each angle function A and each real s there corresponds an integer n such that A(s + 1) - A(s) = 2·7m. (5) However, it is not hard to see (we refer again to [2, p. 463]) that the integer n depends neither on the choice of the angle function A nor on the value of s. It thus depends only on f and e. Its intuitive interpretation is obviously that if x varies one time over S1, then f winds n times about the point e. The assertion (1.13) states that this number n = n(f) equals the winding number u(f(S1), e) as defined in subsection 1.5 (with E = E2). The proof of (1.13) will be given in three steps. Step 1. We consider the special case

fo(x(s))

= g(s) = (1, 27l'ns),

(6)

where n is an integer. Then Ao(s) = 27rns is an angle function for fo, and (5) is true with f = fo. To prove (1.13) in this case, we recall that by definition

(7) for any continuous extension 10 of fo to B(1,0). We define x = (r, ¢), 0 < r -s: 1, 0 -s: ¢ < 27l', we set

10(x)

= (r,n¢) = (r,27rns),

10 as follows:

for

(8)

and for r = 0 we set

10(e) = e.

(9)

Our assertion (1.13) for f = fo is, by (7), equivalent to

d(Jo,B(1,e),0) = n.

(10)

Now 0 is not a regular value for 10 , But we know that the left member of (10) equals d(J0, B(1, e), YO) if Ilyo I is small enough. It is therefore sufficient to prove

d(Jo,B(l,e),Yo)=n

(11)

THE POINCARE-BOHL THEOREM AND APPLICATIONS

86

for Yo = (ro,O) and ro small enough (actually any ro E (0,1) would do). We consider first the case n ¥- O. The roots of the equation

(12)

10(x) = Yo are the points Xi Chapter 2

= (ro, (27ri/n)), i

= 0,1, ...

,n - 1. Now by Definition 2.2 in n-1

d(] 0, B(1, e), Yo) =

I>i'

(13)

i=O

where ji is the index of the linear map D10(Xi; h), and, by subsection 2.3 in Chapter 2, Ji is the sign of the Jacobian Ji of 10 at Xi. Now the Jacobian of the rectangular coordinates Xl = r cos ¢, X2 = r sin ¢ with respect to r, ¢ equals r and thus is positive for r = roo It therefore follows from classical properties of J acobians (see, e.g., [16, p. 171]) that for the computation of ji = sign Ji it does not matter whether we consider 10 as a function of the rectangular or polar coordinates. But in the latter case 10 is given by r' = r, ¢' = n¢ and obviously

Ji

=n

for i

= 0, 1, ... , n -

1.

Thus

Ji

= {

+1 -1

if n > 0, if n < O.

This together with (13) proves assertion (11) if n ¥- O. Now suppose n = o. Then, by (6), 10 maps Sl into the single point (1,0) of Sl, and we obtain a continuous extension 10 of 10 by mapping B(e, 1) into (1,0). The equation 10(x) = e has no roots in O. Therefore d(]o, B(I, e), e) = 0, which proves (11) with n = O. Step II. Let JI(x{s)) = gl(S) = (1, 1li1(S)), where 1li1{S) is a one-valued function satisfying iii 1 (s + 1) - iii 1 (s) = 27rn (14)

< s < 00. For 0 ::; t ::; 1 we set llit(s) = (1 - t)27rn + tIli1(S) and h(x{s)) = gt{s) = (1, llit{s)). From (14) we see that llit(s + 1) - llit(s) = 27rn.

for

-00

This shows that It is for each t E [0,1] a one-valued continuous map of Sl into Sl E E2 - e. It therefore follows from Theorem 1.7 that u(JI{Sl),e) = u(fO(Sl), 0). But here the right member equals n by Step I. We thus proved that U(h(Sl),O) = n. Step III. The general case. Let I(x(s)) = g{s) = (p(s), lIi(s)), where p(s) is as in (4) and where iii (s) is an angle function for I· Since p is continuous and I?eriodic, there exist constants m and M such that 0 < m ::; p( s) ::; M for all s. Obviously we can choose m and M in such a way that also 1 E [m, M]. Consequently, if Pt(s) = (1 - t)p(s) + t ·1, then 0 < m ::; rt{s) ::; M for all t E [0,1] and for -00 < s < 00. If moreover It(x(s)) = (Pt{s), lIi{s)) it follows from Theorem 1. 7 that u(f(Sl), 0) = u(h (S1), 0). But the right member equals n by Step II. This proves our assertion (1.13).

NOTES

87

2. Proof of the assertion in subsection 1.11. Let Yo be a point in the bounded open convex set 0 in the Banach space E. Let h be a ray issuing from Yo and let YI be a point of h different from Yo· Then, by definition, h consists of the points h(t) = (1- t)yO + tYI, 0 ~ t < 00. Now h(t) EO for t small enough since h(O) = Yo E O. On the other hand, Ilh(t)11 -+ 00 as t -+ 00. Thus h(t) lies in the exterior of 0 for big enough t. Thus the least upper bound of those t for which h(t) E 0 is a finite positive number to. Then obviously h(to) EO and h(to) ~ o. Therefore h(to) E a~. lt remains to show that the intersection of h(t) with ao is unique. For this it will be sufficient to prove (A): If h(to) E a~, then h(t) EO for 0 ~ t < to. Indeed if h(to) and h(tt} are in ao with to 1:- tt, we may suppose 0 < tl < to. But then, by (A), h(tl) E 0 which contradicts the assumption h(tt} E a~. To prove (A) we note first that there exists a 8 > 0 such that

B(yo, 8) c 0,

(15)

since Yo is a point in the open set O. Now let 0 < tl < to. We have to show that h(tt} C 0, i.e., that h(tt} is the center of a ball lying in O. We will prove indeed that (16) B(h(tt}, 8(to - ttl/to)

co.

Now if y is an element of this ball, then

Ily -

h(tt}11 < 8(to - ttl/to.

(17)

From the definition of h(t), we see by elementary calculation that

h(tt} = yo(to - ttl/to + h(to)ttfto. Substituting this in (17) we obtain lIyto/(to - td - h(to)td(to -

td -

Yo II

< 8.

This shows that

z = yto/(to - til - h(tO)tl/(to - ttl E B(yo, 8). Therefore, by (15), z E 0 C: rl. But

y

= z(to -

td/to + h(to}ttfto.

We thus see that y is a convex linear combination of the points z and h(to) which lie in n. Since the convexity of 0 implies that of n (see, e.g., [18, p. 413]), it follows that YEn. This proves (16).

CHAPTER 5

The Product Theorem and Some of Its Consequences §1. The product theorem

1.1. The product theorem establishes a connection between the degrees of two L.-S. maps f, g and the degree of composite map gf. In the case that the maps are also Gil, the theorem is essentially a consequence of the chain rule for differentials (see §1.14) and the product theorem in subsection 6.1 of Appendix A for indices of linear nonsingular L.-S. maps. The proof for the general case requires an approximation procedure. 1.2. STATEMENT OF THE PRODUCT THEOREM. Let 0 be a bounded open set in the Banach space E, let f be an L.-8. map ~ E, and let g be an L.-8. map fen) ~ E. Let {G/3} be the set of the bounded components of E - f(aO) where f3 E B, an index set (§§1.33, 1.34). Finally let zo be a point of E satisfying

n

zo ¢ gf(aO).

(1.1)

Then (i) the degree d(g, G/3, zo) exists for all f3 E B for which G/3 C fen); (ii) either the degree d(gf, 0, zo) equals zero (which, in particular, is the case if the equation gf(x) = zo

(1.2)

has no solutions in 0), or there exists a finite subset BO of B such that d(gf, 0, zo)

=

2: d(g, G/3, zo) . d(J, 0, Y/3), (1.3)

/3EBO

for f3 E B O

G/3 C f(O)

with Y/3 E G/3, and where all terms in the sum are different from zero. We will show later (see subsection 1.5) that the product theorem is a consequence of Theorem 1.4 below. In preparation of the latter theorem we state the following lemma. 88

THE PRODUCT THEOREM

89

1.3. LEMMA. Let E,O, and f be as in subsection 1.2. Let Y be a bounded open connected subset of E for which

f(O)

C

Y

(1.4)

and dist(J(O), BY) > O.

Let 9 be an L.-8. map Y (1.1), satisfies

---+

(1.5)

E, and let zo be a point of E which, in addition to zo ¢ g(ay).

(1.6)

Finally let {Ya } be the set of components of Y - f(aO) where a E A, an index set (see §§1.33, 1.34). Then (a) aYa c f(aO) u ay for all a E A; (1. 7)

(b) if y E Y is a root of the equation g(y) = zo,

(1.8)

then (1.9)

(c) if A' is the subset of those a E A for which Ya contains a root y E Y of (1.8), then A' is finite; (d) the degree d(g, Ya , zo) exists for all a E A; (e) let AO be the subset of those a E A' for which Ya contains a root y E J(O) oj (1.8). It is asserted that iJ (1.2) has a root in 0, then AO is not empty. Assertion (a) follows directly from the definition of components. Proof of (b). That a solution y of (1.8) does not lie on ay is the assumption (1.6). Suppose now that y E f(aO) with y satisfying (1.8). Then y = f(x) for some x E ao. But then gJ(x) = g{y) = Zo which contradicts the assumption (1.1). Proof of (c). By part (iii) of Lemma 10 in Chapter 1 the set 8 of solutions y E Y of (1.8) is compact, and, by (b), 8 E Y - J(aO) = Ua Y a . Thus the Ya form an open covering of the compact set 8. Consequently there exists a finite subcovering, i.e., there are a p E A, p = 1,2, ... ,r, such that S C U~=l Yap. If we denote by A' the subset of A whose elements are al, a2,"" a p , then Y a has no point in common with 8 if a ¢ A' since for such an a Ya n Yap would not be empty for some a p . This would contradict the fact that the components Ya are disjoint. Proof of (d). We have to show that Zo ¢ g(aYa). But this follows directly from assertions (a) and (b) together with the definition of A'. Proof of (e). If Xo E 0 is a root of (1.2), then Yo = f(xo) is a root in Y of (1.8). Therefore, by (b) it lies in some Yao ' Then ao E A' by definition of A'. Again by definition ao E Ao since Yo = f{xo) E f(rI). PROOF.

90

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

1.4. THEOREM. Making the same assumptions and using the same notations as in Lemma 1.3, we assert: if (1.2) has solutions in 0, then

d(gf, 0, zo)

=

L

d(g, Y zo) . d(f, 0, YQ), Q ,

(1.10)

QEAo

where YQ E YQ. If (1.2) has no solutions in 0, the right member of (1.10) is to be replaced by zero. 1.5. Before proving Theorem 1.4, we show that it implies Theorem 1.2. Let the assumptions of 1.2 be satisfied. Since the assertion of that theorem is obvious if (1.2) has no solutions, we suppose that this equation does have solutions. It remains to prove assertions (i) and (ii) of Theorem 1.2. Now the map g(y) = y - G(y) is defined on the closed bounded set f(ri). Since G is completely continuous, the set G(f(ri)) is bounded. Thus there exists a bounded convex set 6 such that

G(y) c

6

(1.11)

for y E fen). Now, by §1.31, G can be extended to a completely continuous map E ---+ E such that (1.11) holds for all y E E if the extended map is again denoted by G. Since 6 is bounded we see that for some positive M

IIG(y)11 < M

(1.12)

for all y E E. Now g(y) = y - G(y) is an L.-S. extension to E of the originally given g. From (1.12) we see that

Ilg(y)11

~

IIYII-IIG(y)11 > Ilyll- M.

(1.13)

Now let RI be an upper bound for Ilf(x)ll, x E ri, and let R be a number satisfying R > RI . (1.14) R > M + IlzolI, Then with Y = B(O, R) all assumptions of Lemma 1.3-and thus of Theorem l.4-are satisfied: that (1.4) and (1.5) hold is obvious, and (1.6) follows from (1.14) and (1.13) since Ilyll = R for y E 8Y. Thus, by Theorem 1.4, the formula (1.10) holds. Now since Y is a ball, it follows from part (iv) of §1.34 that each Y a E A, is either a bounded component of E - f(80) or Y = Y n Coo, (1.15) Q ,

Q

where Coo denotes the unbounded component of E - f(80). Now it is possible that (1.15) is true for some a E AO.I But for such a

d(f, 0, YQ) = 0,

yo: E Ao:.

(1.16)

Indeed, since f(O) is .!?ounded, there exists an MI s~ch that the equation f(x) = if Ilyll > MI' Now Coo certamly contains a point Y with IIYIiI > M. Then d(f, 0, yd = O. But d(f, 0, Yo:) = d(f, 0, yd since YOt and YI lie in the same component (viz., Coo) of E - f(80). This proves (1.16), and we

Y has no solution in

°

THE PRODUCT THEOREM

91

see that the only nonzero terms in the sum in (1.10) are among those for which Ya is a bounded component Cf3 of E - f(80). Thus, again, if B = {,B E A I Cf3 is bounded}, then (i) follows from assertion (d) of Lemma 1.3, and (1.10) may be written as

d(gf, 0, zo) =

L

d(g, Cf3 , zo) . d(f, 0, Y(3), f3EB' where B' is a finite subset of Band Yf3 E Cf3. Now if all terms in the sum are zero, then the degree at the left equals zero. Otherwise let BO be the subset of these ,B E B' for which d(g, Cf3, zo)· d(f, 0, Y(3) 1= O. Then the equality in (1.3) follows. To see that also the inclusion in (1.3) is true, we note that d(f, 0, Y(3) 1= 0 for all Yf3 E Cf3 if,B E BO. Consequently, the equation f(x) = Yf3 has a root x E O. This proves the asserted inclusion. (Thus in (1.3) only the original 9 defined on f(11) appears, not in its extension which was used only for the proof.) 1.6. Starting the proof of Theorem 1.4, we note first that the assertion of that theorem in the case that (1.2) has no roots is an obvious one. We therefore assume that (1.2) has a solution. Then by part (e) of Lemma 1.3 the set AO is not empty, but, as a subset of A', it is finite by part (c) of that lemma. We note next that for each a E A the degree d(f, 0, y) is constant for Y E Ya . We therefore define d(f, 0, Ya ) by d(f, 0, Ya ) = d(f, 0, y),

(1.17)

and, restricting a to A 0 , we denote by KO the set of integers k such that

(1.18 ) for at least one a E AO. Moreover, for fixed k E KO we denote by Dk the union of those Ya with a E AO for which (1.18) holds. Since this is a finite union, it follows easily from part (d) of Lemma 1.3 that Zo rt. g(8D k ), i.e., that the degree d(g, Dk, zo) exists for k E KO and that by the sum theorem (1.19) where ak varies over those a E AO for which (1.18) holds. Now if Xo is a root of (1.2), then Yo = f(xo) E Dk for some k E KO, and for that k

Xo E Ok = f- 1 (Dk) = {x E

01

f(x) E Dd·

Since Dk is open and f continuous it follows that Ok is open. We thus see that every root of (1.2) lies in UkEKO Ok. Therefore

D(gf, 0, zo) = d (9f,

U Ok, zo) .

(1.20)

kEKO

Now the Ok are disjoint since the Dk are disjoint. In order to apply the sum theorem to the right member of (1.20), we have to show that

(1.21)

92

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

Suppose (1.21) to be wrong. Then some j; E

aO k

(1.22)

is a root of (1.2). Therefore j; E 0 1 for some l E KO. Now l I:- k by (1.22) since no point of aOk belongs to the open set Ok. But 0 1 is open. Therefore there exists some neighborhood N of j; which is contained in 0 1. On the other hand, N contains points of Ok as follows from (1.22). Thus Ok n 01 is not empty. This contradicts the fact that k I:- l and therefore that Ok and 0 1 are disjoint. We are now able to apply the sum theorem and see from (1.20) that

d(gf, 0, zo) =

I.: d(gf, Ok, ZO)·

( 1.23)

kEKo

1. 7. For fixed k E KO we now turn to the computation of d(gf, Ok, ZO). We do this first under the assumptions, added to the previous ones, that f and g, and therefore gf, are Gil maps and that Zo is a regular value for gf. Then the number of roots in Ok of (1.2) is finite, and so is the number of images under f of these roots. Let Yl, Y2, ... , Ys be these images and, for a = 1,2, ... , s, let Xul, Xu 2, ... , xu{3(u) be the roots of

f(x) = Then the X ur , a = 1,2, ... , s, the Yu are solutions of (1.8). Now the differential

T

(1.24)

Yu·

= 1,2, ... , {3(a), are the roots in Ok of (1.2) and

Dgf(x; h)

(1.25)

is nonsingular for every root x of (1.2) since Zo is a regular value. We claim that also the differentials (a)

Df(x; h),

(b)

Dg(y; fI)

are nonsingular for every root x of (1.2) and for y by the chain rule for differentials

= Yu,

Dgf(x; h) = Dg(f(x); Df(x; h)).

(1.26)

a

= 1,2, ... , s.

Indeed (1.27)

Suppose now that the differential (1.26a) is singular, i.e., that there exists an h I:= 0. (Recall that the differentials are linear 1.-S. maps.) Then, by (1.27), the differential Dgf(x; h) = 0 for this h and thus is singular against assumption. Suppose next that the differential (1.26b) is singular for y = Yu· Then Dg(y; flo) = (j for some flo I:- e. Now from the nonsingularity of D f(x; h) just proved, we conclude that the equation Df(x; h) = flo has a unique solution h = ho· Obviously ho I:- (j. But then Dgf(x; ho) = (j by (1.27), and we arrive again at a contradiction to the nonsingularity of D f g( x; h). Now since the differentials (1.26) with x = Xur and y = Yu are nonsingular L.-S. maps, the indices j of the linear maps given by these differentials are defined. Therefore, by the definition of the degree at a regular value, by the chain

osuch that Df(x; h)

THE PRODUCT THEOREM

93

rule (1.27), by (1.24), and by the product theorem 6.1 for indices in Appendix A, we see that 8 (3(U) d(gJ, Ok, Zo) = j(DgJ(xUT ; h)) u=lT=l 8 (3(U) = j(Dg(f(xUT ); D J(X UT ; h))) u=lT=l 8 (3(U) = Lj(Dg(Yu;77)) . j(DJ(xUT ;h)) u=lT=l 8 (3(U) = Lj(Dg(Yu;77)) Lj(DJ(xUT;h)). u=l T=l But here the inner sum equals d(f, Ok, Yu) since for fixed a the XUT are the roots in Ok of (1.24). But y = J(XUT ) E J(Ok) = Dk. Therefore d(f, Ok, Yu) = k, and we see from the above formula that

LL

LL

L

s

d(gJ,Ok,ZO) = k·

L

j(Dg(yu; 77))·

(1.28)

u=l IT KO contains k = 0, we see that

d(gJ, 00, zo) =

o.

(1.29)

Now let k '" O. We know that Y1,Y2, ... ,Ys are roots in Dk of (1.8). We assert that they are the only ones. Indeed, let y E Dk be a root of (1.8). Then, by definition of Dk, d(f, 0, y) = k '" O. Therefore the equation J(x) = y has a root ~ in O. Since I(x) = y E Dk, it follows that x E 1- 1(y) = Ok. Moreover gl(~) = g(y) = ZOo Thus ~ is a root in Ok of (1.2), and therefore x = X UT for some a E (1, ... ,8) and T E (1, ... , ,8(a». But then y = I(x) = l(xUT ) = Yu by (1.24). From the assertion just proved, the sum in (1.28) equals d(g, Dk, zo). Thus

d(gJ, Ok, ZO) = k . d(g, Dk, ZO).

(1.30)

From this equality together with (1.23), we obtain

D(gJ, 0, zo)

=

E

k· d(g, Die;, ZO).

(1.31)

keKO

Here the sum in the right member is a re-ordering of tht> sum of the right member of the aaaertion (1.10) as is seen from (1.18) and (1.19). We thus proved Theorem 1.4 under the additional assumption made at the beginning of this section. To complete the proof of that theorem, it remains to show that (1.31) holds without these additional assumptions. 1.8. Pro%/ (1.31) without the assumption that Zo is a regular value 0/ gJ but .eill V1ith the assumption that g and J are Gil maps. It will be sufficient to

94

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

prove (1.30). As proved in subsection 1.6 the sets gf(aO k ) and g(aDk) do not contain Z00 Therefore there exists a positive r such that the ball B(zo, r) has no point in common with these sets. Taking into account Definition 4.3 in Chapter 2 and Theorem 4.2 in Chapter 2, this implies that for z E B(zo, r)

d(g, D k , z) = d(g, Dk, zo),

d(gf, Ok, z) = d(gf, Ok, zo).

(1.32)

Now by Theorem 4.1 in Chapter 2 we can choose a point Zl E B(zo,r) such that is a regular value for gf. Therefore by subsection 1.7 the relation (1.30) holds with Zo replaced by Zl. But then, by (1.32), it also holds for ZOo 1.9. It remains to prove (1.31) without the additional assumption that f and g are Gil maps. We start by defining for each k = 0, ±1, ±2, ... , the set Llk by Zl

Llk

= {y E Y

- f(aO)

I d(f,O,y) = k}.

(1.33)

For some k the set Llk may be the empty set 0. We define

d(g, Llk, zo) = 0 if Llk = 0.

(1.34)

Moreover, as will be verified in the Notes,2 the following relations hold:

k· d(g, Llk, zo) = 0 if Llk f=- 0 and k ~ K O;

(1.35)

k . d(g, Llk, zo) = k . d(g, Dk, zo)

(1.36)

for k E KO

(see subsection 1.6 for the definition of KO and Dk). It follows from (1.34)-(1.36) that our assertion (1.31) is equivalent to

d(gf, 0, zo) =

E k· d(g, Llk, zo),

(1.37)

k

where the summation is extended over all integers k. The proof of (1.37) will be based on the following approximation lemma whose proof will be given in the Notes. 3 1.10. ApPROXIMATION LEMMA. With the notation of subsection 1.9 and Lemma 1.3 we assume that the maps f and g satisfy the assumptions of that lemma. Moreover, we assume that (1.37) is satisfied. Then there exist positive numbers 1 and 8 of the following property: if f* and g* are L.-8. maps ~ E and Y ~ E resp. and satisfy the inequalities

n

Ilg*(y) - g(y)11 < 1

( 1.38)

11f*(x) - f(x)11 < 8

( 1.39)

for all y E Y and for x E n, then (i) f* and g* satisfy the assumptions of Lemma 1.3 with f and g replaced by f* and g*, and (ii) d(g* f*, 0, zo) = k . d{g*, Die, zo), (1.40)

L k

where K~ and Die are defined for f*, g* as KO and Dk were defined for f, g.

THE PRODUCT THEOREM

1.11. AN EXTENSION THEOREM. We use the assumptions and notations of subsection 1.9 and of Lemma 1.3. Moreover, as in subsection 1.1 of Chapter 3, we denote by C(O) and C(Y) the space of completely continuous maps F: 0 ---+ E and G: Y ---+ E resp. Let C 1 (r1) and C 1 (y) be linear subsets of CeO) and C(Y) resp. Suppose the product formula (1.10) holds for all couples It: ---+ E and gt: Y ---+ E for which Fl(X) = x - It (x) and Gt(y) = Y - gl(Y) are elements of C t (r1) and C1(y) resp. Then (1.10) holds for all couples f,g for which F(x) = x - f(x) and g(y) = y - G(y) lie in C1(O) and CdY) resp.

n

This theorem is an immediate consequence of the approximation lemma in subsection 1.10 and the fact that (1.37) implies (1.10). 1.12. If C2 (n) = C(n) n C"(O) for every bounded open set [} E E, then, by subsections 1.11 and 1.8, Theorem 1.4 holds for any couple f, g for which F(x) = x - f(x) and G(y) = y - g(y) lie in C2 (O) and in C2 (Y) resp. Thus Theorem 1.4 has been proved for all those Banach spaces E for which (1.41 )

for all bounded open subsets [} of E. (1.41) is true in particular for finitedimensional Banach spaces. 1.13. We know from subsection 2.2 in Chapter 3 that the property (1.41) does not hold in all Banach spaces. In subsection 1.14 we will prove our product formula for finite layer maps in arbitrary Banach spaces, and application of the extension theorem 1.11 will then, in subsection 1.15, yield the product formula (1.10) in the generality stated. 1.14. In addition to the assumptions made for our product formula (1.10) we suppose in the present section that I and 9 are finite layer maps. Then there exists a subspace E" of E of finite dimension n such that not only are I and 9 layer maps with respect to E" but also (1.42)

r

We set 0" = 0 n E", Y" = Y n En and denote by and (gl)" the restrictioll of / and g/ resp. to Likewise gn denotes the restriction of 9 to yn. No,", f(x) E Y by (1.4) and F(x) = x - I(x) c E". Therefore

nne

rex) E Y" c En for x E

nne

Moreover gRey) E E" for y E since.G(,I)

= 11- g(y) E En.

vn,

It follows that for x E (2n

(gl)"(x)

= gr(x) = gn rex).

Moreover (see 1.4»

r(l'F) c yn.

(1.43)

""--_._- .....

Since 8nn

c 80. and 8yn

C

8Y, we see from (1.1), (1.6), and (1.45) that Zo ~ gn r(8n n )

(1.47)

and ( 1.48)

Since it is clear that (1.5) implies that dist(r(rF), 8yn) > 0, we see that the assumptions of Lemma 1.3 (and therefore of our product formula) are satisfied for the continuous finite-dimensional maps In: On ---+ En, gn: yn ---+ En. Therefore (1.10) holds with I and g replaced by rand gn resp. of subsection 1.14. Since (1.10) is equivalent to (1.37), we see that d(gn

r, nn, zo) = L k . d(gn, ~k' zo),

(1.49)

k

where (cf. (1.33)) ~k = {y E yn I d(r,o.n,y) = k}.

(1.50)

It remains to prove (1.37) for the given layer maps I and g. Now from the definition in subsection 3.4 of Chapter 3 of the degree for finite layer maps and from (1.45) we see that d(gl, 0., zo)

= d{(gf)n, o.n, zo) = d{gn r, o.n, zo),

d(f, 0., y) = d(r, o.n, y)

d(g, ilk, Zo) = d{gn, ilk, ZO),

for y E yn,

(1.51)

where ilk is defined by (1.33) and where

ilk = ilk n En.

(1.52)

Now by (1.49) and (1.51) d(gl, 0., ZO) =

E k· d(gn, ~k' zo),

(1.53)

while by the last of the equalities (1.51) our assertion (1.37) reads d(gl, 0., zo)

=

E k . d(gn, ilk, zo).

This together with (1.53) shows that (1.37) will be proved once it is shown that

~k Now, since

=

ilk'

I is a layer map with respect to

( 1.54)

En,

d(f,n,y) = d(r,nn,y)

(1.55)

if y E En and y ~ j(80.). Now let y E ~k' Then by (1.50) the right member of (1.55) equals k. But then the left member equals k, which by (1.33) means that y E ilk. Since y E En, it follows from (1.52) that y E ilk' We thus proved Vk C ilk' Suppose now y E ilk' Then by (1.52), (1.55) and (1.51) y E {y E Y

I d(f,o.,y) =

k} nEn

= {y E yn I d(r,o.n,y) = k} = ~k'

This finishes the proof of (1.37), and therefore of (1.10) for finite layer maps.

THE INVARIANCE OF THE DOMAIN

97

1.15. The product formula (1.10) follows now from 1.13 and the second Leray8chauder lemma in subsection 4.3 of Chapter 3 upon application of the extension theorem in subsection 1.11, with C 1 (D) and C 1 (Y) being the set of completely continuous maps into E whose domains are 0 and Y resp. and whose ranges are contained in finite-dimensional subs paces of E. §2. The invariance of the domain 2.1. As is well known the property of a set to be open is not invariant under a continuous map. For instance Y = f (x) = x2 maps the open interval (-1, + 1) onto the interval 0 ::; Y < 1 which is not open. However, Brouwer proved in 1912 for finite-dimensional spaces that openness is invariant under continuous maps which have a continuous inverse. His theorem is know as the theorem of the invariance of the domain (Gebietsinvarianz). Leray [34] generalized this theorem to L.-8. maps in Banach spaces. He based his proof on the product theorem 1.4. The proof of Leray's generalization is the main object of this chapter. We first need the following theorem. 2.2. THEOREM. Let 0 be a bounded open set in the Banach space E, and

let f be an L.-S. map which maps I1 onto f(I1) in a one-to-one fashion. Then d(f, 0, y) = ±1

for y E f(O).

(2.1)

PROOF. We note first of all that since f is one-to-one on 0, ayE f(O) does not lie in f(aO). Therefore the left member of (2.1) is defined for the y indicated. We note next that by assertion (iv) of §1.1O the assumptions of our theorem imply that f has an inverse 9 which is an L.-8. map: f(O) onto O. Thus

g(y)

= y - G(y),

Y E f(O),

where G is completely continuous. Now gf(zo) = Zo E

n

(2.2)

for every ZO E O. This shows that assumption (1.1) of the product theorem 1.2 is satisfied for every Zo E O. Consequently by this theorem the equality (1.3) holds where by (2.2) the left member equals one. We know that only a finite number of the terms on the right member of (1.3) are different from zero, say for (3 = 1,2, ... , TO. Then TO

1 = ~ d(g, C/3, ZO) . d(f, 0, Y/3),

(2.3)

/3=1

where Y/3 E C/3, and

d(g, C!3, zo) . d(f, 0, Y/3) From the first factor being that

f.

f. 0

for (3 = 1,2, ... ,TO·

(2.4)

0, we conclude that there exists a y!3 E C/3 such

g(y!3) = Zo,

(2.5)

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

98

and from the second factor in (2.4) being =f. 0, we conclude the existence of an x{3 E for which (2.6) I(X{3) = Y{3·

°

From the last two equalities we see that Zo = gl(x{3) = x{3 and therefore by (2.6) (2.7) But since the C{3 are disjoint, the inclusion I(zo) E C{3 determines f3 uniquely, say f3 = 1, and since (2.7) is a consequence of (2.4), we see that ro = 1. Thus, by (2.3) and (2.7),

1 = d(g, C 1 , zo) . d(f, 0, yt} = d(g, C 1 , zo) . d(f, 0, I(zo)). This proves assertion (2.1) since y = I(zo) varies over 1(0) as Zo varies over 0. THEOREM OF THE INVARIANCE OF THE DOMAIN. Under the assumption 01 Theorem 2.2 it is asserted that (i) 1(0) is open, and (ii) 2.3.

(2.8)

1(80) = 81(0).

PROOF. (i). If Yo E 1(0), we have to prove the existence of a neighborhood N(yo) of Yo such that

(2.9)

N(yo) E f(O).

Now, as shown at the beginning of the proof for Theorem 2.2, Yo ¢: I (80). Therefore there exists a neighborhood N(yo) of Yo which does not intersect 1(80). We know that then

d(f,O,y) = d(f,O,yo)

(2.10)

for y E N(yo). But here, by Theorem 2.2, the right member is different from zero. Consequently so is its left member. But this implies that for y E N(yo) the equation I(x) = y has a solution x E 0, i.e., that y E 1(0). (ii) We note first that (2.11) Indeed the inclusion I(n) c 1(0) follows from the continuity of f. On the other hand, f(O) C 1(0.) = 1(0) since f(O) is closed, I being an L.-S. map. Now since is open, n = u 80, and therefore 1(0) = 1(0) U 1(80). But by assertion (i) 1(0) also is open. Therefore f(O) = 1(0) U 8f(0). Using (2.11) we see that

°

°

1(0) U 1(80) = 1(0) U 81(0). Since I is one-to-one on n this implies the assertion (2.8). 2.4. REMARK. An obvious consequence of Theorem 2.3 is the following assertion A: if the assumptions of that theorem are satisfied, if Xo E and if Yo = f(xo), then there exist neighborhoods U(xo) and V(yo) of Xo and Yo resp. such that for y E V(yo) the equation f(x) = Y has one and only one solution x E U(xo). Comparing this assertion with the inversion theorem in §1.21 we note: (i) the latter requires a differentiability assumption while assertion A does

°

THE INVARIANCE OF THE DOMAIN

99

not; (ii) the assumption for A that I is one-to-one is in the inversion theorem replaced by the assumption that the differential of the L.-S. map I at Xo is a oneto-one map. We also note that the "Fredholm alternative" for linear L.-S. maps (cf. part (iv) of Lemma 12 in Chapter 1) asserts: "one-to-one implies existence." Assertion A asserts this implication for not necessarily linear L.-S. maps (in the small). 2.5. COROLLARY TO THEOREMS 2.2 AND 2.3. Under the assumptions and with the notation 01 Theorem 1.2, it is asserted (i) i/O is connected, il I is one-to-one on 0 and Zo E gl(O), then d(gl, 0, zo)

= d(g, 1(0), zo) . d(J, 0, y) = ±d(g, 1(0), zo),

(2.12)

where y E I(n). (ii) Let g: I(n) --> g(J(O)) be one-to-one, let Zo E g(J(O)), and let Yo be the unique root 01 (1.8). Then d(gl, 0, zo)

II,

= ±d(J, 0, yo).

(2.13)

in particular,

(2.14) then d(gl, 0, zo) = d(g, C{3o' zo) . d(J, 0, Yo) lor some unique

/30 E B.

(2.15)

Finally d(J, 0, Yo)

=0

(2.16)

il Yo E (

U C(3) uC

oo .

(2.17)

/3EB- B o

PROOF. (i) Since 0 is connected, so is f(O). Moreover, by Theorem 2.3,

n

is open and (2.8) holds. It follows that 1(0) is one of the bounded components of E - l(aO), say f(O) = C{3o' Suppose first that d(g, f(O), zo) =I O. Then for y E 1(0) d(g, 1(0), zo) . d(J, 0, y) =I 0, (2.18)

since by Theorem 2.2 d(J, 0, y) = ±1.

(2.19)

This shows that the product (2.18) is, by definition of BO, an element of the sum of the right member of (1.3). -On the other hand, i~ is the only term in that sum since for y E C{3 with /3 -=i f30 the equation f(x) = y has no solution in O. Therefore for y E C{3 with /3 -=i f30 d(J, 0, y)

=0

(2.20)

(if a C{3 =I C{3o = 1(0) exists). This proves the first part of (2.12) under our assumption d(g, 1(0), zo) =I O.

100

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

Suppose now

d(g, /(0), zo) = o. This equality together with (2.20) shows that for all

(2.21)

/3 E B

d(g, C(3, zo) . d(f, 0, y(3) = 0

(2.22)

with y(3 E C(3. Therefore the set BO is empty, and

d(g/, 0, zo) = 0

(2.23)

by Theorem 1.2. But (2.23) and (2.21) show that the first part of (2.12) is again true. The second part of (2.12) follows now from Theorem 2.2. Proo/ 0/ (ii). Since Yo rt. /(80) and since the components of E - /(80) are disjoint, there are two cases: (a) Yo E C(3o for some unique /30 E B; (b) Yo E Coo. In case (a) the equation (1.8) has no solution y E C(3 for /3 f. /30, and therefore

d(g, C(3, zo) = 0

(2.24)

/3 f. /30, Now if /30 E BO the sum in (1.3) consists of exactly one term, viz. /3 = /30, and (1.3) is identical with assertion (2.15). But if /30 rt. BO then, because

for

of (2.24), BO is empty. Therefore by Theorem 1.2

d(g/, 0, zo) = 0,

(2.25)

and, by definition of BO, (2.26) d(g, C(3o' zo) . d(f, 0, Yo) = 0 since Yo E C(3o' But by (2.26) and (2.25) the equality (2.15) again holds. This finishes the proof of (2.15). But this equality implies (2.13) since d(g, C(3o' Yo) = ±1 by Theorem 2.2. In case (b) the equality (1.8) has no solution in any C(3. Thus we have d(g, /C(3, zo) = 0 for all /3 E B. Therefore BO is empty, and we see from Theorem 1.2 that (2.25) holds. But the right member of (2.13) also equals zero as is seen by the argument given for the proof of (1.16). Finally, the proof of (2.17) is contained in the preceding arguments. 2.6. Let 0 be an open (not necessarily bounded) subset of E. Let ¢: n ---+ E be an L.-S. map which is one-to-one. Then, for any bounded open set 0 for which nco, d(¢, 0, Yo) = ±1 (2.27) by Theorem 2.2 provided Yo E ¢(O). We call ¢ a coordinate transformation (in 0) if the right member of (2.27) is independent of the choice of 0 and of the point Yo E ¢(O). An example of a coordinate transform~tion in E = 0 is ¢ = l, where l is a linear nonsingular L.-S. map E ---+ E. In thiS case for y E ¢(O)

d(¢, 0, y) = j(¢), (2.28) where j(¢) is the index of ¢ (see subsection 1.4 of Chapter 2). In a generalization ~f the linear case, we define for any coordinate transformation ¢ (in 0) the index J(¢) = i(¢, 0) by (2.28). Note that Ij(¢)1 = 1 by (2.27).

THE INVARIANCE OF THE DOMAIN

101

2.7. LEMMA. If ¢ is a coordinate transformation in 0, then ¢-l is a coordinate transformation in ¢( 0) and (2.29) PROOF. Let 0 be a bounded open set whose closure since ¢-l¢ is the identity map on 0,

n is a subset of O. Then

d(¢-l¢, 0, xo) = +1 for Xo E O. On the other hand, by part (ii) of Corollary 2.5 (with g = ¢-l, that for Xo E 0 and Yo E ¢(O)

(2.30)

1= ¢),

we see

d(¢-l¢, 0, xo) = d(¢-l, ¢(O), xo) . d(¢, 0, yo). Thus by (2.30) and (2.28) 1 = d(¢-l, ¢(O), xo) . j(¢).

(2.31)

Since ¢(O) varies over all bounded open subsets of ¢(O) as 0 varies over all open bounded subsets of 0, (2.31) shows that ¢-l is a coordinate transformation in ¢(O) and that the first factor on the right member of (2.31) equals j(¢-l). Consequently (2.31) implies (2.29). 2.8. THEOREM. The Leray-Schauder degree is invariant under a coordinate transformation. More precisely, let 0 be a bounded open connected set in E, let I be an L.-S. map n ---. E, let 0 be an open set which contains all the bounded components of E - l(aO) and for which

nco,

1(0)

c 0,

(2.32)

let Yo be a point in E satislying Yo ~ f(80),

(2.33)

and let t/Jo be a coordinate transformation in 0 for which ¢(O) c O. Then

d(f, 0, Yo) = d(4)f¢-l, 4>(0), ¢(Yo)).

(2.34)

PROOF. Obviously

I(x)

= Yo

(2.35)

has a solution x in 0 if and only if

¢/¢-l(y)

= 4>(yo)

(2.36)

has a solution y E ¢(O). This shows that (2.34) is true if (2.35) has no solution in 0, for then both members of that equality are zero. Let us assume then that (2.35) has a solution in 0, i.e., that

Yo E 1(0).

(2.37)

Now with 0 the set ¢(O) is connected, and by Theorem 2.3 the latter set is also open. Therefore we can apply part (i) of Corollary 2.5 to obtain

d(f¢-l, ¢(O), Yo) = d(j, ¢-1¢(0), Yo) . d(4)-l, ¢(O), xo),

102

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

where Xo E

¢-1¢(0) = 0

(2.37)

and where d(¢-I,¢(O),xo) =j(¢-I) by Lemma 2.7. Thus

d(f¢-I, ¢(O), Yo) = d(f, 0, Yo) . j(¢-I).

(2.38)

To continue the proof of (2.34) we note that with /I = J¢-1 the right member of that equality equals d(¢Jb ¢(O), ¢(yo)). To compute this degree we apply part (ii) of Corollary 2.5 (with g = ¢ and with J replaced by /I). Now by Theorem 2.3 and by definition of /I, Jd8¢(0)) = /I (¢(80)) = J(80). Thus E - /I(8¢(0)) = E - J(80).

(2.39)

We first consider the case that Yo is not contained in the unbounded component C:x, of the set (2.39). Then Yo is contained in some bounded component C{3o and we see from (2.15) (with the proper substitutions) that

d(¢Jl' ¢(O), ¢(yo)) = d( ¢, C{3o' ¢(yo)) . d(/I, ¢(O), yo)· Here the first factor of the right member equals J.(¢) and the second factor is given by (2.38). Thus

d(¢J¢-I, ¢(O), ¢(yo)) = j(¢) . d(f, 0, Yo) . j(¢-I), and the assertion (2.34) follows from Lemma 2.7. We now consider the case that

¢(Yo) E C:x,.

(2.40)

We use (2.13) (with the proper substitutions) and (2.38) to obtain

d(¢/I, ¢(O), ¢(yo)) = ±d(/I, ¢(O), Yo) = ±d(f, 0, Yo) . J·(¢-I).

(2.41)

But from (2.17) we see that (2.40) implies that d(f, 0, Yo) = O. Therefore by (2.41)

d(¢/I, ¢(O), ¢(yo)) = d(f, 0, Yo) = 0, which by definition of /I proves the assertion (2.34).

2.9. THEOREM. Let 0 be an open bounded connected set in E, let 0- = {x EEl -x EO}, and let J be an L.-8. map 0 U 0- -+ E. Let /I (x) = J( -x). Then d( - II, 0-, -yo) = d(f, 0, yo). (2.42) PROOF. If we set ¢( x) = - x for every x E E, then the assertion (2.42) becomes identical with the equality (2.34) of Theorem ~.8, one of whose assumptions was that ¢ is an L.-8. map. Now ¢(x) = -x IS L.-8. if and only if E is finite dimensional. Thus for finite-dimensional spaces ~h~ assertion (2.42) is a direct consequence of that theorem. To prove the valldlty of (2.42) for arbitrary Banach spaces we note first that -II is L.-8., for if f(x) = x - F(x) then -fl(X) = -J(-x) = x + F(-x). Now let £ be a positive number which is

THE JORDAN-LERAY THEOREM

103

smaller than the distance from Yo to 1(80.) and from -Yo to /(80.-). Then by subsection 5.5 in Chapter 3 there exists a subspace En of E of finite dimension n containing Yo (and therefore -yo) of the following property: if o'n = 0. n En and 0.;;- = 0.- n En, and if and If are the restrictions of I and It to o'n and 0.;;- resp., then

r

d( -

It, 0.-, -Yo) = d( - If, 0.;;-, -yo).

But by the above argument the right members of these equalities are equal to each other. This proves (2.42).

2.10. AN EXISTENCE THEOREM. Let that I is one-to-one and that lim

x--+oo

I

11/(x)11 =

be an L.-S. map E 00.

-t

E. Suppose

(2.43)

Then I(E) = E. PROOF. The only subsets of E which are closed and open are E and the empty set. Since I(E) is not empty, it will be sufficient to prove: (a) I(E) is closed; (b) I(E) is open. Prool 01 (a). Let Yo be a point of the closure of I(E). We have to prove the existence of a point Xo such that

(2.44) Yo = I(xo). Now by assumption there exists a sequence of points Yl. Y2, ... in I(E) such that lim Yi = Yo

(2.45)

0--+00

and points

Xl, X2, •.•

such that

i = 1,2, .... (2.46) Yi = I(xi), The Yi are bounded by (2.45), and by (2.43) this implies that the Xi are bounded. Since F(x) = x - I(x) is completely continuous, it follows that for some subsequence nl! n2.· .. of the integers the points F(xnJ converge to some point Fo. Since xni = Yni + F(x ni ) by (2.46), we see from (2.45) that the x ni converge to the point Xo = Yo + Fo. It is now clear from (2.46) and (2.45) (applied to the subsequence ~) that Xo satisfies (2.44). Prool 01 (b). Let Yo E I( E). We have to show the existence of a neighborhood N(yo) of Yo such that N(yo) c I(E). (2.47) Now by assumption there exists a unique point Xo E E such that Yo = I(xo). Then let 0 0 be a bounded open subset of E containing Xo and let 10 be the restriction of f to Then, by Theorem 2.3, fo(Oo) is open. But Yo E 10(0.0) = 1(00) c feE). This proves (2.47) with N(Yo) = f(Oo).

no.

