Nonlinear Functional Analysis

Nonlinear Functional Analysis

Rajendra Akerkar

Narosa Publishing House New Delhi

Madras Bombay Calcutta London

Rajendra Akerkar Department of Computer Studies Chh. Shahu Central Institute of Business Education & Research University Road, Kolhapur, India

Copyright © 1999 Narosa Publishing House

NAROSA PUBLISHING HOUSE 6 Community Centre, Panchsheel Park, New Delhi 110 017 22 Daryaganj, Prakash Deep, Delhi Medical Association Road, New Delhi 110 002 35-36 Greams Road, Thousand Lights, Madras 600 006 306 Shiv Centre, D.B.C. Sector 17, K.U. Bazar P.O., New Mumbai 400 705 2F-2G Shivam Chambers, 53 Syed Amir Ali Avenue, Calcutta 700 019 3 Henrietta Street, Covent Garden, London WC2E 8LU, UK

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publishers. All export rights for this book vest exclusively with Narosa Publishing House. Unauthorised export is a violation of Copyright Law and is subject to legal action. ISBN

81-7319.230-8

Published by N.K. Mehra for Narosa Publishing House, 6 Community Centre, Panchsheel Park, New Delhi 110 017 and printed at Replika Press Pvt. Ltd., Delhi 110 040 (India).

PREFACE This book presents the central ideas of applicable functional analysis in a vivid and straightforward fashion with a minimum of fuss and formality. The book was developed while teaching an upper-division course

in non-linear functional analysis. My intention was to give the background for the solution of nonlinear equations in Banach Spaces, and this is at least one intention of applicable functional analysis. This course is designed for a one-semester introduction at post-graduate

level. However, the material can easily be expanded to fill a two semester-course. To clarify what I taught, I wrote down each delivered lecture. The

prerequisites for this text are basic theory on Analysis and Linear Functional Analysis. Any student with a certain amount of mathematical

maturity will be able to read the book. The material covered is more or less prerequisite for the students doing research in applicable mathematics. This text could thus be used for an M.Phil. course in the mathematics. The preparation of this manuscript was possible due to the excellent facilities available at the Technomathematics Research Foundation, Kolhapur. I thank my colleagues and friends for their comments and help. I specially thank Mrs. Achala Sabne for the excellent job of preparing the camera ready text. Most of all, I would like to express my deepest gratitude to Rupali, my wife, in whose space and time this book was written. R. AKERKAR

CONTENTS Preface

1. Contraction Banach's Fixed Point Theorem 1.2 The Resolvent Operator 1.3 The Theorem of the Local Homeomorphism 1.1

2. Differential Calculus in Banach Spaces

V

1 1

9 11

17

The Derivative

17

2.2 Higher Derivatives

28 36

2.1

2.3 Partial Derivatives

3. Newton's Method

39

4. The Implicit Function Theorem

47

5. Fixed Point Theorems

55

5.1 The Brouwer Fixed Point Theorem

5.2 The Schauder Fixed Point Theorem

6. Set Contractions and Darbo's Fixed Point Theorem Measures of Noncompactness

55 60

65

6.2 Condensing Maps

65 72

7. The Topological Degree

77

6.1

Axiomatic Definitions of the Brouwer Degree in R" 7.2 Applications of the Brouwer Degree 7.3 The Leray-Schauder Degree 7.4 Borsuk's Antipodal Theorem 7.5 Compact Linear Operators 7.1

77 80 87 92 99

x

Contents

8. Bifurcation Theory 8.1

An Example

8.2 Local Bifurcation 8.3 Bifurcation and Stability 8.4 Global Bifurcation

9. Exercises and Hints References Index

105

105 110 116 123

129

153 155

Chapter 1 CONTRACTION 1.1

Banach's Fixed Point Theorem

Let (X, d), (Y, d) be metric spaces. A mapping F : X -+ Y is said to be Lipschitz continuous, if there exists a constant k > 0, such that for all x1, x3 E X d(F(xi), F(xs)) < k.d(xi, X2)-

F is called a contraction, if for all x1i x2 E X, x1

,-

x2

d(F(xi),F(x2)) < d(xi,xs). F is called a strict or a k-contraction, if F is Lipschitz continuous with a Lipschitz constant k < 1.