§3. The Jordan-Leray theorem 3.1. Let K be a circle of positive radius r in the plane E2. and let J C E2 be a homeomorphic image of K, i.e., a one-to-one continuous image of K. Then

104

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

J is called a Jordan curve. The original Jordan theorem states: if J is a Jordan curve, then the set E2 - J consists of exactly two components (see §1.33 for the definition of component). It is obvious that the set E2 - K has exactly two components, namely the bounded component {x E E2 I I x II < r} and the unbounded component {x E E2 I I xii > r} (if we suppose the center of K to be the zero point () of E2). Therefore the Jordan theorem may be stated as follows: the number of components of E2 - K is invariant under a homeomorphic change of K. Theorem 3.2 below, due to Leray, is a generalization in two directions: E2 is replaced by an arbitrary Banach space E, and the circle K by an arbitrary closed bounded set in E. The admitted homeomorphisms are L.-8. maps. 3.2. THE JORDAN-LERAY THEOREM. Let K and J be closed bounded sets in the Banach space E. Suppose there exists a one-to-one L.-S. map of K onto J. Then the cardinal number of components of E - J equals the cardinal number of components of E - K. PROOF. Let A and B be index sets which are in one-to-one correspondence with sets of bounded components of E - K and E - J resp. The bounded component of E - K corresponding to the element 0: of A will be denoted by Ka:, and the bounded component of E - J corresponding to the element /3 of B will be denoted by J(3. The unbounded components of E - K and E - J will be denoted by Koo and J oo resp. We have to prove

c(B)

= c(A)

(3.1)

where, as in §1.39, c(S) denotes the cardinal number of the set S. For the proof of (3.1) we introduce the vector spaces V and W generated by A and B, resp. (see §1.36). Now by §1.37, A and B are bases for V and W resp. Therefore by §1.39 the assertion (3.1) will be proved once it is shown that

V and Ware linearly isomorphic. We will construct linear maps g. : W

g·f*=Iv,

->

V and

r :V

f·g· = Iw,

->

(3.2) W such that

(3.3)

where Iv and Iw denote the identities on V and W resp. By §1.35 the existence g. implies (3.2). of such 3.3. The definition of rand g. . We recall our assumption that there exists a one-to-one L.-8. map f: K onto J. By part (iv) of §1.1O this implies that 9 = f- 1 is an L.-8. map J onto K. By Lemma 31 in Chapter 1 there exist L.-8. extensions of f and 9 to all of E. We choose one pair of such extensions and call them again f and 9 resp. Before defining f* we need to verify the following facts (i)-(iii): (i) the degree d(f, Ka:, y) exists for every Y E E - J; (ii) d(f,Ka:,Y/3) = const. for Y/3 E J{3, d(f,Kcr.,Yoo) = const. for Yoo E Joo ;

r,

THE JORDAN-LERAY THEOREM

105

(i) and (ii) allow us to define

d(f, Ko,J/3) = d(f, Ko, Y/3) d(f,Ko,Joo ) = d(f,Ko.,Yoo)

for Y/3 E J/3, for Yoo E J oo ;

(3.4)

(iii ) (3.5) and at most a finite number of Ko contain a root of I(x) = Y/3 for Y/3 E J/3, and therefore d(f, K o , J/3) = 0 except for at most a finite number of 0: E A. For the proof of assertion (i) we have to show that Y E l(aKo ) only if Y E J. Now aKo E K; therefore l(aKo ) E I(K) = J. Assertion (ii) follows from subsection 4.4 of Chapter 3. Proof of (iii). (3.5) follows from an argument used earlier: we can choose Y E J 00 in such a way that f (x) = Y has no solution in K 0., and therefore d(f, K o , y) = O. The remaining part of (iii) follows from an argument similar to the ones used in the proof of part (c) of Lemma 1.3. Suppose for some Y/3 E J/3 the equation I(x) = Y/3 has at least one solution Xi E Ko.; where 0:1. 0:2, •.. is an infinite subsequence of elements of A. Since the Ko; are disjoint, Xi "# Xj for i "# i Now the infinite sequence Xl, X2, .•. is bounded since it is contained in the union of the bounded components of E - K (cf. assertion (iii) of §1.34). Since I is L.-S., it follows that the sequence contains a convergent subsequence. Let Xo be the limit of such a subsequence. Then Y/3 = I(xo). Now Xo lies either in E - K or K. If Xo E E - K, then Xo E Ko.o for some 0:0 E A. But then infinitely many Xi E Koo and therefore Ko.; n Koo is not empty for infinitely many O:c E A, which leads to a contradiction since the Ko are disjoint. If Xo E K then Y/3 = I(xo) E I(K) = J. This also is impossible since Y/3 E J/3 C E - J. This finishes the proof of (iii). It follows that for each (3 E B the sum

Eo:' d(f, Ko., Jf3) oEA

is finite and therefore an element of the vector space V generated by A. Since B is a base for the vector space W generated by B, we can define a linear map g* (W): W -+ V as follows: for (3 E B we set g*({3) =

L o:.d(f,Ko ,J/3)

(3.6)

oEA

and extend this map linearly to all of W, i.e., if w = L~ Ti{3i with (3i E B and with Ti real, we set n

g*(w) =

L Tig*({3i).

(3.7)

i=l

If we denote by (I), (ti), (iii) the assertions obtained from (i), (ii), (iii) by replacing A, K , K 0, V, v , I by B , J , J a, W, w , g resp. and use the definition obtained from (3.4) by the same replacement, we see from our definitions and assumptions

106

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

that (i), (IT), (iii) are true. Consequently we obtain a linear map f* : V setting for a E A f*(a) = (3. d(g, J(3, K

L

Q ,)

-+

W by

(3.8)

(3EB

and extending this map linearly from A to V. 3.4. Proof of the relations (3.3). We will prove the first of these relations. The second one then follows by the replacements defined above. Now for a' E A we see from (3.8), (3.6) and (3.7) that

g*

f*(a') = L

g*((3) . d(g, J(3, Ko;l)

=

(3EB

La L o;EA

d(g, J(3, Ko;l) . d(f, Ko;, J(3).

(3EB

Since A is a base for V and since g* f* is linear, it follows that the first of the assertions (3.3) is equivalent to the equality

L

d(g, J(3, Ko;l) . d(f, Ko;, J(3)

=

(3.9)

80;0;1,

(3EB

where

80;0;1

denotes the Kronecker delta. To prove this equality we note first that (3.10)

Indeed, gf is the identity on K and, therefore, on aKo; since aKo; C K. Thus the identity fa on Ka is an 1.-S. extension of the restriction of gf to aKa. Therefore by Theorem 1.3 in Chapter 4

d(gf, Ka, x)

= d(Ia, Ka, x) = {~

~~ ~: ~::

This proves (3.10) since Ka and Kal are disjoint for a' "I a. We now apply the product theorem 1.2 to the left member of (3.10) to obtain

L

d(g,Ka,"pKa l )' d(f,Ka,Ka-y) = oaa "

(3.11)

Here r a is an index set which is in one-to-one correspondence with the bounded components Ka-y of E - f(aKa). On account of (3.11) the proof of our assertion (3.9) is reduced to verifying that

L d(g, J(3, Ka l) . d(f, Ka, J(3) = 2: d(g, Ka-y, Ka l ) . d(f, Ka, Ka-y). (3EB

(3.12)

-yEr"

To establish this relation between the components J (3 of E - J and the components Ka-y of E - f(aKa) we prove first that either or

J(3 E Ka-y for some unique 1 Era, J(3 E K aoo,

(3.13)

where Kaoo is the unbounded component of E - f(aKoJ as noticed before (see the paragraph following (3.10)) aKa E K, and therefore f( 8Ka) C f(K) = J. Consequently,

J(3

C

E- J

= J oo u

(u

(3EB

J(3)

C

E - f( 8Ka)

= Kaoo U

(

U Ka-y) . -YEr"

THE JORDAN-LERAY THEOREM

107

Since J{3 is an open connected set in E - f(8Ko;), the assertion (3.13) now follows from the definition of the Ko;oo and Ko;-y as maximal open connected sets in E - f(8Ka) (3.13) allows us to re-order the J{3 (for fixed 0:) by putting together those which are contained in a fixed component of E - f(8Ko;): let Boo = {/3 E B I J{3 E Ko;oo}, and for I E r let B-y = {/3 E B I J{3 E Ko;-y}. Then 0;

B = Boo U (

U B-y) . -yEr",

Our next step in proving (3.12) consists in establishing the following relations:

d(f,Ko;,J{3) = {d(f,Ko;,Ko;-y) for /3EB-y, d(f, Ko;, Ko;oo) = 0 for /3 E Boo, and, for x E E - K

d(g, Ko;-y, x) =

L

(3.14)

(3.15)

d(g, J{3, x),

(3EB~

where

d(g, J{3, x)

(3.16)

= 0

except for at most a finite number of /3 E B-y. PROOF. Let x E J{3.

(3.17)

Then, by definition, d(f, Ko;, J(3) = d(f, Ko;, x). Now if /3 E B-y then, by (3.13), J{3 E Ko;-y and, by (3.17), x E Ko;-y. Therefore d(f, Ko;, x) = d(f, Ko;, Ko;-y). This proves the first equality in (3.14), and the left part of the second equality in (3.14) follows the same way. The right part of the latter equality follows from the fact that we can choose ayE KO/oo of norm so large that the equation f(x) = y has no root in K 0/. For the proof of (3.15) we note first: if x E E - K and if is a root of

e

g(e) = x,

(3.18)

then (3.19)

e

e

e

since for E J, x = g(O c g(J) = K. By (3.19) E Joo U (U{3EB J/3). If is a root of (3.18) which lies in Ko;-y, then since the J aoo and J/3 are disjoint, by (3.13) lies in some J{3 which is a subset of KO/-y. Thus E U{3EB J{3. Let us now denote by B* the set of those /3 E B-y for which J/3 contains a ro"'ot of (3.18). The set B; is finUe. The proof for this is quite similar to the one for the second part of aSsertion (iii) and will therefore be omitted. It follows that

e

e

d(g, Ko;-y, x) = d (g,

U J{3, x) .

{3EB~

108

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

Application of the sum theorem yields the assertion (3.15) since (3.16) holds for all (3 ~ B~. To finally prove (3.12) we note that by {3.15} with x E Ko:l, and by {3.14} the right member of this equality equals

L L -yEr",

d(g, J{3, Ko:l) . d(f, K o , J{3}.

(3.20)

{3EB~

Here the double summation is extended over all (3 E B - Boo· But if (3 E Boo (Le., J{3 E Ko: oo ), then by (3.14), d(f, Ko, J{3} = O. Thus the sum (3.20) equals the left member of (3.12).

§4. Notes to Chapter 5 Notes to §1. 1. This note refers to the paragraph containing (1.15). It may happen that for an Xo EO which is a root of (I)

gJ(x) = zo

the point Yo = J(xo} lies in the unbounded component Coo of E - J(80}. EXAMPLE. Let E be the two-dimensional Euclidean plane, and let 0 E E be the open circular disc with center 0 and radius 3. Let J: -+ E be defined as follows: J is the identity map for Ilxll ~ 2. For 2 < Ilxll ~ 3 we set

n

J{x} =

xllxll- I (4 - Ilxll}.

(2)

Then J{80} is the circle Ilxll = 1. The bounded component C I of E - J(80) is the open disc 0 ~ Ilxll < 1, and the unbounded component Coo of that set consists of those x E E for which Ilxll > 1. Let g be the identity map on E, and let zo be a point satisfying 1 < Ilzoll < 2. The roots of (I) then are Xo = zo and Xl = zollzoll-I(4 -llzolI}, and for i = 0,1, Yi = J(xd = gJ(Xi} = Zo E Coo. Note that by the argument given in the text d(f, 0, y} = 0 for Y E Coo. In particular o = d(f, 0, zo) = jo + jl where jo and jl are the indices at Xo and Xl. Thus jl = -jo =-1.

The reader will also note the geometric interpretation of (2): the "ring" 2 ~ Ilxll ~ 3 is folded at the rim Ilxll = 2 into the disc IIxll ~ 2, and that J(80) -:p

8J(0}. 2. Proof of (1.35) and {1.36}. These relations are obviously true if k =

o.

Suppose then k -:P

o.

Now if k ~ KO and

d(f, 0, Yo}

{3}

= k,

(4)

then by definition a ~ AO. We claim

a¢A'.

{5}

NOTES

109

Indeed, (3) and (4) imply that y E f(O) for every y E Yo. But this would imply that a E AO if (5) were not true. Now (5) thus proved implies (1.35) since ~k is the union of Yo satisfying (4). Since Dk is an open subset of ~k, it will for the proof of (1.36) be sufficient to show that if y E ~k is a root of

g(y) = Zo,

(6)

then y E Dk • Now y E Yo C ~k for some a E A. But by (6), a E A'. Moreover d(f,O,y) = k. By (3) this implies that y E f(O). It follows that a E AO. Thus y E Dk as we wanted to show.

3. Proof of the approximation lemma 1.10. Proof of assertion (i). Since g, f, and gf are L.-S. maps, we see from (1.1) and (1.6) that there exists a positive 1 such that dist(zo, gf(80)) U g(8Y) > 21.

(7)

We assert first that if 1 satisfies (7) and if the L.-S. map g* satisfies (1.38), then no roots of the equation g*(y) = Zo (8) lie on f(80) U g(8Y). Indeed, by (1.38)

Ilg(f(x)) - g*(f(x))11 < 1

(9)

for all x E 11. In particular for x E 80 we see from (7) and (9) that

IIg*(f(x)) - zoll ~ Ilg(f(x)) - zoll-llg*(f(x)) - g(f(x))11 > 1·

(10)

On the other hand, we see from (7) and (1.38) that for y E 8Y

IIg·(y) - zoll

~

Ilg(y) - zoll-lIg*(y) - g(y)11 > 1.

(11)

(10) and (11) together show that, as asserted, the set (g*)-l(ZO) is disjoint from the set f(aO) U ay. Since the first of these sets is compact (see Lemma 10 in Chapter 1) and the second one is closed, there exists a 0 > 0 such that

dist«g·)-l(ZO), f(aO) U g(ay)) > 20.

(12)

Because of (1.5) we may choose 6 in such a way that, in addition to (12), dist(f(l1), ay)

> 26.

(13)

To prove assertion (i) of our lemma we have to show that (1.1) and (1.6)-(1.8; remain valid if f and g are replaced by rand g. reap., provided that (1.38) ane (1.39) are satisfied. To prove that

Zo ¢ g·I*(80), we suppose this relation to be wrong. Then Zo y* f*(x). Then

=

111*(x) - f(x)1I = Ily·

-

(1{

= g* r(x) for some x E a~.

f(x)1I > 26

by (12) since y* E (g*)-l(ZO). But (15) contradicts (1.39).

Le'

(15

110

THE PRODUCT THEOREM AND SOME OF ITS CONSEQUENCES

That reO) c Y and that dist(f*(O), ay) > 0 (cf. (1.4) and (1.5)) follow from (13) and (1.39). Finally, to prove the relation corresponding to (1.6), we note that for any iJ E ay IIzo - g*(iJ)11 2:: Ilzo - g(iJ)ll- Ilg(iJ) - g*(iJ)11

> 0 by (7) and (1.38).

We now turn to the proof of assertion (ii) of Lemma 1.10. Since (1.37) is true by assumption, the assertion (1.40) will be proved once it is shown that

d(g* r, 0, zo) = d(gl, 0, zo)

(16)

and that for each integer k

k· d(g*, Ll k, zo)

=

k . d(g, Llk, zo).

For the proof of (16) we note that by (9) and (7) for

(17)

x E ao

IlgI(x) - g* l(x)11 < dist(zo, gl(80)) S Ilzo - gl(x)ll· By the Rouche theorem 1.9 in Chapter 4 this implies d(gl, 0, zo) = d(g* 1,0, zo). It therefore remains to prove that

d(g* r, 0, zo) = d(g* 1,0, zo).

(18)

Now if for t E [0,1] we set It(x) = (1 - t)/(x) + tr (x), then (18) will be proved once it is shown that (a) g!t is L.-S. for every t E [0,1]' and (b) Zo ~ g* It (80) for t E [0,1]. (19) Now with I(x) = x-F(x), g*(y) calculation shows that

= y-C*(y) and f*(x) = x-F*(x), elementary

g*(!t(x)) = x - [(1- t)F(x) + tF*(x) + C*(ft(x))], which obviously proves (a). For the proof of (b) let us suppose that Zo = g* !to (x) for some to E [0,1] and some x E 80. Then Yo = Ito (x) E (g*)-l(ZO). Therefore dist((g*)-1(zo),/(80)) S Il/to(x) - l(x)11

s tollr(x) -

But this contradicts (12). Thus (b) is proved. Turning to the proof of (17) we set R* = (g*)-l(zO) and claim first that

d(r,O,y)

= d(f,O,y)

= {y

l(x)11 < 8. E Y I g*(y)

= zo} (20)

for y E R*. To prove this we note that for such y and for x E 80 by (12) and (1.39) Ily - I(x)1I 2:: dist(R*, 1(80)) > 11/(x) - r(x)ll. (21) This inequality proves (20) by the Rouche theorem. Furthermore we will need the equality

(22) We prove first that

(23)

NOTES

111

This inclusion is trivial if the left member is empty. Now let y E R* n Llk. Then dU, 0, y) = k since y E Ll k • But also (20) holds since y E R*. It follows that dU*, 0, y) = k. Thus y E Ll k. This proves (23). To prove that (24) we may again assume the left member contains an element y. Now dU*, 0, y) = k. But again (20) holds, y being an element of R*. Thus dU, 0, y) = k, i.e., y E Ll k . This proves (24), and (22) is established. To prove (17) we may assume k i= O. We will show that (25) and that d(y*, Ll k, zo) = d(y, Llk, zo).

(26)

We first consider the case that Llk is empty. Then the right member of (25) and both members of (26) are zero due to our convention (1.34). But by (22), R* n Llk is empty if Llk is empty. By definition of R* this implies that the equation y*(y) = Zo has no roots in Ll k. It follows that the left member of (25) is also zero. This proves (25) and (26) for empty Ll k . Suppose Llk not to be empty. Now R* has no point in common with BLlk since BLlk C J(BO) n BY. Since Llk is open and R* is compact, it follows that there exists an open set Vk such that (27) Similarly one sees that there exists an open set

V: such that

Llk n R* c V: c Ll k.

(28)

Now for every couple of open sets Vk and V: satisfying (27) and (28) d(y*, Llk, zo)

= d(g*, Vk, zo),

d(g*, Ll k , zo)

= d(g*, V:, zo).

(29)

But by (22) the left members of (27) and (28) are equal. Consequently these two inclusions remain valid if Vk and V: are both replaced by their intersection Vk n V:, and the equations (29) also remain valid under this replacement. Since with this replacement the right members of (29) become identical, the asserted equality (25) follows. To finally prove (26) we note that, for any y E J(BO) u BY, Ily - zoll ~ dist(zo, J(BO) U BY). Therefore by (7) and (1.38) Ily - zoll > IIg(y) - g*(y)ll·

(30)

This inequality holds in particular if y lies in the subset BLlk of !(BO) U BY. Therefore (30) implies (26) by the theorem of Rouche. .

CHAPTER 6

The Finite-Dimensional Case §1. Some elementary prerequisites 1.1. DEFINITION. Let En be a Banach space of finite dimension n. Then r + 1 points ao, al , ... ,aT in En are called dependent or independent according to whether the r points al - ao, a2 - ao, ... ,aT - ao are linearly dependent or linearly independent. (We assume the reader to be familiar with the notions of linear dependence and linear independence of a finite number of points in a vector space, with the notion of the linear space spanned by a finite number of elements in such a space, and with the notion of a base for a finite-dimensional vector space.) 1.2. REMARK. In Definition 1.1 the index zero was distinguished. But we would have obtained an equivalent definition by considering the r points

for any integer i S; r. I These points span an r-dimensional the points ao, aI, ... , aT are independent. It is easy to see independent of i. Thus a set of r + 1 independent points uniquely an ET. 1.3. REMARK. If r > n, any r + 1 points ao, aI,"" aT in (cf. 1.5).

subspace of En if that this space is in En determines

En are dependent

1.4. LEMMA. The points ao, al, ... ,aT of En are independent if and only the relatio'l'ts T LaiAi = B,

if

(1.1 )

i=O

imply that

AO = Al = ... = AT = O. Notes. 2

(1.2)

The proof will be given in the 1.5. DEFINITION. An r-space in En is an r-dimensional linear subspace of En. An r-dimensional plane, shortly an r-plane, ITT in En is the translate of an r-subspace ET of En. If ao E ITT and if bl, b2,"" bT are a base for ET, then the 112

SOME ELEMENTARY PREREQUISITES

113

points x of I1T are given by T

X = ao

+ LbiQi,

Qi

real.

(1.3)

i=l

We sometimes call the bi also a base for I1T.

1.6. LEMMA. The points ao, al, ... ,aT with r :::; n of En are independent if and only if they do not lie in a plane of dimension less than r. The proof will be given in the Notes. 3 1.7. DEFINITION. Let I1T and 11 8 be r- and s-planes in En resp. The hull H of I1T and 11 8 is defined as the plane of smallest dimension containing I1T and 118. The intersection I1T n 11 8 (which is obviously a plane unless empty) will be denoted by K. The dimensions of Hand K will be denoted by h and k resp.

1.8. LEMMA. Using the notations of subsection 1.7 we assert: if K is not empty, then (1.4) k + h = r + s. The proof will be given in Notes. 4 1.9. DEFINITION. A finite or countably infinite set of points in En is said to be in general position if for r :::; n any r + 1 points of S are independent (cf. Definition 1.1). 1.10. Note that by Lemma 1.6 the points of S are in general position if for r :::; n no r + 1 points of S lie in a plane of dimension < r. Note also that in El any set of disjoint points not containing () is in general position.

1.11. THEOREM. Let bo, bl , ... , bT be independent points in En (n 2 1), and let ao, al, . . . be a finite or infinite sequence of not necessarily different points in En. Let E be a given positive number. Then there exist points a~, a~, ... in En such that i = 0, 1,2, ... , lIa~ - at II < E, (1.5) and BUck that the points

bo, b1 , ••• , b %, a~, a~, . .. are T ,

in general position.

PROOF. We give the proof for T = 1. H n = 1 we choose for a~ a point different from bo and bl which satisfies (1.5) for i = O. For a~ we choose a point different from bo, b.. ab which satisfies (1.5) for i = 1. Going on this way we obtain points %, a~, a~, ... of the desired properties. Now let n > 1. We choose for ab a point in B(ao, E) which does not lie in the on~plane (straight line) containing bo and b1 • Then bo, bl , ab are in general II I· h post·t·Ion. Suppose we have chosen ao, ai' ... , aN -1 m l uc a way that the points

(1.6) are in general position and such that (1.5) is satisfied for i = 0,1, ... , N -1. We now consider the points (1.7)

114

THE FINITE-DIMENSIONAL CASE

where x is different from the points (1.6) and satisfies

x E B(aN,e).

(1.8)

We have to show that x can be chosen to satisfy (in addition to (1.8)) the following condition C: for any integer r ::; min(n, N +2) and for any choice of r different points of the set (1.6), say Co, Cl, ... , CT-l, the r + 1 points Co, CI, ... , CT-I, x are independent, i.e., they do not lie in a plane of dimension::; r - 1 (cf. Lemma 1.6). In other words, condition C says that x does not lie in the (r - I)-plane II(co, CI, ... , cT-tl in which the independent points Co, Cl, ... , CT-I lie. Now, as r and the cp vary subject to the above conditions, there are only a finite number of these (r -I)-planes. But since r -1 < n a finite number of (r -I)-planes does not cover a ball in En (see the Notes). 5 Therefore there exists a point x satisfying condition C and (1.8). Taking such a point for aN finishes the induction proof of our theorem. 1.12. LEMMA. Let aO,al, ... ,aT be r assert: (i) the points

+1

independent points in En. We

T

X

= LPiai

(1.9)

i=O

vary over an r-plane lIT as the real numbers i vary subJect to the condition T

LPi = 1;

(1.10)

i=O

(ii) the Pi in the representation (1.9) are uniquely determined. PROOF. (i) r

Po = 1 - LPi

by (1.10).

i=l

Hence by (1.9) T

X

= ao + LPi(ai - ao).

(1.11)

i=l

Here the sum varies over the vector space whose base is the ai - ao, a2 - ao, . .. ,aT - ao as the real numbers PI, P2, ... ,PT vary arbitrarily. This proves assertion (i) (cf. 1.5). (ii) The ai - ao (i = 1,2, ... ,r) are linearly independent. Therefore PI, P2, ... , PT are uniquely determined by (1.11). But then Po is uniquely determined by (1.10). 1.13. DEFINITION. Let ao, al,.··, aT be as in subsection 1.12, and let lIT be the r-plane determined by (1.9) and (1.10). The subset of IIr consisting of those points for which Pi ~ 0, i = 0, 1, ... ,r, (1.12) is called an r-simplex and will be denot~ by lu r! = lao, a;,.:., arl. The points ao, at, ... , aT are called the vertices of 100 I· The sImplex 100 lIS a closed set. The

SOME ELEMENTARY PREREQUISITES

115

open set u r - BuB is called an open simplex. For r ~ 1 and p = 0, 1, ... , r, the r points ao, ... , ap-1, ap+b"" a r are vertices of an (r - 1}-simplex lul~-l called the pth (r -1}-face of lul r. Correspondingly if we omit two of the vertices of lul r we obtain an (r - 2}-simplex called an (r - 2}-face of lul r, and so on. Note that 81ul r+ 1 is the union of the (r - 1}-simplices lul~-l, p = 0,1, ... , r. 1.14. LEMMA. The simplex

lurl = lao, a1,···, arl

(1.13)

is the closed convex hull (see §§1.28, 1.23) of the set V of its vertices.

For a proof see the Notes. 6 1.15. By Definition 1.13 the simplex (1.13) is an r-dimensional set in En and (1.14) where io, i 1 , ... ,ir is a permutation of the numbers 0, 1, ... ,r. We now define as follows the notion of an oriented simplex in which the order of the vertices plays a role. 1.16. Orientation 0/ a simplex. Let ao, a1, ... , ar be r + 1 independent points in En where r ::; n. We consider first the case r ~ 1. Then let (1.15) be a fixed permutation of 0,1, ... , r. Then the oriented simplex (1.16) is the point set (1.14) (or what is the same (1.13}) together with the ordering of the vertices given by the permutation (1.15) or an even permutation thereof. Thus (ajo' ajl' ... ,aj.) = (aio' ail' ... ,ai.) if io, i1, ... ,J~ is an even permutation of io, it, ... ,ir . In contrast the simplex luTI defined in subsection 1.13 is sometimes referred to as an absolute simplex. Every absolute simplex luTI gives rise to two oriented simplices. H ITTI = luTI but u T and TT have different orientation, we write TT = -uT • Now let r = 0. Then luol consists of a single point ao. To introduce an orientation also in this case, we associate the symbols + and - to ao and call +ao and -00 the positive and negative orientation of 10'°1. We make the convention -(+oo) = -ao, -( -00) = +ao· For r ~ 1 the oriented (r - I)-faces 0';-1 (p = 0,1, ... , r) of the oriented simplex O'T = (oo,al, ... ,aT ) are defined as follows: if r ~ 2 then U;-1 = (-l)P(ao,at, ... ,ap-t,ap+l, ... ,aT ). H r = 1, i.e.. O'T = (aO,a1), we set 0'8 = +a1 and O'~ = -ao. 1.17. DEFINITION. Let ET and EP be linear spaces of finite dimension r and p resp. An affine map A: ET -+ EP is a map of the form A(x) = c + l(x), where c is a point of EP and l a linear map ET -+ EP called the linear part of A. A is called nonsingular if l is nonsingular. By determinant A we mean determinant I. (Note that c and l are uniquely determined by A.)

THE FINITE-DIMENSIONAL CASE

116

1.18. LEMMA. Let ao, al, ... ,aT be independent points in En. Let the plane ITT E E and the numbers Po, Pi, . .. , PT be as in Lemma 1.12. Let A be an affine map with domain En. Then for x E ITT T

A(x) =

2: A(ai)Pi.

(1.16)

i=O

PROOF. By Definition 1.17, by (1.9) and (1.10)

A(x) =

C

~Pi + l (~aiPi) = ~(C + l(ai))Pi = ~A(ai)Pi.

1.19. Let En and Ef be linear spaces of dimension n. Let ao, al, ... ,an be independent points in En and let bo, bl , ... ,bn be arbitrary, not necessarily different points in Ef. Then (a) there exists an affine map A: En -+ Ef such that

i = 0, 1, ... , n;

(1.17)

(b) the affine map A is uniquely determined by the conditions (1.17). The proof will be given in the Notes. 7 1.20. In addition to the assumptions of subsection 1.19 we suppose that the points bo, bl , ... ,bn are independent. Then the affine map A determined by (1.17) is nonsingular and maps the simplex lanl = lao, al, ... ,ani onto the simplex lanl = Ibo,bl, ... ,bnl. For proof see the Notes. 8 1.21. DEFINITION. Let an = (ao, al, ... , an) and Tn = (b o, bl , ... , bn ) be two n-simplices in En. Let A be the uniquely determined affine map which satisfies (1.17). We then say that an and Tn have the same or opposite orientation according to whether det A is positive or negative. (For the definition of det A see subsection 1.17; note that det A :f since, by subsection 1.20, A is nonsingular.) 1.22. It follows from Definition 1.21 that a single oriented n-simplex an in En determines what is the same orientation for every n-simplex ITnl of En. We therefore say: an induces an orientation in En. If, in particular, el, e2, ... ,en is a base (in this order) for En, then an = (f), el, e2, ... , en) is an oriented simplex. Thus an ordered base in En determines an orientation of En, and we see from Definition 1.21 that if ei, e2, ... ,e~ is another base for En, then the two bases determine the same orientation if and only if the unique linear map l which maps ei into i = 1,2, ... , n, has a positive determinant. If an orientation for En is given, then an oriented simplex Tn is called positive or negative according to whether the orientation of En induced by Tn agrees with the given one or not, and the terms positive and negative basis are defined correspondingly. 1.23. Let A( x) = c + l (x) be a nonsingular affine map En -+ En, let Yo be an arbitrary point in En, let Xo E. E~ be ~he unique solution of A(x) = Yo, and let 0 be an open bounded domam m E containing Yo. Then, by Definition 2.2 in Chapter 2, d(A, 0, Yo) = j(DA(xo; h)). But DA(xo; h) = l(h). Therefore

°

<,

MAPPINGS BETWEEN SPACES OF THE SAME DIMENSION

11~

d(A, 0, YO) equals j(l). Thus the degree d(A, 0, Yo) is seen to be independent 0 Yo and 0 (provided that Xo = A-I(yO) EO) and will be denoted by j(A). Ther j(A) = j(l). 1.24. DEFINITION. Let rr n - l be an (n - I)-plane in En and let Xl, X2, ... Xn be coordinates in En for which rr n - l is given by Xn = c, a constant. Ther the two open sets {x = (x!, ... ,xn ) EEl Xn > c} and {x = (Xl, ... ,Xn ) E E I Xn < c} are called the halfspaces into which rr n - l decomposes En (shortly halfspaces with respect to rr n - l ). 1.25. Let lun-ll = I(ao,a!, ... ,an-dl be an (n-l)-simplex in En. Then (i; the plane of smallest dimension containing lul n - l is an (n - I)-plane rrn-l; (ii; let HI and H2 be the two halfspaces of En with respect to rr n - l , and let an and bn be points in HI and H2 resp. Then the simplices uf' = (ao, a!, ... , an-I, an) and u~ = (ao, al,' .. ,an-I, bn ) have different orientation. (iii) If En is oriented. then one of these simplices is positive and the other is negative. Since ao, aI, ... ,an-l are n independent points, assertion (i) is obvious from (1.3) with bi = ai - ao and r = n - 1. Assertion (iii) is an obvious consequencE of (ii) and subsection 1.22. The proof of (ii) will be given in the Notes. 9 1.26. If lunl = I(ao,a!, ... ,an)1 is an n-simplex in En, then the (point set-) boundary alunl is obviously the union of the (n - I)-simplices i = 0, 1, ... ,no

The (n - I)-face uf- l = (aO,al, ... ,ai-l,ai+I, ... ,an ) is called a positive or negative face of the oriented simplex un = (ao, a!, ... , an) according to whether i is even or odd. §2. The degree for mappings between finite-dimensional spaces of the same dimension

2.1. A Leray-Schauder map I in a Banach space E is a completely continuous perturbation of the identity map on this space. It therefore was defined only for maps whose domain and range lie in the same space. However, as noted in subsection 2.4 of Chapter 3, the Leray-Schauder maps are identical with continuous maps if E is finite dimensional. This fact makes it easy to extend the degree theory to maps between different spaces if these spaces have the same finite dimensions. 2.2. Let EI' and E~ be two oriented spaces of finite dimension n (see subsection 1.22). Let 0 be a bounded open set in EI', and let I be a continuous map --.~. We want to define a degree d(f,O,Yo) for every Yo E E~ which satisfies the condition (2.1) YO ¢ l(aO)

n

To this end we consider positive bases (see subsection 1.22) bl'~"'" bn and

PI, f32, ... , Pn for En and En resp; Let h be the linear (and therefore continuous) map which maps th~ point ~ = L~ TJif3i of E~ into the point X = L~ TJibi of Ef·

118

THE FINITE-DIMENSIONAL CASE

Then the map It = hI is continuous and maps 11 into E 1. Moreover, since h is one-to-one, the relation (2.1) implies that

h(yo)

~

hl(80).

(2.2)

Therefore the degree d(hl, 0, h(yo)) is defined. Moreover it can be verified that this degree does not depend on the special choices of the positive bases b1,b2, ... ,bn and f31,(h, ... ,f3n (see the Notes).l0 Thus the following definition is legitimate. 2.3. DEFINITION. With the notations and assumptions of subsection 2.2 we define (2.3) d(f, 0, YO) = d(hf, 0, h(yo)). 2.4. DEFINITION. Let Ef,E2 ,0 be as in subsection 2.2. Let f: 11 -+ E2 be a C' map, i.e., f E C'(O). In analogy to §1.17 we define: the point Yo E E2 is called a regular value for I if the equation

f(x) = Yo

(2.4)

either has no solution or if for every solution x of (2.4) the linear map lx(h) = D I (x; h) is nonsingular (equivalently, if det lx :I 0). 2.5. If the point Yo of E2 is a regular value for f and satisfies (2.1), then the equation (1.4) has at most a finite number of solutions. The proof is quite analogous to the one given for part (a) of Lemma 22 in Chapter 1 and will therefore be omitted. 2.6. LEMMA. Let E,?, E 2, I, and h be as in Definition 2.3. Let It = hI. Suppose IE C'(n). Then (i) D!t(x;TJ) = h(DI(x;TJ)); (ii) det D It (x; TJ) = det h . det D I(x; h); (iii) il det D It (x; TJ) :I 0, then sign det D It (x; TJ) = sign det D f(x; h); (iv) Yo is a regular value lor I if and only il h(yo) is a regular value for It. PROOF. (i) follows from the chain rule for differentials together with the fact that h is linear. (ii) is a consequence of (i), and (iii) follows from (ii) together with the fact that det h > O. (2.5)

Finally (iv) follows from (ii) and (2.5) together with the fact that x is a root of (2.4) if and only if x is a root of the equation

It(X) = h(yo).

(2.6)

2.7. LEMMA. Let the assumptions of Definition 2.3 be satisfied. Then (i) if the equation (2.4) has no roots, then d(f, 0, YO) = 0; (ii) iI, in addition, IE C'(O), ilyo is a regular value lor I, and ilx!, X2, .•. , X T are the roots 01(2.4), then T

d(f, 0, Yo) =

E sign det DI(xp; .). p=!

(2.7)

MAPPINGS BETWEEN SPACES OF THE SAME DIMENSION

119

PROOF. (i) If (2.4) has no roots, then (2.6) has no roots. Consequently the right member of (2.3) equals zero. (ii) As above let It = hf. Then by Definition 2.3 and by (2.3) in Chapter 2 r

d(f, 0, Yo)

=

d(hf, 0, h(yo))

=

L j(DIt (x

r

p ; .))

=

p=l

L sign det DIt (x

p ; .).

p=l

But here the right member equals by assertion (iii) of Lemma 2.6 the right member of our assertion (2.7). 2.8. LEMMA. The assertions (ii)-(vii) of Lemma 2.4 of Chapter 2 are true for the degree defined in subsection 2.3. Assertion (i) of that lemma is replaced by the following one: if h is as in subsection 2.2 and l = h -1, then if Yo E l(O), d(l, 0, Yo) - 0 if Yo tf- 1(0).

_{I

The proof of this lemma is omitted since it follows trivially from Definition 2.3, from the corresponding assertions of Lemma 2.4 of Chapter 2, and from subsection 2.4 of Chapter 3. The proof of the following theorem is also omitted since it is an obvious consequence of Definition 2.3 and subsections 2.4 of Chapter 3 and 1. 9 of Chapter

4. 2.9. THEOREM. Let Er,E,!], and 0 be as in subsection 2.2. It is asserted: (a) Let h(x) = f(x, t) be a continuous map 0 x [0, 1] -4 E'!]. Let Yo be a point in E'!] satisfying Yo tf- ft{(30) for all t E [0,1]. Then d(ft, 0, Yo) is independent

oft. (b) Let f be a continuous map 0 -4 E'!]. Let y = y(t), 0 ~ t ~ 1, be a continuous curve in E2 which does not intersect f(aO). Then d(f, 0, y(t)) is independent of t. (c) Theorem of Rouchi. Let I and g be continuous maps n -4 E 2 . Suppose that the point Yo E EJ] does not lie on l(aO). Suppose moreover that, lor all xEaO, IIf(x) - g(x)1I < IIf(x) - yoll· Then d(g, 0, Yo) = d(f, 0, yo). We remark that the latter equality holds in particular if f(x) = g(x) for x E a~. This allows us to define the following. 2.10. DEFINITION. Let fo be a continuous map ao -4 E'!]. Let Yo be a point in EJ] not lying on fo(aO). Then the winding number u l fo(aO), Yo) is defined by

u(fo(aO), yo) = d(f, 0, Yo), where f is an arbitrary continuous extension of fo to n (cf. subsection 1.5 in Chapter 4). The next lemma will be used in the proof of the product theorem in subsection 2.12.

120

THE FINITE-DIMENSIONAL CASE

2.11. LEMMA. Let E'l,E~,E!3 be three oriented Banach spaces of finite dimension n. Let bi , f3i, i = 1,2, ... ,n, and h be as in subsection 2.2. Let fh be a bounded open domain in E~, and let ¢ be a continuous map fh into E!3. Then d(¢h- 1, hfh, zo) = d(¢, fh, zo)

for Zo E E3 - ¢(afh).

(2.8)

PROOF. Since the continuous map ¢ can be uniformly approximated by a map in C'(02) we may, by part (a) of Lemma 2.9, assume that ¢ E C'(O), and by part (b) of that lemma together with the Sard-Smale theorem in subsection 4.1 of Chapter 2 and part (iv) of Theorem 2.6 we may assume that Zo is a regular value for ¢. Now y E O2 is a root of

¢(y) = Zo

(2.9)

if and only if x = h(y) satisfies (2.10) We consider first the case that (2.9) has no roots in O2 . Then (2.10) has no roots x E h(fh). This proves (2.8), both members of this equality being zero. Now suppose Yl. Y2,· .. ,Yr are the roots of (2.9). Then xp = h(yp), p = 1,2, ... , r, are the roots of (2.10), and, by Lemma 2.7 r

d(¢, O2 , Zo) =

L sign det D¢(yp;.)

(2.11)

p=l

and

r

d(¢h- 1, h0 2, zo) =

L sign det D¢h-1(x p; .).

(2.12)

p=l

But by the chain rule and by the linearity of h- 1

detD¢h-1(x p;·) = detD¢(yp;.). deth- 1 . From this and (2.5) we see that the right members of (2.11) and (2.12) are termwise equal.

2.12. THE PRODUCT THEOREM. (A) Let E'l, E~, E!3 be as in Lemma 2.11. Let 0 be a bounded open set in E'l, let f be a continuous map -+ E~, and let 9 be a continuous map f (0) -+ E!3. Let {C/3} be the set of bounded components of E~ - f(aO), where f3 E B, an index set. Finally, let Zo be a point of E!3 not lying on gf(aO). Then

n

d(gf, 0, zo) =

L d(g, C/3, zo) . d(J, 0, Y/3),

(2.13)

/3

where Y/3 is an arbitrary point of C/3 and where the summation is extended over a finite number of f3 E B. (Note that all C/3 occurring in (2.13) are contained in f(O) since d(J, 0, Y/3) = 0 for Y/3 ¢ f(O)·) (B) Let En1, En 2, En 3, 0 and f be as in (A). Let Y be . an open bounded se t·zn E~ for which (1.4) in Chapter 5 (and therefore (1.5) tn Chapter 5) holds. Let 9

MAPPINGS BETWEEN SPACES OF THE SAME DIMENSION

121

be a continuous map Y ~ E'3. Let Zo be a point of E'3 for which (1.1) and (1.6), both in Chapter 5, hold. Let {Yo,} be the set of components of Y - f( an), where a E A, an index set. Then provided that the equation gf(x) = Zo has solutions in 0. d(gf, 0., zo) = d(g, Ya, zo) . d(f, 0., Ya),

2:

aEA

where Ya E Ya and where only a finite number of terms in the sum are different from zero. (The degree at the left obviously equals zero if 9 f (x) = zo has no solutions. ) PROOF. (A) For i = 1,2, ... , n let bi and (3i be as in subsection 2.11, and

Er

let "Ii, "12, ... ,"In be a positive base for E'3. Let hi be the linear map EJ] ~ which maps (3i into bi, and h2 the linear map E'3 ~ EJ] which maps "Ii into (3i. Then hlh2 is the linear map E'3 ~ which maps "Ii ~ bi. Therefore by Definition 2.3 d(gf, 0., zo) = d(hlh2gf, 0., hlh2Z0)' (2.14)

Er

We define for x Erland Y E hd(rl)

ft(x)

= hd(x),

= hlh2ghl1(y),

(2.15)

d(gf, 0., zo) = d(gdl' 0., xo).

(2.16)

gl(Y)

Then by (2.14)

Er,

we may apply the product Since the maps ft and gl both take place in theorem in subsection 1.2 of Chapter 5 to the right member of (2.16) to obtain

d(gdl' 0., xo) =

2: d(gl' C~, xo) . d(ft, 0., Y~),

(2.17)

(3

where the C~ are bounded components of EI' - h(ao.), where where the sum is a finite one. Now by (2.15) and Lemma 2.11

Y~ E C~,

and

d(gt, C~, xo)

= d(hlh2gh-1, C~, xo) = d(h 1h2g, hl1C~, xo). (2.18) But hllC~ = C{3 is a bounded component of E2 - !(ao.). Therefore by (2.18), (2.15), and Definition 2.3,

d(gl' C~, xo)

= d(h1h2g, C{3, h 1 h2ZO) = d(g, C{3, zo).

(2.19)

Now d(ft, 0., Y~) = d(hd, 0., h 1 h l l 14), and if Vf3 E C~ then Y{3 = hl1y~ E C{3. Thus

d(ft, 0., Y~) = d(hd, 0., hlY(3)

= d(J, 0., Y(3)·

(2.20)

The assertion (2.13) now follows from (2.16), (2.17), (1.19) and (2.20). Assertion (B) follows from Theorem 1.4 of Chapter 5 in the same way assertion (A) followed from Theorem 1.2 of Chapter 5. Its proof is therefore omitted.

THE FINITE-DIMENSIONAL CASE

122

2.13. THEOREM. Let 0 be a bounded open set in Ef, and let I be a continuous map 0 - t E'2 which is one-to-one. Then d(f,O,y) = ±1 lor y E 1(0). PROOF. The right member of Definition 2.3 equals ±1 by Theorem 2.2 of Chapter 5.

2.14. THEOREM (INVARIANCE OF THE DOMAIN). Under the assumptions 01 Theorem 2.13, 1(0) is open and l(aO) = al(O). PROOF. The map 11 = hd of Definition 2.3 has by Theorem 2.3 of Chapter 5 the properties asserted for f. This implies that I = h I I It has these properties.

2.15. THEOREM. Let Ef, E'2,E!3, O,I,g,zo be as in Theorem 2.12. It is then asserted: (i) i/O is connected, il I is one-to-one on 0, and il Zo E gl(O), then d(gl, 0, zo)

= d(g, 1(0), zo) . d(f, 0, y) = ±d(g, 1(0), zo),

(2.21)

where y E 1(0); (ii) let g: 1(0) - t gl(O) be one-to-one, let Zo E g(f(O)), and let Yo be the unique root 01 g(y) = Zo. Then d(gl, 0, zo) = ±d(f, 0, yo).

(2.22)

Yo ¢ Coo,

(2.23)

II, in particular, the unbounded component 01 E'2 - l(aO), then d(gl, 0, zo) = d(g, Cf3o' zo) . d(f, 0, Yo)

(2.24)

lor some unique (30 E B. Finally d(f, 0, Yo) = 0

(2.25)

il Yo E (

U C(3) U Coo

(2.26)

f3EB-f3o

PROOF. We use the definitions C2.15) and formula (2.16). To the right member of the latter we apply part Ci) of sUbsection 2.5 in Chapter 5. Using Definition 2.3, Lemma 2.11, and C2.20) we obtain assertion (i). Assertion Cii) is proved similarly by reduction to part (ii) of Theorem 2.5 in Chapter 5.

§3. Simplicial mappings

n

3.1. Again let 0 be a bou~ded open set in Ef, fa c?ntinuous map - t E'2, and Yo a point of E'2 for whlch the ~egree .d(f, 0, Yo) IS defined. Formerly we approximated the continuous I by a differentiable map It which has the £ 11 . d C··) ·f . 0 owmg two properties: (i) deft, 0, YO) = d(f, 0, Yo ) ,an 11 1 Yo IS a regular value for It,

SIMPLICIAL MAPPINGS

123

then the equation h(x} = Yo has at most a finite number of solutions (see §1.22). There is another class of mappings which has these two properties, the simplicial ones defined below. That these enjoy property (ii) (with the proper definition of "regular value") will be shown in the present section. For the proof of property (i) given in §6.5 we will need "subdivisions" treated in §6.4. Historically the approach to finite-dimensional degree theory via simplicial maps was the first one used (see [8, 2]). 3.2. DEFINITION. Let En be a Banach space of finite dimension n, and let v be a nonnegative integer ~ n. A subset IKI of En is called a simplicial set of dimension v if it is the union of a finite number of v-simplices (see Definition 1.13) subject to the condition that for v ~ 1 the intersection of two of these simplices is either empty or a common (v - 1}-face of the two simplices. If En is oriented and v = n, we orient the simplicial set IKnl by giving each of its n-simplices the positive orientation (see subsection 1.22). The thus oriented simplicial set will be denoted by Kn.

IKnl be a simplicial set of dimension n in En. Then the of IKnl, v ~ n, form a simplical set of dimension v, called

3.3. LEMMA. Let

union of the v-faces For a proof see the Notes.!l

IKvl.