If X C Y, F : X - Y, then t E X is called a fixed point

of F, if flt) =.t. If an equation

H(x) = y

(1.1)

is to be solved, where H : U -- X is a continuous mapping from a subset U of a normed space X into X, then this equation can be transformed in a fixed point problem : 1

Nonlinear Functional Analysis

2

Let T : X -- X be an injective (linear) operator, then (1.1) is equivalent to

TH(x) = Ty x = x - TH(x) + Ty hence

x = F(x)

(1.2)

where F(x) = x - TH(x) + Ty. The (unique) fixed point x of (1.2) is a (the unique) solution of (1.1), since T Is injective. T can be chosen, such that some fixed point principles are applicable. Now we will start with the most important fixed point theorem.

Theorem 1.1 (Banach's Fixed Point Principle) Let X be a complete metric space. Let F : X -+ X be a k-contraction with 0 < k < 1, i.e. V x1, x2 E X d(F(xl), F(x2)) < k.d(x1, x2) Then the following hold 10 There exists a fixed point x of F. 20 30

40

This fixed point is unique.

If xo E X is arbitrarily chosen, then the sequence (xn), defined by xn = F(xn_1) converges to x.

For all n the error estimate is true d(xn, ±) <

k

1-k

d(xn, xn-1) <

kn

1-k

d(x1, xo).

Contraction

3

Proof : For xo E X we have

d(xn+l,xn) = d(F(xn),F(xn-1)) <_ k.d(xn)xn-1) < ... < kn.d(x1, xo) and

d(xn+j+l, xn)

:5

d(xn+j+1, xn+j) + ... + d(xn+1, xn)

< (0+1 + ... + k).d(xn, xn-1) k

< 1

- k'

d(xn,xn-1)

kn

1- kd(x1, xo) Since k < 1, the sequence (xn) is a Cauchy sequence. Since X is complete, lira xn = I exists. By continuity of F, we have F(x) = lim F(xn) = lim xn+1

= i,

hence 1 is a fixed point of F. If i is a fixed point of F, then

d(i, x) = d(F(i), F(l)) < k.d(i,1) implies x = x (k is less than 1!) and we obtain the error estimates by

d(i,xn) <

lim d(xn+j-1, xn) H0

<

kn 1

k 1

k.d(xn,xn-1)

k.d(xl,xo) 13

This theorem meets all requirements for a useful mathematical statement: Existence, Uniqueness, Construction and Error Estimate.

Nonlinear Functional Analysis

4

If not necessarily F itself, but almost all iterates

Fn =

FoFn-1

are kn-contractions, we obtain the following result.

Theorem 1.2 Let X be a complete metric space, for F : X X we assume: There exists a sequence (kn) of positive reals, such that for

allx,yEX d(Fnx, F"y) < kn.d(x, y)

Ekn < 00Then F has a unique fixed point ±, and 1 = limxn = limF' (xo) with

d(xn, x) < > kj.d(xl, xo) j>n

Proof : This proof is analogous to the proof of Theorem 1.1.

d(xn+j+1, xn)

:5

d(xn+j+1, xn+j) +

+ d(xn+l, xn)

< d(Fn+jx1, F`+jxo) + ... + d(F`xl, Fnxo)

< (k,,+... + kn).d(x1, xo) Thus, (xn) is a Cauchy sequence. Let x = lim xn, then F(x) = lim F(xn) = lim xn±1 = i, i.e. 1 is fixed point; if 1 is a fixed point of F, so x is a fixed point for all Fn, hence

d(F`i, Fnx) < kn.d(2, x), implies x = i, since kn < 1 for almost all n and d(i, xn) < 1i m d(xn+j+1, xn) < E kj.d(xl, xo). j>n

C)

Contraction

5

If F is just a contraction, then F does not necessarily have a fixed point: Let X = 10, oo) and F : X -+ X be defined by

F(x) = x +

1

x+l.

F(x) = x + z+l+ # x, but

F(x) - F(y) =

i.e. 1 -

(1+47

y) = 1-

)2

(1 +

(x - Y)

< 1, thus, if x # y, I F(x) - F(y)I < I x - yl.

If we additionally assume that (X, d) is a compact metric space,

then we obtain the following result.

Theorem 1.3 Let X be a compact metric space, F : X --> X a contraction. Then F has a unique fixed point and 1 with = 1imxn,

xn = F(xn-1), x0 E X .

Proof : Since X is compact, the sequence (F(xn)) has a convergent subsequence (F(xn, )). Let

= j-oo lim F(xn, ), then F(2) = l lip F(xn,+1).