3.4. Let Ef be oriented, and let IKnl be a simplicial set in Ef of dimension n. Sin;e by definition each simplex is closed, it follows that IKnl is closed and that IKnl = IKnl- 81Knl is open and bounded. Consequently if EJ] is an n-dimensional oriented Banach space and iqhe map f: IKnl ---. EJ] is continuous, then by Definition 2.3 the degree d(f, IKnl, Yo} is defined for any Yo E EJ] provided that (3.1) 3.5. DEFINITION. (a) Let E1', IK1'I, and EJ] be as in subsection 3.4. Then a continuous map f: IK1'I ---. EJ] is called simplicial (with respect to IKrl) if for each simplex lunl of IKfl the restriction of ! to lunl equals the restriction Aun of an affine map A: Ef ---. EJ] to lunl. The simplicial map! is called nondegenerate if all these affine maps are nonsingular (cf. subsection 1.17). (b) The corresponding definition holds if we deal with spaces Efl, E"2 2 , whose respective dimensions nl, n2 are not necessarily equal. (c) Let IK~ll and IK"221 be simplicial sets in E~l and E"22 resp. Then a map f: IK~ll ---. IK"221 is called simplicial with respect to (K~l, K"22) if it has the following two properties: (i) if VO, Vl, ... 'Vnl are the vertices of a simplex U~l of Kn l , then the points Wo = f(vo), Wl = f(vd"",wnl = !(vnl ) are .,ertices of a simplex 0'"2 2 of K"2 2, in other words, the points wo, Wl> ... ,wnl are vertices of a v-face of 0'"2 2 (v :5 min nl> n2); (ii) the restriction of ! to a simplex Ufl of Kil is the restriction of the unique affine map mapping Vi ---. Wi, i = 0,1, ... ,nl> (cf. subsection 1.19). Obviously f is continuous. 3.6. DEFINITION. Let 0'1',0'2', ... ,0': be the simplices whose union is IKnl C Ei'. Let f be a continuous map Er ---. EJ]. Then a point Yo E EJ] is called a

124

regular value for

THE FINITE-DIMENSIONAL CASE

f (with respect to IKn I) if the equation f(x)

=

Yo

(3.2)

has no roots lying on 8a~ for p = 1,2, ... , r. 3.7. With the above notations we assume that f is a simplicial nondegenerate map IKn I ...... E,!], that the simplices a~, p = 1, 2, ... , r, are positive, and that Yo is a regular value for f (with respect to IKn I). Then (i) each solution x of (3.2) lies in the interior of one of the simplices lar!, Ian

... , la;:l; (ii) each la~l, p = 1,2, ... , r, contains at most one solution of (3.2); (iii) the number s of solutions of (3.2) is at most r. (iv) Suppose s > O. Denoting the roots of (3.2) by XI,X2, ... ,X s , we may assume that xp E a~, p = 1,2, ... , s. Let p be the number of those p .:; s for which f(a~) is a positive simplex in E,!], and let q be the number of those p for which f(a~) is a negative simplex of E'!]. Then o (3.3) d(f, IKnl, Yo) = p - q. PROOF. (i) follows directly from the definition of regularity of Yo; (ii) follows from the assumption that f is not degenerate and that therefore the affine map Ap, which is the restriction of f to a~, is nonsingular. (iii) follows from (ii). To prove (iv) we note first that (2.7) holds with 0 = IKnl- 81K n l and with r replaced by s. Now for x E a~, det D f(x; .) = det Ap. But for p = 1,2, ... , s, det Ap equals + 1 if f (a~) = Ap (a~) is a positive simplex, and -1 if that simplex is a negative one. We finish this section by some remarks and definitions which will be useful later on. 3.8. Let Em and En+I be Banach spaces of finite dimension m and n + 1 resp. with n 2 m. Let IKml be a simplicial set in Em and a n + I be an (n + 1)-simplex in En+ I. Suppose thc;,t to every vertex v of Km is assigned a vertex W = ¢( v) of a n+I. Then there exists a unique map ?J: Km ...... 8a n+I which is simplicial with respect to (Km,a n+I ) (see part (c) of subsection 3.5), and which is an extension of ¢. Indeed, let Vo, VI, ... , Vm be the vertices of an m-simplex 01' of Km, and let wo, WI, .. ·, Wv be the different ones among the vertices <1>( vo), ¢( V d, ... , ¢( v m ) of a n+ I. Now v .:; m :s n. Therefore the points wo, WI, ... ,wv are the vertices of a v-face a V of 8a n+1 . Thus the assignment Vi ...... <1>( Vi) determines a unique affine map a1' ...... a V E 8a n+I (cf. subsection 1.19). In this way we obtain the unique map having the required properties. 3.9. Let Kn be a simplicial set in En. We want to define a certain covering of Kn. If an is an n-simplex of Kn and uo, UI, ... , Un are its vertices then the (n - I)-face of an opposite to the vertex Ui is defined as the (n - {)-simplex (UO,UI, ... ,Ui-I,Ui+I, ... ,Un ). Now let_vI,v2, ... ,VN be the vertices of Kn. Then for i = 1,2, ... , N we denote by yi the union of those simplices of Kn having Vi as a vertex, and by Ui the set Ui minus the (n-l)-faces opposite to Vi

ON SUBDIVISIONS

125

of the n-simplices of (h The set Ui is called the (open) star of Vi and denoted by st(Vi, Kn). Ui is open (with respect to IKnl). Thus the Ui, i = I,2, ... ,N, form an open finite covering of the compact set IKnl. It is well known that this implies the existence of a positive b such that each subset S of IKnl of diameter less than b is contained in one of the sets Ui (see, e.g., [2, p. 100, Satz IV]). 3.10. If an+! is an (n + I)-simplex in En+l, we define a covering of oa n+1 in a way similar to the one used in the preceding section: let Va, VI, ... ,Vn+1 be the vertices of an+!. Then for i = 0,1, ... ,n + 1 let Ui be the union of all n-simplices of oa n+1 which have Vi as vertex. The set Ui = st(Vi' a n+!) is then defined as the set Ui minus the (n - I)-faces opposite to Vi of the n-simplices belonging to Ui. Again the sets Ui, i = 0, 1, ... ,n + 1, form a finite covering of oan+1 which is open with respect to oa n+l .

§4. On subdivisions 4.1. DEFINITION. Let En be a Banach space of finite dimension n. If a V = (ao,al, ... ,av), ~ v ~ n, is a v-simplex in En, then the point

°

(4.1) is called the barycenter of a V. If Kn is a simplicial set of dimension n in En and if KV is the simplicial set consisting of the v-faces of the n-simplices of Kn (1 ~ v ~ n) (see subsection 3.3), we will, by induction on v, define the barycentric subdivision BKV of KV: BIKol = IKol (IKOI is the set of vertices of IKnl). Now suppose BKv-1 has been defined for some integer v in the interval [1, n]. Then let aV be a v-simplex of KV, and let a v - I be a (v - I)-face of a V. Then by the induction assumption Ba v - 1 is defined. Now let bo, bl, ... , bv - 1 be one of the (v - I)-simplices of Ba v - 1 • Now if (3 is the barycenter of a V, the v-simplex (3, bo, .. . , bv- 1 is defined as a v-simplex of BKv, and every v-simplex of BKv is defined in this way. Thus BKn is a finite union of n-simplices of En. If En is oriented, we give each n-simplex of BKn the positive orientation. 4.2. LEMMA. BKn is a simplicial set as will be verified in the Notes. 12

4.3. DEFINITION. Using Definition 4.1 and Lemma 4.2 we define for every nonnegative integer r the rth barycentric subdivision Br Kn of the simplicial set Kn by setting BOKn = Kn, BIKn = BKn, and for r ~ 2, BrKn = Bl(Br-1Kn). IT En is oriented, we again orient BrKn by giving each of its n-simplices the positive orientation. 4.4. LEMMA. Let lanl be an n-simplex in En, and let la~1 be a simplex of the rth barycentric subdivision of Ian I. Then

(4.2)

THE FINITE-DIMENSIONAL CASE

126

where for any bounded set SeEn 8(S)

=

sup x',x"ES

lx' - x"I = diameter of S.

For a proof see the Notes. 13

4.5. LEMMA. Let Kn be a simplicial set in En, and let c be a positive number. Then there exists an integer rl ::::: 1 such that

b(lali) < c

(4.3)

for every n-simplex larl of the simplicial set IKfl = 8T 1IKnl· PROOF. Since IKnl contains only a finite number of simplices and since [n(n + I)-It tends to zero as r ---; 00, the lemma is an immediate consequence of Lemma 4.4. 4.6. Simplicial decomposition 01 a cube. Let Xl, X2, ... , Xn be the coordinates of the point x E En (with respect to a given base). Let al,a2, ... ,an be the coordinates of a given point a E En, and let l be a positive number. Then the cube Qn = Qn(a, 2l) of center a and side length 2l is defined as the set of those x E En whose coordinates satisfy the inequalities

lX-t -

a-It < _ l,

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

(4.4)

It is not hard to see that Qn can be subdivided into n-simplices in such a way that the union of these simplices is a simplicial set Kn which as a point set equals Qn (see the Notes).14

§5. Simplicial approximations 5.1. THEOREM. Let IKnl be a simplicial set of dimension n in Er. Let f be a continuous map IKn I ---; E¥, and let c be a positive number. Then there exist a positive integer rl and a map ¢: IKnl---; E¥ of the following properties: (a) ¢ is simplical with respect to IKfl = BT1IKni (see Definition 3.5); (b) iful,u2, ... ,uN are the vertices of IKfl, then the points WI = ¢(ut}, W2 = ¢(U2)' ... ,WN = ¢(UN) of E¥ are in general position; (c) Ilf(x) - ¢(x)11 < c for all x E IKnl· PROOF. The domain IKn I of the continuous map f is a closed and bounded subset of Ei'. Consequently f is uniformly continuous on IKnl, and there exists a 1 > 0 such that Ilf(x') - f(x")11 < ciS (5.1) for any couple of points

x', x"

of IKnl for which

Ilx' - xliii < 1· Now by Lemma 4.5 there exists a positive integer diameter

lall < 1

(5.2) rl

such that (5.3)

SIMPLICIAL APPROXIMATIONS

for every simplex

127

lUll of IKr I = Br, Kn. Now let

= 1,2, ... , N, (5.4) where as above the Ui are the vertices of IKr I. By Theorem 1.11 there exist points W1, W2, ... , WN in E¥ which are in general position and satisfy the inequalities Vi

= /(Ui),

i

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

(5.5)

We define i = 1,2, ... , N.

(5.6)

Now if lurl = lao,a1, ... ,ani is an n-simplex of IKrl, then the ai are a subset of the Ui and the ¢(ai) are a subset of the Wi· Therefore the points bo = ¢(ao), b1 = ¢(a1)' .. . , bn = ¢(an ) are in general position and are thus vertices of an n-simplex Irf I in E¥. For x E lur I we define ¢( x) as the unique affine map determined by (5.7) (see subsection 1.20). Defining ¢ in this way for every n-simplex of IKnl, we obtain a map having the properties (a) and (b) of our assertion. To prove assertion (c) let I¢?I be as above and let x be an arbitrary point of lurl. Then

II/(x) - /(ao)11 + II/(ao) - ¢(ao)11 + 11¢(ao) - ¢(x)ll· (5.8) Since x and ao lie in the simplex lurl whose diameter is, by (5.3), less than " it follows from (5.1) that 11/(8) - /(ao)11 < c/5. But also II/(ao) - ¢(ao)11 < c/5 II/(x) - ¢(x)11

~

as follows from (5.5) since /(ao) = Vi for some i and ¢(ao) = Wi for the same i. We now see from (5.8) that assertion (c) will be proved once it is shown that for

x E lurl

11¢(x) - ¢(ao)11 < 3c/5.

(5.9)

Now by Definition 1.13 n

n

EPi = 1, Pi ~ 0,

X = EPiai, i=O

(5.1O)

i=O

and by Lemma 1.18

n

¢(X) = LPi¢(lli). i=O

Therefore

n

¢(X) - ¢(ao) = LPi(¢(ai) - ¢(ao)).

(5.11)

i=O

But

1I(lli) - (ao) II ~ 1I¢(ai) - f(ai) II + II/(ad - f(ao)1I + II/(ao) - ¢(ao)ll.

(5.12)

Here the first and the last term of the right member are each less than c/5 as is seen from (5.4)-(5.6), and the middle term has the same bound as seen from (~.1)-(5.3). The assertion (5.9) follows now from (5.11) and (5.12) in conjunction wIth the properties of the Pi stated in (5.10).

THE FINITE-DIMENSIONAL CASE

128

5.2. COROLLARY TO THEOREM 5.1. (i) The map 1> is nondegenerate in the sense of Definition 3.5. (ii) If Yo, Y1 are two disjoint points of EJ], then assertion (b) of our theorem may be replaced by the assertion that the points Yo, Y1, W1, W2, ... , W N are in general position. The corollary is an obvious consequence of Theorem 5.1 and Theorem 1.11. 5.3. THEOREM. Let IKnl and f be as in Theorem 5.1. Let Yo be a point of EJ] satisfying (5.13) Then there exist a positive integer r1 and a map 1>: IKn I ~ EJ] which is a nondegenerate simplicial map with respect to BT1IKni and for which ,

,

d(f, K n , Yo) = d(1), K n , Yo),

,

(5.14)

where Kn = IKnl- 81K n l. PROOF. The assumption (5.13) implies the existence of a positive c; for which (5.15) Ilyo - f(x)11 > c; for x E 81K n l· Now let 1> be a map having the properties (a), (b), (c) of Theorem 5.1 with an c; satisfying (5.15). Then by property (c)

Ilf(x) - 1>(x) I < Ilyo - f(x)11 (5.16) for x E 81K n l. But by the theorem of Rouche (see subsection 2.9) the inequality (5.16) implies the assertion (5.14). §6. Notes to Chapter 6 Notes to §1.

1. Proof of Remark 1.2. X1,X2,···,X r T

Suppose, e.g., i = 1.

Then for any numbers

T

I>J(aj - ao) = l:)xj(aj - ad - xJ(ao - ad] j=l j=l T

T

= (ao - a1) 2)-Xj) + I>j(aj - ad. j=l j=2 This identity shows that the r elements aJ - ao, j = 1,2, ... , r, are linearly dependent if and only if the r elements aj - a1, j = 0,2, ... , r, are linearly dependent. 2. Proof of Lemma 1.4. Suppose the points ai are independent, and suppose that (1.1) holds. Subtracting the second equality (1.1) mUltiplied by ao from the first one, we see that T

2)ai - aO»\i i=l

= O.

NOTES

Since the ai - ao are linearly independent, this implies that Al = >'2 = ... = Ar = O. Therefore by the second equality (1.1) also AO = O. Thus (1.2) is satisfied. Suppose now that (1.1) implies (1.2). By Definition 1.1 we have to show that the ai - ao are linearly independent or that the relation r

LJ.li(ai - ao) = 0,

J.li

real,

(1)

i=l

implies that

= J.l2 = ... = J.lr = O.

J.l1

(2)

Now (1) may be written as ao

r

r

i=l

i=l

L( -J.li) + L aiJ.li = O.

Then both relations (1.2) are satisfied with r

>'i

AO = - LJ.lJ' j=l

= J.li

for i = 1,2, ... ,r.

(3)

Then, by our assumption, (1.2) is satisfied with AO, AI, ... , Ar given by (3). But (1.2) together with (3) implies the assertion (2).

3. Proof of Lemma 1.6. We will show that the points ao, aI, ... ,ar are dependent if and only if they lie in a plane of dimension less than r. Suppose that ao, aI, ... ,ar are dependent. Then by definition the points a1 - ao, a2 - ao,· .. ,ar - ao are linearly dependent, i.e., r

EJ.li(ai - ao) i=l

=0

for some J.li not all zero. Suppose, e.g., that J.lr

(4)

# O. Then

r-1

a r = 0.0

+L

Ai(ai - ao),

i=l

This shows that the points 0.0, aI, ... ,ar lie in the plane II which is the translate by ao of the subspace of En spanned by the r -1 elements a1 - ao, ... , ar-l - ao. This subspace is of dimension ~ r - 1 < r. Thus II is of dimension < r. Now suppose ao, ... ,ar lie in a plane Ilk of dimension k ~ r - 1. Then there are only k among the elements a1 -an, ... , ar-o.o which are linearly independent. Suppose they are at - ao, ... ,ak - ao. Then k

ak+1 - ao

= 2: J.li(ai -

0.0)

i=l

f?r some J.l.i. This implies that the elements al - 0.0, a2 - 0.0, ... , ar - ao are linearly dependent. Consequently points ao, al, ... , a r are dependent.

THE FINITE-DIMENSIONAL CASE

130

4. For the proof of (1.4) we will make use of the following well-known fact of linear algebra: if vn is a vector space of dimension n and if b1 , ~, ... ,br are arbitrary r < n linearly independent elements of V n , then there exist elements br +1, ... ,bn of vn such that the b1 , b2 , ... ,bn form a base for En. Now let ko be an element of K, and let Xl. ... ,X r , Yl, ... ,Ys, Zl, ... ,Zk be bases for ITT, ITs, and K resp. (see 1.5). Then the elements x of ITT, Y of ITs, Z of K may be written as k

T

X

= ko

+L

Qi Xi,

Z

i=l

i=l

where the

Qi,

= ko

+

L

li Z i,

i=l

f3i, li are real numbers. We consider first the special case that s = k.

(5)

In this case ITs = KeITT = Hand r = h. This together with assertion (1.4). Suppose now s -# k. Then s > k. Since K C ITs remark made at the beginning of this proof choose Yl. Y2, ... , Ys that Yi = Zi for i = 1,2, ... ,k. Since KeITT it is clear that with r + s - k elements

(5) proves the we can by the in such a way this choice the

(6) span H - K. Thus we obtain the inequality h~r+s-k

(7)

for the dimension h of H, and our assertion (1.4) will be proved once it is shown that the equality sign holds in (7), i.e., that the elements (6) are linearly independent. Now a linear relation T

L i=l

S

AiXi

=

L

(8)

J.LiYi

i=k+l

implies that all J.Li = O. For otherwise the right member of (8), let's call it y, would be different from (} since the Yi are linearly independent. Now Y + ko E ITs n ITT = K by (8). But this contradicts the definition of Yk+l. Yk+2, ... , Ys' Thus indeed J.Ll = J.L2 = ... = J.Lr = 0, and therefore the left member of (8) equals zero. But this implies that all Ai = 0 since the Xi are linearly independent. We thus proved that (8) implies that all Ai and J.Li = 0, i.e., we proved the linear independence of the elements (6). 5. That the union of a finite number of (n - l)-planes in En does not contain an open ball of En is a consequence of Baire's theorem [18, p. 20], since obviously a single (n - 1)-plane does not contain an open ball of en.

6. Proof of Lemma 1.14. We have to prove: (a) IUTI is a convex set; (b) if Q is a convex set containing the points 110, al, ... ,aT! then

IU I c T

Q.

NOTES

r

131

r

(9)

LPi = Lqi = 1, i=O

i=O

and Pi 2: 0,

qi 2: 0,

i = 0, 1, ... , T.

(10)

We have to show that (11) for all t E [0,1]. Now r

(1 - t)x + ty = L Tiai,

(12)

i=O

where Ti = (1 - t)pi

+ tqi,

and therefore by (9)

r

r

L Ti

= (1 -

r

t) LPi

i=O

+ t 2: qi = 1,

i=O

and, by (10), Ti 2: 0. This proves (11). (b) We will prove by induction for containing the points ao, at, ... , ak, then

i=O

°: ;

k ::;

T

that if Qk is a convex set (13)

°

The assertion is trivial for k = since laol = ao is contained in any set containing ao. Suppose then the assertion (13) is true if k is replaced by k-1. We will prove it for k. Now the convex set Qk certainly contains the points aO,al, ... ,ak-l· Therefore by the induction assumption

lak-11 = lao,al, ... ,ak-ll We have to show that if x E lakl, then

C

Qk.

(14)

XEQk.

(15)

By (14) this is clear if x E lak-11. Suppose now that k

x = 2:Piai

E

lakl-lak-11.

i=O

Then Pk

I- 0,

Pi 2: fl.

(16)

i=O

Consider first the case that Pk = 1. Then Po = PI = ... = Pk-l x = ak E Qk. Thus (15) is true in this case. Suppose Pk I- 1. Then

k-l x

=L i=O

=

0 and

k-l Piai

+ Pkak= (1 -

Pk)

2: 8i a i=O

i

+ Pkak,

(17)

THE FINITE-DIMENSIONAL CASE

132

where Si

= pi/(1 - Pk)

~

O.

Therefore by (16) L~':~ Si = 1. This shows that the point Xo = L~':~ Siai lies in luk-ll, and therefore by (14) in Qk. Since ak E Qk the assertion (15) now follows from (17) since Qk is convex.

7. Proof of 1.19. By our assumptions the plane IIn spanned by ao, at, ... , an is the whole space En. Therefore every x E En has the unique representation (1.9) (with r = n) subject to the condition (1.10). For x given by (1.9) we set n

A(x)

(18)

L biPi.

=

i=O

Since the Pi are uniquely determined by x, (18) implies (1.17). We now verify that the A given by (18) is affine: from (1.9) and (1.10) (with r = n) we see that n

X = ao

+ L(ai -

(19)

aO)Pi,

i=l

and, taking into account (18), that n

A(x) = bo + L:(bi - bO)Pi.

(20)

i=l

Since the ai - ao form a base for En, there exists a unique linear map I: En such that l(ai - ao) = bi - boo Then

---+

~(bi - bo)Pi = I (~(ai - ao)Pi) , and we see from (19)-(21) that A(x) = bo -l(ao) To prove the uniqueness assertion (b) let Cl

=

+ l(x).

Ef

(21)

Thus A is affine.

const. and it linear,

(22)

be an affine map En ---+ Ef satisfying (1.17) with A replaced by AI. If x is again given by (1.9) (with r = n), then n

Al(X) =

Cl

+ LPiit(ad

(23)

i=l

and, in particular, bj = Al (aj) = Cl + it (aj) for i = 0, 1, ... , n. Multiplying by Pj, adding over i, and taking into account that LPj = 1, we see that

t )=1

bjpj =

Cl

+ it

(t

a),pj) =

Cl

+ II(X)

= AI(X).

1=0

Comparison with (18) shows that Al(X) = A(x) as asserted.

NOTES

133

8. Proof oj 1.20. By Definition 1.12 the point x given by (1.9) (subject to (1.10) and with r = n) lies in lunl if and only if Pi 2 o. But the same condition is necessary and sufficient so that the point A{x} given by (1.16) with A(ai) = bi lies in the simplex Irnl. That A is not singular follows from the fact (shown in the Notes 7) that the linear part l of A maps ai - ao into bi - bo for i = 1,2, ... ,n. 9. Proof of part (ii) of 1.25. Every x E En has the unique representation n X

(24)

= LPiai, i=O

i=O

Since HI and H2 are convex, it is clear that two points of En - rr n - 1 given in the form (24) lie in the same or different halfspaces Hl and H2 according to whether the coefficients of an in (24) for these points have the same sign or not. H we assume Pn > 0 for x = an, then since an and bn lie in different halfspaces we see that if n

{25}

bn = Laiqi, i=O

then qn < O.

(26)

Now to prove our assertion (ii) we have to show that the affine map A which maps the vertices of uf onto the vertices of u2' (in the given order) has a negative determinant (cf. Definition 1.21). Now by (1.19) for x given by (24) n-l

y = A(x) = L

Piai

+ Pnbn,

i=O

or by (25). n-l

y

= A(x) =

L(Pi

+ qiPn)ai + Pnqnan.

i=O

The linear part of A is then given by n-l

y - ao = L(Pi

+ qiPn)(~ - an) + Pnqn{an -

ao).

(27)

i=l

Now Pl,P2, ... ,Pn are the coordinates of x - ao with respect to the base al - an, a2 - an,· .. , an - ao. If we denote by Yt. Y2, ... , Yn the coordinates of y - ao with respect to the same basis, we see from (27) that the matrix of the corresponding coordinate transformation has only zeros below the main diagonal, has ones in the first n - 1 elements of that diagonal, and qn as its nt'l element. Its value is therefore qn. By (26) this finishes the proof.

Note. to §2. 1~. Proof of the independence of d(hJ, 0, h(yo)) of the chosen base (see subsectIon 2.2). In addition to the notations used in subsection 2.2, let b~, b~, ... , b~

THE FINITE-DIMENSIONAL CASE

134

and ,8~, ,82, ... ,,8~ denote positive bases for Ef and E!f resp., and let h' be the linear map E!f ---+ Ef which maps the point y = L 1/~,8~ E E!f into the point x = L1/~b~ E Ef. We have to show that d(h'I, O, h'(yo)) = d(hl,O,h(yo)), or that (28) d(gh,O,zo) = d(h,O,xo), where

h = hi,

9 = h'h- 1 ,

xo = h(yo),

zo = g(xo).

(29)

To prove (28) we use part (ii) of Corollary 2.5 in Chapter 5 with I replaced by h and Yo by xo· With this replacement we see from (2.13) and (2.16), both in Chapter 5, that if Xo lies in the unbounded component of Ef - h (80), then (28) holds, both members being zero. But if Xo lies in a bounded component C(3o of Ef - h(80), then by (2.15) in Chapter 5

d(gh, 0, zo)

=

d(g, C(3o' zo)· d(h, 0, xo),

and for the proof of (28) it remains to show that d(g,C(3o'zo) = +1, i.e., by (29) that the determinant of the linear map 9 = h'h- 1 is positive. Now since b1 , b2 , ... ,bn is a base for Ef, we may write n

j = 1,2, ... , n,

bj = I>]kbk, k=1

where the CJk are real numbers. But since the bi and the b~ are both positive bases for Ef, it follows from subsection 1.22 that the determinant ICI of the matrix C of the CiJ is positive. Similarly n

,8J

=

L l]k,8k'

j = 1,2, ... , n,

k=1

where the determinant If! of the matrix r of the IJk is positive. Now if 1/1,1/2, ... , 1/n are the coordinates of x E Ef with respect to the base b1 , b2 , ... ,bn and if 1/~ , 1/2, ... ,1/~ are the coordinates with respect to the same base of the point x' = g( x) = h' h -1 (x), then direct computation shows that 1/;

=L

1/i

L likCkJ = L 1/i(rC)iJ,

j

= 1,2, ... , n,

(30)

k

where (rC)i] denotes the term in the ith row and the jth column of the matrix rc. Therefore the determinant of the linear map h' h- 1 is positive, the determinants of rand C being positive. Notes to §3. 11. Proof of Lemma 3.3. If lavi = laio,ail'··· ,ad is a v-face of the simplex lanl = lao,al, ... ,ani, then the point n X

= LPiai, i=O

n

I: Pi = 1, Pi ~ 0, i=O

(21)

NOTES

of lanl belongs to lavi if and only if Pi = 0 for all indices i not contained in the set i o, iI, ... , iv . Noting that the representation (21) of a point x E an is unique (Lemma 1.12), we conclude that if lal-'I = lai'o ,ai',· .. , ai',. I is a Jl-face of 1 Ian I, then the intersection la v n al-' I is empty if the index sets io, iI, ... ,i" and ib, ii, ... ,i~ have no point in common. If they have common elements, we may assume these are io = ib, il = ii, ... , it = i~ with 0 ::; t ::; min(Jl, v). Then lavi n lal-'I is the common face laio, ail'···' ait I of lavi and Ia I-' I·

Notes to §4. 12. Proof of Lemma 4.2. By Definition 3.2 we have to show that the intersection of two n-simplices of BKn is either empty or a common face of the two simplices. For this purpose we will prove the following assertion Av for v = 0,1,2, ... , n: if 81' and 82 are two v-simplices of BKn (i.e., v-faces of n-simplices of BKn), then either A~ : 81' n 82 is empty, or A~ : 81' n 82 is a common face of 81' and 82· Ao is trivially true since 8? and 8g are vertices of Kn. Suppose then that Ao, AI, ... , A v - 1 are true for some v ::; n. We want to prove Av. Now for i = 1,2, the simplex 8i is of the form (22) where (3i is the barycenter of a v-simplex a1' of Kn and where (bb, bi,···, b~_d = ai- 1 is a (v - I)-simplex of the barycentric subdivision of a (v - I)-face of a V (i.e., of aaf). But by induction assumption either

n a2v-I a v-I 1

.

IS

empty,

(23)

or (24) is a common J1,-face of see that

ar-

v

1

v

and

a1 n a2 =

{

a2- 1 with 0 ::; JL ::; v the empty set (3'

1. Now in case (23) we

if (3' =f:. (32, if (3' = (32.

This proves the assertion Av in case (23). Now suppose (24) takes place. In this case we may assume the indexing to be such that

ul-' Then by (24)

= (bb = b~, bi = b~, ... , b~ =

b!).

THE FINITE-DIMENSIONAL CASE

136

13. The proof of Lemma 4.4 is based on the following. LEMMA

A. Let lanl = lao,al, ... ,anl. Then 8(la n l) = sup Ilai - ajll.

(25)

i,j=O,l, ... ,n

PROOF. Since ai E lanl for i = 0,1, ... , n, it is clear that the right member of (25) is not greater than the left member. Thus it remains to prove

8(la n l) :::;

Ilai - ajll.

sup

(26)

i,j=O,l, ... ,n

Now if (26) were not true, there would exist a 80 such that for i, y" = 0,1, ... ,n (27) To show that this inequality leads to a contradiction, we will show that it implies

8(la n l) > 80 > Ilx -

yll

for all x, y E lanl,

(28)

which indeed leads to the contradiction

8(la n l) > 80 2:

Ilx -

sup

yll

=

8(la n l).

x,yElunl

To prove that (27) implies (28), we note that the first parts of these inequalities are identical. It thus remains to show that (27) implies the second part of (28). To show this we choose an arbitrary but fixed i among the integers 0,1, ... , n. Then, by (27), aj E B(ai,80 ) for every y" = 0,1, ... , n. Thus the ball B(ai,80 ) is a convex set containing all vertices of lanl. By Lemma 1.14 this ball contains lanl. Thus Ilx-ail! < 80 for every x E lanl, or ai E B(x, 80 ) for each such x. But ai was an arbitrary vertex of lanl. Therefore for fixed x E lanl the convex set B(x,80 ) contains all vertices of lanl. So, again by Lemma 1.14, lanl E B(x, 80 ), This obviously implies the second part of (28), and Lemma A is proved. 'furning to the proof of (4.2) we note that it will be sufficient to prove the assertion for r = 1 since the general case then follows by iteration. Then let af = (bo, bl , ... , bn ) be a simplex of BKn. To prove (4.2) for r = 1 it will, by Lemma A, be sufficient to show that

Ilbi -

b)" II

n:

:::;

1 8 (la n l)

(29)

for i, j = 0,1, ... , n, i i= j. If as above lanl = lao, al, ... , ani, it is no loss of generality to assume that bo = ao,

bl = (ao + ad/2, ... ,

bi = (ao + al + '" + ai)/(i + 1), ...

Then for i =I- y"

bi - bj = i

~ 1 [t a 0=0

=

"+11 t

t

0=0

o -

(i + 1) . J

t

~ 1 (J=o t a{J]

[ao - J.+1 1 {J=O a{J]

=i

~ 1 t j! 1 {J=O t(a o- a{J). 0=0

(30)

NOTES

Now i ¥- j and we may suppose j > i. The double sum at the right member of (30) contains formally (i + I)U + 1) terms and exactly i + 1 of these equal zero. Thus the sum contains (i + l)(j + 1) - (i + 1) = (i + 1)j nonzero terms. Therefore we see from (30) and Lemma A that Ilbi

-

bJil ::; (i

~\~(?! 1) 8(IO"nl) =

j!

1 8(IO"nl).

(31)

Now j/(j + 1) ::; n/(n+ 1) since j ::; n. Therefore (31) implies the assertion (29). 14. To obtain a simplicial decomposition of Qn (cf. subsection 4.6), we first define r-faces of Qn for r = n, n - 1, ... , 1, O. The n-face of Qn is defined as Qn. To define (n - I)-faces we write (4.4) in the form Xi - ai } ai - Xi

<_ l,

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

(32)

Then a point X belongs to an (n -1 )-face if and only if for its coordinates at least one of the inequalities (32) is replaced by equality: e.g., the set of X satisfying Xn = an

+ l,

Xi - ai ::; ai - Xi ::;

l'} l,

c i = 1,2, ... ,n - 1, lor

is an (n - I)-face. It is an (n - I)-cube of side length 2l and center (aI, a2,"" an-I, an

+ l)

and lies in the plane Xn = an + l of En. The number of (n - I)-faces is 2n, and their union is obviously the boundary of Qn. One obtains the (n - 2)-faces by replacing exactly two of the inequalities (32) by equalities and so on. Each r-face is a cube in some r-plane of En and its boundaries consist of (r -I)-faces. The desired subdivision of Qn into simplices is constructed as follows by induction. The subdivision of a O-face Q O (i.e., a vertex of Qn) is Q O• Suppose now that for some r ::; n - 1 the r-faces of Qn are subdivided into r-simplices, and let Qr+ 1 be an (r + 1)-face of Qn. The boundary of Qr+ 1 consists of a finite number of r-faces of Qn, say Q~, Q2' ... , Q~. Let Vo be the center of Qr+ l. Then the simplices of the subdivision of Qr+l are of the form (vo, Vl,"" Vr+l), where (Vi! V2, . .. ,vr+d is a simplex of the simplicial subdivision of one of the cubes Q~, ... , Q~. In this way we obtain a decomposition of Qr+l into simplices. The proof that the union of the simplices of the subdivision is a simplicial set will be omitted (cf. the corresponding proof in Note 12; cf. also [2, part 8 of Chapter m, §2]).

CHAPTER 7

On Spheres Elementary properties and orientation of spheres. Degree of mappings between spheres 1.1. DEFINITION. Let En+1 be the (n + I)-dimensional Euclidean space whose points are the (n + I)-tuples x = (Xl, X2, ... , xn+d of real numbers and whose norm Ilxll is given by n+1 ( 1.1) IIxl1 2 = xI = (x, x),

2: i=l

where (x, y) denotes the usual scalar product of the elements x, y of En +1. Let c = (C1, C2, ... , Cn+1) be a point of En+1 and let r be a positive number. Then the n-sphere sn(c, r) of center c and radius r is defined by sn(c,r) = 8B n+1(c,r), (1.2) where as usual Bn+1(c,r) denotes the (open) ball with center c and radius r. Equivalently,

sn(c, r)

= {x E

E n+1

I%: (Xi - Ci)2 = r2} .

(1.3)

The sphere sn(fJ, 1) is called the unit sphere in En+1 and is often denoted by sn. (Note that for n 2:: 1 the sphere sn(c,r) is a connected set, while SO(c,r) consists of the points c - r, c + r in E1 with r > 0 and is thus disconnected.) 1.2. LEMMA. If 8 i8 an integer < n, then the intersection of sn(c, r) with an (8 + 1)-plane fP+1 is either empty or a single point or an s-sphere.

For a proof see the Notes. 1 1.3. If the (8 + 1)-plane 11 8 + 1 of Lemma 1.2 contains the center c of sn(c, r), then the intersection sn(c, r) n 11 8 + 1 (which by that lemma is an 8-sphere) is called a "great sphere on sn(c, r)." If s == 1, the great sphere is a great circle. If s == 0, then the O-sphere sn(c, r) n IT1 consists of the two points in which the line 111 through c intersects sn(c, r). These two points are called antipodal to each other. 138

ELEMENTARY PROPERTIES AND ORIENTATION

139

1.4. LEMMA. Let N be a point of sn(c, r). Then the intersection of all great circles through N consists of N and its antipode S. 2

1.5. DEFINITION. Let x be a point of sn(c, r). Then the tangent plane Tx to sn(c, r) at x is defined as the set of all t E En+l for which t - x is orthogonal to x-c. Obviously Tx is an n-plane containing x. The tangent space E,!: at x is the subspace of En+l of which Tx is a translate, i.e., y E E,!: if and only if t = y + x E Tx. 1.6. Stereographic projection. Let no and no be two antipodal points on sn(c, r). Let x be a point of sn(c, r) different from no, and let III be the I-plane (straight line) determined by no and x. Then the pair of these two points is the intersection sn(c, r) n III (cf. Lemma 1.2), and III intersects the tangent plane Tn o in a single point t = t(x) (as follows, e.g., from Lemma 1.8 of Chapter 6). Conversely, if t is a given point on Tn o' then the straight line determined by no and t intersects sn(c, r) in no and in exactly one point x # no. We thus obtain a one-to-one map of sn(c, r) - {no} onto Tn o. This map is called the stereographic projection of sn(c, r) with pole no. It is easily seen to be continuous (see the Notes).3 The stereographic projection will be the basis for extending the notion of the degree for mappings between linear spaces to mappings between spheres. 1. 7. Orientation 0/ spheres. Let En+ I be an oriented Euclidean (n + 1)space. Orienting a sphere sn(c, r) in En+l means to give an orientation to the tangent space E,!: to sn+! (c, r) for each x in that sphere. We define the orientation of E,!: induced by the one of En+ I as follows: let Tn be an n-simplex in T;- with vertices to = x, tt, t2, ... , tn. Then the points bl = tl - to, b2 = t2 -to, ... , bn = tn -to form a base for E,!:, and the points bo = c- to, bl , ... , bn form a base for En+!. We then call the base bt, b2 , ... , bn a positive base for E; if and only if bo, bt, . .. ,bn form a positive base for En+l (cf. subsection 1.22 in Chapter 6). This determines an orientation of E; provided it can be shown that this definition does not depend on the particular choice of the points t I, t2, ... , tn. This verification will be given in the Notes. 4 The orientation of Tx is defined as that of E;. When it is necessary to distinguish between the oriented and unoriented Tx, we denote the former by Tx. The orientation of sn(x, r) defined above will be referred to as its positive orientation. We conclude this section by the following remarks. (a) The simplex (to, tt, . .. ,tn ) is a positive simplex of Tx if the simplex (to, c, tl, .. . , tn) is a positive simplex of En+!. This follows from subsections 1.22 and 1.26 of Chapter 6. (b) In our definition for the orientation ofTx , the center c of sn(c,r) = aBn+!(c,r) !ray be replaced by an arbitrary point of B(c, r) - {x}. This follows easily from (a) and subsection 1.25 of Chapter 6. (c) Let Xl and X2 be points of sn(c, r); let 0 1 be a bounded open set in Tzu and let f be a continuous map of fIl -+ Tx2 • Let t2 be a point in TX2 satisfying t2 fI. /(aod. Then the degree d(f, 01, t2) is defined as the degree of the corresponding map between the tangent spaces to sn(c, r) at Xl and X2 resp.

140

ON SPHERES

1.8. LEMMA. Let En+1 be an oriented (n

t6

+ I)-Euclidean space. Let tli and Let b6 = c - tli and b6 = c - t6·

be points on the sphere Sn(c,r) C En+1. Let r12 be a rotation of En+1 onto itself which preserves the orientation of En+1 and for which (1.4) Then r12(Etol) = E t02,

(1.5)

where Etlo and Et20 are the oriented tangent spaces (see subsection 1.7) to sn (c, r) at the points tli and t6 resp. Equivalently (1.6)

where Ttl, Tt20 are the corresponding tangent planes. o For a proof see the Notes. 6 1.9. LEMMA. With the notation of Lemma 1.8, let Xl III be an open bounded set in T xl . Then the degree

=

tli and X2

=

t6. Let (1.7)

for every t2 E r12(1lt} C T x2 · The proof is obvious from Lemma 1.8, from Definition 2.3 in Chapter 6 with h = (r12)-1, and from remark (c) in subsection 1.7. 1.10. LEMMA. For i = 0, 1, let ni be a point of sn(c, r), let Si be the antipode of ni, let Ti be the tangent plane at Si, and let (1i be the stereographic projection with pole ni. Let Uo be an open subset of sn(c, r) having a positive distance from no and n1 such that the subsets Il i = (1i(UO) of Ti are well defined. Let (101 = (11(10-1 an d (1.8)

Then, assuming n ;:::: 2, (1.9)

where to E Ilo = (1oUo·

(1.10)

The proof will be given in the Notes. 7 1.11. For i = 1,2, let E tn+1 be an oriented (n •+• I)-space, and let sn(c-1, r·) C t Ef+l where Ci is a point in Ef+1 and ri a posItIve number. Let f be a continuous map Sf = sn (C1' r1) into S~ = S~ (C2' r2). We want to define a degree d(f, Sf, S~). If n = 0 the sphere SO(C1 r1) consists of the two points Xl :::::: Cl - r X = , . 1 d h hOI, 2 Cl +rl of the oriented I-space (X-axIS) E l , an t e sp ere S2(c2,r2) consists of

ELEMENTARY PROPERTIES AND ORIENTATION

141

the points Yl = C2 - r2, Y2 = C2 + r2 of the oriented "y-axis EJ." The degree d(f, SP(cl, rl), S~(C2' r2)) is defined as follows: +1 iff(xd=Yl, f(X2)=Y2, - 1 if f(xd = Y2, f(X2) = Yl, 0 if f(xd = f(X2) = Yl,

o 0 _ { d(f,Sl (cl,rd,S2(c2,r2)) -

o

if f(xd = f(X2) = Y2.

Now let n = 1. If in addition Ef = E~, Cl = C2 = e, rl = r2, we define d(f, Sr, Sf) as the winding number u(f, Sr) discussed in subsection 1.6 of Chapter 4 and Note 1 to §4.1. It is obvious how to formulate the definition if the above additional conditions are not satisfied. If n ~ 2 we will first define a degree d(f,Sr(Cl,rd,yo) where Yo is a point of Sr(C2, r2) and prove later (see Lemma 1.12) that this degree is independent of Yo. We distinguish three cases: Case I: Yo tf- f(Sr(Cl,rt)); Case II: f maps (Cl, rl) into a single point of Sr (C2, r2); Case III: neither of the preceding two conditions holds. In Case I as well as in Case II we define

sn

(1.11) In Case III we note first that there exists a point

fend

i= Yo,

nl

E Sr(Cl, rd such that (1.12)

and the set (1.13) is not empty. If moreover n2 is a point of Sr (C2, r2) different from Yo, then as will be proved in the Notes8 there exists an open subset U1 of Sr(cl,rd which has the following properties: (a)

dist(nll Ud > 0;

(1.14)

For i = 1,2, let 8i be the antipode of ni, let Ti be the tangent plane at 8i, and let t7i be the stereographic projection with pole ni. Then 0 1 = t71(Ud is a bounded open set of Tl and by (1.14c) the projection t72(f(Ud is well defined. Thus (1.15) maps the subset solution of

nt =

t71 (Ud of Tl continuously into T2. Moreover tl E 0 1 is a h(td

if and only if x

= t72(YO)

(1.16)

= t711(tl) E U1 is a solution of f(x) = Yo,

(1.17)

i.e., if (1.18) Now it is easy to see that (1.19)

142

ON SPHERES

(see the Notes).9 Therefore the degree d(h,01,0"2(YO)) is defined. As will be shown in the Notes,10 this degree is independent of the choice of n1 and U1 provided that n2 # Yo, and that (1.12) and (1.14) hold. Therefore we may define

(1.20) 1.12. LEMMA. The degree d(f,Sf(cl,rd,Yo) is independent of the choice of the point Yo E S2(c2, r2). PROOF. If f maps Sf (C1, rd into a single point of S2 (C2, r2), i.e., in Case II of subsection 1.11, then, by definition, (1.11) holds for all Yo E S2(c2,r2). Suppose now that f(Sf(c1,rd) is not a single point. Since the sphere is compact, it will be sufficient to prove that to every Yo E S2 (C2, r2) there exists a neighborhood V(yo) C S2(C2, r2) such that

d(f, Sr(C1, rd, y) = const.

for y E V(yo).

(1.21)

Suppose first that for some Yo E S2 (C2, r2) the equation

f(x) = Yo

(1.22)

has no solutions, Le., there are some Yo in Case I (see subsection 1.11). It is clear that such a Yo has a neighborhood V(yo) such that the equation f(x) = y for y E V(yo) has no solution since f(Sf(cl, rt}) is closed and is therefore a positive distance from Yo. Consequently every y E V(yo) is in Case I, and thus (1.21) holds for y E V(yo}, with the constant being zero. Finally consider Case III. Then Yo is an image point, and d(f, Sf, Yo) is given by (1.20). If Xo E Ul is a root of (1.22), then tl = O"l(XO) is a root of h(tt} = 0"2(YO) and 0"2(YO) satisfies (1.19). This implies that 0"2(YO) E O2 = 0"2(Ut} C T2. Consequently there exists a neighborhood Wo of 0"2 (Yo) which does not intersect h(aOt}. Thus d(h, 01. w) = d(/t, 0 1, 0"2 (Yo)) for all wE Wo, and, by definition (1.20), the relation (1.21) will be true with V = 0"2 1 (WO). Lemma 1.12 allows us to define the following. 1.13. DEFINITION. For n ~ 2 the degree d(f, Sf(Cl, rd, S2(C2, r2)) is defined by the equality d(f,Sf(c1,rd,S2(c2,r2)) = d(f,Sf(cl,rt},yo) where Yo is an arbitrary point of S2(r2, C2). §2. Properties of the degree d(f, Sr, S2) We use the notations of the preceding section and assume n ~ 2. For St(Ci, ri) we will often simply write St. 2.1. LEMMA. If f(Sf) is a proper subset of S2' then d(f, 8r, 8 2) = o. PROOF. By assumption there exists a YO E 8 2 such that Yo ~ f(8r). Therefore, by definition, (1.11) holds. Our assertion follows now from Definition 1.13.