If i L F(x), there exist disjoint closed neighbourhoods U of x and V of F(i). The mapping P:UxV

R, P(x, y) =

F(y)) ,y) d(Fdx),

(

6

Nonlinear Functional Analysis

is continuous, and attains its maximum k < 1. Let p E N, such

that for j > p F(xn,) E U, F(xn,+l) E V. Then d(F(xn,+2), F(xn,+1)) < k.d(F(xn,+1), F(xn,)) and

d(F(xn), F(x -}1 )) <- d(F'(xm), F(xm+1))

for n > m. Hence for j > p d(F(xn,), F(xni+1))

d(F(xni-1+1), F(xni-i+2))

< k.d(F(xn!-1), F(xn, +1)) < ... < O-P+l.d(F(xn,+l), F(xna+2))

< k'-P.d(F(xn,), F(xnp+1)) Therefore

d(x, F(x)) = im d(F(xnj ), F(xn,+1)) = 0.

This contradiction shows that x is a fixed point of F. The uniqueness follows from the contraction property. Finally, we will show :k = lim xn. This follows from d(x, F(xn,+1)) = d(F' (x), F'(F(xn, ))) <_ d(x, F(xn, ))

0 The following example shows that there exist contractions on compact spaces, which are not strict: 2 Let F : [-1, 01 -- [-1, 0] be defined by F(x) = x + x.

Contraction

7

Then

F(x)-F(y) = x+x2-y-y2=x-y+(x+y)(x-y) (1+x+y)(x-y). If x # y, then 11 + x + y] < 1, but there is no k < 1, such that I F(x) - F(y)I < klx - yI. As an application of Banach's fixed point theorem we will consider the following nonlinear Volterra integral equation x(t)

- f k(t,T,x(T))dT = y(t).

(*)

Assume the function

k: [0, 1]x[0,1]xR-+R is continuous and fulfills the following Lipschitz condition: there is a ry > 0, such that for all t, T E [0,1], r, s E 7Z

(k(t,T,r) - k(t,T, s)I < ryIr - sI. Then the mapping F, defined by

F(x)(t) =1 t k(t,T,z(T))dr, maps C[0,11 into C[O,1]. As the complete metric space we choose (X, d) = (C[0,1], dry) with dry(xi, x2) = max Ixi(t) - x2(t)I

a-try.

la-2,

Then F: X -- X and F is a 2-contraction. d,y(F(xi), F(x2))

= max IF(xi(t)) (xi(t)) - F(x2(t))I max 1 t

Ik(t,T,xi(T)) - k(t,T,x2(T))Idr.e-2'

Nonlinear Functional Analysis

8

maxf

t

'YI xl (T) - x2 (T) I

d-y(xi, x2).-y. max

f

e-27

t

dT.e-2

e2''Tdr.e-27t

o
d7(xi, x2)-7-- max a

- 2,yt

(e

2'yt

- 1)

I d,(xl, x2) By Banach's fixed point theorem we have the following result. The Volterra integral equation (*) has for every continuous function y a unique continuous solution x. Especially, we obtain the theorem of Picard - Lindelof : The initial value problem y' = f (t, y), y(O) = rl with Lipschitz continuous f has a unique solution, since ( is equivalent to

*

y(t) = 77 + Jot f (T, y(T))dr.

In general operator F is defined on a subset U of the complete

metric space X or is a k-contraction only on a subset of X. In such cases it is an additional problem to find a subset Uo C U with the properties: F maps Uo into Uo, and Uo itself is a complete metric space. A sufficient condition for such a situation is described in the following result.

Theorem 1.4 Let (X, d) be a complete metric space. Let U C X and F: U -+ X be a k - contraction with k < 1. Let xi E U, x2 = F(xi) E U, r = - d(xi, x2). Let the closed ball Uo := B(x2ir) C U. Then: 1. F maps Uo into Uo.

Contraction

9

2. F has a unique fixed point x E Uo and the sequence (xn) = (F(xn-I)) converges to x.

Proof : The closed subset Uo = B(x2i r) is complete. Let x E Uo. Then

d(F(x, x2)) = d(F(x), F(xi)) < k.d(x, xl) < k(d(x, x2) + d(x2, xl))

< k(1

k

k

+ 1).d(xl, x2))

< r hence F : Uo --+ Uo and Theorem 1.1 applies.

1.2

The Resolvent Operator

Let X be a Banach space and U C X. Let F : U - X be a continuous mapping. Let V C X be a subset, such that for all y E V the equation

x - F(x) = y has a unique solution x. Then x can be represented by

x=y-R(y) and the mapping R : V -+ X is said to be the resolvent oper-

ator to F. In the case, where F is a contraction with Lipschitz constant k < 1, the resolvent operator exists and is Lipschitz continuous, too.