PROPERTIES OF d(f, S1', S~)

143

2.2. THEOREM. II the continuous map I: Sr ---- Sr is one-to-one, then (i) d(f,Sr,Sr) = ±1, and (ii) I is onto. PROOF. Let Yo E I(Sr). Since I is one-to-one, neither of the Cases I or II takes place (see subsection 1.11). We are thus in Case III. Therefore d(f, sr, Sr) = d(f,Sr,Yo) (Definition 1.13), and by (1.20), d(f,Sr,Yo) = d{h,01,(J2(YO)) where h is defined by (1.15). But the right member of this equality equals ±1 since the "plane" mapping h is obviously also one-to-one. This proves assertion (i). If assertion (ii) were not true then, by Lemma 2.1, d(f, sr, Sr) = 0 which contradicts assertion (i). 2.3. On the index j(f;xo) 01 an i80lated 80lution Xo 0/ (1.17). We recall first the second part of (1.20), where Ul is an open subset of Sr which has the property (among others) that it contains all roots of (1.17) (see (1.14) and part (b) of (1.14)). Suppose now that there exists a point Xo in Sr which is an isolated root of (1.17), i.e., there exists a neighborhood of Xo in which Xo is the only root. Such neighborhood contains an open set Uo such that (2.1)

and therefore (Jdxo) E (Jl(UO) C (Jl(Ud = 0 1 . It is clear from the sum theorem for the "plane" map h that d(h, (Jl(UO), (J2(YO)) does not depend on the particular choice of Uo provided that (2.1) is satisfied and that Xo is the only root of (1.17) which lies in Uo. For such Uo we define (cf. (1.20)) (2.2)

We denote this number by j(f; xo) and call it the index of Xo as a solution of (1.17). If in particular this equation has only a finite number of roots, say Xl, X2, ... , X r , and if for p = 1,2, ... , r the open set UP contains xp and is contained in Ul, and if the UP are disjoint, then r

d(h,0'1(Ud,(J2(YO)) = Ld(/I,(Jl(UP),U2(YO)) p=l as follows from the sum theorem. Therefore by (1.20) and (2.2)

d(f, Sf, YO) =

r

r

p=l

p=l

L d(f, UP, Yo) = L

j(Jj xp).

(2.3)

2.4. THEOREM. Let It (x) = I(x, t) be a continuous map Sr x [0,1] ---- Sr.

Then d(!t,Sr,S~) is independent olt. The proof will be given in the Notes.u 2.5. THE PRODUCT THEOREM. For i = 1,2,3, let Ei+ l be an oriented (n ~ 1)-8pace, and let SF' = Si(Ci, ri) be the sphere in Ei+1 with center Ci and

radzus Ti· Let I and g be continuous maps Sr ---- Sf and Sf ---- Sf resp. Then d(gl, Sf, Sf) = d(f, Sf, S2) . d(g, S2' Sf)·

(2.4)

ON SPHERES

144

PROOF. Let

nl be a point of 8r, and (2.5)

Suppose first that at least one of the following conditions holds: (a) / (8r) = n2; (b) g(8r) = n3. In either case g/(8r) = n3, and, therefore, the left member of (2.4) equals zero. But in case (a) the first factor and in case (b) the second factor of the right member of (2.4) equals zero. Thus (2.4) holds in either case. Suppose now that neither of the conditions (a) and (b) holds. We next prove (2.4) for the special case that

(2.6)

g/(8r) = n3·

In this case the left member of (2.4) equals zero. Now we excluded case (b). Therefore g-l(n3) is a proper subset of 8r, and since by (2.6), /(8r) c g-l(n3) the set /(81) is a proper subset of 8r. Thus, by Lemma 2.1, d(f,8 1,8r) = 0, and (2.4) is true, both of its members being zero. Suppose finally that neither (a), (b), nor (2.6) holds. It is then easy to verify (see the Notes l2 ) that there exists an Xo E 8 1 of the following property: if

= /(xo),

Zo

/-I(yO) ~ nl, /(gf)-I(ZO) C g-I(ZO),

(b)

Yo

= g(yo),

(2.7)

then (a)

(c)

g-I(ZO) ~ n2, (d) /-l(yO) c (gf)-I(zO).

(2.8)

For i = 1,2,3, let Si be the antipode to ni, let Ti be the tangent plane to 8r at Si, and let Ui be the stereographic projection of 8r with pole ni. It is then clear from the proof given in Note 8 for (1.14) (see also (1.13)) that a D2 of the following property can be chosen: if

U2 =

U

B(y, D2),

(2.9)

yEg-l(ZO)

then (a)

dist(n2, U2 ) > 0;

(c)

n3

rt g(U2 ).

(2.10)

Moreover (cf. (1.20)) (2.11) where -1

gl = U3g u 2

(2.12)

•

In the same way we see that if Dl is a small enough positive number and U1 =

U

B(Xl,Dt},

(2.13)

XE(gf)-l(ZO)

then (a)

dist(nl, Ud > 0;

(c)

n3 ~ gf(Ud. (2.14)

PROPERTIES OF

dU. Sf. s!f)

Moreover (2.15) where (2.16) with

II =0'2fO'I 1.

(2.17) But because of part (c) of (2.8) and the closedness of g-1 (zo), we see from (2.9) and (2.13) that Ih can be chosen in such a way that feUd c

u2.

(2.18)

Finally we note that the U1 thus defined satisfies (1.14). Indeed, part (a) of (1.14) follows from part (a) of (2.14); part (b) follows from part (b) of (2.14) because of part (d) of (2.8) (and (1.13)); if part (c) of (1.14) were not true, then n2 = feud for some U1 C U1 · But then, by (2.5), n3 = g(n2) = gf(ud c gf(Ut} which contradicts part (c) of (2.14). But (1.14) thus proved implies that

d(J,Sr,Yo) = d(II,0'1(Ud,0'2(Yo)). (2.19) It now follows from Definition 1.13 and from (2.11), (2.25), (2.19) that our assertion (2.4) is equivalent to d(gdt, O'lUt, 0'3 (Zo)) = d(g1,0'2U2,0'3(ZO))' d(II,0'1Ut,0'2(YO».

(2.20)

To prove this equality we want to apply the product theorem in subsection 2.12 of Chapter 6. We have to verify that the assumptions of that theorem are satisfied with n = O'lUt, Y = 0'2U2, and with f and g replaced by II and gl resp., and with Zo replaced by 0'3(ZO), we have to show that

II(O'lUt} C 0'2U2, (b) 0'3(ZO) rt. g(O(0'2U2))' (c) 0'3(ZO) rt. gl!1(OO'lUt}. Now (a) follows from (2.17) and (2.18). Suppose now (b) were not true. Then 0'3(ZO) E gl(8(0'2U2» = gl(0'2(8U2», and, by (2.12), 0'3(ZO) E 0'3(g(8U2 which implies that ZO E g(8U2)' This however contradicts part (b) of (2.10). Finally (c) follows similarly from (2.12), (2.17) and part (b) of (1.14) together with (1.13). We are now in a position to apply Theorem 2.12 and obtain (a))

»

d{gl/t,UIU1 ,0'3(ZO» = Ld(gll Yo,0'3(ZO»' d(/t,UIU1. U2{Yo»,

(2.21)

o

where {Yo} is the set of components of U2(U2) - /t(8UIUt}, where 0'2(Yo) E Yo, and where the summation is extended over those Q for which gl (,,) = 0'3 (zo) has a solution" E Yo. Now by (1.20) and by Lemma (1.12)

d(/t,UIU1 ,U2(Yo» = d(J, Sr, Yo) But by the SUID theorem

= d(J, Sr, Yo) = d(/t,0'1Ut, U2(YO».

Ld(gl, Yo, 0'3 (Zo)) = d(gll 0'2U2,U3(ZO»

= d(g, S;, zo).

o

The assertion (2.20) now follows from (2.21), (2.23), and (2.22).

(2.22)

(2.23)

ON SPHERES

146

2.6. for no

THEOREM. Let f and 9 be continuous maps S1' -> Sr and x E S1' are the points f(x) and g(x) of Sr antipodal. Then

(2.24)

d(f,Sr,S2') = d(g,SI,S2'). PROOF. Since Sr Yo and YI of Sr

=

Sr(C2, r2), it is clear that for two nonantipodal points

(1 - t)yO + tYI - C2 = (1 - t)(yO - C2) + t(YI - C2) Therefore

suppose that

i- a

for t E [0,1].

(1 - t)f(x) + tg(x) - C2 f(x, t) = 11(1- t)f(x) + tg(x) _ c211 r2 + C2

is a continuous map defined on SI x [0, 1]. Its range is on Sr since II! (x, t) - c21i = r2, and the assertion (2.24) follows from Theorem 2.4. 2.7. DEFINITION. Let c be a point of En+l, and let M and ~ be subsets of En+l not containing c. Suppose that each ray issuing from c (see §1.23) intersects ~ in exactly one point. Then the central projection 7r = 7r(M,~, c) from c of M on ~ is defined as follows: if m is a point of M, then 7r(m) is the intersection of the ray em with ~. 2.8. LEMMA. Let En+1 be oriented, and let So and SI be positively oriented spheres in En+l with common center c and with radii ro and rl resp. Let 7r1 be the central projection from c of So on SI. Then

d(7rI' So, SI) = +1.

(2.25)

For a proof see the Notes. 13 2.9. Let So and S1' be two positively oriented spheres in En+1 of equal radius and with centers Co and CI respectively. Let r: So -> SI be given by r(x) = x + CI - co. Then d(r, So, S1') = +1. The rather obvious proof is omitted. §3. The order of the image of a sphere with respect to a point 3.1. Let Ef+l and E~+I be oriented spaces. Let SI be a positively oriented sphere in Ef+l, and let CI be a point in E~+I. Let f be a continuous map S1' -> E~+I. In subsection 2.10 of Chapter 6 the winding number u = u(f(SI),cd was defined provided that CI rf- f(SI)· (3.1)

In the present chapter we will give another interpretation of the winding number. The assumption (3.1) implies the existence of a positive r2 such that (3.2) Therefore the central projection 11"2 = 7r2(cd with center CI of f(sn) on the sphere Sr = 8B(CI' r2) is defined. :rhus It = 7r2!(S1'). is a contin~ous map Sf -> S!] and its degree d(h, Sf, Sr) IS defined. Moreover It follows immediately

THE ORDER OF THE IMAGE OF A SPHI£K~

... ,

from Lemma 2.8 in conjunction with the product theorem in subsection 2.5 that this degree is independent of the particular choice of an r2 satisfying (3.2). Therefore the following definition is legitimate. 3.2. DEFINITION. With the assumption and notations of subsection 3.1 the order v(f(Sf), cd of f(Sr) with respect to Cl is defined by

v(f(Sr), cd = d( 7r 2f(Sr), S2)'

(3.3)

3.3. LEMMA. v(f(Sr),cd is constant if Cl varies continuously in such a way that (3.1) remains true during the change. The proof will be given in the Notes. 14 The theorem whose proof is the main object of this section is the following.

3.4. THEOREM. With the above notations and assumptions v(f(Sr), cd = u(f(Sr)'cd·

(3.4)

By Definition 2.10 in Chapter 6 of the winding number u the above theorem is equivalent to the following one.

3.5. THEOREM. Let Bf+l be the ball whose boundary is Sr. Let f be a continuous map of Bf+l into E~+ 1 . Then v(f(Sr), cd = d(f, Bf+l, cd

(3.5)

provided that (3.1) holds. 3.6. Our first step to proving Theorem 3.5 consists in formulating an equivalent theorem in which the ball Bf+l is replaced by an (n+1}-simplex ,af+l whose barycenter "I is the center of Bf+l and whose vertices lie on Sf+l = 8Bf+l: let K, be the central projection from "I of Sr on 8,af+l (see subsection 2.7). We extend the definition of K, from Sf to Bf+l as follows: we set K,C"!) = "I; if x E Bf+ 1 - b} and if x is the point in which the ray ;:yX intersects Sf, we set

K,(x)

= "I + 1IK,(x) -

"III (x - "I) .

(3.6)

rl

Then K, is a one-to-one continuous map of Bf+l onto ,an+l. We define a continuous map g: ,an+l ---+ E~+l by setting

g(x)

= f(K,-I(x)),

x E ,an+l, f as in Theorem 3.5.

(3.7)

Then by part (i) of subsection 2.15 in Chapter 6

d(f, Bf+l, cd = d(gK" Bf+l, cd = d(g, pn+l, CI) . d(K" Bf+l, 17), where 17 is an arbitrary point in K,Bf+l = pn+l. Here t;e second factor at the right equals +1 (see the Notes).15 Thus

de!, Bf+l, cd

= d(g, Pf+l, cd,

(3.8)

and the assertion (3.5) is equivalent to

v(f(Sr),cd = d(g,Pf+ 1 ,cd·

(3.9)

ON SPHERES

148

3.7. Proof of (3.9) in a special case. Since g((3n+l) is bounded there exists a positive R2 such that the ball B(O, R 2) E E~+1 contains g((3n+l). Now let Yl be a point in E~+1 for which (3.10) We will prove (3.9) for Cl = Yl by showing that both members of that formula are zero. Since by definition of YI the equation g(x) = Yl has no solution, it is clear that the right member equals zero. To show that the same is true for the left member, we choose an rl > 0 such that

(3.11) Then by Definition 3.2

v(f(Sr),yt} =

d(7rf,Sl'S~),

(3.12)

where Sf = 8B(Yl,rt} and where 7r is the central projection from Yl on Sf. Now f(Sr) = g(8(3n+l) C B(O, R2)' Therefore

7rf(Sr)

C

7rB(O, R2)'

7rB(O,R2) = a proper subset of S~,

(3.13) (3.14)

as follows from (3.11) and (3.10),16 and by (3.13) the same is true of 7rf(Sr). By Lemma 2.1 this implies that the right member of (3.12) equals zero, and we see that the left member of (3.9) (with Cl = yt} equals zero as asserted. 3.B. To prove (3.9) for an arbitrary Yo = Cl satisfying (3.1) or what is the same (3.15) we consider the segment

yet) = (1- t)yO + tYI,

o ~ t ~ 1,

(3.16)

where YI is as in subsection 3.7. Since we proved (3.9) for CI = Yl, that equality will be proved for Cl = Yo once it is shown that both members of (3.9) with Cl replaced by yet) undergo the same change as t varies from 0 to 1, Le., that

v(f(Sr),yo) - v(f(Sl)'yt}

=

d(g,(31+ 1,yo) - d(g, (31+1, yt}.

(3.17)

For those t E [0, 1] for which

yet)

f/: g(8(31+ 1 ),

(3.18)

we denote by 7rt the central projection from yet) on a sphere Sf(y(t)) of center yet) (for which the ball whose boundary is Sf(y(t)) does not intersect g(8(3n+l)). Now in a subinterval of [0, 1] all of whose points t satisfy (3.18), both members of (3.9) with Cl replaced by yet) remain constant (see ~he Notes).17 Thus a change of these members can occur only ift crosses a point t for which yet) E g(8(31+1). If there are no such l, then both members of (3.17) are zero an~ this equality is proved. We therefore assume from now on that ther~ are such t. The difficulty in this case is that there may be infinitely many such t and that each intersection

THE ORDER OF THE IMAGE OF A SPHERE

149

of the segment YOYl with g(B,Bf+l) may be of a complicated nature. To avoid this difficulty we will introduce a simplicial approximation of g. 3.9. A simplicial approximation. We note first that there exists a positive c of the following property: if h is a continuous map ,Bf+l -+ E~+l which satisfies Ig(x) - h(x)1 < c

(3.19)

for all x E ,Bf+l, then neither member of (3.9) changes its value if g is replaced by hand f = g",-l (cf. subsection 3.7). For the right member this follows from Theorem 2.9 of Chapter 6, and for the left member from Lemma 3.3 and Theorem 2.6. Let c be such a number. We know from Theorem 5.1 of Chapter 6 that there exists a barycentric subdivision Kn+l = Br ,Bf+l of ,Bf+l and a map h: IKn+ll = l,Bf+ll -+ E~+l which satisfies (3.19), is simplicial with respect to Kn+l, and is such that the images under h of the vertices of Kn+l together with the points Yo and Yl form a point set of E~+l which is in general position. In other words, for the proof of (3.9) we may assume that g is simplicial with respect to the subdivision Kn+l of ,Bf+l and has the following property: if Uo, Ul, ... , UN are the vertices of Kn+l and if Vi = g(Ui) for i = 0, 1, ... , N, then the points (3.20) Yo, Yl, Vo, Vl,···, VN are in general position. This property has the following consequences C l , C 2 , C3 : iffor some l E [0,1] the point y(l) of the segment (3.16) lies on g(8Kn+l), then (Cd the equation (3.21 ) g(x) = y(t) has in BKn+l one and only one solution; (C 2 ) if this solution lies in the simplex un of BKn+l and if Tn = g(u n ), then y(l) lies in the interior of Tn; (C 3 ) y(l) is the only intersection ofthe ray YoY! with Tn. For a proof of these three properties see the Notes. 18 Since there are only a finite number of n-simplices in g( BKn+ 1 ), it follows that there are only a finite number of t-values in (0,1) for which y(t) E g(B,Bf+1), say tt < t2 < ... < t r • Since we know that t = 0 and t = 1 do not belong to them, the intersection points of the segment YOYI with g( B,Bf+ l) are

o < tl < t2 < ... < tr < 1.

(3.22)

If, for t different from any of the to" a = 1,2, ... , r, rrt is defined as in the lines directly following (3.18), we write for simplicity's sake l(y(t)) and r(y(t)) for the left and right member of (3.9) (with Cl replaced by y(t) resp.). Now the discU88ion in subsection 3.8 showed that the left member of (3.17) equals the sum of the '~umps" of l(y(t)) as t crosses one of the to: and that the right member equals the sum of the corresponding jumps of r(y(t)). Thus the assertion (3.17) will be proved once it is shown that for each to: the jumps of l(y(t)) and r(y(t))

ON SPHERES

150

are the same. More precisely we have to show that for

l{y{t;t)) -l{y{t;;))

= r{y{t;t)) -

0:

= 1,2, ... , r

r{y{t;;)),

(3.23)

where the numbers t;; and t;t are subject to the inequalities to:-l < t;; < to: < t;t < t~+l < ta+l

if we define to = 0 and tr+1 of the equation

= 1.

(3.24)

Now let Xa be the unique solution in 8Kn+l

g{X) = y{to:),

(3.25)

let O'~ be the unique n-simplex of 8Kn+1 which contains XO:, let a~+l be the positively oriented {n + I)-simplex of Kn+1 for which (3.26) and let (3.27) Then let 0:0 be a fixed one of the integers 1,2, ... , r. For this 0:0 the simplex r;::l may be a positive or negative simplex of E~+1. Since the proof of (3.23) is essentially the same in these two cases, we carry it out under the assumption that r;::l is positively oriented. (3.28) We denote by Wl, W2, ••. ,Wn the vertices of r;:o' and by WO, Wl, •.• ,Wn the vertices of r;::l. Moreover, if ITS denotes the n-plane spanned by r;:o we distinguish two cases: (i)

Wo

and Yt+ao lie in the same half space of E~+1 with respect to ITO';

and tt+ao lie in different half spaces of E~+l with respect to ITO'. (3.29) 3.10. Computation of the right member of (3.23) for 0: = 0:0. We will show that + _ _ { + 1 in case (i) of (3.29), (3.30) r(y(to:o)) - r(y{to:o)) - 1 in case (ii) of (3.29). (ii)

Wo

The idea of the proof is to show that as t varies from t- to t+ then as t 0:0 0:0 ' crosses to: the point y(t) enters exactly one simplex of g(8Kn+l), namely, rn+l in case (i), while it leaves that simplex in case (ii). This together with O:tOhe assumption (3.28) will yield the assertion (3.30). Precisely, with the notations employed in subsection 3.9, let Kn+1 be the simplicial set consisting of the (n + I)-simplices of Kn+1. except for a~+1. It then follows from (Cd and (C 2 ) th~t for 0: = 0:0 the equatIon (3.25) has no root x E 8kn+1. Thus every root x E Kn+1 of that equation is an interior p . t f d + om 0 Kn+1. Consequently we can choose t;;;o an tao so near to to:o that d(g,k n+1,y(t;;o))

= d(g,K n+1,y(tto)) = d(g, Kn+1, y(to:)).

(3.31)

THE ORDER OF THE IMAGE OF A SPHERE

Thus, by the sum theorem, d(g, K n+ 1, y(t~o)) = d(g, k n+1, y(t~o))

151

+ d(g,a~:1, y(t~o))'

and this equality remains true if y(t~o) is replaced by y(tto)' It therefore follows from (3.31) and the definition of r(y(t)) that

r(y(tto)) - r(y(t~o)) = d(g, a~:1, y(tt)) - d(g, a~:1, y(t~o))' (3.32) Now let H+ and H- be the two half spaces into which the plane TI8 spanned by 0 divides E'2+1 with y(tto) E H+ and y(t~o) E H-. Then in case (i) of (3.29) the interior of the simplex = g(a~:1) lies in H+. Since by assertion (C 2 ) stated in subsection 3.9 the point y(t oo ) lies in the interior of the simplex 7rJ = g(a~o) E TIn, we may assume that y(tto) E 7::: 1. By assumption (3.28) this implies that the first term ofthe right member of (3.32) equals +1. On the other hand the second term of the right member of that equality equals zero n+1) = 7 00 n+1 E H+ • ) E H- while g(a00 since y(C 00 This proves the first part of (3.30). The second part is proved correspondingly. 3.11. Computation 01 the left member 01 (3.23) Before giving a formal proof for the result (3.48) we make the following intuitive remark: if the ray YOYi intersects at y(t oo ) the simplex 7::0 of 8gKn+1 and if (3.27) and (3.26) hold with a = ao, then on account of the general position of the points (3.20) there are two possibilities as t increasing crosses too: (i) y( t) enters the interior of from the exterior of g(Kn+1); (ii) yet) leaves the interior of and enters the exterior of g(Kn+1). Thus under assumption (3.28) the degree d(g, a;;: 1 ,y(t)) changes from 0 to 1 in case (i) and from 1 to 0 in case (ii). It is then not hard to compute the related degree obtained by replacing 9 by 1Ttg appearing in the right member of (3.46). Therefore our first goal in our formal proof of (3.48) will be to prove (3.46). Again let yet) be given by (3.16), 0 ~ t ~ 1. For any t value for which (3.18) holds or what is the same for which

7::

7:::1

7:::1

yet)

rt.

f(Si),

i.e., t

:f. to,

a = 1,2, ... , r,

7:::1

(3.32)

holds, we denote by S2(y(t)) a sphere with center yet) and radius so small that the closure of the ball whose boundary is S2(y(t)) contains none of the y(t o ). Then by definition (3.33) where 1Tt I(x) denotes the central projection of I(x) on S2(y(t)) with center yet). Now by Definition 1.13 the right member of (?-.33) equals the degree d( 1Tt f,Sr,y*) if y* is an arbitrary point of S2(y(t)). Now if too < t < too+! we choose for y* the point y(t·) which is the intersection of Sr(y(t)) with the segment y(t)Y1. Then too < t < t· < t oo +1 (3.34) (if the radius of S2(t) is chosen small enough), and by (3.33) v(f(Sf),y(t)) = d(1T t /,Si,y(t*)). (3.35)

ON SPHERES

152

Now from our discussion of the root x'" of (3.25) and the relation (3.7) between 9 and /, we see that the equation (3.36)

has on

S1' exactly one root E",

=

11:- 1x"" II:

-1

and if cr", = U"', n

(3.37)

then (3.38)

But from (3.34) we see that (3.39)

if and only if a = ao

+ 1,

ao

+ 2, ... ,ar . Tlt/(O

=

It follows that the roots on

S1' of

y(t*)

(3.40)

are E"'0+1, ... , Er. Therefore we see from (3.35) and subsection 2.3 that r

L

v(f(Sl'),y(t)) =

J(Tl t /; E",)·

(3.41)

"'="'0+ 1 To express the right member in terms of 9 rather than /, we note that the simplices cr~ and 7"{; = g(cr~) are oriented (as boundary simplices of cr~+1 and 7"{;+1 resp.) They therefore induce an orientation in the n-planes P;; c Er+ 1 and Q~ C E~+1 in which they lie resp.) Now y(t",) is an interior point of 7"{; = g( cr~) and the only intersection of YOY1 with 7;;. Let TIt 9 be the central projection (from yet)) on the tangent plane T;' to Sf(y(t)) at y(t*). Then the degree d(Tltg,cr~,y(t*)) is defined and, as will be verified in the Notes,19

J(Tl t /; E",) = d(Tltg, cr~, y(t*)).

(3.42)

Since t satisfies (3.34) we may choose an arbitrary value satisfying tOto < t < t"'0+1. Therefore, by (3.24) we may choose t = tto. Then by (3.41) and (3.42) r

v(f(Sl'),y(tto))

=

L

d(Tlttog,cr~,y(tto)*)·

(3.43)

2: d(Tlt;;og,cr~,y((t~o)*))'

(3.44)

"'="'0+ 1 In the same way one proves r

V(f(Sl'),y(t~o)) =

where t;;o satisfies (3.24) with a replaced by ao and where y((t;;o)*) is the intersection of Sf (y( t;;o )) with the segment y( t;;o), Yl· But for a = ao + 1, ... , r

d(Tlt;tog,cr~,y((tto)*)) = d(Tlt;;og,cr~,y((t;;o)*))

(3.45)

(see the Notes).20 It therefore follows from (3.43), (3.44) and the definition of l(y(t)) (given in the paragraph directly following (3.22) that

l(y(t+)) -l(y(t-)) = -d(Tlt;:;og,cr~o,y((t;;o)*))·

(3.46)

THE ORDER OF THE IMAGE OF A SPHERE

153

Now since t;;o < tno we can apply the assertion (77) of Note 20 with t = t;; and a = ao. Since, moreover, tno < tto we see that for a = ao the cases (i) and (ii) defined in (76) of Note 20 are identical with cases (i) and (ii) resp. (defined in (3.29)). Because of our assumption (3.28) we apply (77) with Cn = +1 to obtain

d(7rt~og'(}"~o,y((t;;o)*)) =

{+-11

~n case case ((~!)' 11.

III

(3.47)

Therefore by (3.46)

+ l(y(tno )) -l(y(t;.;o))

=

{

+1 _ 1

in case (i), in case (ii).

(3.48)

Comparison of this formula with (3.30) proves (3.23). This finishes the proof of Theorems 3.5 and 3.4. 3.12. Interpretation of order and degree as intersection numbers (cf. subsections 2.1-2.6 in Chapter 4). We use the definitions and notations of the previous sections of the present chapter. In addition we state the following definition. 3.13. DEFINITION. A map I: Br+ 1 --+ E~+1 as well as a map f: 8Br+1, Sr --+ E~+1 is called simplicial if 1 is given by (3.7) where 9 is simplicial with respect to the barycentric subdivision Kn+1 of the simplex ;3r+ 1 (see subsection 3.9) and where the points (3.20) are in general position. Now let 1 be a simplicial map Sr --+ E~+l. We consider the intersections of I(Sr) = g(8Kn+l) with the ray YOYt given by

y(t)

=

(1 - t)yO

+ tYI,

0::::: t <

00,

(3.49)

where Yo and Y1 are as in subsection 3.8. If tn is one of the at most finite number of positive t-values for which y(t) intersects I(S1') = g(8Kn+I), then for tt and t;; as in (3.24) the "jump" v(f(Sf),y(tt)) - v(f(Sr),y(t;;)) given by (3.48) is different from zero, actually ±l. However if y(t) is not an intersection point with I(S1') and t- < t < t+ with t+ - t- small enough, then v(f(S1'), y(t+)) v(f(S1'), y(t-)) equals zero. Therefore for any t 2: 0 we define the intersection number i(f(Sr), y(t)) at y(t) of I(S1') with our ray (3.49) by

i(f(Sr), y(t)) = v(f(Sr), y(t+)) - v(f(Sr), y(r)).

(3.50)

The sum over t of all these intersection numbers is called the intersection number of I(S1') with the ray yoy~ and denoted by i(f(S1'), YOy{). Then r

i(f(Sr), YOYt) =

2: v(f(Sl), y(t;t)) - v(f(Sl), y(t;.;)).

(3.51)

n=1

We recall that v(f(S1'), y(t)) is constant in every t interval which contains none of the tQ (see subsection 3.8). From this, from (3.24) and the fact that v(f(S1'),Y1) = 0 (see subsection 3.8), it follows that the right member of (3.51) equals -v(f(Sf), Yo). Thus

v(f(Sl), Yo) = -i(f(Sr), YOyi).

(3.52)

ON SPHERES

154

This represents the order (for a simplicial f) as an intersection number. To obtain a corresponding representation for the degree we need the following definition: let Tn+! be an oriented simplex in the oriented En+!, let Tn+! be its interior, and let Yo be a point in En+!. Then if co denotes one of the symbols + and -, we define the intersection number i(coyo, Tn+!) of the oriented zero simplex coYo (see subsection 1.16 in Chapter 6) with T n + 1 by setting if Yo E Tn+! and if Tn+! is positively oriented, if Yo E Tn+! and if Tn+1 is negatively oriented, if Yo rt. Tn+l.

(3.53)

The intersection numbers are not defined if Yo E UT n +!. But because of the general position of the points (3.20) the numbers (3.54) are defined for all (n+ 1)-simplices O"~+! of Kn+!. Moreover the O-face containing Yo of the oriented I-simplex (YO, yt) is -yo according to the definition given at the end of subsection 1.16 in Chapter 6. Thus co = -1. We now define the intersection number of -yo with g(K1+!) as the sum of the numbers (3.54): (3.55) Q

where we sum over all (n + I)-simplices O"~+! of Kl+!' But for each 0: the restriction of 9 to O"~+! is affine. Therefore we see from (3.53) and the proof of assertion (iv) of subsection 3.7 in Chapter 6 that n+1)) -_ t'{ -yo,g (O"Q

d{ g'O"Qn+l ,Yo. )

(3.56)

Consequently, by (3.55) and the sum theorem for the degree,

i{ -yo, g(Kn+l)) = -d(g, IKf+1l, Yo).

(3.57)

But IKr+ 1 1= ,ar+1 (see subsection 3.6) and, by (3.8), the right member of (3.57) equals -d(f, Br+1, yo). Thus

-d(f, Br+l, Yo)

= -d(g, ,an+!, Yo) = i( -Yo, g(,an+1)),

(3.58)

and we expressed the degree in terms of an intersection number. Now in (3.52) as well as in (3.58) / was supposed to be a simplicial map. Now let / be an arbitrary continuous map satisfying only Yo f1. !(Sr). But we know from subsection 3.9 that for small enough c the values of the numbers v(f(Sr)' Yo) and d(f, Bl+!' Yo) do not change if / is replaced by a continuous! satisfying II/ - /111 < c, and that among such It there are simplicial maps. Thu~ (3.52) and (3.58) hold for continuous! if the left member is replaced for such an

§4. TWO APPROXIMATION LEMMAS

155

approximation it. We therefore define for continuous maps f the intersection numbers i(f(Sf), yoy{) and i( -yo, f(Bf+l)) by setting

i(J(Si),YOYi)

= -v(f(Sr),yo),

i(-yo,f(Bf+1))

= -d(f,Bf+l,yO). (3.59)

We finally note the "duality" formula

i(f(oBf+l), YOYi) = i( -Yo, Bf+1),

(3.60)

which on account of (3.59) follows directly from Theorem 3.5 since Sf = oBf+l.

§4. Two approximation lemmas 4.1. It is often important to approximate a given equation in such a way that the approximating equation has at most a finite number of roots. The lemmas of this chapter deal with such approximations. They will be used later on.

4.2. LEMMA. Let ,8f+l and ,8'2+1 be (n + I)-simplices in Ef+1 and E'2+1 resp. We suppose n 2': 1. Let Yo be a point in the interior of f3'2+ 1 , and let z, z be two points in 0,8'2+1. Let the map : of3f+1 - t 0,8'2+1 be continuous, and let "I be a positive number. Then there exists a continuous map g: 0 f3f+ 1 - t 0,8'2+1 such that (i) for all x E 0,8r+ 1 I(x) - g(x)1 < "I,

(4.1)

and (ii) each of the equations g(x) = z,

g(x) =

z

(4.2)

has at most a finite number of solutions. PROOF. Let Rl be such a positive number that

(8,Bf+1) c B(Yo,Rt}.

1iOZ and ~ resp. for which R < {IY11,

(4.3)

Let Yl and fit be points on the rays I

liitl.

(4.4)

Then if e is a positive number, one sees by arguments quite similar to those used in subsection 3.9 that there is a barycentric subdivision Ki+1 of ,8i+1 and a map '"'(: of3i+1 -+ E'2+1, which is simplicial with respect to oKi+1, which satisfies

I(x) -,",((x)1 < e

(4.5)

for all x E of3i+1, and which has the following property: if UI, ..• ,UN are the vertices of oKn+1 and if VI = '"'((ud, ... ,VN = '"'((UN), then the points

YO,YI, Jil' Vt. ... , VN

(4.6)

are in general position. Moreover, if we choose c < dist(yo, 0,8'2+1),

(4.7)

ON SPHERES

156

then (4.8)

and each of the rays yOZ = YOY{ and ~ = YaY; intersects -y(8Kr+1) in at most a finite number of points. Finally, if y is one of these intersection points, then the equation -y(x) = y (4.9) has (because of the independence of the points (4.6)) exactly one solution x in 8Kr+1, and this solution lies in the interior of one of the n-simplices of 8Kr+1 (cf. condition (Cd on equation (3.21)). It follows from all this that if 1r is the central projection from Yo on 8/3'2+ 1 and if g(x) = 1r(-y(x)), (4.10) then 9 maps 8/3r+1 into 8/3'2+1 and is continuous since the convexity of /3'2+1 implies that 1r(Y) is continuous for y i= Yo (see the Notes).21 Moreover, x is a solution of one of the equations (4.2) if and only if x is a solution of (4.9) with a y as defined above. Thus equations (4.2) have at most a finite number of solutions. It remains to show that for 'YJ > 0 given c can be chosen also in such a way that the inequality (4.1) holds with 9 given by (4.10). Now since 1r(Y) is uniformly continuous on any closed bounded set not containing Yo, there is an c such that for all il E 8/3'2+1

(4.11) if

Iy - ill < c. (4.12) Since Q>(x) E 8/3'2+1 we may apply this with il = Q>(x) and y = -y(x) to see that for such c the inequality (4.5) implies (4.1) with 9 given by (4.10). 4.3. LEMMA. Let Sjl and Sr be spheres in Er+ 1 and E'2+1 resp. Let p and p be two points of Sr which are not antipodal. Let 'f/; be a continuous map Sjl -+ Sr and c > O. Then there exists a continuous map h: Sf -+ sr of the following properties: (i) 11'f/;(x) - h(x)11 < c (4.13) for all x E Sjl; (ii) the equations

h(x) = p,

hex) = p

(4.14)

have at most a finite number of solutions. This lemma is a corollary to the preceding . one. Indeed, let the simplex /3 1n + 1 be as in subsection 3.6, let, as in that sectIOn, Kl = K be the projection from the center of Sjl on 8/3r+1 , and let /3'2+ 1 and K? b~ defined in the corresponding

Sr.

Then from the contmUlty of Kl and K2, we see that way with respect to ¢ = K2'f/;K 11 is a continuous map 8/3i+ 1 -+ 8/3'2+1 and that to a given c > 0 there corresponds an 'YJ > 0 such that for all x E Sf

IIK2"lQ>Kl(X) - K2"lgK(X)11 < c

(4.15)

NOTES

if 9 is a continuous map 8{3~+1

--+

157

8{3'2+! satisfying for all ~ E 8{3~+1

Iit/>(~) - g(~)11

< 17·

(4.16)

We now choose for 9 a map 8{3~+! --+ 8{3'2+! which has the properties (i) and (ii) asserted in Lemma 4.2. Then (4.1), i.e., (4.16), and therefore (4.15) is satisfied. Thus (4.13) holds with h = /\,2"1 g/\'1. But also assertion (ii) of the present lemma is satisfied since x is a solution, e.g., of the first of the equations (4.14)' if and only if ~ = /\,1(X) is a solution of the first of the equations (4.2) with z = /\,2p.

§5. Notes to Chapter 7 Notes to §1. 1. Proof of Lemma 1.2. II s+ 1 is a translate of an (8 + 1)-dimensional subspace of En+! (Definition 1.5 in Chapter 6). It therefore follows from the well-known orthogonality properties of En+l that there exists a Co E lIS+! such that Co - C and x - Co are orthogonal for every x E II S+ 1 . Since x - c = x - Co + Co - c we see that IIx - cll 2 = IIx - coll 2 + IIco - cll 2 for all x E lIS+!. If in particular x E II s + 1 n S(c, r), then IIx - cll 2 = r2. Thus IIx - coll 2 = r2 -Ileo - cll 2 .

(1)

This equation for x has no solution if r < lleo - cll, and the unique solution x = Co if r = IIco - cll. If r > lleo - cll we see from (1) that lIS+! n sn(c,r) = II s + 1 n sn(co, (r2 -Ileo - cll 2)1/2). 2. Proof of Lemma 1.4. By definition a great circle S1 through N is the intersection of sn(c, r) with a two-dimensional plane II2 containing N and c. It therefore contains the antipode S of N. That no third point lies on the intersection of two (not identical) great circles follows from the fact that a rr 2 is uniquely determined by three independent points.

no,

3. With the notations of subsection 1.6 let Xo = to = and let t = O'(x) be the stereographic projection of x with pole no. Then the formulas (3) and (4) below hold, which obviously imply the asserted continuity:

= no + J.L(x -

= 4r2/lix -

(a)

t

no),

(b)

JJ

noll 2,

(3)

(a)

x=no+,X(t-no),

(b)

,X=4r2/IIt-noll2,

(4)

Proof of (3) and (4). It is obvious that (3a) and (4a) hold for some real J.L and 'x. To prove that these numbers have the values given by f3b) and (4b), we note that

(no - e, t - c) = (no - e, t - xo) + (no - c, Xo - c) = _r2, (5) since (no - e, t - xo) = O. Moreover, since IIno - ell 2 = IIx - ell 2 = r2, we see that 2(no - e,no - x) = 2(no - e,no - c) + 2(no - c,e - x) = IIno - cll 2 + 2(no - e,e - x) + lie - xll 2.

ON SPHERES

158

Thus

2(no - c, no - x) = Iino - c + c - xl1 2 = Iino - x11 2 .

(6)

To prove (3b) we subtract c from both members of (3a) and then multiply by the scalar no - c. We then see from (5) and (6) that

_r2 = r2 - (/-L/2)ll n o - x11 2 . This proves (3b). To prove (4b) we note that by (3)

(x - no,t - no) = /-Lllx - nol1 2 = 4r2. Therefore scalar multiplication of (4a) by t - no yields 4r2 = ).llt - no11 2 . This proves (4b). 4. As in subsection 1.7 of Chapter 6 let b1, b2, ... , bn be a base for the tangent space Ex. Suppose, moreover, that bo = c - to, bl , ... , In form a positive base for En+ 1 and that likewise bi, b~, ... ,b~ is a base for E'; and that bb = c to, bi, ... , b~ form a positive base for En+!. We have to show that bl ,.· . , bn and bi, . .. , b~ determine the same orientation of E';, i.e., (see subsection 1.22 of Chapter 6) that if n

bi =

L {3ijbj,

i = 1,2,3, . .. , n,

(7)

j=l

then the determinant ~ n of the {3ij is positive. If we enlarge (7) by adding bb = bo, then (again by subsection 1.22, Chapter 6) the determinant ~n+! of the enlarged system is positive since the bi and bi, i = 0,1, . .. , n, both form a positive base. But obviously ~ n = ~ n+!. 5. It is easy to see that an r01 with the properties assumed in Lemma 1.8 exists: assuming c = () let E2 be the 2-subspace of En+! containing bA and b~, and let bl be a point of E2 orthogonal to bA. Then obviously there exists an orthogonal map (rotation) of E2 onto itself which maps bA into b~ and whose determinant with respect to the base bA, bl is positive. If n = 1 we subject bl to the additional condition that bA, bl form a positive base for E2 = En+1. If n ~ 2 we choose points bl, i = 2,3, ... , n, such that the points bA, bl, b~, ... , b~ form an orthogonal base for En+l which is positive. We extend the map rl2 defined on E2 to En+1 by setting

rl2(bl) = bl,

i = 2,3, ... ,n.

(6)

The linear map determined by these additional conditions has the desired properties.

6. Proof of Lemma 1.8. With the notations of that lemma, let bl bl bl be an orthogonal base for En+l, which is positive. Then bl, b~, . . ~: bi'," b~in; l form a base for the tangent space Et 1 of sn(c r) t t1 orthogonal to bol = C - t 0' 0 , a o.

NOTES

Moreover by the definitions given in subsection 1.7 this base for Et1o is a positive one. On the other hand, since r12 is orthogonal and preserves the orientation, it follows that the b; = r12(bl)' i = 0,1, ... , n, form a positive orthogonal base for En+1. This again implies that r12(bl), r12(b~), ... , r12(b~) form a positive base for the tangent space Et2o to 8(c, r) at t5 = c - b5. This obviously proves the assertion (1.5). The equivalence of (1.6) with (1.5) follows from the definitions given in subsection 1.7.

n

1

7. Proof of Lemma 1.10. 0'01 is a continuous one-to-one map oH10 c To onto c T 1 , and in 1 E 0 1 , then the equation

(7) has one and only one solution to = to E 0 0. So d(O'ol. 0 0, h) any open set 0 satisfying to E 0 C 0 0 c To. Thus

= d(O'ol. 0, td

for

(8) for any open subset U of 8 n (c, r) for which

Xo =

0'0 1(to)

cUe Uo.

(9)

Consequently to prove the assertion (1.9) it will be sufficient to show that d(O'Ol,O'O(U),td = +1

(10)

for an arbitrary U satisfying the above conditions. Now let n 2 be a 2-plane containing c, no, and n1 (and therefore So and sd. Let 8 1 be the great circle 8 n (c, r) n n 2 (cf. subsection 1.3). We first consider the case that (11)

We then choose for U an open set which, in addition to (9), satisfies

un8 1 = 0.

(12)

Let 8J1: x = n(O), 0 ::; 0 ::; 1, be one of the segments of 8 1 which is bounded by no and n1 such that

n(O)

= no,

(13)

Let 8(0) be the antipode of n(O), let To be the tangent plane to sn(c, r) at s(O), and let tTl be the stereographic projection with pole n(O). Then by assumption (12) the stereographic projection tTo(x) is defined for every x E U and 0 E [0,1]. We denote the open subset O'o(U) of To by 00. For 0 E [0,1] we set 0'01

-1 .

= O'ltTo ,

(14)

160

ON SPHERES

Then

O"Ol(tO) = 0"10"0 1(to) = tt E T1 for to E O"o(U), 0"11(t1) = t1 E T1 for tt E O"l(U), (TOt}Ol = 00. (15) Now similar to the construction used in Note 5 we define a rotation ro of En+! with center c which keeps the orientation of En+1, which maps 81 into 80 and for which ro('rt) = To, while for an open set V1 C T1 and for an arbitrary point to E ro(Vt} C T1 (16) Now let rPo = (ro)-l To1 such that, by (15), rPo(Ot) = (ro)-10 0 c T 1. Since then T01 = rorPo, we see from subsection 2.15 in Chapter 6 and (15) that for to E 00 d( T01, 01, to) = d(rorPo, 0 1 , to) = d(ro, rP001, to) . d(rPo, 0 1 , td,

(17)

where t1 is an arbitrary point of rPo(Ot} = (ro)-10 0 c T 1. We now see from (16) with V1 = rPo(Ot) and from (17) that (18) Now let u be an arbitrary point of the open subset U of sn+1(c,r) (cf. (12)), and let t1(0) = (ro)-lO"o(u). Then t1(0) E (ro)-10 0 = (ro)-l To1 (Od = rPo(Od by (15) and definition of rPo, which shows that the right member of (18) equals d( rPo, 0 1 , tt (0)) and also that h (0) ~ rPo (aO t) since rPo is continuous and one-toone while 0 1 is open. Therefore, by Theorem 5.4 in Chapter 3, d( rPO, 0 1 , t1 (0)) is independent of 0 and has for all 0 E [0,1] the value +1 since rP1 is the identity on 0 1 . It now follows from (18) that d(T01,Ol,tO) = +1. By (14) this proves the assertion (1.9) since the first part of (1.9) is obvious. Suppose now the assumption (11) is not satisfied. In this case, using the notations of (7), we choose a positive number P1 such that

(20) if (21) On the other hand to is the unique solution of (7). Therefore there exists a position Po such that if

to E B(to, Po). (22) Thus (20) is true with t1 = O"01(tO) = 0"10"0 1(to) }f to satisfies (22). But since n ~ 2 the condition (22) is satisfied for so~e to = to for which 0"0 1(to) ~ Sf, i.e., for which (11) is true with to replaced by to. Therefore d(O"Ol, Oo,td = 1

(23) for t1 = 0"01 (to). The assertion (1.9) follows now from (20) and (23) since (20) holds for t1 = t 1.

NOTES

161

8. Proof of the existence of a U1 satisfying (1.14). Since f is continuous the set C 1 = f-1(yO) is closed and since, by (1.12), n1 ~ Cl, there exists a positive 151 such that Ilnl - cll > 151 for all c E C 1 • (24) Moreover, since n2 i- Yo and since f is uniformly continuous on Cl, there is a positive 152 such that Ilf(x) - f(e)11 < IIn2 - yoll/2 (25) for all couples x, ewith eE C1, x E Sr(el, rd for which that if 15 = min( 151 ,152 ) and

Ilx - ell < 152.