10

Nonlinear Functional Analysis

Theorem 1.5 Let X be a Banach space and F : X - X be Lipschitz continuous with Lipschitz constant k < 1. Then the resolvent operator R to F exists and has Lipschitz constant i " k

Proof : For every y E X the equation

x - F(x) = y

(1.1)

has a unique solution, since the operator FO : X - X defined by Fo(x) = F(x) + y

is a k-contraction and has a fixed point x E X. So a mapping G : X - X with G(y) = x is defined. Let R(y) = y - G(y), then

x=y-R(y)

(1.2)

is the unique solution of equation (1.1). Let yl, y2 E X and x? - F(x;) = yy. Then IIR(yi)

- R(y2)II = IIxi - yI - (X2 - Y2)II = IIF(x1) - F(x2)II < kllxi

- X211

kllyj - R(y2) - (y2 - R(yi))II klly, - y211 + kIIR(yj) - R(y2)II hence

IIR(yi) - R(y2)II : 1

k k Ilyl - Y211

0 The representation

x = y - R(y) of the solution x of equation (1.1) shows that properties of R determine the structure of the solution. We will illustrate this fact by the following result.

Contraction

11

Theorem 1.6 Let X be a Banach space, F : U - X a continuous operator and R : V - X its resolvent opeator. Let Y C X be a linear subspace, such that Range F C Y. Then Range

RcY. Proof : From equations (1.1) and (1.2) it follows that

R(y) = -F(x) = -F(y - R(y)). Thus Range R C Range (-F) C Y.

0 Theorem 1.7 Let SI C R" be an open subset, X = C(SI). Let k : SI x SI x R - R, such that D' k(., r, ) exists for all a E N" with l al :5 V,7- E 0,e E R. Then

F : C(Q) - C(Q) defined by

(Fx)(t) = jk(t,r,x(r))dr maps C(fl) into the subspaces C"(S2) C C(fl) of v - times differentiable functions. If further the resolvent operator R to F exists, then the solution x of an equation

x - F(x) = y y E C(SI), has the property x - y E C"(SI).

1.3

The Theorem of the Local Homeomorphism

The strongest version of the solution of the equation

Nonlinear Functional Analysis

12

is the construction of the inverse map F-1, such that

x = F-1(y) The existence (and construction) of this inverse map in some cases is possible. In this section we will see, that F is (continuous) invertible, if F can be approached by a linear continuous invertible operator T in the following sense.

Theorem 1.8 Let X, Y be Banach spaces and U = B (xo, r) a closed ball in X with centre xo and radius r. Let F : U -+ Y be a mapping, and let T : X --+ Y a continuous linear homeomorphism, such that there exist q with 0 < q < 1, such that for all

u,vEU, IIF(u) - F(v) - T(u - v)II < gIIT(u - v)II.

(*)

Then

10 F is injective and continuous in U. 20

Let p = r(1- q).IIT-III-1. The closed ball V = B(F(xo), p) with centre F(xo) and radius p is contained in F(U), i.e. V C F(U).

30

F-1 is continuous in V.

40

Let Uo = F-1(V). Then Uo is a closed neighbourhood of xo, and FI Uo : Uo - V is a homeomorphism from Uo onto V.

50

For all yl, y2 E V we have the inequality IIF-lyl

- F-1y2 - T-1(yl -

with p =1 .IITI12

IIT-1II2.

plIT-1(yl y2)11:5

- y2)II

Contraction

13

Proof : Let F(u) = F(v) for some u, v E U. Then (*) implies

1°

IIT(u - v)II <_ gIIT(u - v)II

and since q < 1, Tu = Tv and u = v by the injectivity of T. Let y E V. Then F(x) = y is equivalent to the equations

2°

F(T-'z) = y and Tx = x, and further to z = G(z) where

G(z) = x

- F(T-1z) + V.

We will see that G is a q-contraction on the closed ball Vo = B(Txo, rl IT-' 11-1). Let z1, zz E Vo, zf = Tx1, then

IIG(zl) - G(z2)II = Ilzl - zz - F(T-'zl) + F(T-lzs)II I ITxI - Tx2 - F(x1) + F(x2)II <_ gIIT(xl -.T2)11 = gIlzi -'z2II,

since IIzj = TxoII S rIIT-111-1 and

IIxi - x°II = IIT-1(zi - Txo)II

< IIT

TxoII

< IIT-1II.r.IIT-'II-1 = r, i.e.

x3 E U.