U B(e, 15),

U1 =

We claim

(26)

cEC 1

then U1 satisfies (1.14). Indeed part (a) of this assertion follows from (24), and part (b) is obvious from (26). To prove part (c) we note that for x E U1 we can, on account of (26), find a eE C 1 for which !Ix - ell < 15 < 152 and for which therefore (25) holds. Since f(e) = Yo we see that

IIn2 - f(x)11 2 IIn2 - f(e)II-llf(e) - f(x)11 = IIn2 - yoll-Ilf(e) - f(x)11 2 IIn2 - yoll/2 > O. 9. Proof of (1.19). Suppose that (1.19) is not true, i.e., that (1.16) holds for some tl E a0 1 = O'daud. Then we know that x = 0'1 1 (td E aUl satisfies (1.17) and, therefore, (1.18). Thus x E (aUd n C1 . But this intersection is empty as follows from part (b) of (1.14) since U1 is open and C 1 is closed. 10. Proof of the independence of the degree d(/t, fh, 0'2 (yo» of the choice of n1 and U1. Let n~ be a point of (Cl' rd satisfying

Sr

(27)

Let n~ be a point of Sr(C2, T2) different from Yo, and let Uf be an open subset of Sr(ct,Td which satisfies (1.14) with nl,n2,Ul replaced by n~,n~,Uf resp. For i = 1,2 let 8~ be the antipode of n~, let TI be the tangent plane to Sr(Ci, Ti) at 8~, let O'~ be the stereographic projection with pole n~, and let o~ = O'~ (Uf). Finally let if be the map o~ ---+ T2 defined by fHt~)

= 0';f(O'D- 1 (t~),

t~ E o~

= O'~ (Un.

(28)

We have to prove that (29) Clearly the intersection Uf' = U1 n Uf satisfies (1.14) with U1 replaced by Uf'· In particular part (b) of that relation together with (1.13) shows that

d(!t,0'1(Ut},0'2(YO»

= d(/t,0'1(Un,0'2(YO)).

(30)

ON SPHERES

162

In the same way one sees that

d(fL (1~ (uD, (1~ (yo)) =

d(f~, (1~ (Un, (1~ (yo)).

(31)

It follows from (30) and (31) that for the proof of (29) it will be sufficient to show that that relation holds with U~ and U1 replaced by U~/. In other words we may assume that (32) Now (1~(Ud = ((1~,(111)(11(ud,

(33)

and by (28) and subsection 1.15 f~

= ((1~(121)(121(111(11((1~)-1 = (1~(121 h((11((1~)-1).

(34)

Now from (32)-(34) together with part (i) of subsection 2.15 in Chapter 6 we see that d(f~, (1~ (Ud, (1~(yo)) = d((1~(121 h, (11 (ud, (1~ (yo)) . d((11 ((1~) -1 , (1~ (Ud,

td,

where t1 E (11 (Ut}. But the second factor of the right member equals +1 by Lemma 1.10. Thus

d(fL (1~ (Ut), (1~(yo)) = d((1~(121 h, (11 (Ut), (1~ (Yo)).

(35)

Using part (ii) of subsection 2.15 in Chapter 6 we see that d((1~(121 h,(11(Ut},(1~(yo)) = d((1~(121, h(11U1,(1~(yO))· d(h,(11U1, t2)'

(36)

where t2 is an arbitrary point in h(1lU1 = (121(U). Since Yo E f(U) we may choose t2 = (12(YO). On the other hand, again by Lemma 1.10, the first factor at the right member of (36) equals +1 since (1~(yo) E (1~f(Ut) = (1~(121(f1(11Ud. This together with (36), (35), and (32) proves the assertion (29). Notes to §2.

11. Proof of Lemma 2.4. By Definition 1.13 it will be sufficient to prove that

d(ft, SI' YO) is independent of t for some Yo E S'f.. For this again it will, by the Borel theorem, be sufficient to show that if to E [0,1] then there exists a positive 8 = 8(to) such that (37) if t E [0,1]. It - tol < 8, (38) We consider the three cases I, II, III defined in subsection 1.11. Case I. Yo ¢. fto(Sr) for some Yo E Sr· Then the right member of (37) equals zero (e.g. by Lemma 2.1), and YO has a po~itive distance 2c from fto(SI) since the latter set is closed. We now choose 8 10 such a way that for all x E Sr, Ilft(x) - fto(x)11 < c if (38) is satisfied. For such t

Ilyo - ft(x)11 ~ Ilyo - fto(x)II-llfto(x) - h(x)1I ~ c > 0

NOTES

for all x E Sf. This shows that It (Sf) is a proper subset of Sf. Therefore d(ft, Sf, Yo) = 0 by Lemma 2.1. This proves (37), both members of that equality being zero. Case II. Ito (Sf) is a single point Yl of Sf· Then, for any point Yo E Sf different from Yl, we are in Case I. Case III. In this case there exists an nl E Sf such that

(39) Let

n2

be a point of Sf such that (40)

and for t E [0,1] let

Ct = {x

E

Sf I It(x) = yo}.

(41)

Then we know that there exists an open set U1 C Sf such that (1.14) holds with C1 replaced by Cto and I by Ito, and that

d(fto,Sf,yo) = d(0'2It o O'1 1,U1,0'2(YO))

(42)

(see (1.20) and (1.15)). On the other hand, we see from (39), (40), from part (c) of (1.14) (with I replaced by Ito), and from the uniform continuity of It on Sf x [0,1] that there exists a 151 > 0 such that (43) if

It - tol < 151 .

(44)

Moreover, as will be proved below, 15 1 can be chosen in such a way that (44) implies the inclusion C t CUI. (45) But from (43) and (45) we see that (1.14) is true if I is replaced by It and C 1 by Ct for t satisfying (44). It follows that for such t

d(ft,Sr,yo) = d(0'2!tO'1 1,0'1(Ud,0'2(YO)).

(46)

Now since 0'21tO'1 1 for fixed t is a "plane" map, it follows from Theorem 2.9 of Chapter 6 that there exists a 152 > 0 such that for It - to I < 152 the right members of (46) and (42) are equal. Thus these two equalities show that (38) with 6 = min(61, 62 ) implies (37). It remains to prove that 151 can be chosen in such a way that (44) implies (45). Since Sf is a compact subset of En+l, this assertion is a consequence of the following lemma: let ~ be a compact subset 01 Ei+; and I a continuous map ~ -+ ~+1. Let Yo be a point of E2+1 and CU) = {x E ~ I f(x) = yo}. Let U1 be an open bounded subset of Ei+1 containing CU). Then there exists a positive 'YJ 01 the following property: if the map g: ~ -+ E2+1 satisfies

Ilf(x) - g(x)1I < 'YJ

for all x E ~,

(47)

ON SPHERES

164

then (48)

where C(g)

{x E Sf I g(x) = yo}. Suppose there is no such 'r/. Then there exists a sequence 'r/1,'r/2, ... of positive numbers converging to zero and continuous maps gv such that for all xEL: (49) Ilgv(x) - f(x)11 < 'r/v, =

PROOF.

while there exist points

(50) for which

Xv ¢:. UI .

(51)

Since L: is compact we may assume that lim Xv = Xo,

(52)

v--+oo

where Xo E L:. Now Yo = gv(xv) by (50), and therefore

Ilf(xo) - yoll = Ilf(xo) - gv(xv)11 -s; Ilf(xo) - f(xv)11

+ Ilf(x v ) - gv(xv)ll·

Here the two terms of the right member tend to zero as v -> 00, the first one by (52) and the second one by (49). It follows that f(xo) = Yo, i.e., that

Xo E CU)

CUI.

(53)

Now UI is open. Therefore (53) together with (52) implies that Xv E U I for v big enough. But this contradicts (51). 12. Proof of (2.8). Since (2.6) does not hold, there exists an Xo E Sf such

that

(54) We now verify that for such Xo the assertions (a)-(d) of (2.8) are true: if (a) were not true, then f(nJ) = Yo, and, by (2.5) and (2.7), n3 = gf(nJ) = g(yo) = gf(xo), which contradicts (54). Similarly, if (b) were not true, then g(n2) = zo and n3 = gf(nJ) = zo = gf(xo), which again contradicts (54). For the proof of (c) let X E (gf)-I(ZO). Then gf(x) = Zo, i.e., f(x) E g-l(ZO), which proves (c). Finally, for the proof of (d) let x E f-l(yO)' Then f(x) = Yo and, by (2.7),

gf(x)

= g(yo) = Zo,

i.e., x E (gf)-l(ZO).

13. Proof of Lemma 2.8. The sphere Sr is defined for t = 0 and t = 1. We extend this definition to all t E [0, 1] by denoting by Sr the sphere with center c and radius Tt = (1 - t)ro + trl. By 7rt we denote the central projection from c of S{f on Sr. Now let ao be an arbitrary point on S{f, let at = 7l't(ao) E Sr, and let nt be its antipode. We will show that d(7l'f,S{f,at) = +1.

(55)

NOTES

For t = 1 this equality is, by Definition 1.13, equivalent to our assertion (2.25). In order to prove (55) we consider a positive number 0 < min(rO,rl). If Xt = 7rt(x) for x E So, we see that Ilat-ntll = 2rt > 20. Now if Uo = {x E Sollix-aoll < 8}, then Ilx - noll ~ Ilao - noll - Ilx - aoll = 2ro - Ilx - aoll > 0 (56) for x E Uo. We claim that then (1.14) is satisfied with by no, Uo resp., i.e., that (a)

dist(na, Uo) > 0;

f = 7rt and nl, U1 replaced (c)

nt rf- 7rt(Uo).

(57)

Indeed, (a) follows directly from (56), and (b) from the fact that 7r t 1 (at) = ao E Uo. To prove (c) we note that 7rt(x) is given by 7rt(x) - C = (x - ch(ro)-l, and therefore 7rt (x) - nt = 7rt (x) - 7rt (no) = (x - no)rt (ro) -1. Consequently, by (56), for x E Ua II7rt(x) - ntll ~ 8rt(ro)-1 ~ 8 min(ro, rdro1. This proves (c). Now in subsection 1.11 we showed that (1.14) implies (1.20). Thus by (57) (58) where, for each t E [0,1]' lTt is the stereographic projection of Sr with pole nt on the tangent plane Tt to Sr at the point at. But the degree at the right of (58) is-by remark (c) in subsection 1. 7-equal to the degree of the corresponding mapping between the tangent spaces Eo and E t to So and Sr at ao and at resp. Now E t is the translate of Tt which maps the point at into O. Therefore taking into account the definition of the orientation of tangent spaces we see that E t = Eo (as oriented spaces) since the planes Tt are parallel to To. Thus

(59) where 0 0 = {~ E Eo I ~ + ao E lToUo} and where 4>t is the map 0 0 --+ Eo given by 4>t(~) - at = lTt7rtlTo1(e + ao). But, by the continuity property of the degree, the right member of (59) is independent of t and therefore equals + 1 since 4>0 is the identity on 0 0 . The assertion (55) now follows from (59) and (58).

Notes to §3. 14. Proof of Lemma 3.3. It will be sufficient to show that if Cl is a given point in E2+1 satisfying (3.1), then there exists a 8 > 0 such that v(f(Sf),c) = v(f(Sf),cd

(60)

IC- Cll
(61)

if Now let T2 be a positive number satisfying (3.2). Then obviously B( c, r2!2) C B( CI, T2)

(62)

ON SPHERES

166

if c satisfies (61) with (63) Now it was already established in subsection 3.1 that the right member of (3.3) does not depend on the particular choice of the (small enough) radius of Sf. Therefore (64) where hand,

1l'C l

is the central projection with center CIon Sf(CI, T2/2). On the other

v(f(Sl), C)

=

d(1l'cl,Sr,Sf(c,T2/2)).

(65)

We have to prove that for Ilc - CIII small enough the right members of (64) and (65) are equal. For this purpose we introduce the translation Tc(Y) = Y + CI - c. Then TcSf(c,T2/2) = Sf(Cl,T2/2), and Tc1l'cl maps Sr on Sf(CI,T2/2). From the product theorem of subsection 2.5 in conjunction with Lemma 2.9 we see that d(Tc1l'cl, Sr, Sr(Cl, T2/2)) = d(1l'e!, Sr, Sr(c, T2/2)) . d{ Tc, Sr(c, T2/2), Sr(CI, T2/2))

(66)

= d(1l'e!, Sr, Sr(c, T2/2)).

This holds if 6 and c satisfy (63) and (61) resp. We will show below that {) can be chosen in such a way that (61) implies that also

d(TC1l'e!, Sr, Sr(Cl,T2/2)) = d(1l'cJ,Sr,Sr(Cl,T2/2)).

(67)

But (66) and (67) together show that the right members of (64) and (65) are equal as we wanted to prove. Finally to verify (67) (for Ic - cil small enough) we note that Tc1l'e! and 1l'cJ both map Sr into Sf(cl, T2/2). Therefore by Theorem 2.6 it will for the proof of (67) be sufficient to show that Tc1l'e!(X) tends to 1l'Cl f(x) uniformly for x E Sr as c -> CI. To prove this assertion we note that for Y E I(Sr), 1l'c(y) = c + (y - c)(211Y - cll)-IT2, and therefore Tc1l'c(Y) = CI + (y - c)(211Y - cll)-IT2' Thus

Tc1l'c(Y) -

1l'C l

(y)

=

C Y - CI ) T2 ( lIyY -- cil - Ily - cd 2'

This proves our assertion since (f(x) - c) '1If{x) - cll-l, on account of (3.2), is uniformly continuous on Sr X {Ie - cil ::; T2/ 2}. 15. Proof that

d{K, Br+ l , 11) = +1

(68)

(cf. subsection 3.6). It is sufficient to prove (68) with 11 = "'(. For t E [0,1] and x E Br+1, let Kt{X) be defined by Kt(X) - "'( = (1- t)(K(X) - "'() + t(x - "'(). Then by (3.6): (69)

NOTES

If 8 = dist(8/3r+1, ,), then IIK(x) - ,II ~ 8 and the bracket in (69) is not smaller than min(8rll, 1), and it follows from (69) that Kt(X) i=- , for x E S~+ I = 8 Br+ I. Therefore d( Kt, Br+ I, ,) is independent of t and thus equals d(KI,Br+ I ,,) = +1 since, by (69), KI(X) = x.

16. Proof of (3.14). Since 7rB(O, R 2 ) = 7r8B(O, R 2 ) (as point sets), the assertion is a consequence of the following lemma: for i = 1,2 let ~i be a sphere in En+ I of radius ri and center Yi. Suppose that (70) (i.e., the closures of the balls bounded by ~f and ~2 are disjoint). Let 7r be the central projection from Yl of ~2 on I;f· Then 7r~2 is a proper subset of ~f. PROOF. Let pn be the n-plane through YI which is orthogonal to YI - Y2. Let HI and H2 denote the two open half spaces into which pn decomposes En+l (see subsection 1.24 of Chapter 6). By (70) the sphere ~2 lies in one of them, say in H 2. Since H 2 is convex, all rays Y1X, x E ~2' lie in H 2. Thus no point of 7r~2 lies in HI. But obviously ~f contains points in HI. 17. That the right member of (3.9) (with CI replaced by y(t)) is constant in a t-interval whose elements satisfy (3.20) follows from Theorem 2.9 of Chapter 6. The corresponding assertion for the left member follows from Lemma 3.3, since f(S],) = g(8!3 n+ 1 ). 18. Proof of the properties (Cd-(C 3 ) in subsection 3.9. The proof will be based on the following properties (C 4 ) and (C 5 ). (C 4 ) Let af and a!J be two different n-simplices of 8Kn+l, and let r]' = g(af) and r2 = g(a~). Suppose that

(71) is not empty. Then dim rf2 :::; n - 1,

(72)

where dim r]'2 denotes the dimension of the plane spanned by rI'2. (C 5 ) y(l) does not lie in a simplex rffl of g(8Kn+1) of dimension m :::; n - 1 if I is as in the sentence following (3.20). Proof of (C 4 ). Let bo, bl , ... , bn be the vertices of af, and Co, Cl, ... , Cn be those of a~. Since af and a!J are different, at least one of the vertices of a!J, say Co, is different from all vertices of af· Then the n + 2 points = g(co), /30 = g(bo),·.·,!3n = g(bn ) are in general position since they form a subset of the points (3.20). They therefore span the (n + I)-space E:;+1. We thus see that if III and II2 are the planes spanned by rI' and r2 resp., then their hull H (see Definition 1.7 of Chapter 6) equals E~+l. Suppose now the assertion (72) not to be true. Then dim( r]' n r2) = n, and therefore dim(111 n 112) = n. Thus

,0

dim(IIl

n II 2) + dim H = 2n + 1 > 2n = dim III + dim 112

168

ON SPHERES

in contradiction to Lemma 1.8 of Chapter 6. This proves (C 4 ). Proof of (C 5 ). Suppose the assertion to be wrong and yell E Tm with m ::; n - 1. Then the plane rr T spanned by T m and the ray YaY! is of dimension r ::; m

+ 1 ::; n.

(73)

On the other hand, if va, Vb ... , Vm are the vertices of Tm, then the m+3 ::; n+2 points Yo, Yl, Va, ... , Vm are in general position, being a subset of the points (3.20). Therefore they do not lie in a plane of dimension < m + 2. But these points lie in the plane ITT whose dimension, by (73), is ::; m + 1 < m + 2. This contradiction proves (C 5 ). We now turn to the proofs of assertions (Cd-(C 3 ) of the text. Proof of (Cd. Suppose the equation (3.21) has two roots Xl -I- X2 in aKn+l. For i = 1, 2 let Xi E O'f, where O'f is an n-simplex of aKn+l. Since the restriction of 9 to a single simplex of aKn+l is one-to-one, the simplices O'r and 0'2 are different. Since Xl and X2 satisfy (3.21), the point yell lies in the intersection of Tl = g(O'r) and T2 = g(0'2)· Thus this intersection is not empty and therefore (72) holds by (C 4 ). But this contradicts (C 5 ) since yell is a point of Tf n T2. Proof of (C 2). Let yell E Tn = g(O'n) where un is an n-simplex of aKn+l. Suppose now (C2) not to be true. Then yell E aTn, and yell E Tm, where Tm is an m-simplex of g(aKn+1) of dimension m ::; n - 1. This again contradicts (C 5 ). Proof of (C 3). Suppose yay; has two intersections y(td and y(t2), tl -I- t2, with Tn. Then YOYi lies in the n-plane rrn spanned by Tn. Thus if Va, VI, ... , Vn are the vertices of Tn, then the n + 2 points Yo, Va, ... , Vn lie in the plane rrn of dimension n < n + 1. This contradicts the fact that these points as a subset of the points (3.20) are in general position.

19. Proof of (3.42). Let Uo: = K-lO'~. Then Uo: is a neighborhood of ~o: which contains no other root of (3.40). Therefore j(rrt f, ~o:) = d(rrt f, Uo:, y(t*)). Now if Tf is the tangent plane at ~o: to from its center on Tf, then

Sf

(73a)

and So: the central projection of Sf

d(rrt f, Uo:, y(t*)) = d(rrt fs;;l, flo:,y(t*)), where flo: = so:Uo:. But fs~l = gKS~l. So j(rrtf,~o:) = d(rrtgKs~l,flo:,y(t*)). Since KS;;lflo: = KUo: = O'~, we see from the product theorem that j(rrtf, ~o:) = d(rrtg,O'~,y(t*)). d(KS;;l, flo:, x), where X is a fixed point of O'~. This proves (3.42) since d(KS~l,flo:,x) = +1. (Note that by (2.2) the right member of (73a) is defined as the degree of a map flo: - T!'.) 20. Proof of (3.45). Since t~o and tto are both less than to: for 0: ~ 0:0 + 1 (cf. (3.24)), the equality (3.45) is a consequence of the following assertion: with the notations used in subsection 3.11 the degree d(rrtg,O'~,y(t*)) (74)

NOTES

is independent of t if

o < t < tOt.

(75)

This assertion in turn is a consequence of the following one: let Q~ be the nplane in E~+ 1 which contains the simplex r;; = g( a;;) = (WI, W2, ... , Wn+d. Let r;;+1 = g(a;;+I) = (wo,wl, ... ,wn+d. Let H+ and H- be the two halfspaces of E~+1 with respect to Q~, where we suppose that yet") (and therefore yet)) lies in H;;. As in subsection 3.11 let T: be the tangent plane to S:r(y( t)) at yet"). Finally let COt equal +1 or -1 according to whether r;;+1 is a positive or negative simplex. We distinguish two cases:

(i)

Wo

E

H+,

(ii)

Wo E H-,

(76)

and assert that

d(

7r

t

n (t")) = { - COt g,aOt'y + COt

in case (i), . (.. ) III case 11,

(77)

if t satisfies (75). We present the proof of (77) for case (i) with COt = +1: we note that the simplex (y(t),7rtr;;) = (y(t),7rtwl, ... ,7rtwn+d has the same orientation as the simplex (y(t),r;;) = (y(t),wl, ... ,wn+d. But since yet) E H- and Wo E H+, it follows from subsection 1.25 of Chapter 6 that the simplices (y(t),r;;) and (wo, r;;) = 7;;+1 have different orientations, i.e., by the assumption COt = + 1, it follows that the simplex (y( t), 7;;), and therefore the simplex (y( t), 7rtr;;), is negative. But by subsection 1.7 the negativity of (y(t), 7rtr;;) implies that the simplex 7rtr;; is a negative simplex of the tangent plane T:. Thus d( 7rt, r;;, y( t")) = -1. Since r;; = g(a;;), this proves (77) under the assumptions considered. The remaining assertions follow similarly.

Notes to §4. 21. THEOREM. Let V be a bounded open convex set in a Banach space E, let Yo be a point in V, and let y be a point in E different from Yo. Then (i) the ray yoy intersects the boundaryaV of V in exactly one point y = y(y), and (ii) y(y) is continuous at every point y = Yl ::I Yo· PROOF. Assertion (i) was already proved in Note 2 to Chapter 4. For the

proof of (ii) we note that for y

y=

::I Yo Yo

+ (y -

Yo

) Ily - yoll Ily - yoll·

(78)

If we set

t( ) z =

Ilzll Ilyo + z - yoll'

z::l 0,

(79)

ON SPHERES

170

then by (78) _ Y = Yo

To prove the continuity of y(y) at Y then

Y - Yo

(80)

+ t( Y -Yo )

= Yl -:I Yo

we note that if IIYI - yoll

= 281 , (81)

if

(82) Now t(y - Yo) is bounded away from zero for Y satisfying (82). Indeed, for such y, by (81) Ily - Yo II i: / ( ) t Y - Yo = Ily _ yoll 2 VI TO, where TO is the positive distance of Yo from av. It now follows from (80) that for our continuity proof it will be sufficient to show that t(y - Yo) is continuous at y = Yl. We claim next that this continuity is a consequence of the following "triangular inequality" to be proved below:

(83) where Y2 denotes a point y satisfying (82). Indeed from (83), from the inequality obtained from (83) by interchanging Yl and Y2, and from the definition (79) of t(z), we see that It (Y2 - Yo ) - t (Yl - Yo ) I ~ t (Y2 - Yl ) = II

II Y2 - Yl II + II· Y2 - Yl Yo - Yo

(84)

But Y2 - Yl + Yo E av since we obviously may assume Y2 -:I Yl. Thus the denominator of the right member of (84) is not smaller than TO = dist(yo, av), and (84) implies the continuity of t(y - Yo) at Y = YI. It remains to prove the inequality (83). We follow the proof given by Minkowski in [40, pp. 12, 13]. We note first that for a > 0 t(a(y - Yo)) = at(y - yo),

y

-:I Yo,

(85)

since obviously Yo + a(y - YO) = Yo + (y - yo)· Suppose now that (83) is not true for some y = Y2 satisfying (82). Then there exists a number r for which t(Y2 - Yo) > r > t(Y2 - yd

+ t(YI

- yo).

To arrive at a contradiction, we consider points Y3, Y4, Y5, Y6 defined by Y3 Y5

= Yo + (Yl

= Yo + (Y3 -

- Yo)/r,

Yo)/(l - 1),

Y4

= Y3 + (Y2 - yt}/r, Y6 = Yo + (Y4 - Y3)h,

where 1

= t(Y3 - YO) + t(Y4 - Y3) .

(86)

NOTES

171

From these definitions, from (86), and from (85) it follows that

t(Y3 - Yo)

=

t(Yl - Yo)/r < 1,

+ t(Y4 - Y3) t(Y3 - Yo) + t(Y4 - Y3)

t(Y4 - Yo)

=

t(Y2 - yo)/r > 1,

+ t(Y2 (1/r)[t(Yl - Yo) + t(Y2 -

t(yS - Yo) = t(Y3 - Yo)

= (1/r)[t(Yl - Yo)

yt}l < 1,

t(Y6 - Yo) =

=

yt}l < 1.

Now, by definition, t(y - Yo) = lIy - yoll . (11v - Yo)II)-l. Thus t(y - Yo) > 1 if and only if Y is exterior to V, and t(y - Yo) < 1 if and only if Y is a point of the open convex set V. In particular we see from the above inequalities that Ys and Y6 belong to V and Y4 does not. This however contradicts the convexity of V since Y4 is a convex combination of Ys and Y6, viz., Y4 = (1 - I)Ys + lY6 with 0<1<1.

CHAPTER

8

Some Extension and Homotopy Theorems §1. An extension and a homotopy theorem 1.1. Let E be a Banach space, let 0 be a bounded open connected set in E, let Yo be a point of E, and let f be a g.L.-S. map (see subsection 6.2 of Chapter 3) which maps 80 in E - Yo. Then by Definition 1.5 in Chapter 4 of the winding number (1.1) for every g.-L.8. extension the equation

j

of f to

n.

Consequently if u(f(80), Yo)

j(x) = Yo

#- 0,

then ( 1.2)

has at least one solution x E O. Thus for the existence of a g.L.-S. extension of f mapping into E - Yo, the condition

n

u(f(80), Yo) = 0

(1.3)

is necessary. The following theorem asserts that this condition is also sufficient. 1.2. THEOREM. Let E, 0, Yo, and f be as in subsection 1.1, and suppose that (1.3) is satisfied. Then (i) there exists a g.L.-5. extension of f which maps into E - Yo; (ii) if f is L.-5., then there exists an L.-5. extension of the above property.

n

In the finite-dimensional case this theorem is a special case of H. Hopf's classical extension theorem (see [2, p. 499ff.] and the Hopf papers quoted on p. 621 of [2], cf. also [29, p. 142ff.]). For this reason we refer to Theorem 1.2 as the Hopf extension theorem. 1.3. Let f and 9 be g.L.-S. maps 80 -> E - Yo· We know from Theorem 1.7 of Chapter 4 that if f and 9 are g.L.-S. homotopic, then

u(f(80), yo) = u(g(80), Yo).

(1.4)

Thus this equality is necessary for f and 9 to b~ g.L.-S. homotopic. The following theorem states that this equality is also suffiCIent for the g.L.-S. homotopy of f and 9 if 0 is convex. 172

AN EXTENSION AND HOMOTOPY THEOREM

173

1.4. THEOREM. Let 0 be an open bounded convex set 01 the Banach space E. Let I and 9 be y.L.-S. maps ao -+ E -- Yo. Suppose that (1.4) holds. Then I and 9 are y.L.-S. homotopic in E - Yo·

We refer to this theorem as the Hopf-Krasnoselskil homotopy theorem since in the finite-dimensional case it is a special case of the Hopf theorem quoted in subsection 1.2 and since as already mentioned in the introduction Krasnoselskil was the first to prove the theorem for L.-S. maps in arbitrary Banach spaces. 1.5. We start the proof of Theorems 1.2 and 1.4 by considering the finitedimensional case. For i = 1,2 let Ei+ 1 be a Banach space of finite dimension n + 1, let Oi be a point of Ei+ 1, let Bi+ 1 be a ball with center Oi, and let Sr = aBi+1. Let I be a continuous map Sf -+ S'2 satisfying (1.5) Under this assumption we will prove the following assertions

hn

and

en + 1

for

n = 0,1, ....

1.6. Assertion hn. I is homotopic to a continuous map It which maps Sf into a single point of S'2. 1. 7. Assertion en +1. I can be extended to a continuous map Br+1 -+ S'2. Moreover we will prove 1.8. Assertion e~+l. If I is a continuous map Sf -+ E~+1-02 which satisfies (1.5), then I can be extended to a continuous map Br+ 1 -+ E~+l - O2. For the proof of these assertions the following lemma is basic. 1.9.

LEMMA.

en +1 implies

Let n 2: O. Then (a) h n and en +1 are equivalent, and (b)

e~+l.

(a) Suppose h n is true. Then there exists a continuous I(x, t): SI n X [0,1] -+ S:S such that f(x,O) = f(x) and such that f(x, 1) maps S1" into a single point Y2 of S:s. We now define 1: Bf+1 -+ S:S as follows: we set f (fh) = Y2; if ~ E Bf+1 - O!, then the ray ~ intersects Sr in a unique point x = x( e) and PROOF.

e= (1 -

t)x + tOl,

(1.6)

where t = t(~) is a unique value in [0,1]. We then set l(e) = I(x,t) to obtain the desired extension. Now Suppose en +1 is true. Then there exists a continuous map 7: Bf+l -+ S'2 such that lex) = f(x) for x E Sf. We define a map f(x, t): Sr x [0,1]-+ S'2 as follows: we set f(x, 1) = 1(Od for all xESI'. For 0 :5 t < 1 and XESI' the point (1.6) lies in Bf+1 but is different from 01. We then set I(x, t) = 1( (1- t)x + tOd to obtain the desired homotopy. Proof of (b). Since by assumption 02 rt f(Sr) there exists a ball B~+1 with center (h whose closure does not intersect !(Sr)· Let k be the central projection from (}2 of f(SI') on S'2 = aB~+1, and let 4>(x) = kf(x). Then 4> is a continuous map SI' -+ S'2, and by Definition 3.2 and Theorem 3.4, both in Chapter 7,

u(J(SI'), ( 2) = v(J(Sf), ( 2) = d(4), Sf, S~)

= v(4)(Sf), ( 2 ) =

u(4)(Sf), ( 2 ).

174

SOME EXTENSION AND HOMOTOPY THEOREMS

Thus u{4>{Sr),(l2) = 0 by assumption (1.5), and the assumed validity of en + 1 implies the existence of a continuous extension ¢ of 4> which maps Br+1 into S2. To obtain an extension] of I of the desired properties, we set

r{x) = 11/{x) - (1211 for x E Sr· Since for I{x) :f:. (12 for these x there exists a, (3 such that 0 < a ~ r{x) ~ (3 and since the interval [a, (3] is convex, it follows from §1.31 that r{x) can be extended to a continuous function r{x) which is defined in all of Elt+1 and satisfies the inequality 0 < a ~ r{x) ~ (3. We now define ](x) for each x E Br+ 1 as that ~

point on the ray (l2¢{X) whose distance from (12 has the positive value r{x). 1.10. We turn to the proof of assertions hn, en+1 , e~+1, n ~ o. Prool 01 hO• From the definition of the winding number and by definition of the degree d(f, Sp, S~) given in subsection 1.11 of Chapter 7 it follows that u(f{Sp), (12) = d(f, Sp, S~) and that the right member of this equality equals zero if and only if I maps Sp into one of the two points of which S~ consists. Thus in this case (1.5) implies that I maps Sp into a single point of S~. Prool 01 hI. We refer to subsection 1.6 of Chapter 4 and Note 1 to Chapter 4. If rl and r2 are the radii of Sf and Si resp. and if 4>1 and 4>2 are polar angles with the poles (II and (12 resp., then the argument given in Step II of that note shows that the map I: Sf - Si is homotopic to a map I: (rl, 4>d - (r2,4>2) where 4>2 = 4>1 multiplied by an integer whose value, by (1.13) of Chapter 4, equals u(f(Sr), (12). Thus, by assumption (1.5), 10 maps Sr into the single point (r2, 0) of S2. Prool 01 hn lor n > 1:Suppose h k - l has been proved for some integer k in the interval 2 ~ k ~ n. We will show that then hk is true. This will prove hn since hI has been established. Let I be a continuous map Sf - Sf which satisfies (1.5) with n replaced by k. We have to show that I is homotopic to a map h which maps Sf into a single point of Sf. We note first that it is sufficient to show that I is homotopic to a continuous map h which maps Sf into a proper subset of Sf. Indeed, for such map there exists a point Yo of Sf which is not contained in h(Sf). Let no be the antipode to Yo, and let h be the map which maps Sr into no. Then for no x E Sf are the points h(x) and h(x) antipodal. But this implies that hand h are homotopic (cf. the proof of Theorem 2.6 in Chapter 7). It follows that I is homotopic to h if it is homotopic to h. We note next that if in Lemma 4.3 of Chapter 7 with t/J = I and n = k we choose c less than the diameter of Sf, then (4.13) in Chapter 7 shows that for no x E Sf are the points I{x) and h{x) antipodal. As pointed out above, this implies that h is homotopic to I. Therefore for our proof it is sufficient to assume that I has the property (ii) asserted for h in that lem~a. In other words, using different notation, we may assume that there are two different but not antipodal points Yo and n2 on Sf which have the property that each of the equations (a) I(x) = YO, (b) I(x) = n2 (1.7) has at most a finite number of solutions.

AN EXTENSION AND HOMOTOPY THEOREM

175

We will show that f is homotopic to a map It for which at least one of the points Yo, n2 does not lie in f(Sf), thus proving that the latter set is a proper subset of S~. Now if one of the equations (1.7) has no roots, this is obvious; we simply have to take It = f. Suppose now that each of the equations has roots. In this case we will show that f is homotopic to a map It: Sf --+ S~ for which

Yo

rt. It(Sf)·

(1.8)

Then let Xl. X2, ... , Xr be the points of f- 1(Yo), and n}, n~, . .. , nj be the points of f-1(n2). Let U1 be an open subset of Sf which contains the finite set f-1(yO) and whose closure is disjoint from the finite set f-1(n2). Such U1 satisfies (1.14) in Chapter 7 with n1 = nl and C 1 = f-1(yO). Therefore if the tangent planes T 1,T2, the stereographic projections 0"1,0"2, and 0 1 are defined as in the lines directly following (1.14) in Chapter 7, we see from Definition 1.13 and from (1.20), both also in Chapter 7, that

dU, Sf, S~)

=

dU, Sf, Yo)

=

d(r, 01. Yo),

(1.9)

where r(~) = 0"2fO"11(~),

If Tlo

(1.10)

0 1 = 0"1 (Ud,

= 0"2 (Yo), then the roots of the equation r(~) = Tlo

(1.11)

are the points ~p = 0"1(X p ), P = 1,2, ... ,r, which obviously lie in 0 1. Now let a > 0 be such that (1.12) Bf = B(6, a) C 0 1 . As will be shown in the Notes,1 there exists a homotopy

Jt(€) = J*(x, t): IT! x [0,1)

--+

T2

which has the following properties: (i) fo(e) = r(e)i (ii) Jt(e) = 1*(e) for E ao 1 , t E [0, l)j (iii) all roots of the equation

e

(1.13)

lie in

BY.

Then

dU;, 01. Tlo)

= dU*, 01. Tlo)

(1.14)

as follows, e.g., from subsection 2.9 of Chapter 6 and (ii). Moreover from (iii) and (1.12) together with the sum theorem we see that

dU;, 01. Tlo)

= dU;, B~, Tlo).

(1.15)

Now since fi is homotopic to 1*, it follows that the maps It = 0"2 1fiO"l and f = 0"2 11*0"1 (cf. 1.10) are homotopic maps Sf --+ S~. Therefore

dU;, 01. Tlo) = d(lt, Sf, S~) = dU, Sf, S~).

(1.16)

176

SOME EXTENSION AND HOMOTOPY THEOREMS

But by Definition 3.2 and Theorem 3.4, both in Chapter 7, the right member of this equality equals u(f(Sf), fh) and thus equals zero by assumption (1.5) with n replaced by k. We thus see from (1.16) and (1.15) that d(fi, Bf,1]o) = 0, or what, by definition of the winding number, is the same that ( 1.17) We now recall that h k - 1 is true by our induction assumption. But by Lemma 1.9, h k - 1 implies ek . Therefore (1.17) implies the existence of a continuous map fi: Bf -+ T2 - 1]0 which satisfies fi(~) = fi(~)

We extend the definition of fi to

for ~ E aB~.

(1.18)

IT! by setting

!i(~) = fi(~)

for ~ E 0 1

-

B~.

Because of (1.18) this is a continuous map. Moreover by property (iii) of ft, our extension fi maps 0 1 -+ T2 - 1]0· Finally we see that fi( 0 = f* (~) for x E a0 1 . Thus fi is homotopic to f* (cf. the proof of Theorem 1.3 in Chapter 4). Denoting the extension fi of fi simply by fi, we see that It = a2 1fia1 is a continuous map of the subset U 1 of Sf into S~ - Yo which agrees with f on au1. If we extend the definition of It to all of Sf by setting It(x) = f(x) for x E Sf - U 1, we obviously obtain a continuous map It: Sf -+ Yo which is homotopic to f. 1.11. We proved hn for n ~ 0 and therefore by Lemma 1.9 also assertions en + 1 and e:z+ 1 . Our next goal is to prove the Hopf extension theorem in subsection 1.2. This entails two steps: (I) to show that e:z+ 1 is still true if Bf+1 is replaced by an arbitrary bounded open connected subset on+1 of Ef+1 and Sf by aOf+ 1 ; (II) to show that the theorem thus obtained remains true if the finite-dimensional spaces Ef+1, E'2+ 1 are replaced by an arbitrary Banach space E provided the maps involved are g.L.-S. For both of these steps we will make use of the following theorem.

s:r -

1.12. THE HOMOTOPY EXTENSION THEOREM. Let C 1 be a bounded closed subset of the Banach space E, let Co be a closed subset of C 1, and let Yo be a point of E. Let go and fo be two g.L.-S. maps Co -+ E - Yo. Assume that (i) there exists a g.L.-S. extension 9 of go which maps C1 into E - Yo, and (ii) go and fo are g.L.-S. homotopic in E - Yo· Then there exists a g.L.-S. extension f of fo which maps C 1 into E - Yo. In the finite-dimensional case the theorem remains true if domain and range spaces are different spaces of the same dimension.

The proof will be given in the Notes. 2 We noW turn to the generalization of assertion e:;+l stated as Step I of subsection 1.11. We will need the following lemma.

AN EXTENSION AND HOMOTOPY THEOREM

111

1.13. LEMMA. Let on+1 be a bounded open set in E~+!, let Yo be a point in E'2+ 1, and let f be a continuous map On+ 1 ---.. E'2+ 1 for which

(1.19) Then there exists a map 9 E c'(on+1) which on E'2+1 - Yo, and for which (ii) the equation

ao n + 1 is (i)

homotopic to f in

g(x) = Yo

(1.20)

has at most a finite number of roots. PROOF.

By (1.19) Ilf(x) - Yoll > 4c

for x E ao n+1

(1.21)

for some positive c. Let g1 be a C1-map: On+1 ---.. E'2+1 for which Ilgl(X) - f(x)11 < 2c

for x E On+1.

(1.22)

Such g1 exists by §1.32. By Sard's theorem the ball B(yo, c) contains a point Yo which is a regular value for g1. If g(x) = g1(X) + Yo -- Yo, then equation (1.20) is equivalent to the equation gdx) = Yo, and the differential of 9 equals the one of g1. Consequently Yo is a regular value for 9 which implies that (1.20) has at most a finite number of roots. Moreover by (1.22) Ilg(x) - f(x)11 ~ Ilg(x) - g1(x)11

+ Ilg1(X) - f(x)11

~ Ilyo - Yoll

+ 2c < 3c

for all x E rI. From this inequality and from (1.21) we see that for x E t E [0,1] 11(1 - t)f(x)

+ tg(x) - Yoll

~

111(x) - Yo II

-

ao and

tllg(x) - f(x)11 ~ 4c - 3c.

This proves that the homotopy (1 - t) 1 + tg between 1 and 9 has the asserted properties. 1.14. We are now ready to prove the assertion stated as step I in subsection 1.11. We have to show that if 1 is as in subsection 1.13 and if in addition (1.3) is satisfied with 0 = on+!, then there exists a continuous extension 1 to On+! of the restriction 10 of 1 to ao n+1 such that (1.23) By property (i) of Lemma 1.13 the restriction go of 9 to aon+! is homotopic to 10 in E!l+! - Yo. Suppose now we can prove that go has an extension g to On+! such that Yo rt. g(On+!). Then application of the homot0py extension theorem of subsection 1.12 with Co = ao n+1 and C1 = On+! shows also that 10 has an extension on 1 satisfying (1.23). Thus we have to consider only a map g with the properties (i) and (ii) of Lemma 1.13. In other words it is sufficient for our proof to assume that the equation

fex) = Yo

(1.24)

178

SOME EXTENSION AND HOMOTOPY THEOREMS

has at most a finite number of solutions. But under this assumption we know from Lemma 1 of Note 1 that if Bf+! is a ball in 0 having a positive distance from 80 n+1, then there exists a continuous map 11: Bf+ 1 -+ E'2+ 1 such that all roots of the equation (1.25) ft(x) = Yo lie in Bf+ 1 and such that

ft (x)

= I(x)

for x E 80 n+!.

( 1.26)

The first of these properties implies that

deft, on+!, Yo) = deft, Bf+!, Yo),

( 1.27)

and the second one together with assumption (1.3) that

u(fl(80 n+1 ),yo) =

o.

(1.28)

But by definition of the winding number u, the left member of (1.28) equals the left member of (1.27). Therefore 0 = deft, Bf+l, Yo) = u(ft (8Bf+l), Yo). But the latter equality implies by e~+! (see subsection 1.8) the existence of a continuous map 12: Bf+! -+ E2+! - Yo which agrees with ft on 8Bf+l. Therefore if we set hex) = ft (x) for x E On+! - Bf+l we obtain a continuous map On+l -+ E2+! - Yo. Moreover 12 = It = I on 80. Thus 12 is a continuous extension of the restriction of I to 80. This finishes the proof of the Hopf extension theorem of subsection 1.2 in the finite-dimensional case. 1.15. Proof of Theorem 1.2 for the case that the g.L.-S. map I(x) = >.(x)x - F(x) is a finite lauer map. In this case there exists a subspace En+! of E of finite dimension n + 1 such that not only F(80) E En+! but also by Yo E En+!. Let On+! = 0 n En+l, and let In+l be the restriction of I to 80 n+! = En+! n 80. Then

u(f(80), Yo) = u(fn+d80n+d, yo).

(1.29)

Indeed by Definitions 3.4 and 6.4 of Chapter 3

d(j,O,yo) = d(jn+l,On+!,Yo)

(1.30)

if j is a g.L.-S. extension of I to nand In+l is its restriction to On+!. But by definition of the winding number u, the left and right members of (1.29) equal the left and right members of (1.30) resp. Now by assumption (1.3) the equality (1.29) implies that u(fn+l(On+l), Yo) = O. But by subsection 1.14 this condition implies the existence of a continuous extension 7n+! of In+l which maps Dn+! into E - Yo· We define a map Fn+!: On+! -+ E by the equality

7n+l(X) = A(x)x - Fn+1(x),

(1.31)

where:X is an extension of>. to E which is positive and bounded away from zero (cf. §31 of Chapter 1). Fn+l is continuous and therefore completely continuous

AN EXTENSION AND HOMOTOPY THEOREM

179

on the closed bounded subset On+! of En+ 1. We thus obtain a completely continuous map F 1 : C = On+l U an - En+l by defining

F ( ) = {Fn+l{X), 1 X F{x),

x

E

On+l,

(1.32)

xEan.

Now C is a closed bounded set in E and F{x) lies in the convex set En+l. Therefore by §1.31 there exists a completely continuous extension of Fl which maps E into En+l. We set

1{x) = :X{x)x - F\{x),

(1.33)

where :x and Flare extensions of >. and Fl with the above properties. Then is a g.L.-8. map E, and our theorem will be proved if we can show that

n-

1{x)

# Yo

for all x

En.

1

(1.34)

Now if x E On+1. then (cf. (1.32)) 1{x) equals 1n+!{x) (see (1.31)) and 1n+!{x) Yo. But if the point x of does not lie in On+l, then x and :X{x)x do not lie in En+l, and since F{x) E En+l it follows from (1.33) that 1{x) (j. En+!. However, since Yo E En+! the inequality (1.34) follows again. 1.16. Conclusion of the proof of the Hopf extension theorem 1.2. Since by assumption f is a g.L.-8. map an - E - Yo, there exists a positive c such that

n

#

Ilf{x) -

yoll

~ 2c

(1.35)

for x Ean. On the other hand there exists a finite layer map g defined on an such that (1.36) Ilf{x) - g{x)11 < c. It follows from the last two inequalities that IIg(x)-YolI~c>O,

(1.37)

and from the Rouche theorem (subsection 1.9 in Chapter 4) that u(g(aO), YO) = u(f(an), yo). Consequently u{g(an), Yo) = 0 by assumption (1.3). We now see from subsection 1.15 that g can be extended to a g.L.-S. map g: E - Yo. But on the maps f and g are homotopic in E - Yo since the map (1 - t)f(x) + tg(x) # Yo for t [0,1] and x as follows from (1.35) and (1.36) (cr. the last

an

n-

E

Ean

inequality in subsection 1.13). Therefore application of the homotopy extension theorem shows that with g also f can be extended to a g.L.-S. map 7: n - E-yo. This proves part (i) of Theorem 1.2. To prove (ii) we note that an L.-8. map f may be considered as a g.L.-8. map f{x} = >.(x}x - F(x) with >.(x) = 1. Now in the proof of O} we showed that the E - Yo where the conditions on the extension map 7 given by {1.33} maps X of >. are that >. is continuous, bounded, nonnegative, and bounded away from zero. But if >. = 1, then :x = 1 satisfies all these conditions. This finishes the proof of the Hopf extension theorem. 1.17. For the proof of the Hopf-Krasnoselski'i homotopy theorem 1.4 we will show that it is a consequence of Theorem 1.2 just proved. For this we need the

n_

180

SOME EXTENSION AND HOMOTOPY THEOREMS

following lemma: let VI be a bounded open convex subset

0/ the Banach space E,

let a be a point 0/ VI, let (J be a positive number less than unity, and let Vo = {x EEl a + (x - a)(J-l E Vd.