This shows that G is a contraction. Now we will see, G maps V° into itself. Let z E Vo, zo = Txo, then

14

Nonlinear Functional Analysis

IIG(z)-Txoll

= IIz-F(T-iz)+y

TxoII

< II z - F(T-'z) - (zo

- F(T-izo))II +

+11y - F(T-izo)II)

gllz- zoII+ IIy- F(xo)l IIT-'11-'+ (1 q)r.IIT-ill-i < qr

-

r.IIT-ill-i

=

-

(y E V implies Ily - F(xo)II < (1 111-1). Since Vo is a closed ball in a Banach space, Vo is complete, and by the fixed point theorem of Banach G has a unique fixed point z E Vo. Thus I = T-iz is the unique solution of F(x) = y. 3°

Let yi, y2 E V, zi, z2 E Vo be the fixed points of

-

F(T-'z)

02(z) = z -

F(T-'z)

G1(z) = z

+ yi

+y2

By the properties in 2° we obtain Ilzi - x211 = IIG1(zi) - G2 (Z2)11

< IIG1(zi)-G2(zi)II +IIG2(zi)

-

G2(z2)11

: llyi-y211+gllzi-x211, hence

Ilzi-x211

1

1

For x; = F-i(y?) we obtain I1x1-x211 = 11T

i(zi-x2)11

<

-

i

II1 gl.Ilyi-y2II

Differential Calculus in Banach Spaces

15

and the continuity of F-1 follows from

IIF-1(yi)

- F-i(y2)II < Il1 QIIy1- Y2II,

if

1/1,112 E V.

40

Since V is closed, Uc is closed by continuity of F.

5°

Let

1/1,1/2EV,w=F-i(yi)-F-1(1/2)-T-1(yi-1/2).

Then with xj = F-1(yj) ,

Tw = T(xi - x2) - F(xi) + F(x2) and

IIwIl

IIT-1II.IITwII

<

1IT-1II.q.IIT(xi

F(x2) - T(xi - x2)11

-

x2)11 X211

11T-1112.IITII.1 q

q

q

IIT-1.(yi - 1/2)11

This proves 50.

0

Chapter 2 DIFFERENTIAL CALCULUS IN BANACH SPACES 2.1

The Derivative

In this chapter, X will be a Banach space, X' its dual space, i.e. the space of continuous linear functionals. Let U C X be an open subset of X, xo E U and f : U -+ R a real valued function. There are several possible meanings to the statement,

that x' E X' is the derivative of f at xo.

Definition 2.1 1°

x' is the Gateaux derivative of f at xo,

if V u E X, lim 1 (f (xo + ru) - f (xo) - rx*(u)) = 0 20

x' is the Frechet derivative of f at xo if lim

1

!lull-.o Null

(f (X0 + u) - f(xo) - x*(u)) = 0.

In both cases, we denote by f'(xo) = x' E X' the derivative. Clearly, 20 = 10, 17

Nonlinear Fhnctional Analysis

18

If f is Frechet differentiable at x, then f is continuous at x, since

Ve>O 36>0 dyEX IIyII
IIf(x + y) - f(x)II < EIIyII +

implies the continuity of f at x. On the other hand, let (e2

f : R2 -> R, f (0) = 0, f (e, rl) _

+ 712).

77

Let x' = 0. Then (f (0 + ry)

- f(0) - x'(y)) =

T.T .r2(S2 +712)

hence lim 1 f (ry) = 0

and f (0) = 0 is the Gateaux derivative of f , but f is not continuous at 0. This definition can be extended to map between Banach spaces. Let X, Y be Banach spaces, U C X an open subset

and F:U-'Yamap. Definition 2.2 1° A continuous linear map T : X - Y is the Gateaux derivative of F at x° E U if

+ ru) - F(xo) - rTuII = 0. Vu E X lim 1.IIF(xo r 20 A linear map T : X -+ Y is the Frechet derivative of F

atx°EU if I1lim

I ICI I

I I F(x°

+ u) - F(xo) - Tu11 = 0.

Differential Calculus in Banach Spaces 3°

19

T is the, weak Gateaux derivative of F at xo i ff

Vu E X Vx' E X' lim 1 x'(F(xo + Tu) - F(xo) - TTu) = 0. T

41

T is the weak Frechet derivative of F at xo if

Vx' E X'

lim

1

Hull 40 11uII

x'(F(xo + u) - F(xo) - Tu) = 0.