(1.38)

Then (i) Vo is an open convex subset o/Vl ; (ii) i/c is a given positive number, then there exists a positive number 8 such that Ilxl - Xo I < c i/O < 1 - (J < 8, where xo and Xl are points 0/ avo and aVl resp. which lie on the same ray issuing from a. The proof will be given in the Notes. 3 Here we only remark that if VI is a ball with center a and radius rl, then the Vo defined above is a concentric ball with center a and radius TO = (Jrl. Thus the assertions of the lemma are quite trivial in this case. 1.18. Proof of the Hopf-KrBlno,ellkia homoto,,, theorem 1.4. Let 1 be a g.L.-S. extension of / to n. We assert first that it is possible to choose a positive number (J < 1 with the following property: if VI = 0 and if Vo is the set defined by (1.38), then 0 0 = 0 - Vo contains no root of the equation

1(x)

= yo·

(1.39)

This assertion is trivial if (1.39) has no roots in VI = n, i.e., if (])-l(yO) is empty. Then let (])-l(yO) be not empty. This set is disjoint from aVl = 130 since 1(130) = /(130) is contained in E - yo by assumption, and it is compact by part (iii) of §1.10 and subsection 6.3 of Chapter 3. Therefore (])-l(yO) has a positive distance from aVl, and our assertion follows from part (ii) of Lemma 1.17. If Vo has the above properties, it follows that

d(], VI, YO)

= d(], VO,yo),

and we see from the definition of the winding number u and the fact that / = 1 on aVl that u(](avo), Yo) = u(f(avt}, yo). (1.40) / and g are defined on aVl = 130. We now "transfer" / to avo, i.e., we define a g.L.-S. map h: avo --+ E - Yo as follows: if (as in the proof of Lemma 1.17) Xo and Xi denote points on avo and aVl resp. which lie on the same ray issuing from a, we define h(xo) = /(xd, (1.41 ) i.e.,

h(xo) = /(a + (xo - a)(J-l). We thus obtain a g.L.-S. map b: 1300 = aVl u avo --+ E - yo by setting g(x) for x E avl , b(x) = { hex) for x E avo.

(1.42)

(1.43)

Then, as will be proved below, the assumption (1.4) of our theorem implies that

u(b(aOo), yo)

= O.

(1.44)

AN EXTENSION AND HOMOTOPY THEOREM

lIn

Therefore we see from the Hopf extension theorem 1.2 that a g.L.-S. map ¢: no E - Yo exists for which

,I.,(x)

for x = Xl E aVl = ao, hex) for x = Xo E avo,

--+

= { g(x)

'+'

(1.45)

Now if x E 0 0 there are unique points Xo E avo and Xl E aV1 such that the points xo, x, Xl lie on the same ray issuing from a. Then

x

= (1 -

t)xo + tXl

= (1 -

t)(a + (Xl - a)O)

+ txt,

Denoting the right member here by Xt(Xl), we set '¢t(xt} Since ¢(x) f. Yo for x E no, we see that

o :s: t :s: 1.

= ¢(Xt(xt}) (cf. (1.45)).

'¢t(xt) f. Yo for all Xl E ao and t E [0,1]. Moreover since ¢ is g.L.-S. and since the factor of Xl in Xt(xt} equals (1- t)O + t and thus lies in the interval [0,1] and therefore is bounded away from zero, we see that '¢t(xt) is for each t E [0,1] a g.L.-S. map. Finally we see from (1.45) and (1.41) that

'¢o(xt}

= ¢(xo(xd) = ¢(xo) = h(xo) = J(xd, '¢t{xt} = ¢(xd = g(xd·

We thus see that '¢t(x), x E ao, t E [0,1]' is a homotopy between J(xd and g(xd with the desired properties, and for the proof of our theorem it remains to verify (1.44). For this purpose we set

ht(xo) =

7 (a + (1 ~o t)/+ t) ,

Xo E avo, t E [0,1].

(1.46)

Then by (1.41)

ho(xo)

= ](xt) = f(xt} = h(xo),

(1.47)

Moreover for t Indeed,

x(t)

xo - a

= a + (1 _ t)O + t

E

E

VI

[0,1]. -

Vo

(1.48) 7'i

= "0·

Therefore x{t) is not a root of (1.39). This together with (1.46) proves (1.48). But the latter inequality together with (1.47) shows that

= u(ho{aVo), Yo) = ua{aVo), Yo). and assumption 1.4 ua{avo) , Yo) = u{g{aVI ), yo). u{h{aVo), Yo)

But by (1.40) (1.49)

u(h{aVo), Yo)

= u{g{avt}, Yo).

(1.49) Therefore by (1.50)

To prove (1.44) with b given by (1.43) we define a g.L.-S. map b:V1 --+ E which on Vo is an extension of the map h given on avo and which on 0 0 = VI - Vo is an extension of the b given on aoo. Then by the sum theorem

d(1), VI, Yo)

= de!), Vo, Yo) + de!), 00, yo).

182

SOME EXTENSION AND HOMOTOPY THEOREMS

By definition (1.43) of b and by the definition of winding numbers, we see from this equality that

u(g(aVd, Yo)

= u(h(aVo), Yo) + u(b(ano), Yo).

This equality together with (1.50) proves the assertion (1.44). This finishes the proof of the Hopf-Krasnoselskil homotopy theorem.

§2. Two further extension theorems 2.1. In the finite-dimensional version of the Hopf extension theorem (see subsections 1.5 and 1.7) the domain and range space were of the same dimension n+l and the condition (1.4) was necessary for the desired extension. But, as was also shown by Hopf (see [2, §1.5]), this condition is not necessary if the dimension of the domain space is less than that of the range space. For a more precise statement see Theorem 2.2 below. This theorem will be used to prove Theorem 2.3 which deals with "odd" mappings and which will be useful in Chapter 9.

2.2. THEOREM. Let Ern and E 2+ 1 be Banach spaces of finite dimension m and n + 1 resp. with m ~ n. Let n1 be a bounded open domain in Ern, and let Yo be a point in E2+1. Let f be a continuous map anI --+ E2+1 satisfying

Yo

fI. f(and·

(2.1)

Then there exists a continuous extension 7 of f mapping

nt -+ E n +1 - Yo.

(2.2)

PROOF. The argument used in the last paragraph of the proof for part (b) of Lemma 1.9 shows that it will be sufficient to prove Assertion A. If assumption (2.1) is replaced by

(2.3) where B2+1 is a ball with center Yo, then f can be extended to a continuous map

(2.4) Now let U2+ 1 be an (n + I)-simplex which is inscribed in S~ and which contains Yo as an interior point, and let II: be the central projection from Yo of S~ on au n +1. Then 7 is a continuous map satisfying (2.4) if and only if II:f is a continuous map mapping nt into au2'+1. It is therefore clear that assertion A is equivalent to Assertion B. Let

f: anI

-+

aU2+1

(2.5)

be continuous, and suppose that

m:::;n,

(2.6)

TWO FURTHER EXTENSION

THJ!;UKt;M~

where m and n + 1 are the dimensions defined in the statement of our theorem. Then there exists a continuous extension

(2.7)

Er

Proof of assertion B. Let Q be a closed cube in containing 0 1 in its interior. By §1.30 there exists a continuous extension 9 of f: (2.8)

g(x) = f(x)

for x E

ao 1 .

(2.9)

9 is uniformly continuous on Q. Therefore if r is the distance of Yo from a(1~+ 1 there exists a 8 > 0 such that

Ilg(x') - g(x")11 < r/2

(2.10)

if

Ilx' - xliii < 8,

x', x" E Q.

(2.11)

Now by subsections 4.6 and 4.5 of Chapter 6 we can subdivide Q simplicially in such a way that the diameter of each simplex of the subdivision is less than 8. We denote by Km the set of those simplices which have a point in common with ril, and by km the set of those simplices which have a point in common with ao 1 . Thus if (2.12) then x lies in some simplex T m of km which contains a point for such x Ilx - xii ~ diam Tn < 8.

x of ao 1 • Therefore

Thus (2.11) holds with x' = x, x" = x. Consequently Ilg(x) - g{x)11 < r/2 by (2.1O). Moreover since g(x) = f(x) by (2.9) we see that Ilg(x) - f{x)11 < r/2. But f(x) E a(12+ 1 by (2.5), and therefore by definition of r, Ilf(x) - Yoll ~ r. Consequently IIg(x) - Yo II ~ Ilf(x) - yoll-lIf(x) - g(x)1I ~ r - r/2 > 0 for all x satisfying (2.12). Therefore for such x we can project g(x) centrally from Yo on a(12+1 to obtain a continuous map (2.13) Now g(x) E 8(12+1 for x E 80 1 by (2.9) and (2.5). Therefore h{x) = g(x) = f{x) for these x. Thus h(x) = f(x) for x E 80 t . (2.14) Now Km :) Ill. We therefore see from (2.14) that assertion B will follow from Assertion C. The map (2.13) can be extended to a ccrrtinuous map

h: K m

-+

8(12+1.

(2.15)

For the proof of C we note first that this assertion is nearly trivial if the map (2.13) is simplicial with respect to (km, 8(12+1) in the sense of part (c) of Definition 3.5 in Chapter 6. For if we arbitrarily assign to every vertex v of Km

184

SOME EXTENSION AND HOMOTOPY THEOREMS

which is not a vertex of km a vertex W = h(v} of 0(12'+1 and set h(v} = h(v} if v is a vertex of km, we obtain by the construction given in subsection 3.8 of Chapter 6 an extension h of h which maps Km into 0(12'+ 1 and is simplicial with respect to (Km, 0(12'+ 1 ) and is therefore continuous. The following assertion 0 reduces the case of a continuous h to the case of a simplicial map. Assertion D. There exists a barycentric subdivision Kr of Km of the following property: if kr consists of those simplices of Kr which belong to a subdivision of km, then the continuous map (2.13) is homotopic to a map hi: kr ---+ 0(12'+1 which is simplicial with respect to (Kr, 0(12'+1). Before proving this assertion we show that it implies assertion C. We note first that by 0 there exists a continuous map k(x, t}: Kr x [0,1] ---+ 0(12'+1 such that h(x,O} = h(x} and h(x, 1) = hdx). Now as we just saw, the fact that hi is simplicial with respect to (kr, 0(12'+1) implies the existence of an extension hi: Kr ---+ 0(12'+1 of hi which is simplicial with respect to (Kr, 0(12'+ 1 ) and is therefore a continuous map Kr ---+ E2'+1 - Yo since Yo ~ 0(12'+1. It now follows from the homotopy extension theorem 1.12 that h(x) = h{x,O) can be extended to a continuous map h: Kr ---+ E2'+1 - Yo. It is then clear that h{x) = ~h{x) is a continuous extension of h mapping Kr ---+ 0(12'+1 if K, is the central projection from Yo of h( x} on 0(12'+1. It remains to prove assertion D. Let wo, Wi, . .. , Wn+1 be the vertices of (12'+1. Then the sets Wi = st(Wi, 0(12'+1), i = 0,1, ... , n + 1, defined in subsection 3.9 of Chapter 6 form an open covering of 0(12'+1. This implies-as was already pointed out in that section-that there exists a positive c such that every subset S of 0(12'+ 1 of diameter < c is contained in at least one of the Wi. Since h is continuous, we can choose a 8 > 0 such that

Ilh{x'} - h{x"} II < c

{2.16}

Ilx' - x"ll < 8.

{2.17}

if Now by subsection 4.5 of Chapter 6 there exists a barycentric subdivision

Kr of Km such that the diameter of each simplex of Kr is less than 8/2. If kr consists of those simplices of Kr which are contained in a simplex of . ." .. . Km and if Vi is a vertex of Kr and x', x are pOIhts 10 st{Vi, Kr}, then the inequality (2.17) holds since Ilx' - viii and Ilx" - viii are both less than 8/2 and Ilx' - x"l1 ~ Ilx' - viii + Ilx" - viii- It follows that (2.16) is satisfied and that therefore h{x'} and h(x"} lie in the same star of 0(12'+1, i.e., there is a vertex W· of 0(12'+1 such that h(x'} and h(x") lie in st(Wj, 0(12'+1}. We thus see that if w~ set {2.18} then (2.19)

TWO FURTHER EXTENSION THEOREMS

185

for all vertices Vi of iqn. Using the construction described in subsection 3.8 of Chapter 6, we extend the map (2.18) to a map denoted again by ho which maps kr into 80'2+ 1 and which is simplicial with respect to (kr, 80'2+ 1). We now show that the simplicial map ho is homotopic (on 80'2+ 1 ) to the continuous map h. Let x be a point of kr, let be the simplex of smallest dimension of kr which contains x, and let vo, VI, ... , VI-' be its vertices. Then x E c st(Vi, kr) for i = 0,1, ... , J.L. Therefore for these i by (2.19)

O'r

O'r

h(x) C h(st( Vi, kr)) c st(ho(Vi), 80'2+1).

(2.20)

Now the points ho(vo),ho(vd, ... ,ho(vl-') are vertices of 80'2+ 1 but not necessarily different ones. Therefore they are vertices of a v-face 0'2 of 80'2+ 1, v :::; J.L, and we see from (1. 20) that

n I-'

h(x) =

st(hO(Vi),80'2'+I) = 0'2'

(2.21)

t=u

But also ho (x) E 0'2 since ho is simplicial and the ho (vd are vertices of 0'2' From this, from (2.21), and from the fact that 0'2 is convex, it follows that

(1 - t)ho(x)

+ th(x) E 0'2 E 0'2'+1

for 0 :::; t :::; 1. This establishes the desired homotopy between hand ho.

2.3. THEOREM. Let om be a bounded open subset 01 Em whose closure does not contain the zero point flt 01 Em (e.g., a "ring" bounded by two spheres with center Od. We assume moreover that om is symmetric with respect to 01 , i.e., that it contains the point -x il it contains x. Let n 2: m and let I be a continuous map 80 m --+ En+1 - O2 , where O2 is the zero point 01 En+1. We assume that I is odd with respect to 81 , i. e., that f( -x) = - f(x) lor all x Then

E

80 m

(2.22)

•

f can be extended to a continuous odd map rIm

--+

En+! - 82 •

PROOF. For fixed n we prove the theorem by induction for m = 1,2, ... , n. Let m = 1. By our assumption on 0 1 we may write 0 1 = o~ u O~, where o~ = {x E 0 1 I x > O} and o~ = {x E 0 1 I x < O}. Since n + 1 > 1 and 02 1(80~), application of Theorem 2.2 to n~ shows that the restriction of I to 8n~ can be extended to a continuous map 7: rI~ --+ E2+1 - O2 • Now if xE then -x E Therefore we obtain an odd continuous extension which maps 1 --+ E2+1 - O2 by setting l(x) = -1( -x) for x E o~. Now suppose our assertion has been proved for an ~nteger m - 1 :::; n - 1. We want to prove it for m. Let Er, om, and I satisfy our assumptions, let Xt.X2, ... ,Xm be coordinates for Em, let E m- 1 be the subspace of Em given by Xm = 0, and let Im-1 be the restriction of I to 80m n Em-I. By induction assumption fm-1 can be extended to an odd continuous map

rt

n_, n

n+.

,-

.7'im

m-1. U

n E m - 1 --+ E 2n + 1 - O2 •

(2.23)

SOME EXTENSION AND HOMOTOPY THEOREMS

186

We extend this definition by setting

=f

m-l

(x) = {im-l(X) f(x)

for x E om n Em-l, for x E 80m .

(2.24)

1

Then is a continuous odd map whose range lies in E'2+1 - (h. Now let E"t and E"!! be the two half spaces of Em characterized by Xm > 0 and Xm < 0 resp., and let 0+ = om n E+., O~ = om n E"!!. Then the restriction of 1 to 80+ maps the latter set into E'2+ l - ()2. Since m :::; n we see from Theorem 2.2 that this E'2+1 - ()2. Since om is symmetric with map can be extended to a map]: respect to (), we simply have to define 1(x) = - f ( - x) for x E 80~ to obtain a map defined on om with the asserted properties.

n+ ---.

§3. Notes to Chapter 8

1. The existence of a homotopy with the properties asserted in the lines between (1.12) and (1.13) is an obvious consequence of the following lemma. LEMMA 1. Let Ef and E~ be Banach spaces of finite dimension k. Let 0 be a bounded connected open set in Ef, let f be a continuous map 0 ---. E~, and let Yo be a point of E~. We assume that

Yo ~ f(80)

(1)

and that the number r of solutions of f(x) = Yo

(2)

is finite. Let B be a ball whose closure lies in O. Then there exists a continuous map g: 0 ---. E~ of the following properties: (i) all solutions of the equation g(x) = Yo

(3)

lie in B and their number is r; (ii) g(x) = f(x)

for x E 80.

(4)

(Note that condition (ii) implies the existence of a homotopy connecting f and g whose values on 80 are the same as those of f. This is clear from the proof given for Theorem 1.3 in Chapter 4.) The proof of this lemma will be based on Lemmas 2 and 3 below. LEMMA 2. Let c be a point of Ef, and let a be a positive number. Let f be a continuous map B(c, a) ---. E~. Let the point Yo of E~ satisfy

Yo

f!. f(8B(c, a)),

(5)

and let Cl be an arbitrary point in B(c, a). Suppose that the equation (2) has exactly one solution Co in B(c, a).

NOTES

~UI

Then there exists a continuous map ¢: B(c, a) (i) Cl is the only root of the equation

--+

E~ such that

(6)

¢(x) = Yo;

(ii) ¢(x) = f(x)

for x E BB(c, a).

(7)

PROOF. We set

(8)

¢(ct} = Yo

and define ¢ on BB(c, a) by (7). Now let x be a point of B(c, a) different from Then the ray Cl?t intersects BB(c, a) in exactly one point x = x(x); this point depends continuously on x, and Cl.

x = (1 -

t)Cl

+ tx,

(9)

where t is a uniquely determined number in the interval (0,1). We then define ¢( x) by setting ¢(x) = (1 - t)yO + tf(x). (10) It is clear from the above definitions that ¢: B(c, a) --+ E~ is continuous, that (7) is satisfied, and that Cl is a root of (6). It remains to verify that Cl is the only root of (8). To do this we note that by (10) a root x E B(c,a) of (6) satisfies

() = ¢(x) -

Yo

= t(f(x) -

yo).

(11)

But since x E BB(c, a) we see from (5) and (7) that f(x) - Yo -=J B. Thus, t = 0 by (11), and therefore x = Cl by (9). LEMMA 3. We use the notation and assumptions of Lemma 1. Let Xl> X2, ... , XT be the roots of equation (2), and let Xl be a point of 0 different from these roots. Then there exists a map 1j;: n --+ E~ and an open subset 0 1 of 0 having the following properties: (i) Xl and Xl lie in flt; (ii) 0 1 has a positive distance from BO as well as from the points X2, X3, . .. , Xr ; (iii) the roots of the equation

1j;(X) = Yo

(12)

are the points X!, X2, X3, ... , Xr ; (iv) 1j;(x) = f(x) for x 0 1.

En -

PROOF. Since 11 is open and connected there exists <.1 simple polygon Q c 11 connecting Xl and Xl (see §§1.24 and 1.25) and having a positive distance 28 1 from 811. Obviously Q can be chosen in such a way that it has a positive distance 282 from the points X2, X3, ... ' Xr . Let 8 = min(8l,82). We consider the balls B(q,8) as q varies over Q. Since the polygon Q is compact, it contains a finite number of points ql = Xl, Q2,q3, ... ,qN-l> QN = Xl (monotone with

188

SOME EXTENSION AND HOMOTOPY THEOREMS

respect to the arc length on Q counted from xt} such that the intersections I", = Q n B(q"" 8), v = 1,2, ... , N, cover Q and such that the intersections I", n 1",+1, v = 1,2, ... , v - 1, are not empty. Then the set N

01 =

U B(q",,8)

(13)

",=1

is an open subset of 0 which contains Xl and Xl and whose distance from ao and the points X2, X3,' .. ,Xr is not smaller than 8. It thus satisfies assertions (i) and (ii). It follows that Xl = ql is the only root of (2) which lies in B(Xl,8). Now let X12 be a point of It nI2. Then X12 E 11 nB(Xl,8). Therefore by Lemma 2 there exists a continuous map W12: B(Xl' 8) ----> E~ such that X12 is the only root of the equation W12(X) =

(14)

Yo

and such that W12(X) = f(x) for X E aB(Xl, 8). We therefore obtain a continuous map W12: 0 ----> E~ if we define W12(X)

= f(x)

(15)

for X E n - B(Xl,8). Since B(x1,8) = B(ql,8) c 0 1 by (13), the equality (15) holds for X En - 0 1 , and the roots in n of (14) are X12, X2, X3,"" Xr . Moreover X12 E 0 1 . Thus conditions (i)-(iv) are satisfied with Xl replaced by X12 and W by W12. We repeat this process by replacing the point Xl = q1 E It by the point X12 E 11 n 12 c 12 and choosing a point X23 C 12 n 13 c 13. Going on this way, we obtain a point XN-1,N E IN-1 n IN and a map WN-1,N: ----> E~ such that conditions (i)-(iv) are satisfied with Xl replaced by XN-l,N and Wby WN-1,N. Now XN-1,N and Xl both lie in IN and thus in B(XN-l,N,8). Therefore by again employing Lemma 2 we obtain from WN-l,N a map Wsatisfying assertions (i)-(iv) by using the argument used above to obtain W12 from f. Proof of Lemma 1. Let Xl, X2,"" Xr be the roots of equation (2). Let B c 0 be a ball which has a positive distance from ao, and let Xl, X2, ... ,xr be points in B. Then by Lemma 2 there exists a continuous map W1: 11 ----> E~ which agrees with f on ao and is such that the roots of W1 (x) = Yo are Xl, X2, ... , Xr . Again by Lemma 2 there exists a continuous map W2: n ----> E~ which on ao agrees with W1, and therefore with f, and is such that the roots of W2 (x) = Yo are Xl, X2, X3,· .. , Xr· Going on this way, we obtain a map g = Wr satisfying the assertion of Lemma 1.

n

2. Proof of the homotopy extension theorem 1.12. Let fo{x) = AO{X)X - Fo{x) and go{x) = 1L0(X)X - Go{x), where Fo and Go are ~ompletely continuous maps Co ----> E and where AO and 1L0 are real-valued contmuous functions defined on

189

NOTES

Co and satisfying ;\o(x)

0< mo $ { J.lo(x) $ Mo,

x E Co,

(16)

for some constants mo and Mo. By assumption, there is a completely continuous map Ho(x, t) and a continuous map v(x, t) which maps Co X [0,1] into E and the reals resp. These two maps have the following properties (17)-( 20).

Ho(x,O)

=

Ho(x, 1) = Fo(x),

Go(x),

v(x,O) = J.l(x),

v(x, 1) = ;\(x),

o< m

x

E

Co,

x E Co,

$ v(x, t) $ M

(17) (18)

(19)

for some constants m, M.

ho(x, t) = v(x, t)x - Ho(x, t) =I- Yo,

(x, t) E Co x [0,1].

(20)

Also by assumption there exists a completely continuous extension Go of Go to C l , and a continuous extension Ji of J.l to C such that

o< m

(21)

$ Ji( x) $ M,

for some constants m and M and such that (22)

?/o(x) = Ji(x)x - Go(x) =I- Yo, We now consider the set Z = C1

X

[0,1] c E x [0,1]

and its subset

Zo = (C 1

X

0) U (Co x [0,1]).

(If as in subsection 1.4, and also in subsection 1.14, C 1 = 0, Co = an, we may think of Z as a cylinder over 0, with C 1 x 0 as its base and Co x [0,1] as its "lateral surface.") We define a completely continuous map H1(X, t): Zo --+ E by setting

H1(X,0) = Go(x) for x E C1 X 0, H 1(x, t) = Ho(x, t) for (x, t) E Co x [0,1],

(23)

and define a real-valued continuous function J.l1(X, t) on Zo by setting

J.ll(X,O) = Ji(x), x E Cl , J.ll(X, t) = v(x, t) for (x, t) E Co x rO,l].

(24)

Then by (21) and (19) 0< m1 $ J.ll(X,t) $ Ml, for some constants

ml

(x,t) E Zo,

(25)

and M 1 • If we define

hI (x, t) = J.ll (x, t)x

- HI (x, t),

(x, t) E Zo,

(26)

190

SOME EXTENSION AND HOMOTOPY THEOREMS

we see from (22)-(24) and (20) that

h 1(x,0) = go(x) for x E C 1, h1 (x, t) = ho(x, t)

for (x, t) E Co

X

[0,1]'

(27)

and from (20) and (22) that

h1 (x, t) =J: Yo

for (x, t) E Zoo

(28)

Now Zo is a closed set. Therefore, by §1.31, H1 (x, t) has a completely continuous extension to Z. We denote one such extension again by H1 (x, t). Again by §1.31, III (x, t) has a continuous extension to Z satisfying (25) since the interval [m1' M 1] is convex. Calling one such extension again III (x, t), we obtain a g.L.-S. map h1 (x, t) = III (x, t)x - H1 (x, t) defined on all of Z. Now by (27), (20), (18), and (17) (29) h 1(x, 1) = >.(x)x - Fo(x) = lo(x) for x E Co. Thus h 1(x, 1), x E C1, is an extension of 10. Therefore if h 1 (x, 1) =J: Yo

for x E C 1 ,

(30)

then hex, 1) is an extension of 10 of the desired property. In general, however, (30) will not be true for all x E C1 . We therefore construct a g.L.-S. map I(x, t): Z -+ E which agrees with h1 (x, t) on Zo but which satisfies the condition

I(x, t) =J: Yo

for all (x, t) E Z.

(31)

For this purpose we consider the set Xo = {x E C1 I h1 (x, t) = Yo for some t E [0, I]}. Now by (28) the closed sets Xo and Zo are disjoint. As is well known, this implies the existence of a real-valued continuous function p with the properties p(x) = 1 for x E Zo, p(x) = 0 for x E Xo, and 0::; p(x) ::; 1 (see e.g. [18, p. 15]). We now set for (x, t) E Z

I(x, t) = 1l1(X, tp(x))x - H1(x, tp(x)). Then (31) holds. Indeed suppose I(xo, to) = Yo for some (xo, to) E Z. Now Xo E Xo and therefore p( xo) = O. Thus

Yo = I(xo, to) = III (xo, O)xo - H1 (xo, 0)

= h1 (xo, 0).

But this contradicts (28) since (xo,O) E C1 X 0 E ZOo

3. Prool 01 the lemma in subsection 1.17. That Vo is an open subset of V1 is obvious. To prove the convexity of Vo we have to show that if xg and X6 are points of Vo and if Xb = (1 - t)xg + tX6, then

xb E Vo

for t E [0,1],

(32)

NOTES

191

i.e., by the definition of Vo that a + (xb - a)O-l E Vl . Now elementary computation shows that a + (x& - a)O-l = a + [(1 - t)x8 + tX6 - a]O-l

+ (x8 - a)O-l] + t[a + (X6 - a)O-l]. But since xg and xA are points of Vo, it follows that a + (xb - a)O-l E = (1- t)[a

Vl for i = 0,1, and since Vl is convex the right member of the last equality lies in VI. Therefore so does the left member. But this implies (32). This proves assertion (i). To prove assertion (ii), let Xo and Xl be points on avo and aVl resp. which lie on the same ray issuing from a. Then Xo = a + 0(X1 - a) or Xo - Xl = (1 - O)(a - xd. This obviously proves (ii) since a and Xl are points of the bounded set VI.

CHAPTER 9

The Borsuk Theorem and Some of Its Consequences §1. The Borsuk theorem 1.1. THEOREM. 1 Let E be a Banach space and 0 a bounded open subset of E which contains the zero point () of E and which is symmetric with respect to (). Let f: ao ---+ E be an L. -8. map which is odd with respect to (). We suppose moreover that (1.1 ) () fi. f(aO). Then the winding number u(f(aO), ()) is odd. (For the meaning of the terms symmetric and odd with respect to (), see the statement of Theorem 2.3 in Chapter 8.) The following theorem is equivalent to Theorem 1.1.

1.2. THEOREM. Let E and 0 be as in Theorem 1.1. Let f be an L.-8. map which is odd with respect to n, i.e., for which

n ---+ E

f(-x) = -f(x)

(1.2)

for all x E n. We suppose moreover that (1.1) is satisfied. Then the degree d(f, 0, ()) is an odd number. 1.3. Proof of the equivalence of the two preceding theorems. 1.1 implies Theorem 1.2 is obvious, since by definition d(f, 0, ()) = u(f(aO), ())

n

That Theorem (1.3)

for any L.-S. map I defined on and satisfying (1.1). To prove the converse implication, we note that an f satisfying the assumptions of Theorem 1.1 has an L.-S. extension h to n which is odd. Indeed if h is an arbitrary L.-S. extension to we only have to set h (x) = (h(x) - h( -x))/2. Then (1.3) holds with f replaced by h on the left, while d(h, 0, ()) is odd by Theorem 1.2. 1.4. It is instructive to note that if in addition to the assumptions of Theorem 1.2 we suppose that I E C'(O) and that () is a regular value for I, then that theorem is a nearly immediate consequence of the fact that the differential of an

n,

192

THE BORSUK THEOREM

.llM

odd map is even (see §1.22b). To see this we note first that 0 is a root of the equation f(x) = 0, (1.4) since f(O) = f( -0) = - f(O) by (1.2). Now if 0 is the only root of (1.4), then, being a regular value for f, d(f, 0, 0) = j(f,O) = ±1 by subsection 2.1 of Chapter 2. Suppose now that 0 is not the only root of (1.4). We know that the regularity of 0 implies that the number of roots is finite, and (1.2) implies that, along with x, -x also is a root of (1.4). Thus we may denote the roots of that equation by 0, Xl,-Xl, ... ,Xn-Xr , where r is a positive integer. Then by Definition 2.2 in Chapter 2 of the degree

o

r

d(f,O,O) = j(Df(O; .))

+ L[j(Df(xp; .)) + j(Df(-xp; .))].

(1.5)

p=l

But, as already mentioned, (1.2) implies that Df(-xp;') = Df(xp;')' This shows that the right member of (1.5) is an odd integer since each indexj(D f(x p ; .) = ±1. 2 1.5. Turning to the proof of Theorem 1.2 without the additional assumptions made in subsection 1.4, we consider first the finite-dimensional case: E = En+! , a space of finite dimension n + 1. Following J. T. Schwartz (see [51, 111.33]) we start with a special case in which the assertion of our theorem is rather obvious.

°

1.6. THEOREM. Let = on+l be a bounded open subset of En+! which contains 0 and is symmetric with respect to O. Let 9 be a continuous map n ---+ En+l which satisfies the relations (1.1) and (1.2) with f replaced by g. Moreover we assume that there exists a ball Bn+ 1 with center 0 whose closure is contained in on+ 1 and for which g(x) = x (1.6) for x E Bn+!. Finally we assume the existence of a subspace En of En+! such that g(x) #- 0 for x E En n (on+! - Bn+l). (1.7) Under these assumptions we assert that d(g, on+! ,0) is odd. PROOF. Since (1.1) holds with that

d(g, on+! , 6)

f = 9 we see from

= d(g, Bn+!, 0) + d(g, on+! -

Bn+!, 6)

(1.6) and the sum theorem

= 1 + d(g, on+! -

Bn+!, 0).

It remains to show that here the second term of the ri~ht member is an even integer. Now if H+.+! and H':!:.+!are the half spaces into which En divides En+l, then the assumption (1.7) allows us by application ofthe sum theorem to obtain the relation

d(g, On+! - Bn+!, 6)

= d(g, H+.+! n (on+! _ BMl), 0) + d(g, H:':.+l n (on+! -

Bn+!), 6).

194

THE BORSUK THEOREM AND SOME OF ITS CONSEQUENCES

But by Theorem 2.9 in Chapter 5 the two degrees on the right are equal, since the transformation x --+ -x maps

and since -g(-x) = g(x). 1.7. To prove Theorem 1.2 with E = En+l without the additional assumptions of Theorem 1.6 we recall that if f and g are continuous maps On+ 1 --+ En+ 1 , if (1.1) is satisfied with 0 = on+l, and if

g(x) = f(x)

for x E aO n+1 ,

(1.8)

then d(g, on+l, 0) = d(f, on+! , 0). It is therefore clear from Theorem 1.6 that Theorem 1.2 (with E = En+!) is a consequence of the following theorem. 1.S. THEOREM. Let 0 = on+l be a subset of E = En+l satisfying the assumptions made in Theorem 1.2. Let f be a continuous map aon+! --+ En+! - 0 which is odd with respect to O. Then there exists an extension g of f to On+! which satisfies the assumptions of Theorem 1.6. PROOF. For x E aon+! we define g( x) by (1.8), and if Bn+ 1 is as in Theorem 1.6 we set g(x) = x for x E Bn+!. (1.9) Then g(x) is continuous and odd. Moreover if (1.10) then

g(x)i-O forXEaU n + 1 •

(1.11)

If, furthermore, En is an n-dimensional subspace of En+l and un = U n+1 n En and if for a moment gn denotes the restriction of g to aU n , then the assumptions of Theorem 2.3 in Chapter 8 are satisfied if in that theorem we set m = n, om = un, and replace f by gn. It follows that gn can be extended to a continuous odd map Un --+ En+! - O. We denote this extension again by g. Now let U+.+! = un+! n H+.+l and U~+l = U n +1 n H~+l where the half spaces H+.+l and H~+! are as in the proof of Theorem 1.6. Then au+.+! belongs to the domain of g and g( au+.+!) =I- O. Therefore g can be extended to a continuous map U+.+! --+ En+!. If we now define g(x) = -g(-x) for x E U~+!, then g is continuous and odd on Un+! while g(x) = f(x) on aon+! and g(x) = x on aBn+!. Taking into account definition (1.9) we see that g is an extension of f to On+! of the asserted properties. This finishes the proof of Theorem 1.2 in the finite-dimensional case. The following lemma will allow us to reduce the proof of this theorem for the general Banach space case to the finite-dimensional one.

SOME CONSEQUENCES

195

°

1.9. LEMMA. Let and f(x) = x-F(x) satisfy the assumptions of Theorem 1.2. Then to every c > 0 there exists a subspace En of E of finite dimension n and a layer map A(X) = x - A(x) (1.12)

with respect to En which, in addition to satisfying the inequality Ilf(x) - A(x)11 = IIF(x) - A(x)11 < c

( 1.13)

for all x E fl, is an odd map.3 PROOF. Since F is completely continuous there exists according to the second Leray-Schauder Lemma 4.3 in Chapter 3 an En and a layer map J.L(x) = x-M(x) with respect to En such that

IIF(x) - M(x)11 < c

(1.14)

for all x E fl. If we set

A(x) = !(M(x) - M( -x)), then the map A defined by (1.12) is odd. Moreover, since F is odd by assumption,

F(x) - A(x) =

! [F(x) -

M(x) - (F( -x) - M( -x))].

This equality together with (1.14) proves (1.13). 1.10. Conclusion 0/ the pro%/ Theorem 1.2. By (1.1) 0 has a positive distance from /(80.). Let c be a positive number less than this distance. With such an c let En and A be as in Lemma 1.9, and let An be the restriction of A to o.n = 0. n En. Then by subsections 5.5 and 5.6 of Chapter 3

d(f, 0, 0) = d(An, on, 0). Here the right member is an odd number since An is odd, and our theorem is proved in the finite-dimensional case.

§2. Some consequences of the Borsuk theorem In this section 0. always denotes a subset of a Banach space E which is bounded, open, and symmetric with respect to the zero point 0 of E which it contains.

n

2.1. THEOREM. Let f be an L.-8. map -+ E which satisfies (1.1). If, moreover, (1.2) holds for x E 80, then the equation

f(x) = 0

(2.1)

has at least one root x EO. PROOF. By definition of the winding number

d(f, 0, 0)

= u(f{80.) , 0).

Here the right member is an odd number by Theorem 1.1. Thus d(f, 0, 0) which implies the existence of a root of (2.1).

(2.2)

i= 0

196

2.2. x E 80

THE BORSUK THEOREM AND SOME OF ITS CONSEQUENCES LEMMA. Let 1/; be an L.-S. map 80 ..... E - O. Suppose that for no do the vectors1/;(x) and 1/;( -x) have the same direction, i.e., that

1/;(x)

-I- )..1/;( -x)

(2.3)

for all x E 80 and all ).. > O. Then u(1/;(80), 0) is odd. PROOF. g(x) = (1/;(x) -1/;(-x))/2 is odd. Moreover g(x) -I- 0 for all x E 80 since otherwise 1/;(x) = 1/;( -x) for some x E 80 which contradicts the assumption (2.3). Thus by Theorem 1.2, u(g(80), 0) is odd, and our assertion will be proved once it is shown that u(1/;(80), 0) = u(g(80), 0). (2.4) To prove this equality we set

h(x, t) = (1 - t)g(x)

+ t1/;(x),

and show that

h(x, t)

-I-

t E [0,1]'

°

(2.5) (2.6)

for all x E ao and t E [0,1]. This relation is obvious for t = 0 since h(x, 0) = g(x). Suppose now that h(x, t) = 0 for some x E 80 and some t in the interval o < t ~ 1. We then see from (2.5) and from the definition of g by elementary calculations that

1-t 1/;(x) = 1 + t t/J( -x).

Now this relation contradicts (2.3) if 0 < t < 1, and if t = 1 it contradicts the assumption that 1/; maps 80 into E - O. We thus proved (2.6). But by Theorem 1.8 in Chapter 4 the relation (2.6) implies (2.4). 2.3. THEOREM. Let 1/; be an L.-S. map n ..... E which on 80 satisfies the assumption of Lemma 2.2. Then the equation 1/;(x) = 0 has at least one solution xE O.

PROOF. The theorem follows from Lemma 2.2 the same way Theorem 2.1 followed from Theorem 1.1. 2.4. THEOREM. Suppose that El is a subspace of E of codimension 1, i.e., that there exists an element eo -I- 0 in E such that every x E E has the unique

representation

(2.7) where el EEl and where).. is a real number. Let f be an odd L.-8. map 80 ..... E 1 . Then the equation f(x) = 0 (2.8) has at least one solution x E 80. PROOF. Suppose the assertion to be wrong. Then

f(80) eEl - 0 c E - 0,

(2.9)

SOME CONSEQUENCES

197

and we see from Theorem 1.1 that u(J( ao), 0)

i- o.

(2.10)

Thus by Theorem 2.2 in Chapter 4 every ray issuing from 0 intersects f(aO). But the ray x = Aeo, A ~ 0, does not intersect f(aO) as is seen from (2.9). This contradiction proves the theorem. 2.5. BORSUK- ULAM THEOREM. 4 Let El be as in subsection 2.4. Let g(x) = x - G(x) be an L.-S. map ao --4 E 1 . Then the equation

g(x) = g( -x)

(2.11)

has at least one solution x E ao. (Note: the theorem shows that El contains no one-to-one L.-S. image of ao.) PROOF. Let

f(x)

= g(x) -2 g(-x) = x _

G(x) -2G( -x).

Then f satisfies the assumptions of Theorem 2.4. Therefore, the equation (2.8) has a solution. But this equation is equivalent to equation (2.11). 2.6. COROLLARY TO THEOREM 2.5. Let U be an arbitrary open set in E. Then El contains no one-to-one L.-S. image of u. PROOF. Obviously we may suppose that 0 E U. Assume then that there exists a one-to-one L.-S. map g such that g(U) EEl. Let 0 be a bounded open subset of U which contains 0 and is odd with respect to 0 (e.g., a ball with center oand with a small enough radius). Then the restriction of g to ao satisfies the assumption of Theorem 2.5. Therefore (2.11) holds for some x E ao. This contradicts our assumption that g is one-to-one. 2.7. THEOREM. Let f be an L.-S. map E --4 E. We assume that for every bounded set (3 E E the inverse f-1«(3) = {x EEl f(x) c .B} is bounded. We assume moreover that there exists a monotone sequence of positive numbers r1, r2, ... tending to 00 such that if Si is the sphere with center 0 and radius ri, then for i = 1,2, ... one of the following conditions (a) and (b) is satisfied: (a) f is odd on Si, (b) f satisfies (2.3) with 1/J = f on Si. Then for every Yo E E the equation

f(x) = Yo

(2.12)

has a solution. PROOF. Since the set (30 consisting of the single point 0 is bounded, it follows from our assumptions that f-1(O) is bounded. Therefore there exists an io such that

198

THE BORSUK THEOREM AND SOME OF ITS CONSEQUENCES

and we see from Theorem 1.1 in case (a) and from Lemma 2.2 in case (b) that u(f(Si),{}) is odd for i ~ i o. Therefore in either case

d(f, B(Ti' 0), 0) = u(f(Si), 0) f 0 for i

~

i o.

(2.13)

Now let Yo be a given point in E. Then the set (3 = {y EEl y = tyo, 0 ~ t ~ 1} is bounded. Consequently there exists an i1 ~ io such that 1- 1 «(3) lies in the open ball B(O, TiJ and thus contains no point of Sil = BB(O, TiJ. In other words, tyo ct. I(SiJ for 0 ~ t ~ 1. Therefore d(f,B(TillO),yO) = d(f,B(TillO),O). Since (2.13) holds for i = it, we see that d(f, B(Till 0), Yo) "# O. But this implies that (1.12) has a root x E B(Til' 0). §3. Notes to Chapter 9

Notes to §1. 1. Theorem 1.1 (and the equivalent Theorem 1.2) is referred to as Borsuk's theorem. In his original paper [7] Borsuk dealt with the case that [} = Bn+ 1, a ball in En+!, and proved without the use of degree theory that under condition (1.2) a continuous map I: sn = BBn+! -+ En+1 - 0 cannot be extended to a continuous map Bn+! -+ En+l - 0 (Borsuk's "Antipodensatz z"). Obviously (see subsection 1.1 in Chapter 8) this theorem is a consequence of Theorem 1.1 with [} = Bn+!, proved by H. Hopf (see [2, p. 483]), since it implies that u(f(sn),O) f O. The generalization ofthe latter theorem to Banach spaces (i.e., Theorem 1.10) was given by Krasnoselskil [31, p. 124], while the generalization to Banach spaces of the original Antipodensatz (and of related theorems) was given by Granas [24, Chapter IV].

2. For the finite-dimensional case the argument of subsection 1.4 was used by Eisenack-Fenske [19, p. 106]. 3. Lemma 1.9 is taken from [24, p. 41]. 4. In the finite-dimensional case with [} a ball, Theorem 2.5 was conjectured by Ulam and proved by Borsuk [7, p. 178]. (See also [2, p. 486]). The generalization to the Banach space case with [} a ball was given by Granas [24, p. 45].

ApPENDIX A

The Linear Homotopy Theorem §1. Motivation for the theorem and the method of proof 1.1. Let E be a Banach space, and let

l(x)

=

x - L(x)

(1.1)

be a nonsingular L.-8. map, E ---> E. We want to define an index j(l) which indicates how often E is covered by l(E). 8ince l is a nonsingular L.-8. map, it maps, according to §§1.11 and 1.12, E onto E in a one-to-one fashion. Therefore we should define j(l) as a number of absolute value 1. In order to decide for which l to define j(l) as +1 and for which as -1, we consider the finite-dimensional case. Let E = En be an oriented space of finite dimension n (see subsections 1.21 and 1.22 in Chapter 6 for the concept of orientation). It is then natural to define j(l) as +1 if l preserves the orientation (as does, e.g., the identity map I) and as -1 otherwise. By the sections just quoted, this is equivalent to defining .(l) J

= { + 1 if det l > 0, _ 1

if det l < 0,

(1.2)

where det l denotes the determinant of l. (Note that det l i: 0 since l is supposed to be nonsingular.) Now definition (1.2) in the form stated cannot be used in general Banach spaces since it depends on determinants; however, it can be reformulated in a way to become meaningful for Banach spaces. This is possible on account of the following facts (a) and (b) which are well known in linear algebra, being a consequence of the existence of the Jordan normal form for n x n matrices (see, e.g., [26, §58 and §77]): (a) if L has real eigenvalues greater than 1, say Al > A2 > ... > Ar > 1, and if for p = 1, ... , r, p,p = p,(Ap) is the generalized multiplicity of Ap , i.e., the dimension of the subspace EP of En which consists of those x E En for which

(ApI - L)mx =

(J

for some positive integer m, then

.

SIgn

de t l

I-'P., = (1)E:=1 -

199

(1.3)

200

THE LINEAR HOMOTOPY THEOREM

(b) if L has no real eigenvalues greater than 1, then sign det 1 =

+ 1.

( 1.4 )

It follows from (1.3) and (1.4) that definition (1.2) is equivalent to

j(l) = { (_l)L:=l#lP

+1

in case (a), in case (b).

(1.5)

As is well known and as will be seen in the following sections, the right member of (1.5) makes sense for an arbitrary Banach space E if 1 is a nonsingular linear L.-S. map E - t E. We note that Definition 1.6 in Chapter 2 appears as a theorem in the original Leray-Schauder theory (see [36, II. 11]). In §2 of this appendix we recall background material from the spectral theory oflinear operators M, referring for proofs mainly to the presentation given in [18, Chapter VII]. Since in general the spectrum of M contains complex numbers, we deal here with a complex Banach space Z. In order to apply the theory to the given real Banach space E, we have to "complexify" E, i.e., embed it into a complex Banach space Z. This is done in §3. §4 discusses the index j(l) in E as defined by (1.5) and also its definition for the complexification Z of E. Finally, in §5, the linear homotopy theorem (cf. subsection 1.3 in Chapter 2) is proved. §2. Background material from the spectral theory in a complex Banach space Z 2.1. DEFINITION. Let Z be a complex Banach space, let M be a linear bounded operator on Z, and let I denote the identity map on Z. Then the set of complex numbers A for which the inverse R>.(M) of U - M exists as a bounded operator Z - t Z is called the resolvent set of M and will be denoted by p(M). The complement u(M) of p(M) in the A-plane is called the spectrum of M. u(M) is closed and bounded. A subset uo(M) of u(M) is called a spectral set of M if it is open and closed with respect to u(M) [18, pp. 566,567,572]. 2.2. For fixed M the domain of R>.(M) as a function of A is the set p(M), and its range is a subset of the family of bounded linear operators on Z. This family is a Banach space with the linear operations defined in the obvious way and with the norm defined [18, pp. 475, 477] by

IIMII

=

sup

Ilzll::;!

IIM(z)ll·

(2.1)

This Banach space will be denoted by Zl, and for M fixed we consider R>.(M) as a map p(M) - t Zl. The topology of Zl induced by the norm (2.1) is called the uniform operator topology. 2.3. An example for a spectral subset of the spectrum u(M) is u(M) itself, and so is every isolated point of u(M). 2.4. LEMMA. p(M) is open, and u(M) is a nonempty closed bounded set. R>.(M) as a function of A is analytic at each A E p(M), i.e., at such A the

THE LINEAR HOMOTOPY THEOREM

201

derivative of R), (M) with respect to >. exists. Obviously 00

L

IIMllnl>'I-(n+l)

n=O

converges if

1>'1> IIMII·

(2.2)

Moreover for such >. (2.3) For a proof of this lemma see [18, pp. 566, 567]. 2.5. LEMMA. Let Ll be a closed not necessarily bounded subset of p{M). Then R),{M) is bounded on Ll. PROOF. On the bounded closed set

PII>'I ::; IIMII + I} the boundedness of PI 1>'1 > II Mil + I} it follows

R),{M) follows from its analyticity. But on the set

from (2.3). 2.6. DEFINITION. 1{M) denotes the family of all complex-valued functions f which are defined and analytic in some open neighborhood U of u{M). (U may depend on f.) 2.7. Let Uo be a spectral set of M, and let Vo be an open set in the complex >.-plane containing Uo but no point of u - Uo where u = u(M). Let f E 1{M) and let Uf be the domain of definition of f. Let f 0 and r I be two rectifiable Jordan curves lying in Vo n Uf oriented in the counterclockwise sense and such that Uo lies in the open bounded sets whose boundaries are ro and r l . Then (2.4) We note that the integrands in (2.4) are operator-valued. For the definition of such integrals and for the proof of the "Cauchy theorem" (2.4), we refer the reader to [18, pp. 224-227]. 2.8. DEFINITION. For f E 1{M) we define f{M) by setting

1 . f f{Il)R",{M) dll, f{M) = -2 7r~ lr

(2.5)

where f is a rectifiable Jordan curve of the following properties: (i) u{M) lies in the bounded open domain whose boundary is f; (ii) r is contained in the domain UJ of definition for f; (iii) r is oriented in the clJunterclockwise sense. The uniqueness of Definition 2.5 follows from Lemma 2.7. 2.9. LEMMA. Let r be a rectifiable Jordan curve satisfying conditions (i), (ii) and (iii) of subsection 2.8. Then

M =

~

( IlR",(M) dll,

2nlr

(2.6)

THE LINEAR HOMOTOPY THEOREM

202

1. 1= -2 1rl

irf RJL(M) dJ.l.

(2.7)

If, moreover, f and 9 are elements of J(M) with domains Uf and Ug resp. and if, in addition to the above requirements, r E Uf n U g , then

(2.8) For the proof we refer the reader to the proof of Theorem 10, p. 468 in [18], which contains the assertions of the present lemma. 2.10. DEFINITION. A bounded linear map P: Z -+ Z is called a projection if p2 = P. (2.9)

2.11. LEMMA. Let 0'0 be a spectral set of M. Let Then the operator Pua defined by

1. Pua = -2 1f2

f RJL(M) dJ.l ira

ro

be as in Lemma 2.7.

(2.10)

is a projection.

PROOF. Let 0'0 = a - 0'0. By the definition of spectral sets, there exist bounded open sets Va, Va", Uo, U(j such that 0'0 C Va C Va C Uo, 0'0 eVa' c V~ c U(j with U o and U~ disjoint and such that ro = avo and r' = avo are rectifiable Jordan curves. Let r = ro u r·. Then (2.8) holds if we set

J.l E UO, J.lE U(j.

(2.11)

But with this choice of f and 9 we see from (2.10) that the right member of (2.8) equals Pua since f(J.l) = g(J.l) = 0 for J.l E r and taking account also of (2.5), we see that f(M) = g(M) = Pua · Thus the left member of (2.8) equals P;a'

o,

2.12. LEMMA. Let 0'0 and 0'1 be two disjoint spectral sets for M. Let ro and Pua be defined as in subsection 2.11, and let r 1 and PUI be defined correspondingly 11Jith respect to 0'1. Then

Pua . PUI = PUI • Pua = where

(}IX

= (} for every x

(}1

(2.12)

E E.

PROOF. Since 0'0,0'1, and 0'2 = a - (0'1 U ( 2) are closed bounded sets there exist for i = 0, I, 2 bounded open sets Vi, Ui such that (Ti C Vi C Vie i with the sets U 1, V 2, U 3 being pairwise disjoint and such that r i = aVi is a rectifiable Jordan curve which we assume to be oriented in the counterclockwise sense. If we set () { I for J.l E U1 , gJ.l= o for J.l E Uo U U2 ,

U

THE LINEAR HOMOTOPY THEOREM

then with

r

=

203

r 1 ur 2 ur 3

1n irr f(J.L)RIJ-(M) dJ.L = f(M),

1. PO'o = -2

PO'l

1. = -2 1ft

irr g(J.L)RIJ-(M) dJ.L = g(M),

and application of (2.8) yields the assertion (2.12) since f(J.L) . g(J.L) = 0 for all J.L E r. 2.13. LEMMA. Let 0'0 be a spectral set of M, let PO'o be the projection (2.10),

and let Zo be the range of PO'o' i.e., Zo spectrum of the restriction of M to Zo.

=

PO'oZ, Then MZo C Zo and 0'0 is the

For a proof see [18, pp. 574, 575]. 2.14. LEMMA. The restriction MO'o of M to Zo = PO'oZ is given by

MO'o = where

ro

r

iro

J.LRIJ-(M) dJ.L

(2.13)

is as in subsection 2.7.

PROOF. Let r be as in Lemma 2.9, and let Vo and V be the bounded open sets whose boundaries are ro and r resp. Then M is given by (2.6), i.e., by (2.5) with f(J.L) = J.L. Now z = PO'o(z) if and only if z E Zo = PO'oZ, Therefore M(z) = MPO'o(z) for z E Zoo But PO'o is given by (2.10) or what is the same 1. PO'o = -2 1ft

irr g(J.L)RIJ-(M) dJ.L,

(2.14)

with g(J.L) as in (2.11). The assertion (2.13) follows now from (2.5) and (2.14) on application of (2.8). 2.15. So far we assumed of the linear operator M only boundedness or what is the same continuity. In the remainder of this section we require complete continuity, i.e., we make the additional assumption that for any bounded set (3 C E the closure of M((3) is compact. The definitions and assertions stated below are classical. Proofs may be found, e.g., in [18, p. 577ff.]. 2.16. The spectrum a(M) of a completely continuous linear operator M on E is at most countable. If a(M) contains infinitely many points, then 0 is the only accumulation point of a(M). It follows that each Ao E a(M) which is different from 0 is a spectral set 0'0 of M. For the projection PO'o defined by (2.10) we write: PAO = ~ RIJ-(M) dJ.L. (2.15) 21ft

r

iro

For ro we may and will take a circle with center Ao and radius smaller than the distance of AO from the rest of a(M). ro is supposed to be oriented in the counterclockwise sense. We recall that the points of a(M) are also referred to as eigenvalues of M.

204

THE LINEAR HOMOTOPY THEOREM

2.17. Let Ao E a(M) and Ao

i= O.

P>'oZ = {z E Z

Then

I (Ao! -

M)n = O}

(2.16)

for some positive integer n. The dimension v( Ao) of the linear subspace P>'o Z of Z is finite and is called the generalized multiplicity of Ao as an eigenvalue of M. 2.18. Let Ao E a(M) and Ao i= O. Then the equation

Aoz-M(z)=B

(2.17)

has nontrivial solutions (i.e., solutions z i= 0) called eigenelements of M to the eigenvalue Ao. These eigenelements obviously form a linear subspace of E. The dimension r(Ao) of this subspace satisfies the inequality 1 :S r(Ao) :S v(Ao) and is called the multiplicity of Ao as an eigenvalue of M.

2.19. LEMMA. Let M be a completely continuous linear map Z --+ Z and suppose that the linear map m = 1- M mapping Z onto Z is one-to-one. Then the inverse n = m- l exists and is of the form m(O = ~ - N(O, where N is completely continuous on Z. PROOF. By assumption m is a linear, continuous, and therefore bounded, one-to-one map of Z onto Z. By a well-known theorem (see, e.g., [18, p. 57]) these properties imply that the linear map n = m- l is bounded. Now if ~ = z - M(z), then n(O = z = ~ + M(z) = ~ + M(n(~)). Since n is bounded, i.e., continuous, and M is completely continuous, the map N(~) = -M(n(~)) is completely continuous.

§3. The complexification Z of a real Banach space E 3.1. The construction of Z from E is analogous to the construction of complex numbers as ordered pairs of real numbers: the elements z of Z are ordered pairs (x, y) of points of the real Banach space E. If Zi = (Xi, Yi), i = 1,2, are two elements of Z, then their sum is defined by Zl + Z2 = (Xl + X2, YI + Y2), and if C = 0 + i(3, 0, (3 real, is a complex number, then CZI is defined by (3.1) Finally if as usual II . II denotes the norm in E, we set for the point z Ilzlll = sup Ilxcos
= (x, y)

E Z

(3.2)


3.2. In the case that E is the real line, then Z is the complex plane. The reader may verify that then Ilzlll is the absolute value of the complex number z. 3.3. The number Ilzlll defined by (3.2) is a norm, i.e., for points Z,ZI,Z2 in Z and complex number c . Ilzlll ;::: 0 (3.3) with the equality holding if and only if z = 0, the zero element of Z, and

Ilzl

+ z211 :S

Ilztll

+ Ilz211,

Ilczlll = lel·llzll·

(3.4) (3.5)

THE LINEAR HOMOTOPY THEOREM

205

3.4. With the norm Ilzll! the linear space Z is complete and thus a Banach space. The verification of the assertions contained in subsections 3.3 and 3.4 is of course based on the corresponding properties of the Banach space E of which Z is the complexification. We leave this verification to the reader. 3.5. It is easily verified from our definitions that II(x,edll! = Ilxll. Thus the linear one-to-one map E into Z given by

x

--+

(x, e)

(3.6)

is norm preserving. Identifying x E E with (x, e) E Z, we consider E as a subset of Z. This identification also allows us to write Ilzll instead of Ilzll! without ambiguity. 3.6. From the multiplication rule (2.1) we see that (e,y) = i(y,e) for any y E E where as usual i = (0,1), the imaginary unit of the complex plane. Therefore z = (x, y) = (x, e) + (e, y) = (x, e) + i(y, e), and by our above identification we see that every z E Z may be written uniquely in the form z = x

+ iy,

(3.7)

x E E, y E E.

3.7. For later use we note the inequality

~(llxll

+ Ilyll) :S

Ilzll :S Ilxll

+ Ilyll

for z

= x + iy.

(3.8)

The right part of this inequality is a restatement of the triangle inequality (3.4). To prove the left part we note that Ilxll = IlxcosO + ysinOl1 :S supq, Ilxcos¢ + ysin¢11 = Ilzll by (3.2). In the corresponding way we see that Ilyll :S Ilzll· Thus Ilxll + Ilyll :S 211zll· 3.8. DEFINITION. Let L be a linear bounded operator E --+ E. Then the map M: Z --+ Z defined for z = x + iy by

M{z) = L{x) + iL{y)

(3.9)

is called the complex extension of L. 3.9. LEMMA. (i) M is linear; (ii) M is bounded; (iii) M is completely continuous if and only if L is completely continuous; (iv) if L is continuous, then m = 1 - M is nonsingular if and only if 1 = 1 - L is not singular (1 denotes the identity map in the respective space); (v) m(x) = 0 has a solution z '" e if and only if l(x) = e has a solution x '" e; (vi) m is not singular (i.e., m(Z) = Z) if and only ifm is one-to-one, i.e., if z = e is the only root ofm(z) = O. PROOF. The proof of (i) consists of an obvious verification. Proof of (ii). Let J.L be a bound for L. Then by (3.9), (3.4), and (3.8) IIM(z)11 :S IIL(x)11

+ IIL(y)11 :S J.L(llxll + lIyll) :S 2J.LllzlI·

Proof of (iii). Suppose L is completely continuous. Then, by (ii), M is continuous and it remains to show that if Zn = Xn + iYn where Xn and Yn are

THE LINEAR HOMOTOPY THEOREM

206

bounded sequences in E, then the M(znJ converge for some subsequence Zn, of the Zn. Now by (3.8) the boundedness of the Zn implies the boundedness of the Xn and Yn. It therefore follows from the complete continuity of L that for some subsequence ni of the integers the L(x n,) and L(Yni) converge. But then for zni = x ni + iYni the sequence M(znJ = L(xnJ + iL(YnJ converges. The converse is obvious since L is a restriction of M to E. Proof of (iv): If Z = x + iy, ~ = ~ + iTJ, with x, Y,~, TJ elements of E, then the equation m(z) = ~ is equivalent to the couple of equations l(x) = ~, l(y) = TJ. This obviously implies that I maps E onto E if and only if m maps Z onto Z. We obtain a proof of assertion (v) if in the proof of (iv) we set ~ = ~ = TJ = (). Finally assertion (vi) follows from assertions (iv) and (v) in conjunction with part (iv) of Lemma 12 in Chapter 1, the latter being valid also for complex Banach spaces. 3.10. DEFINITION. With E considered as subset of Z (cf. subsection 3.5), the elements of E are called the real elements of Z. If Z = x + iy with x, Y real, the element z = x - iy is called conjugate to Z and the map Z ----> z is called a conjugation. A subspace Zl of Z is called invariant under conjugation if Z E Zl implies that z E Zl. 3.11. LEMMA. Let M be a bounded linear map Z ----> Z. Then M is the complex extension of a linear bounded map L: E ----> E if and only if for all Z E Z

M(z)

=

M(z).

(3.10)

PROOF. The necessity is obvious from Definition 3.8. Then let M be a linear bounded map Z ----> Z satisfying (3.10). Now for each x E E, M(x) is an element of Z. Therefore

M(x) = Ldx) + iL 2(x),

(3.11)

where L 1 (x) and L2(X) are elements of E. It will be sufficient to show that L 2 (x) = (), for then M will be the complex extension of L 1 . Now by (3.11) for Z = x + iy

M(z)

=

M{x) + iM{y)

This holds for all

Z

=

Ldx) - L2{y) + i{L2{x) + Ldy)).

(3.12)

E Z. Therefore

(3.13) On the other hand, we see from (3.12) that (3.14) By assumption (3.10) the left members of (3.13) and (3.14) are equal. Comparing

the imaginary parts of the latter equalities, we see that L2(X) L 2 {x) = () as we wanted to prove.

=

-L 2(x), i.e.,

THE LINEAR HOMOTOPY THEOREM

207

3.12. LEMMA. Let Z1 be a linear subspace of Z and let E1 be the set of real elements in Zl. It is asserted: (i) if Z1 is invariant under conjugation, then for each z = x + iy E Z1 the elements x and y belong to E 1; (ii) if, besides satisfying the assumption of (i), Z1 is finite dimensional, then there exists a base for Z1 which consists of elements of E 1; (iii) a finite-dimensional subspace Z1 of Z is invariant under conjugation if and only if there exists a projection P: Z onto Z1, such that P(z) = P(z)

for all z E

z.

(3.15)

PROOF. (i) If z = x+iy E Z1(X,y real), then by assumption z = x-iy E Z1. Consequently x = (z + z)/2 and y = (z - z)/2i lie in Z1 n E = E 1. Proof of (ii). E1 is obviously a linear subspace of the real space E. Moreover, if b1, b2 , ••. ,br are linearly independent elements of this subspace, they are also linearly independent as elements of Z1. For if r

I>jbj = (), j=1

with

Cj

= aj

+ i{3j,

aj, {3j real, then r

r

'~JJ " a ·b . + i'" {3 b· - () ~JJ-' j=1 j=1

and thus

r

r

L ajbj j=1

=

L {3jbj j=1

= ().

Therefore aj = {3j = 0, j = 1,2, ... , r, on account of the independence of the bj as elements of E 1 • Thus Cj = aj + i{3j = o. This proves the independence of the bj as elements of Z1. It follows that the subspace E1 of E is finite dimensional since Z1 is finite dimensional. Then let b1 , ... ,bn be a base for E 1 • Since the bj are independent also as elements of Zl. it remains to show that they span Z1. Now let z = x + iy E Z1. Then there exist real numbers aj, {3j such that n

n

y = L{3jbj , j=1

X = Lajbj , j=1

and thus

(3.16)

n Z

= X + iy = L(aj

+ i{3j)bj.

j=1 Proof of (iii). The coefficients aj in the first sum in (3.16) are obviously linear in x for x in the finite-dimensional subspace E1 of E. They are therefore also continuous (see, e.g., [18, p. 245]). Consequently, by the Hahn-Banach theorem, these bounded linear functions aj = aj(x) can be extended to bounded

208

THE LINEAR HOMOTOPY THEOREM

linear functions A)(x) defined for all x E E. We now define for j continuous linear functionals C) on Z hy setting for z = x + iy

= 1,2, ... ,n (3.17)

Then

CJ(Z) = C)(z)

(3.18)

since the A)(x) are real-valued. We now set n

P(z)

=

L Cj(z)bJ,

(3.19)

)=1

where the b) form a real base for Zl. We then see from (3.18) that (3.15) is satisfied. We now show that P is a projection, i.e., that (2.9) is satisfied. Since the b) are real, we see from (3.19) and (3.17) that

P'(z)

~ P(P(z)) ~

t.

b,Cj

(~C'(Z)b') ~

t. ~ bj

C,(z)Cj(b,).

(3.20)

Since bl EEl, C)(bt) = A)(bt) = D:J(bt). But D:)(bt) = 0 if j =I- l, and 1 for j = l as is seen from (3.16) with x = bl. Therefore 1 if j = l, CJ(bt) = { 0 if j =I- l,

and we see from (3.20) and (3.19) that n

p 2 (z) = ~ b)C)(z) = P(z). )=1

This finishes the proof that the invariance of Zl under conjugation implies the existence of a projection P satisfying (3.15). Conversely if such projection P exists, then for z E Zl we see that z = P(z), and therefore by (3.15) that Z = P(z) = P(z) which shows that z E Zl.

§4. On the index J. of linear nonsingular L.-S. maps on complex and real Banach spaces 4.1. DEFINITION.

Let Z be a complex Banach space. Let

m(z) = z - M(z)

(4.1 )

be a nonsingular L.-S. map Z ----> Z. Then with R(A) denoting the real part of the complex number A, the index j(m) of m is defined as follows: if M has no eigenvalues A with R(A) > 1 we set j(m) = 1. If M does have eigenalues A with R(A) 2 1, we denote them by Al, A2, ... ' A8 and define j(m) = (_1)2::=1

Vi,

(4.2)

where Vi = V(Ai) is the generalized multiplicity of Ai as an eigenvalue of M (see SUbsection 2.17 for the definition of "generalized multiplicity," and see §1

THE LINEAR HOMOTOPY THEOREM

for a motivation for the definition of j(m)). Note that by subsection 2.16 the number of eigenvalues A of M with R(A) ~ 1 is finite and by subsection 2.17 the generalized multiplicity by V(A) of an eigenvalue A of M is finite. 4.2. DEFINITION. Let E be a real Banach space, and let

l(x) = x - L(x)

(4.3)

be a nonsingular L.-S. map E ---+ E. An eigenvalue AO of L is then a real number such that for some point x E E different from f)

L(x) = AoX.

(4.4)

If Z is the complexification of E and M the complex extension of L, it is clear from §3 that AO is an eigenvalue of L if and only if Ao is a real eigenvalue of M. We define the generalized multiplicity of an eigenvalue Ao f:. 0 of L as the dimension 11-( AO) of the space (4.5)

for some integer n ~ 1. I1-(Ao) is finite, for it is clear that I1-(AO) S v(Ao), the generalized multiplicity of Ao as an eigenvalue of M (see subsection 2.17; actually I1-(AO) = V(AO) as we will see in subsection 4.11). 4.3. DEFINITION. Let land L be as in subsection 4.2. We define the index j(l) to be 1 if L has no eigenvalues> 1. If L has eigenvalues> 1, we denote them by A1 > A2 > ... > As and define the index j(l) by (4.6)

where I1-p = I1-(A p), P = 1,2, ... , s. Note that this definition agrees with (1.5), which was shown to agree in the finite-dimensional case with (1.2). Note also that A = 1 is not an eigenvalue of L since l is not singular. 4.4. DEFINITION. A linear map l of the real Banach space E into itself is said to be type 1- if it can be constructed as follows: let E1 be a one-dimensional subspace of E and let E2 be a complementary subspace such that every x E E can be written in the form (cf. Lemma 4 of Chapter 1) (4.7)

Then l maps x = Xl + X2 into -Xl + X2 (cf. §8 of the introduction). Maps of type 1- in a complex Banach space Z are defined correspondingly. Note that l(x) = x - 2X1, i.e., l(x) is ofthe form (4.3) with L(x) = 2X1. Since Xl lies in the one-dimensional space E1, L(x) is completely continuous. Thus l is an L.-S. map. Since obviously l maps E onto E, the map l is not singular. Thus j(l) is defined for a map l of type 1-. 4.5. LEMMA.

. _ { +1 J(l) -

_ 1

if l.= I, the identity map, if l is of type 1-.

(4.8)

210

THE LINEAR HOMOTOPY THEOREM

PROOF. If I = I then the L(x) in (4.3) equals () for all x. Therefore (4.4) for x i' () is satisfied only with AO = o. Thus L has no eigenvalues 2: 1 which by Definition 4.3 proves the first part of assertion (4.8). Suppose now I is of the 1- type. Then with the notation used in (4.7), l(x) = -Xl + X2 = X - 2Xl, i.e., L(x) = 2Xl by (4.3) and the eigenvalue equation (4.4) reads 2Xl = AOX = AO(XI + X2)· This implies X2 = () and AO = 2. Thus 2 is the only eigenvalue of L, and j(l) = (_1)1'(2). (4.9) Here /1(2) is the dimension of the space given by (4.5) with AO = 2 and L(x) = 2Xl. Now for i = 1,2 let Pi be the projection x -+ Xi (cf. (4.7)). Then AoI - L = 2( PI + P2) - 2Pl = 2P2· Thus, by (4.5), E2 consists of those x for which x= 0 for some integer n 2: 1. But for these integers = P 2 since P2 is a projection, and we see that E2 = {x EEl P2(x) = O} = El. Thus /1(2) = dimE l = 1. This together with (4.9) proves the second part of the assertion (4.8). Since L.-S. homotopy (see subsection 1.2 of Chapter 2) is transitive, it follows from Lemma 4.5 that the linear homotopy theorem (stated in subsection 1.3 of Chapter 2) is equivalent to the following one.

pr

pr

4.6. THEOREM. Let lo and It be two linear nonsingular L.-S. maps E -+ E. Then lo and it are linearly L.-S. homotopic (see Definition 1.2 in Chapter 2) if and only if j(lo)

= j(It).

(4.10)

Now for questions of continuity it is preferable to deal with the complexification Z of E and the complex extension M of the operator L given by (4.3). For if M changes continuously, a real eigenvalue of M may become complex. But in this case the eigenvalue of L as an operator on E "disappears" since the concept of a complex eigenvalue makes no sense in the real space E. We therefore state the following theorem whose proof will be given in §5. 4.7. THEOREM. Let lo and it be as in Theorem 4.6. Let Z be the complexification of E, and for i = 0, 1 let Mi be the complex extension of Li = I - Ii on E, and let mi(z) = Z - Mi(Z). It is asserted that the equality j(mo) = j(md

(4.11 )

(see Definition 4.1) holds if and only if there exists an L. -So linear homotopy m(x, t) = Z - M(z, t), 0::; t ::; 1, between mo(z) = m(z, 0) and mt{z) = m(z, 1) which satisfies for z E Z and t E [0,1]

M(z, t)

= M(z, t).

The remainder of §4 is devoted to showing 4.8. THEOREM. Theorems 4.6 and 4.7 are equivalent. The proof of this theorem requires some lemmas.

(4.12)

THE LINEAR HOMOTOPY THEOREM

211

Let E and Z be as in Theorems 4.7 and 4.8. Let L be a continuous linear map E -+ E, and let M be its complex extension. Then (i) the spectrum u(M) of M is symmetric with respect to the real axis of the complex A-plane, and (ii) for A E p(M) 4.9.

LEMMA.

R>:(M)(z) = R>.(M)(z)

(4.13)

(see Definition 2.1). PROOF. (i) u(M) and the resolvent set p(M) are disjoint subsets of the A-plane whose union is the whole A-plane. Therefore assertion (i) is equivalent to the assertion that p( M) is symmetric to the real A-axis. Let AO E p( M). We have to prove (4.14) Xo E p(M).

Now by the definition of R>.(M) (4.15)

(AoI - M)R>.o(M)(z) = z for all z E Z. But by (3.10) for u E Z

(AoI - M)(u) = (XoI - M)(u). Applying this relation with u

= R>.o(M)(z) we see from (4.15) that

(XoI - M)R>.o(M)(z) = z. Replacing z here by

z we see that (XoI - M)R>.o(M)(z) = z

for all z E Z. This shows that

S(z) = R>.o(M)(z)

(4.16)

is a right inverse of (XoI - M). But interchanging the factors in (4.15) we conclude that S (z) is also a left inverse of >'01 - M. Thus S is an inverse of >'0 1 - M. Since S is bounded, the assertion (4.14) follows by definition of p(M). Proof of (ii). By definition, RXo(M) is the inverse of >'01 - M. Therefore by (4.16)

RXo(M)(z) = S(z) = R>.o(M)(z). This proves assertion (ii). 4.10. LEMMA. Let E, Z, L, M be as in Lemma 4.9. Let >'0 be an isolated point of the spectrum u(M) of M such that by Lemma 4.9 >'0 is also an isolated point of u(M). Let Co be a counterclockwise-oriented circle with center >'0 and radius ro where ro is such that the closure of the circular disk B(>'o, ro) contains no point of u(M) other than Ao. Let c~ be the counterclockwise-oriented circle with center >'0 and radius ro. Let P>'o and PXo be the linear operators defined by

P>'o

= ~ { Rp.(M) dJ.l, 27l"Z leo

Px = f-= 0

'In

{ Rp.(M) dJ.l.

le~

(4.17)

THE LINEAR HOMOTOPY THEOREM

212

Then the ranges Z>'o = P>'o Z and Z-Xo = P-XoZ are conjugate to each other, i.e., if z varies over Z>'o' then z varies over Z-xo' PROOF. >'0 and Xo are spectral sets (see subsections 2.1 and 2.3). Therefore, by subsection 2.11, P>'o and P>"o are projections. We will show first our lemma follows from the relation P>"o (z)

= P>'o (z) for all z E Z.

(4.18)

Indeed, since P>'o is a projection, z E Z >'0 = P>'o Z if and only if z = P>'o (z) or z = P>'o (z), or, by (4.18), z = P>"o (z). But since P>"o is a projection, the last equality is a necesary and sufficient condition for z to be an element of P>"oZ = Z>"o· Thus for the proof of our lemma it remains to verify (4.18). For this purpose we set J.l = Xo + roeic/> and>' = 11 = >'0 + roe-ic/> such that dJ.l = iroeic/> d¢ and d>' = -iroe-ic/>d¢. We then see from (4.17) and (4.13) that

P>.. (z) o

=

21 11'

10r

27r

r 1I'k 1I'Z

10r

27r

R>.(M)(z)roeic/> d¢

------~------------------

11

= --2'

= 21 11'

27r

= 21

RJl(M)(z)roeic/> d¢

R>.(M)(z)roe-ic/>d¢

-co

nkr

1. = --2

11

R>.(M)(z) d>' = -2' 1I'Z

Co

27r

R>.(M)(z)(-iroe-ic/»d¢

-

R>.(M)(z) d>' = P>'o (z).

This proves (4.18), and thus our lemma. 4.11. COROLLARY TO LEMM~ 4.10. (i) The spaces Z>'o and Z>"o have the same dimension; thus 1.1(>'0) = 1.1(>'0) (cf. subsection 2.17). (ii) If >'0 is real, then Z>'o is invariant under conjugation. (iii) If >'0 is real, then the generalized multiplicity 1.1(>'0) of >'0 as an eigenvalue of M equals the generalized multiplicity J.l(>'o) of >'0 as an eigenvalue of L (cf. subsections 2.17 and 4.2). PROOF. Assertions (i) and (ii) are obvious consequences of Lemma 4.10. As to assertion (iii) we noted already in subsection 4.2 that J.l(>'o) < 1.1(>'0)' It remains to prove that (4.19) Now it follows from assertion (ii) of our corollary and Lemma 3.12 that Z>'o = P>'o Z has a base b1 , b2 , .•• ,br consisting of elements of E. Then

r = 1.1(>'0), since rand 1.1(>'0) both equal the dimension of Z>'o (cf. sUbsection 2.17). Since each bj E Z>'0' we have (also by subsection 2.17)

(>'ol - M)nbj

= (),

j

= 1,2, ... ,1.1(>'0)

(4.20)

for som.e n. But by definition (3.9) of the exten!ion M of.L, we see that M(b j ) = L(bj ) since bj E E. Thus by (4.20) p.ol - L) bj = 0, J = 1,2,···,1.1('\0). The

THE LINEAR HOMOTOPY THEOREM

213

asserted inequality (4.19) now follows from the definition of {l{.>.o) given in 4.2 since '>'0 is real. 4.12. LEMMA. We use the notation of Lemma 4.10. We assume now that the linear map l{x) = x - L{x) is a nonsingular L.-8. map E -> E. Let m(z) be its complex extension. Then j(m) = j(l). (4.21) PROOF. By Lemma 3.9 the nonsingularity of l implies that m(z) = z - M(z) is a nonsingular L.-S. map Z -> Z. Thus both members of (4.21) are defined. To prove their equality we note first: if '>'0 is a complex eigenvalue of M, then so is Ao (see Lemma 4.9), and v(.>.o) = v(Ao) by part (i) of Corollary 4.11. Thus v(.>.o) + v(Ao) is an even number. Therefore in Definition 4.1 (see in particular (4.2)) we may omit those Vi = V('>'i) for which '>'i is complex. But for a real eigenvalue'>'i of M we know from part (i) of the corollary 4.11 that V('>'i) = {l('>'d, and the definition of j(l) in subsection 4.3 (see in particular (4.6)) shows that j(m) = j(l). 4.13. We are now ready to prove Theorem 4.8. In the present section we show that Theorem 4.7 implies Theorem 4.6. (a) Suppose (4.10) is satisfied. We have to show that lo and Ii are linearly L.-S. homotopic. Let mo and ml be the complex extensions of lo and it resp. Then, by Lemma 4.12, the equality (4.10) implies (4.11). Therefore, by Theorem 4.7 there exists a homotopy m(z, t) = z-M(z, t) of the properties asserted in that theorem. But by Lemma 3.11 the relation (4.12) implies that for each t E [0,1] the map M(z, t) is the complex extension of a bounded linear map L(x, t): E -> E which by Lemma 3.9 is completely continuous and such that l(x, t) = x - L(x, t) is nonsingular. This l{x, t) gives the desired homotopy between 10 and ll' (b) Assume now that lo and it are linearly L.-S. homotopic. We have to prove (4.1O). By assumption there exists a linear L.-S. homotopy l(x, t) = x - L(x, t) connecting 10 with it. For each t E [0, 1] let m{x, t) = z-M(z, t) be the complex extension of l(x, t). Then m(z, t) is an L.-S. homotopy connecting mo(z) = m{z,O) with ml = m(z, 1) which by Lemma 3.11 satisfies (4.12). The assertion (4.10) follows now from Theorem 4.7 in conjunction with Lemma 4.12. 4.14. In this section we show that Theorem 4.6 implies Theorem 4.7. (a) For i = 0,1 let mi(z) = Z - Mi(Z) be a linear nonsingular L.-S. map Z -> Z which is the complex extension of the linear nonsingular L.-S. map li{X) = X-Li(X): E -> E, and suppose that (4.11) holds. Now by Lemma 4.12 the relation (4.11) implies (4.10). Consequently by Theorem 4.J there exists a linear L.-S. homotopy l(x, t) = x-L(x, t) connecting 10 = l(x, 0) with Ii = l(x, 1). Then the complex extension m(z, t) = Z - M{z, t) of l(x, t) is a linear L.-S. homotopy which connects 1110 with ml and which, by Lemma 3.11, satisfies (4.12). (b) Again, for i = 0,1, let mi{z) = Z - Mi(Z) be a nonsingular linear L.-S. map Z -> Z which is the extension of a linear nonsingular L.-S. map li(X) = x - Li(X): E -> E. Suppose now there exists a linear L.-S. homotopy which

214

THE LINEAR HOMOTOPY THEOREM

connects mo with ml and satisfies (4.12). From the latter relation and from Lemma 3.11 we conclude that there exists a linear 1.-S. homotopy which connects 10 and h and that therefore by Theorem 4.6, j(lo) = j(h). The asserted relation (4.11) follows now from Lemma 4.12.

§5. Proof of the linear homotopy theorem 5.1. The linear homotopy theorem (see subsection 1.3 of Chapter 2) is, as already stated in the last paragraph of subsection 4.5, equivalent to Theorem 4.6, which in turn, by Theorem 4.8, is equivalent to Theorem 4.7. It is thus sufficient to prove Theorem 4.7. We start by proving that if a homotopy mt(z) = m(x, t) of the properties stated in that theorem exists, then (4.11) is true. To do this we will show that j(md is constant for t E [0,1]. Since j(md is integer-valued it will be sufficient to prove the "continuity" of that number or, more precisely, to prove the following theorem. 5.2. THEOREM. Let Z be the complexification of the real Banach space E, and let mo(z) = z - Mo(z) be the complex extension of the nonsingular linear £.-8. map lo(x) = x - Lo(x) mapping E into E. Then there exists a positive number eo of the following property: ifm(z) = z-M(z) is the complex extension of a linear nonsingular L.-8. map l: E -+ E and if

11m - moll

=

11M - Moll < e < eo,

(5.1)

then j(m) = j(mo).

(5.2)

For the proof we need the following well-known elementary lemma.

5.3. LEMMA. Let Z be an arbitrary complex Banach space, and let no be a bounded linear map Z -+ Z which has a bounded everywhere-defined inverse no 1 Then every linear bounded map n: Z -+ Z satisfying

lin - noll < Ilno111- 1

(5.3)

has a bounded everywhere-defined inverse. (As the following proof shows, the lemma is also valid in a real Banach space E.)

PROOF. Suppose first that no = I. Then the assumption (5.3) reads

lin - III < 1,

(5.4)

from which it easily follows that (5.5) i=O

converges (with (I - n)O = /). Moreover, writing n nnl = nln = I, and thus nl = n- 1 . In the general case we see from (5.3) that

III - nolnll = IInol(no -

n)1I ::;

=I

- (I - n), one sees that

Iinollilino - nil < 1.

(5.6)

THE LINEAR HOMOTOPY THEOREM

215

Thus (5.4) is satisfied if n is replaced by n o 1n. Thus n2 = (n o 1n)-1 exists and (n2nO l)n = n2 (no 1 n) = I. This shows that n2nO 1 is a left inverse of n. Similarly, we see from the inequality 111- nn 1 1 < 1, which is proved the same way as (5.6), that n has a left inverse. But the existence of a left and a right inverse proves the existence of a unique inverse. 5.4. PROOF OF THEOREM 5.2. We distinguish two cases: (A) Mo has no eigenvalues A with real part R(>.) ~ 1; (B) Mo does have such eigenvalues. Proof in case (A). We define the subset A = A(Mo) of the >.-plane by A = P I R(>.) ~ I}. Then A is a closed subset of p(Mo). Therefore by Lemma 2.5 there exists a positive number /10 such that

o

IIR,,(Mo)11 < /10/2

(5.7)

for>. E A.

We now claim that if M is a linear completely continuous map Z

---+

Z satisfying (5.8)

then

A C p(M).

(5.9)

Indeed if no(z) = AZ - Mo(z) and n(z) = AZ - M(z) with>. E A, then by (5.8), by (5.7), and by the definition of R" (5.10)

But by Lemma 5.3 this inequality implies that n -1 = R" (M) exists, as a bounded operator, i.e., that>. E p(M) as asserted. It is now easy to see that (5.8) implies the asserted equality (5.2). Indeed j(mo) = 1 by Definition 4.1. But by definition of A we see from (5.9) that no eigenvalue>. of M has a real part 2 1. Thus, again by Definition 4.1, j(m) = 1. Proof in case (B). a(Mo) is closed and bounded (see subsection 2.1). Moreover by Lemma 3.9 Mo is completely continuous and therefore (see subsection 2.16) zero is the only accumulation point of a(M). It follows that a(Mo) contains only a finite number of points A with R(>') ~ 1. Moreover, if we denote them by A?, >.g, ... , >.~, then there exists a positive number r of the following properties: if for i = 1,2, ... , s, di is the open circular disk with center >'? and radius r, then (a) >.? is the only eigenvalue of Mo in the c~osed disk di ; (b) the di are disjoint; (c) for those i for which R(>'?) > 1 the disk di is contained i~ the open half plane R (>.) > 1; (d) for those i for which >.? is not real the disk di does not intersect the real axis; (e) for those>. E a(Mo) for which R(>.) < 1 we have R(>.) < 1- r; (f) 0 < r < 1. In our present case (B) we define 8

A = A(Mo) =

P I R(A) ~

1- r} -

Ud

j .

(5.11)

j=l

Then A is a closed subset of p(Mo). As in case (A) we conclude from this the existence of a positive /10 such that (5.7) holds and for completely continuous linear M satisfying (5.8) the inclusion (5.9) holds.

216

THE LINEAR HOMOTOPY THEOREM

To prove that (5.1) with small enough co implies the assertion (5.2), we choose our notation in such a way that ).~, ).g, ... , ).~ are real and ).~+ 1 , ).~+2' ... , ).~ are complex, 0 ::::; r ::::; 8 (r = 0 means that there are no real numbers among the ).~, and r = 8 that all of them are real; in any case 8 ~ 1 by assumption). Since M o is the complex extension of a completely continuous operator L o: E --+ E, we see from Lemma 4.9 and the corollary to Lemma 4.11 that if), is one of the complex eigenvalues ).~+1, ... ,).8 ofMo, then so is >: and vO().) = vo (>:) , where vo().) denotes the generalized multiplicity of ). as an eigenvalue of Mo. Consequently by definition (4.2) of j(m) j(mo)

= (-1)2::=1 v('\~),

(5.12)

where it is understood that the sum stands for zero if r = O. Suppose now that M is the complex extension of a linear completely continuous operator L: E --+ E and that M satisfies (5.8). Then (5.9) is true and therefore by definition (5.11) of Ll s r s

a(M) n PI R()') ~ 1- r} c

U dj = U dj + U

j=1

j=1

dj .

j=r+1

Now by property (d) of the radius r of the disks dj, these disks do not intersect the real axis for j = r + 1, ... ,8. Thus the eigenvalues of M contained in these disks are complex, and the argument used to establish (5.12) shows that j(m) = (-l)2: v (,\),

where the sum is extended over those eigenvalues). of M which lie in one of the dj for j = 1,2, ... , r and where v().) denotes the generalized multiplicity of >. as an eigenvalue of M. Thus if ).{, ).~, ... ,).t.3 are those eigenvalues of M which lie in dj for j = 1,2, ... ,r, then j(m)

=

".

"'j

(,\j)

(-l)L..j=1 L..i=1 v i .

(5.13)

Comparison of (5.12) with (5.13) shows that our assertion (5.2) will be proved once it is shown that rj

v(>.~) =

L v(>.1)'

j = 1,2, ... , r.

(5.14)

i=1

We recall that here the multiplicity v at the left refers to M o while the v's at the right refer to M. Now by subsection 2.17 and (2.15)

v(>.J) = dim pJ Z, where

pO = J

~ 21l't

r

ladj

j = 1,2, ... , r,

Rj.t(Mo)dJ1.

(5.15) (5.16)

with the circle adj oriented in the counterclockwise sense. In the same way we see that i = 1,2, ... ,rj, (5.17)

THE LINEAR HOMOTOPY THEOREM

where 1 Pj. = -2' 7l't

In ad~

(5.18)

R I1 (M) dJL

and where di is a circular disk with center closures Cf;. are disjoint and lie in di . The assertion (5.14) now reads

Ai

217

and a radius so small that the

T)

dimPJoZ = LdimPj Z,

j = 1,2, ... , r.

(5.19)

j = 1,2, ... ,r,

(5.20)

i=l

If we set pi =

f r

R I1 (M) dJL,

7l't } ad)

we see from the "Cauchy theorem" (2.4) that Tj

PJ(Z)

=

LP1Z

(5.21)

i=1

and from this equality together with Lemma 2.12 that pi Z is the direct sum of the spaces Pi Z, PdZ, ... , PlZ, and that therefore )

Tj

(5.22)

dimPJZ = '~ " dimPJZ t' i=l

Thus the assertion (5.19) may be written as dimPJ Z = dim pi Z,

J' = 1,2, ... ,r.

(5.23)

We will prove that if JLo is as in (5.7), if C1 = (JLO/2)-1 (cf. (5.8)), if C2

=

JLo 2 J=~i~",r IIPJII-l,

(5.24)

and if co = min(cI/2, c2),

(5.25)

then (5.1) implies (5.23) (and thus (5.2)). For the proof we need two lemmas; the first is related to Lemma 5.3 while the second is due to J. Schwartz [50, p. 424J.

5.5. LEMMA. Let no(z) be as in Lemma 5.3, and let n be a linear bounded map Z ---+ Z satisfying (5.26) Then n -1 exists and satisfies

Iln- 1 - no 111 < 211no111211n - noll· 5.6. LEMMA. Let Qo and Q1 be projections Z

---+

(5.27)

Z. Suppose that

IIQ1 - Qoll ::; ~IIQoll-1. Then the dimension 80 of QoZ equals the dimension 81 of Q1Z,

(5.28)

THE LINEAR HOMOTOPY THEOREM

218

5.7. We postpone the proof of the two preceding lemmas and prove that (5.1) implies (5.23). Let no{z) = f..lZ - Mo{z) and n{z) = f..lZ - M{z). Then by (5.8) and (5.7) Iln(z) - no(z)11 = IIM(z) - Mo(z)11 < ~(f..lO/2)-1

<

~IIRJl(Mo)ll-l = ~llnolll-l,

i.e., (5.26) is satisfied. Therefore by Lemma 5.5 the inequality (5.27) holds:

IIRJl(M) - RJl(Mo)II < 21I RJl{Mo)1121IM - Moll· Thus by (5.1), (5.25) and (5.24)

IIRJl(M) - RJl(Mo)ll < 21I RJl(Mo)112f..lo2 . min IIP?II- l , J=1,2, ... ,r and by (5.7)

IIRJl(M) - RJl{Mo)11 < ~ . min IIPJoll-l. J=1.2, .... r But by (5.20) and (5.16) Ilpi - p?11 =

2~ lifad 1

j

(5.29)

RJl(M) - RJl{Mo) df..lll

[21<

~ 271' 10

IIRJl{M) - RJl(Mo)llrlei"'l d¢.

P?

Therefore by (5.29) (recalling that 0 < r < 1) we see that liP} II < ~ IIPJo 11- 1 . But by Lemma 5.6 this inequality implies the assertion (5.23). It remains to prove Lemmas 5.5 and 5.6. 5.S. PROOF OF LEMMA 5.5. That the inequality (5.26) implies the existence of n- l follows from Lemma 5.3. Moreover it follows easily from the proof for that lemma (see in particular (5.5) and (5.6)) that 00

n-lno

= (noln)-l = L(I - nolnF j=O

or 00

-l . n -1 = ""(I ~ -no- ln) jno j=o Thus 00

n- 1 - nol =

L(I - noln)jnol = (I -

00

noln)

j=l

L(/ - noln)knol. k=O

Therefore 00

Iln- 1 - nolll ~ Iinolll· III - nolnll .

L

III -

nolnllk

k=O _ IInolll·III - nolnll < IInolll211no - nil 1 - III - no lnll - 1 - IInOlll . IIno - nil'

THE LINEAR HOMOTOPY THEOREM

219

This inequality together with the assumption (5.26) proves the assertion (5.27). 5.9. PROOF OF LEMMA 5.6. We will establish the inequalities IIQo - QoQtI! } < 1 IIQ1 - Q1QoII ' and then show that they imply our lemma. Since

Q5 =

(5.30) Qo we see that

IIQo - QoQtI! = IIQo(Qo - Qdll ::; IIQoll·IIQo - Q111· By assumption (5.28) this inequality implies the first of the two inequalities (5.30). To prove the second one, we note that IIQol1 ~ 1 since Qo is the identity on its range QoZ. Using this and again the assumption (5.28), we see that

+ Q1 - Qoll ::; IIQol1 + IIQ1 - Qoll IIQol1 + 1/(21IQoll) ::; IIQol1 + IIQoll/2.

IIQtI! = IIQo ::; Thus

(5.31)

IIQ111 < 211QolI· Since Q~

= Q1 we see that IIQ1 - Q1QoII = IIQ1(Q1 - Qo)1I ::; IIQdl·IIQ1 - Qoll·

But if follows from (5.31) and from (5.28) that here the right member is less than 1. This proves the second of the equalities (5.30). To derive the lemma from (5.30), we set Zo = QoZ, Zl = Q1Z, and denote by (QoQdo the restriction of QOQ1 to Zoo Then (QoQdo maps Zo ----t Zo, while Qo restricted to Zo is the identity on Z00 Therefore the first of the inequalities (5.30) implies by Lemma 5.3 that (QoQdo has an inverse on Z00 Thus Zo c range of(QoQdo c range of QOQ1 C range ofQ!, and Zo C Q1 Z = Zl. Therefore 80 = dim Zo ::; 81 = dim Z1. In a similar way one derives from the second of the inequalities (5.30) that 81 ::; 80 . Thus 81 = 80 as asserted. 5.10. We finished the proof that the equality (4.11) is necessary for the existence of the homotopy described in Theorem 4.7. We now turn to the sufficiency proof. We recall that A = 1 is not an eigenvalue of the complex extension Mo of Lo since mo = 1- Mo is not singular. We consider first the special case that Mo has no real eigenvalues> 1. In this case it follows from Lemma 4.12 and Definition 4.3 that j(mo) = j(lo) = 1 = j(I). Therefore in the special case considered we have to prove

5.11. LEMMA. Let Mo satisfy the assumption of Theorem 4.7. In addition it is assumed that Mo has no real eigenvalues A > 1. Then there exists a linear L.-8. homotopy which connects mo with I and which satisfies (4.12). PROOF. For t E [0,1] we set m(z, t) = z - (1 - t)Mo(z). That M(z, t) = (1 - t)Mo(z) satisfies (4.12) is clear fiom Lemma 3.11. It remains to show that mt(z) = m(z, t) is nonsingular for all t E [0,1]. For t = and t = 1

°

THE LINEAR HOMOTOPY THEOREM

220

the nonsingularity is obvious since m(z,O) = mo(z) and m(z, 1) = I. Suppose then mto(z) to be singular for some to in the open interval (0,1). Then Zo (1- to)Mo(zo) = Zo for some Zo i- 0, or M(zo) = Aozo with Ao = (1- to)-1 > 1. This contradicts our assumption that Mo has no real eigenvalues> 1. 5.12. We suppose now that Mo has real eigenvalues greater than 1. We denote them by A1 > A2 > ... > A8 • Let 0"0 = O"o(Mo) be their union, and let 0"1 = 0"1 (Mo) be the complement of 0"0 in O"(Mo): (5.32) Let (5.33) be the projections defined in Lemma 2.11 (cf. also subsection 2.12), and let

zo = PoZ,

(5.34)

Then as will be shown in subsection 5.13

z

=

Zo

+ Z1

(direct sum)

(5.35)

with Zo being finite-dimensional. Corresponding to the decomposition (5.34) we set (5.36) M8=poMo, We will show that

Mo =

(~8 ~J)

(5.37)

(see subsection 5.15) and that 0"0 is the spectrum of M8 and 0"1 the spectrum of MJ (see subsection 5.16). Thus MJ has no real eigenvalues> 1, and Lemma 5.11 may be applied to ma = h -MJ, where II denotes the identity map on ZI. Discussion of the "finite-dimensional" part M8 of Mo will begin in subsection 5.20. 5.13. PROOF OF (5.35). Let Vo and VI be open sets in the A-plane containing 0"0 and 0"1 resp. with Vo and V 1 disjoint. Let r 0 = avo and r 1 = av1, where ro and r 1 are supposed to be rectifiable Jordan curves oriented in the counterclockwise sense. Then by subsections 2.11 and 2.12, by (5.33) and by Lemma 2.9

I = Po

+ PI = ~

r

Rp.(Mo) djl, 2m jr This shows that every z E Z can be represented by Z=ZO+Zl, The uniqueness of this representation follows from (2.12). To establish (5.37) we prove the following lemma.

(5.38)

(5.39)

THE LINEAR HOMOTOPY THEOREM

221

5.14. LEMMA.

(5.40) 1. = -2

MJ

7n

r

ir 1

J.LRJJ.(Mo} dJ.L = MoP!,

(5.41)

where M8 and MJ are the operators defined in (5.36). Moreover, for z E Z,

+ MJ(z},

(5.42)

MJ(zo} = B,

(5.43)

Mo(z} = M8(zo} with Zo, Zl as in (5.39). PROOF. Let

f(J.L} = {01 for J.L E Vo, for J.L E Vl, where Vo and Vl are as in subsection 5.13. Then by (2.10) and (5.33)

Po

1. = -2

r

1n iro

RJJ.(Mo} dJ.L

r

1. = -2

1n irO+r 1

f(J.L}RJJ.(Mo} dJ.L,

and from (5.32) and Lemma 2.9 we see that

1. Mo = -2 1rZ

Therefore from (2.8) with g(J.L}

PoMo

1. = MoPo = -2 1rZ

r

irOurl

r

irO+r 1

= J.L

J.LRJJ.(Mo} dJ.L.

on Vo U V 1

f(J.L}J.LRJJ.(Mo} dJ.L

1. = -2 1rZ

r

iro

J.LRJJ.(Mo} dJ.L. (5.44)

By definition (5.36) this equality proves (5.40). Assertion (5.41) is proved correspondingly. To prove (5.42) we note that by (5.39) and (5.34)

Mo(z) = Mo(zo

+ zd = Mo(zo) + Mo(zt}

= Mo(Poz) + MO(PlZ) =

MoPo(ZO) + MOPl(zt}.

By (5.40) and (5.41) this proves (5.42). Finally by (5.36), (5.44), and (2.12)

M8(zd

= PoMo(zd

= MoPo(zt}

= MOPOPl(z) = O.

This proves the first of the relations (5.43). The second one follows similarly. 5.15. PROOF OF (5.37). Since PoMo = MoPo by (5.44) and since PlMo MOPl is established the same way,

Mo(z)

~ (~:~~~;;) ~ (~~~:~:;) .

Therefore by the direct decomposition (5.39)

Mo(z) = (MoPO(ZO) MOPl(zo)

MOPO(Zt}). MoPl(zt}

=

THE LINEAR HOMOTOPY THEOREM

222

This relation proves (5.37) as we see from (5.40), (5.41), and (5.43). 5.16. Proof that (10 is the spectrum of M8 and (11 is the spectrum of MJ. It follows from (5.42) and (5.43) that, restricted to ZO, Mo = M8, and, restricted to ZI, Mo = MJ. Our assertion now follows from subsection 2.13. This finishes the proof of the assertions contained in subsection 5.12. We next prove

5.17. LEMMA. (i) Zo and ZI are invariant under conjugation (see Definition 3.10); (ii) Zo is finite dimensional. PROOF. (i) Let Aj, j = 1,2, ... , s, be as in subsection 5.12. Let dj be a circular disk with center Aj of radius so small that the closures dj are disjoint and lie in Vo (cf. subsection 5.13). Let the boundary adj of dj be oriented in the counterclockwise sense, and let

Qj

1. = -2 7n

r

ladj

RI'(Mo) dJ.L,

j

= 1,2, ... , s.

(5.45)

Since the A)' are the only eigenvalues of Mo in Vo, it follows from the Cauchy theorem 2.7 and from 5.12 that (5.46)

and

8

E QjZ.

Zo = PoZ =

(5.47)

j=1

Since each >'j is a spectral set for M o, each Qj is a projection by Lemma 2.11, and by the Corollary 4.11 to Lemma 4.10 each of the spaces QjZ is invariant under conjugation since >'j is real. Thus we see from (5.47) that Zo is invariant under conjugation. Since Z also is invariant under conjugation, being the complexification of a real Banach space E, it now follows that ZI also is invariant under conjugation. (ii) By Lemma 2.12 QiQj = 0 for i ':I j. It follows that the decomposition (5.47) is direct. Therefore 8

dimZo = ~dimQ)·Zo. j=1

But by 2.17 the dimension of QjZO is the finite number I/(>"j). Thus 8

dim Zo = ~ 1/(>")-).

(5.48)

j=1

5.1S. Let 10 and II be the identity maps on Zo and ZI resp., and let, as always, I denote the identity map on Z. Then (5.49)

THE LINEAR HOMOTOPY THEOREM

223

where rn8 = 10 -

Mg,

rnA = It - MJ.

(5.50)

This is an immediate consequence of (5.37). 5.19. LEMMA. There exists an L.-S. homotopy rn6(zl' t) = Zl - MJ(Zl, t), Zl E Zl, t E [0,1], connecting rnA with 11 and satisfying

MJ(z,t) = MJ(z,t). PROOF. The lemma will be proved once it is shown that MJ satisfies the assumption made in Theorem 5.11 for Mo. That is, we have to verify: (i) Zl is the complexification of a real Banach space E 1·; (ii)rnA = 11 - MJ and It are the complex extensions of linear nonsingular L.-S. maps on E 1 ; (iii) has no real eigenvalues> 1. Now (i) follows from the fact that Zl is invariant under conjugation (Lemma 5.17). The assertion of (ii), that It is the complex extension of the identity on E 1 , is obvious. To prove that MJ is the complex extension of a linear continuous map LA on E 1 , it is by Lemma 3.11 sufficient to prove the relation

MP

MJ(z) = MJ(z),

z E Zl.

(5.51)

But this relation follows from the fact that MJ(z) = Mo(z) for z E Zl, by (5.41) and that Mo(z) = Mo(z) since Mo is by assumption the extension of a linear continuous map on E. That LA is completely continuous follows from Lemma 3.9 since MJ(z) = Mo(z) is completely continuous by assumption. Finally, that lA = 11 - LA is nonsingular follows also from Lemma 3.9 since otherwise, by that lemma, rnA and therefore rno would be singular. Finally that assertion (iii) is true was already noticed in the lines directly following (5.37). 5.20. Our next goal is to prove that rno = 1- Mo is L.-S. homotopic (with a homotopy satisfying (4.12)) to either I or to a map of type [- (see Definition 4.4). It follows from subsection 5.19 and (5.49) that rno is L.-S. homotopic (with the condition (4.12)) to

rno ~ (~8 ~),

mg ~ 10 -

Mg.

(5.52)

It will therefore be sufficient to prove the corresponding assertion for the map

rn8 which maps the finite-dimensional space Zo onto Zo (cf. Lemma 5.17). Our first step in this direction will be to prove 5.21. LEMMA. There exists a real base b1 , b2 , ... ,bN uf Zo such that j = 1,2, .. . ,N,

(5.53)

where N is the dimension of Zo (cf. (5.48)). PROOF. As shown in the proof of part (ii) of Lemma 5.17 the decomposition of Zo given by (5.47) is direct. Therefore a basis for Zo is obtained by putting

THE LINEAR HOMOTOPY THEOREM

224

together the bases for each of the spaces QjZO, j = 1,2, ... , s. But by subsection 2.17 and by assertion (iii) of Corollary 4.11 (5.54)

Now let >'0 be one of the >'j, and let Qo be the corresponding Qj. QoZo is invariant under conjugation (see Corollary 4.11). It follows that QoZo is the complexification of a real Banach space Eo of the same dimension, and every base of Eo is a real base of Zo (cf. Lemma 3.12). Therefore if Lo is the linear map Eo -+ Eo of which the map M8 (restricted to QoZo) is the complex extension, it will for the proof of our lemma be sufficient to show that there exists a base b1 , •.. , bNo of Eo such that j

= 1,2, ... ,No,

(5.55)

where

lo(x) = x - Lo(x),

x

E

Eo.

Let nl be the "index of nilpotency" of the map QoZo

(5.56) -+

QoZo given by (5.57)

l:>.o = >'oIo - Lo,

i.e., the smallest positive integer n for which (4.5) holds for all x E QoZo. Then by a well-known theorem of linear algebra (see, e.g., [26, p. 111]) there exist integers r, n2, n3, ... ,nr and elements /31. /32, ... ,/3r such that nl 2: n2 2: ... 2: nr and such that the elements

/31, l:>.0/31. ... ,l~~ -1/31, ... ,/3p, i>"o/3p, ... ,l~:-l /3p, ... ,/3r, l:>.o/3r, ... ,l~~ -1 /3r (5.58)

form a base for Eo, while

l'n>"01 /3 1 -- ln'>"02 /32 --

... -ln - '>"0 /3r -r

() .

(5.59)

Thus Eo is the direct sum of spaces Eop with bases

1 /3P , ••• , /3n p --'>"0 ln p-l/3p. /3P1 = /3P' /3P2 --'>"0

(5.60)

We then see from (5.58) that l:>.0/3~

=/3;, lAO/3;

=/3~, ... ,lAo/3;p-l =/3;p,

(5.61)

while by (5.59) (5.62)

But from (5.56) and (5.57)

lo = (1- >'o)Io + LAo, and therefore from (5.61), (5.62), lo/3~ = (1 - >'0)/3~

+ /3;, ... , lo/3;p-l

=

(1 - >'O)/3;p-l

lof3;p = (1 - Ao)f3;p.

+ f3;p, (5.63)

THE LINEAR HOMOTOPY THEOREM We now define for x E Eop and t E [0,1] a homotopy It =

Itf3! = [(1- Ao)(l- t) Itf3~p-l Itf3~P

225

lex, t)

by setting

+ (-l)t]f3! + (1- t)f3;,

= [(1- Ao)(l- t) + (-l)t]f3~p-l + (1 -

t)f3~P,

(5.64)

= [(1- Ao)(l- t) + (-I)t]f3~p.

Obviously (5.63) and (5.64) agree for t

= 0, while for t = 1 by

(5.64)

j = 1,2, ... ,np '

(5.65)

Recalling that Ao > 1 we see that [(1 - Ao)(1 - t) + (-I)t] is the linear convex combination of the points -1 and 1 - Ao < 0, and therefore does not contain the point O. This shows that It is not singular for all t E [0,1]. Thus 10 and it are homotopic and (5.65) proves our lemma 5.22. We will now prove the assertion stated in subsection 5.20 as our "next goal." By subsection 5.20 and Lemma 5.21 this assertion obviously follows from the following lemma.

5.23. LEMMA. Let Zo be a complex Banach space of finite dimension N which is the complexification of a real Banach space Eo. Let m8 be a nonsingular linear map Zo - Zoo Suppose there exists a real base b1 , b2, ... , bN in Zo for which (5.54) holds. Then m8 is L.-S. homotopic if (4.12) holds to the identity I if N is even and to a map of type 1- if N is odd. PROOF. If N is even, we set N = 2p and define m8(x, t) for t E [0,1] and for i = 1, 2, ... ,p by

m8(b2i-l, t) = b2i-l cos(l - t)7r - b2i sin(1 - t)7r, m8(b2i' t) = b2i - l sin(l - t)7r + b2i cos(l - t)7r.

(5.66)

Then m8(z,0) = m8(z) by (5.54), while m8(z,1) sends bj into bj for j = 1,2, ... ,N, i.e., m8(z,1) is the identity map. Moreover m8(z, t) is not singular, the determinant of (5.66) being 1. If N is odd, we change our notation by denoting the given base of Zo by bo, bt, ... ,bN. We then set m8(bo, t) = -bo, while m8(bj , t) for J' = 1,2, ... ,N = 2p is given by (5.66). We thus obtain a nonsingular homotopy m8(z, t) connecting the map m8(z) given by m8(b3') = -bj for j = 0, 1, ... ,N with the map given by m8(bo, 1) = -bo, m8(bj, 1) = bj, j = 1,2, ... , N. The latter map is by definition of the 1- type. 5.23. We thus proved the assertion of subsection 5.20. It is clear that the identity map I on Z is not homotopic to a map m of type 1- since for such m the equality J'(m) = j(I) would hold (see 5.10). But by (4.21) and Lemma 4.5 this leads to the contradiction +1 = -: 1. Thus to finish the proof of Theorem 4.7 it remains to prove that two maps of the 1- type are L.-8. homotopic.

THE LINEAR HOMOTOPY THEOREM

226

5.24. LEMMA. Let mo(z) = z - Mo(z) and Ml(z) = z - Ml(Z) be linear L.-S. maps Z -- Z of type ]- which are complex extensions of the real maps 1o and h resp. Then there exists an L.-S. homotopy m(z, t) = z-M(z, t) connecting mo with ml and satisfying the condition (4.12). PROOF. Again let E denote the real Banach space of which Z is the complexification. 1o and h are then L.-S. linear maps E -- E of type ]-. It is easily seen that it will be sufficient to show that 1o and II are L.-S. homotopic. By Definition 4.4 there exists for i = 0, 1 a direct decomposition E = Ef E~, where is a one-dimensional subspace, and there exists a corresponding unique representation for x E E

+

Ei

such that

lo(x)

= -x~

+ xg = x -

2x~,

Now let eO and e l be unit elements in

h(x) = x - 2x~.

(5.67)

E? and Ef resp.:

Ileo I = Ilelll = 1.

(5.68)

Then by (5.67)

lo( eO) = _eo,

h (e l )

lo(x) = x - 2a(x)eO,

= _e l

,

h(x) = x - 2,8(x)e l

(5.69) ,

where a and ,8 are real-valued continuous functionals on E satisfying

a( eO) = ,8( e l ) = 1.

(5.70)

For the proof that the maps (5.69) are L.-S. homotopic, we distinguish three cases:

(A) eO = elj (B) eO = -elj (C) eO and e l are linearly independent. Case (A). We set

It(x)

=

(1 - t)a(x)

+ t,8(x),

tE[O,I],

(5.71)

and (5.72)

It is clear that for t = a and t = 1 this definition agrees with the one given by (5.69). It remains to verify that It is nonsingular for each t E [O,IJ. If this were not true, then ltex) = for some I E [O,IJ and x E E different from 0, i.e., by (5.72), x = aeo for some real a i:- o. Since it is linear, we see that It(eO) = 0, and therefore from (5.72) that eO(l - 2,t(eO)) = O. Thus

°

(5.72a) But ~ince eO

(1- t)a(eO)

=_ e l

by assumption, we see from (5.70) and (5.71) that It(e O) = = 1 in contradiction to (5.72a).

+ t/3(e l )

227

THE LINEAR HOMOTOPY THEOREM

Case (B). This case reduces to case (A) since, by (5.69), ll(X) = x - 2,8(x)eO with ,8(x) = j3(-x), and since ,8(eO) = j3(-eO) = j3(e 1) = l. Case (C). Let E2 be the two-dimensional subspace of E spanned by eO and e1 , and let E3 be a complementary subspace to E2 such that every x E E has the unique representation (5.73) Now let

rt

be the rotation in E2 determined by

rt(eO) = eO cos(t7r/2)

rt (e 1 ) =

+ e 1 sin(t7r/2),

-eo sin( t7r /2)

+ e 1 cos( t7r /2).

O~t~l.

(5.74)

We note that

r1(e o ) = e 1 . We extend

rt

(5.75)

to a map rt on E by setting, for x E E and t E [0,1],

rt(x) = x - P2(X) + j\(P2(x)),

(5.76)

where P2 is the projection x = X2 + X3 ---+ X2 (cf. (5.73)). This is indeed an extension of rt since x = P2(X) for x E E 2. (5.76) is obviously an L.-8. map. To show that it is nonsingular we will prove that r;l exists: let y = rt(x). 8ince pi = P2 and since rt(P2(x)) E E 2, we see from (5.76) that P2(Y) = rt(P2(x)), and since rt obviously has an inverse, it follows that (5.77) On the other hand, if P3 is the projection x = X2 + X3 ---+ X3, we see from (5.76) that P3(x) = P3(Y). Thus by (5.77) x = P2(X) + P3(x) = r;1(p2(y)) + P3(Y) which shows that the inverse r; 1 exists. We therefore may define for x E E and t E [0, 1J l(x, t) = x - 2o:(rt 1 (x))rt(eO) (5.78) with 0: as in (5.69). Then

l(x,O) = lo(x),

(5.79)

since ro is the identity map. Moreover by (5.75)

l(x, 1)

=x-

20:(rl1(x))e 1 = x - 20:1 (x)e 1 ,

(5.80)

where (5.81 ) To show that lo(x) and l(x, 1) are 1.-8. homotopic we have, because of (5.80), only to show that l(x, t) is not singular. If this were not true, then by (5.78)

x = 20:(r;1(x))rt(eO) for some x E E with x =f. () and some t E [0,1]. Thus x = o. Substitution in (5.82) and cancelling a shows that 1 = 2o:(eO ) which contradicts (5.70). Now we want to prove that lo and (5.70)) are L.-8. homotopic. Since we just proved that

(5.82) art(eO) with a =f. 20:(r;lrt(eo)) = and it (see (5.69) lo and l(·,I) are

228

THE LINEAR HOMOTOPY THEOREM

L.-S. homotopic, it remains to show that h and l(., 1) are L.-S. homotopic. But comparing (5.80) with the definition (5.69) of II we see that we are in case (A) provided that a1(e 1) = 1. (5.83) Now by (5.81), (5.75) and (5.70), a1(e 1) = a(rl1(e 1)) = a(rl1(e 1)) = a(e O) = 1. This finishes the proof of the linear homotopy theorem.

§6. Two multiplication theorems for the indices 6.1. THEOREM. Let E be a real Banach space, and lo and II be two nonsingular L.-8. maps E ~ E. Then (6.1)

j(lold = j(lo) . j(h).

PROOF. Let I be the identity map on E and let l- be a fixed L.-S. map of type 1- on E. Then by Lemma 4.5 j(I) = +1,

(6.2)

By the linear homotopy theorem each of the maps lo and h is L.-S. homotopic to exactly one of the maps I and l-. Denoting L.-S. homotopy by the symbol "~,, there are four cases:

(i) (ii)

lo lo lo lo

(iii) (iv)

~

I,

h

~I;

~

I,

~

~

l- , l-,

II II II

~

~ ~

l-; I; l-.

(6.3)

Then loll

~

I· I = I

loh

~

I . l- = l-

loh

~

l- 1= l-

in case (iii),

loh

~

l-l- = I

in case (iv).

in case (i), in case (ii),

By Theorem 4.6 and by (6.2) this implies j(loh) = +1 j(lold

= -1 in case (ii),

j(lold = -1 j(lold

in case (i), (6.4)

in case (iii),

= +1 in case (iv).

On the other hand we see from (6.3), from Theorem 4.6 and from (6.2) that j(lo) = 1, j(lo) = 1, j(lo) = -1, j(lo) = -1,

j(ld = 1 j(h) = -1 j(ld = +1 j(h) = -1

in in in in

case case case case

(i),

(ii), (iii), (iv).

Comparison of (6.4) with (6.5) makes the assertion (6.1) evident.

(6.5)

THE LINEAR HOMOTOPY THEOREM

6.2. THEOREM. Let l be a nonsingular L.-8. map E such that every h E E has the unique representation

-+

229

E. Let E

=

El

+E2 (6.6)

Let

It

and l2 be the restrictions of l to El and E2 resp., and suppose that l(h)

=

(It(h d

o

0) .

l2(h 2 )

(6.7)

Then j(l)

= j(It)

. j(l2)'

(6.8)

It is clear that It and l2 are linear nonsingular L.-8. maps El E2 resp. Thus the right member of (6.8) is defined. Now let

PROOF.

and E

-+

ml =

C~ ~),

-+

E

(6.9)

where It and 12 are the identity maps on El and E2 resp. Then l = mlm2. Therefore, by Theorem 6.1, j(l) = j(mt) . j(m2)' It remains to verify that

(6.10)

j(lt) = j(mt),

By Definition 4.3 of the index j, it will for the proof of the first of these equalities be sufficient to show: (i) a real number>' i=- 0 is an eigenvalue of Ml = 1- ml if and only if>. is an eigenvalue of Ll = II - ll; (ii) for such an eigenvalue>. i=- 0

(6.11) where J.,Lh (A) and J.,Lrnl (A) denote the generalized multiplicity of A as an eigenvalue of Ll and Ml resp. (see Definition 4.2). Proof of (i). Let>. i=- 0 be an eigenvalue of Ml = I - mI· Then there exists an h i=- 0 such that (I - mt)h = >'h or by (6.9) and (6.6) such that (6.12) and Oh 2 = >.h 2. This implies h2 = B since>. i=- O. Thus h = hI, and (6.12) shows that>. is an eigenvalue of L 1 . The converse follows even more easily. Proof of (ii). Let>. i=- 0 be an eigenvalue of Ll and therefore, by (i), an eigenvalue of Ml = I - mI' By (6.9) >.I - Ml = (

>.II - Ll 0

and for any positive integer n (>.I _ M )n = 1

((Aft -0 Lt)n

0 ) >'h'

0) .

>.nI2

230

THE LINEAR HOMOTOPY THEOREM

From this, from (6.6) and from>. -# 0 it follows that for h E E, (>..J - Mdnh = (j if and only if (>'h - Ldnhl = (j and h2 = (j. This implies (6.11) by Definition 4.2, and therefore, by Definition 4.3, proves the first of our assertions (6.10). The second one is proved correspondingly.

6.3. COROLLARY TO THEOREM 6.2. Let E, E I , E 2, h, ht, h2 be as in Theorem 6.2, and for i = 1,2 let 7ri be the projection h = hI + h2 ~ hi. Let m be a nonsingular L.-8. map E ~ E. Let ml denote the restriction of 7rlm to E I , and let m2 denote the restriction of 7r2m to E 2 . We assume (6.13) Then j(m) = j(md . j(m2).

PROOF. Let ml2(h 2 )

(6.14)

= 7rlm(h2). Then

m(h)

=

(ml(h d (j

m l2 (h 2 )). m2(h 2 )

(6.15)

We note first that m2 is nonsingular, for otherwise m2(h 2 ) = 0 for some h2 -# (j. Then m(h) = 0 for h = h2 since then hl = (j and ml2(h 2 ) = 7rlm(h2) = (j. This contradicts the assumed nonsingularity of m. Similarly one sees that ml is not singular. Now let for t E [O,lJ mt(h)

= (

ml~hd

(6.16)

We prove that for each t this map is not singular by showing that for arbitrary

k E E the equation (6.17) has a solution. Now if k = kl (6.17) is equivalent to

+ k2

with kl E Ei, then by (6.16) the equation

Since ml and m2 are nonsingular, this system obviously has a solution. We thus see from (6.16) and (6.15) that m = m l is L.-S. homotopic to the map 0_ (ml(hd m (j

(j)

m2(h2)·

Consequently by Theorem 4.6, j(m) = j(mO). But j(mO) Theorem 6.2. This proves the assertion (6.14).

= j(mt} . j(m2)

by

ApPENDIX

B

Proof of the Sard-Smale Theorem 4.4 of Chapter 2 1. THE THEOREM OF SARD. Let Rn and Rm be real Euclidean spaces of finite dimension nand m resp. Let U be an open subset of Rn, and let f E CT (U): U ----t Rm, where r is a positive integer satisfying

r > max(O, n - m).

(1)

Then the set of critical values of f (see Definition 17 in Chapter 1) is of Lebesgue measure O.

For a proof of and further literature on Sard's theorem we refer the reader to [1, §15]. 2. COROLLARY TO SARD'S THEOREM. Let En and Em be (real) Banach spaces of finite dimension nand m resp. Let [2 be a bounded open set in En and f E CT(O) (cf. §1.16) where r is a positive integer satisfying (1). Then the set of points in Em which are regular values for f (see §1.17) is dense in Em.

This is an obvious consequence of Theorem 1 since on the one hand a subset of Rm of measure 0 contains no open set and therefore its complement contains a point in every open subset of Rm and is thus dense in Rffl, while on the other hand every finite-dimensional Banach space is linearly isomorphic to a Euclidean space of the same dimension (see, e.g., [lS, p. 245]). 3. The proof given below for the Sard-Smale theorem is an adaptation to the simpler Banach space case of the proof given in [1, §16] for Smale's generalization of Sard's theorem to certain Banach manifolds. Using the preceding corollary we will prove a "local version" of the Sard-Smale theorem in Lemma 4 for maps of a very special form. The lemma in §5 will allow us to show that the local theorem holds also for the more general maps treated in §6. The latter result implies the Sard-Smale theorem as shown in §7. 4. LEMMA. Let III, K, and K* be Banach spaces, where K is finite dimensional and where dimK* = dimK - p, (2)

231

232

PROOF OF THE SARD-SMALE THEOREM 4.4 OF CHAPTER 2

with p being a positive integer. Let TI and direct products

~

be the Banach spaces defined as the (3)

(cf. §1.3). Let V be a bounded open subset ofTI. Let'l/J be a CS-map V where s is an integer satisfying

---+

(4)

s;:::p+l. We suppose moreover that the map 'l/J: V

---+

L:

W = 'l/J(V) is of the special form

(5) where (6)

and where the map X: V ---+ K* is completely continuous. Now if <;0 is a point of V, then there exists an open neighborhood Vo of <;0 such that (7) VocVocV and such that the set R(i/Jo) of those points of L: which are regular values for the restriction 1fio of'l/J to Vo is dense and open. Let <;0 = <;P + kO, <;P E TIl, kO E K. Since <;0 is an element of the open set V, there obviously exist positive numbers Cl and C2 such that the "rectangular" neighborhood Vo = VI X V2 satisfies (7) if PROOF.

VI = {<;l

E

TIl I 11<;1 - <;PII < cI},

V2 = {k

E

K I Ilk - k011 < c2}.

(8)

We will show first that the set R('l/Jo) of regular values for the restriction 'l/Jo of 'l/J to Vo is dense in ~. For this purpose we consider an arbitrary point (j = ~ I + k* in L: where ~ 1 E TIl and k* E K*. We have to show that every neighborhood N of (j contains a point which is a regular value for 'l/Jo. Now this is obvious if the equation (9) has no solution in Vo since then, by definition, (j is a regular value of 'l/Jo. Suppose now (9) has a solution ~l + k, ~l E VI, k E V2· Then by (5) and (9), ~l = ?" 1 and

x(?"l+k)=k*. Now for ~l

E

VI we define a map X~l : V2 X~l (k)

C

(10)

K into K* by setting

= X(?"l + k).

(11)

From (2) and (4) we see that this map satisfies the assumption of the Corollary 2 (with Em = K*, En = K). Therefore by that corollary every neighborhood N2 C K* of k* contains a regular value k* for X~l· We will show that for such k* the point a = ~l + k* E L: is a regular value for 'l/Jo· This will obviously prove our assertion that every neighborhood N of (j = ?" 1 + k* contains a regular value for'l/Jo. Now if the equation X~l (k) = k* (12)

PROOF OF THE SARD-SMALE THEOREM 4.4 OF CHAPTER 2

233

has no solution k C V2 , then the equation (13) has no solution ~ E Vo as is seen from (5) and (11). Thus in this case ~1 + k* is a regular value for 'l/Jo. Suppose now (12) has a solution k E V2 . Then since k* is a regular value for X~l' the differential DX~l (k; x:), x: E K, is nonsingular. On the other hand, ~ = ~ 1 + k is the corresponding solution of (13), and with h = A + x:, A E III, x: E K, we see from (5) that

D¢ok;h)

~ (DX~;A) DX~(;K)).

(14)

But DX(~; x:) = DX~l (k; x:) by (11). This shows that the differential (14) is not singular and therefore that ~ 1 + k* is a regular value for 'l/Jo for every solution ~ E VI X V2 of (13). This finishes the proof that R('l/Jo) is dense in E. Now let Uo be a rectangular neighborhood of ~o whose closure U 0 is a subset of Vo. As follows directly from the definition of "regular value," the relation U 0 c Vo implies that the set of regular values for the restriction 1[;0 of 'I/J to U 0 contains the set R ('l/Jo) and is therefore dense in E. In other words, changing our notation, we can choose the rectangular neighborhood Vo of ~o satisfying (7) in such a way that R(1[;o) is dense in E. To prove our lemma it remains to show that R(1[;o) is open or, what is the same, that its complement, the set 8(1[;0) of singular values for 1[;0' is closed. Now it follows directly from the definition of "singular value" that 8(1[;0) = 1[;0(8t} if 8 1 denotes the set of singular points of 1[;0 in Vo. But to prove that 8 1 is closed, it will be sufficient to show that 8 1 is closed as follows from the special form (5) of'I/J (cf. the proof of part (ii) of §1.1O). The proof that 8 1 is closed is quite similar to the proof of part (ii) of Lemma 19 in Chapter 1: let ~1, ~2, .•. be a convergent sequence of points in 8 1 . We have to prove that ~ = lim ~i E 8 1 . ,~oo

Since ~i E 8 1 C V 0 and since V 0 is a closed set, we see that ~EVoCV.

Now if ~ ~ 8 1 the differential D'I/J(~; h) would not be singular, i.e., ~ C V would be a regular point for 'I/J. But then, by the continuity of the differential and by Lemma 5.3 of Appendix A, all points of some neighborhood of ~ would be regular points for 'I/J. This contradicts the fact that every neighborhood of ~ contains infinitely many of the singular points ~i. 5. LEMMA. Let II be a Banach space, let E be a subspace of II of finite codimension p, let Z be an open bounded subset of II, and let ¢ = ¢(z) be an

234

PROOF OF THE SARD-SMALE THEOREM 4.4 OF CHAPTER 2

L.-S. map Z -+ E. We suppose that ¢ E GS(z) for some integer s :::- 1. Let Za be a point of Z and la(T}) = D¢(za;T}). (15) Moreover, let K denote the kernel ofla, i.e., K = {T} E II Ila(T}) = O}. Then the following assertions (i) and (ii) are made: (i) K is finite dimensional, and there exists a subspace III of II such that

II = IIl-+-K.

(16)

Moreover if R c E is the range of la, then there exists a finite-dimensional subspace K* of E with the properties:

(a)

(b)

E = R-+-K*;

dimK* = dimK - p.

(17)

Thus every y E E has the unique representation

r E R, k* E K*,

( 18)

where 7rl and 7r2 are the projections y -+ rand y -+ k* resp. (ii) Let I; be the direct sum of III and K* (see part (ii) of§1.3): I;

= III -+- K*

(19)

such that each point a E I; has the unique representation a

= al + a2 = 7rf(a) + 7rr(a),

al

E

III,

a2 E K*,

(20)

where 7rf and 7rr are the projections a -+ al and a -+ a2 resp. It is asserted: there exists a linear L.-S. isomorphism h of E onto I; and an L.-S. GS isomorphism hI of some neighborhood U C Z c II of Za onto some neighborhood V C II of h1(Za) such that the map

'l/J(~)

=

h¢(hl1(~)),

\ E V,

(21)

is of the form (5) where \, \1, \2 and X are as described in the lines following (5). PROOF. (i) That K is finite dimensional follows from Lemmas 12 and 18 in Chapter 1, and the existence of a III satisfying (16) follows from Lemma 4 in Chapter 1. Since ¢ is an L.-S. map whose domain is the subset Z of II and whose range lies in the subspace E of II, it may also be considered as an 1.-S. map Z -+ II. Therefore, by Lemma 12 of Chapter 1 there exists a finite-dimensional subspace Ki of II such that II = R-i-Ki· (22) Now En R = R since R C E. Therefore intersecting both members of (22) with E we see that the finite-dimensional space K* = E n Ki satisfies (17 a). Since p is the codimension of E in II, comparison of (22) with (17a) shows that dimK* = dimKi - p. But by §1.12, dimKi = dimK. This proves (17b). Proof of (ii). We note first that the restriction of la to III maps III onto R. Indeed, if r is a given element of R, there exists by definition of R a Z E II such that la(z) = r. But by (16)

Z = Z1

+ k,

(23)

PROOF OF THE SARD-SMALE THEOREM 4.4 OF CHAPTER 2

235

8ince loCk) = () by definition of K, we see that lo(zt} = r. Thus 10 as a map III -+ R is a nonsingular L.-8. map III onto R. It follows from the argument given for the proof of part (v) of §1.12 that this map has an inverse m: R -+ III which is a nonsingular 1.-8. map. Now every y E E has the decomposition (18). If for r = 7rl(Y), k* = 7r2(Y) we define hey) = mer)

+ k*,

(24)

we obtain a map of E onto the space E defined by (19). 8ince mer) is linear and 1.-8. and since K* is finite dimensional, it follows that h is a linear nonsingular 1.-8. map. Using the decomposition (16) we set for z = Zl + k E Z, Zl E Ill, k E K, (25)

8ince ¢(z) E E, it follows from (18) that 7rl¢(Z) E R. Therefore the first term of the right member of (25) is an element of Ill, and hl(z) E II by (16). Moreover since ¢ and mare L.-8. mappings and since K is finite dimensional, we see from (25) that hl is an L.-8. map. We prove next that hl maps some open neighborhood U of Zo onto some open neighborhood V of hl(zo) and that hll E CS(V). For this purpose it will by the inversion theorem in §1.21 be sufficient to show that the differential of hl at Zo is the identity map on II, i.e., that (26)

for all rJ = rJl + '" E II, rJl E Ill, '" E K (cf. (16)). Now since m and 7rl are linear, we see from (25) that Dhl (z; rJ)

= m7rl D¢(z; rJ) + '"

(27)

(28)

for all Z E Z. But for Z = Zo we see from (15) and from (27) that D¢(zo;rJ) = 10(rJ) = lo(rJd since", E K and, therefore, lo("') = (). Since R is the range of lo, we see from (18) that 7r l D¢(zo,rJ) = 7rlIO(rJd = 10(rJd· But 10(rJd is the restriction of lo to II 1 and, by definition, m is its inverse. Therefore m7rl D¢( zo; rJ) = m1o(rJd = rJl. Thus by (28) Dhl(zo;rJ) = TIl + "', which by (27) proves (26). Then let U and V be sets of the properties stated above. We set (29)

If Z varies over U, then ~ varies over V. 8ince ~ E II we may write ~ = ~l ~l E Ill, ~2 E K (cf. (16)). Then by (25) and (20) ~2 =

Now if

tP(~)

= h¢hll(~),

k E K.

~ E V,

then with z = hll(.;) by (24), (18) and (30)

tP(.;)

= h¢(z) = m7rl¢(z) + 7r2¢(Z) = ~l + 7r2¢hll(~).

+ ~2, (30) (31)

236

PROOF OF THE SARD-SMALE THEOREM

4.4

OF CHAPTER

2

With X(~) = 1r24>hll(~) this equality shows that 'l/; is of the asserted form (5). Obviously X( ~) is bounded and lies in K*. Since K* is finite dimensional, we see that X is completely continuous. Thus assertion (ii) of Lemma 4 is proved. 6. LEMMA. With the notation used in Lemma 5 we suppose that the assumptions of that lemma are satisfied. In addition we strengthen the assumption that 4> E CS (Z) with s ~ 1 by requiring that s~p+l.

(32)

It is asserted that corresponding to each point Zo E Z there exists an open set Uo which in addition to satisfying zOEUoCUoCZ

(33)

has the following property: if¢o is the restriction of 4> to Vo, then the set R(¢o) of points in E which are regular values for ¢o is dense and open.

PROOF. It follows from Lemma 5 together with the assumption (32) that all assumptions of Lemma 4 are satisfied. Consequently, the point

(34) has an open neighborhood Vo with the properties asserted in that lemma. We claim that then the set

(35) has the properties asserted in the present lemma. In the first place Uo is open and U 0 = hl1 (V 0) as follows from Theorem 2.3 in Chapter 5. Next the following two assertions hold: (a) for Yo E E the equation 4>( z) = Yo has a solution Z E U 0 if and only if for 0"0 = h(yo) E E the equation 'l/;(~) = 0"0 has a solution ~ E Vo; (b) for Zo E U 0 the differential D4>( Zo; .) is not singular if and only if the differential D'l/; (h I (zo); .) is not singular. (a) follows directly from the relation (21) between 'l/; and 4> and the properties of h and hI. (b) holds for the same reasons in conjunction with the chain rule. But (a) and (b) together imply that R(¢o) = h- l R(if;o)· Now R(00 ) is dense and open by Lemma 4. Therefore h- l R(00 ) is dense and open since h- l is a linear one-to-one 1.-S. map E onto E. 7. Prool 01 the Sard-Smale theorem 4.4 0/ Chapter 2. Let Yo be an arbitrary point of E. We have to prove that every neighborhood (in E) of Yo contains a regular value for 4>. Now if the equation

Yo

E

E - 4>(8Z),

(36)

has no solution, then, by definition, Yo is a regular value for 4> and every neighborhood of Yo contains a regular value, e.g., Yo. Suppose now that the set Co of roots of (36) is not empty. Co is compact by Lemma 10 in Chapter 1, and every Zo has a neighborhood Uo of the property

PROOF OF THE SARD-SMALE THEOREM 4.4 OF CHAPTER 2

237

asserted in Lemma 6. As Zo varies over Co, these neighborhoods form an open covering of this compact set. Consequently there exists a finite subcovering, say, UJ,UJ, ... ,Uti such that (cf. (33)) for i = 1, 2, ... , q q

[J,io C

U i0 C Z·,

Co

c

U= 2:Uo.

(37)

i=l

If ¢i is the restriction of ¢ to Ub, then by Lemma 6 the set R(¢i) of regular values for ¢i is dense and open in E. Consequently by Baire's density theorem (see e.g. [2; p. 108, Theorem V']) the set

n q

~=

R(¢i)

(38)

i=l

is dense in E, and therefore in every neighborhood of Yo. We show next the existence of an open neighborhood No of Yo such that (39) Indeed, otherwise, there would exist a sequence of positive numbers 8v converging to 0, of points Yv E B(yo, 8v ), and of points Xv such that

Xv EZ,

¢(Xv) = Yv,

(40)

but

Xv

tt U.

(41)

Now since the Yv converge to Yo and since ¢ is an 1.-S. map, it is easy to see that a subsequence of the Xv converges to a point Xo which by (40) satisfies ¢(xo) = Yo, i.e., (42) But by (41)

Xo

tt U

(43)

since U is open. But by (42) and (37), Xo E Co C U which contradicts (43). Then let No be a neighborhood of Yo satisfying (39), and let ~l = ~ n No. ~l is not empty since ~ is dense and No open in E. We now show that every point Yl of ~l is a regular value for ¢. If the equation

¢(z) = Yl

(44)

has no solutions, there is nothing to prove. But if it has a solution, then every solution lies in U by (39), and therefore, by (37), in some of the sets U&, say for i = 1,2, ... , ql with i ::::; ql ::::; q. Consider now for such i a solution z E U&. Then ¢(z) = ¢i(Z) = Yl E ~ = n;=l R(¢j) c R(¢i), and D(z; h) is not singular. This finishes the proof that every point of No n ~ is a regular value for , and therefore it finishes the proof of the Sard-Smale theorem since an No satisfying (39) can be chosen as a subset of a given neighborhood of Yo·

References

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Index Affine map, 115 Antipodal points, 138

Convex hull, 22 set, 21 Coordinate transformation, 100, 101 CT-map,19 Critical point, 19 value, 19

Ball, 13 Banach space, 13 Barycenter, 125 Barycentric subdivision, 125 Base of a vector space generated by a set, 24 Bilinear map, 18 Borsuk theorem on odd maps, 192 Antipodensatz, 198 Borsuk-Ulam theorem, 198

Differential, 18 Direct sum, 13 summand, 13 Dugundji extension theorem, 22

Cauchy theorem (for operator-valued integrands), 201 Central projection, 146, 147 Chain rule for differentials, 18 Codimension, 14 Co-kernel, 17 Compact set, 15 Complementary space, 13 Complete continuity, 15, 203 Complex extension of a linear "real" operator, 205 Complexification of a real Banach space, 204 Component of an open set, 23 Conjugation, 206 Connected set, 21

Eigenelement, 204 Eigenvalue: definition, 203 generalized multiplicity of, 204 multiplicity of, 204 Extension theorem for odd maps, 185 Face of a simplex, 115 Finite layer map, 28 l(M), definitiOl. of, 201 Fredholm alternative, 17 General position of a point set, 113 Generating set for a vector space, 24 Great circle, great sphere, 138 241

242

INDEX

Half-space, 117 Homotopy extension theorem, 176 Hopf extension theorem, 172 Hopf-Krasnoselskil homotopy theorem, 173 Hull of planes in En, 113

1- type map, 209 Independence of points in En, 112 Index of a linear nonsingular L.-S. map on a complex Banach space, 208 on a real Banach space, 27, 209 of an isolated root for mappings between spheres, 143 Intersection interpretation of degree and winding number, 84, 153 Invariance of domain, theorem of, 98 Inversion theorem, 20 Jordan-Leray theorem, 104 Kernel of a linear map, 17 Krasnoselskil's lemma, 19 Layer map, 28 Leray-Schauder's first lemma, 64 second lemma, 66 Leray-Schauder (L.-S.) maps, 15 generalized (g.1.-S.), 69 Linear homotopy theorem, 26, 210 1.-S. homotopy, 26 Mazur lemma, 22 Multiplication theorem for indices, 228 Nonsingular linear map, 17 Order with respect to a point, 147 Orientation of En, 116 of a simplex, 115, 116 of a sphere, 139 ABCDEFGHIJ -

89876

Poincare-Bohl theorem, 76, 78 Polygon, simple polygon, 21 Precompact, 22 Product of Banach spaces, 13 Projection, 14, 202 Ray, 21 Real elements of a complexified Banach space, 204 Regular linear map, 17 point, 19 value, 19 Resolvent set, 200 Rouche theorem, 78 Sard theorem, 231 Sard-Smale theorem, 40 Segment, 21 Simplex, 114 Simplicial approximation, 126 map, 123, 153 set, 123 Singular linear map, 17 point, 19 value, 19 (2-) Smooth Banach space, 63 Spectral set, 200 Spectrum, 200 Sphere, 138 Star, 125 domain, 79 Stereographic projection, 139 Subspace, 13 Sum theorem, 28 Tangent space, tangent plane, 139 'Transversal, 22 Vector space generated by a set, 24 Winding number, 77