We denote by F'(xo) the derivative of F at xo. (Weakly) Frechet differentiable maps are (weakly) continuous, but there are weakly continuous maps, which are not continuous,

e.g. F. [0,11 -+co

A(t)

0

F(t) = (fis(t)) F(O) = 0 _1 It -'

linear and continuous elsewhere

(a) limn.., f (t) = 0, since fn (t) = 0, if n > 1 hence (fis(t)) E co. (b) I IF(0 + h) - F(O)II = maxis I fn (h) I =1, h =

,

hence

F is not continuous at 0.

(c) co = 4. Let x' = (Cn) E ll. Choose for e > 0 an integer no, such that E ISn I <_ E,

n>nO

then

1x`(F(h))I = I Ftnfn(h)1 <- I E Enfn(h)I+ E 1tnI. n:Sno

n>np

The finite sum of continuous functions is continuous

hence, if IhI < - , then Ix'(F(h))I < e, and F is weakly continuous at 0.

20

Nonlinear Functional Analysis

Remark 2.1 Let F : U -+ Y be continuous and Frechet differentiable at xo, then F(xo) is a continuous linear map. Let > 0, b > 0 such that if llull < b < 1

IIF(xo + u) - F(xo)ll < 2 and

IIF(xo + u) - F(xo) -- F'(xo)ul1 < 2.1Iu11- 2. Then

IIF'(xo)ull < E if Iluli < 6.

This is the continuity of F'(xo) at the origin, hence everywhere.

Remark 2.2 If F is Frechet differentiable at xo, then there exist -y > 0, 5 > 0, such that for all x E U

llx - xoll <- b * II F(x) - F(xo)II <-

xoll

Let cpF(x) = F(x) - F(xo) - F'(xo)(x - xo), then II F(x) - F(xo)Il <-

xo11 + I1(PF(x)II

Since, lim)WF( 0)11 = 0, there is a b > 0, such that llx - xoll < 6

IIWF(x)II < IIx - xohI. Thus II F(x) - F(xo)1I <_ (1 + IIF'(xo)II).Ilx -- xoll

Examples 1. The constant mapping F(x) = a for x E U has the derivative F'(x) = 0. 2. Let F : X -- Y be a continuous linear map. Then

F(x + u) - F(x) - F(u) = 0. Hence F'(x) = F for every x E X.

Differential Calculus in Banach Spaces

21

3. Let k : [0,1] x [0,1] x R --- R be continuous and as k(t1, t2, t3) be continuous for t1, t2 E [0, 1], t3 E R. Let F : C[O, 1] -- C[O, 1] be defined by

F(x)(t) = f k(t,T, x(T))dr, 1

then F is in x E C[0,1] Fr chet differentiable and F'(x) is the linear operator F'(x) : C[O,1] - C[O,1]

F'(x)u(t) =

j

k(t,T,x(T)).u(T)dT.

This can be seen by the following.

Let t3 E R, for c > 0 choose 8 > 0, such that, if IhI < 8, then Ik(t,T,t3 + h) - k(t,T,t3) -

3k(t,T,t3).hI C EIhI.

If t3 = x(-r), then

I k(t, T, x(t)+h)-k(t, r, x(T))-

3

k(t, r, x(T))

.

h{ < EIhI.

Let u E C[0,1] with Jul < S, then Iu(T)I < a and 1

Ik(t, T, x(r) + u(r)) - k(t, T, x(r))

-

3k(t,T,x(T)),u(r)Idr,

<Ellull 4. Let f : R -' R be differentiable and

F:C[O,1]--C[O,1] F(x)(t) = f (x(t))

22

Nonlinear Functional Analysis

Then F'(x) : Cf0,11-- CEO, 1)

F'(x)u(t) = i.e. the Frechet derivative of F is the multiplication operator, which multiplies each continuous function u by fox. 5. Let X, Y, Z be Banach spaces and the Banach space X x Y endowed with the norm II(u,v)II = max(Ilull, IIvil) Let

B : X x Y - Z be a continuous bilinear map, i.e. Sup

IBI I =

II B(x, y)II < oo.

I I x I I <_ 1

IIvII<_1

Let (xo, yo) E X x Y. Then IIB(u,v)II thus

lim

IIB(u, v)ll = 0.

II(u,v)ll Therefore B is Frechet differentiable and B'(xo, yo) is the linear map II(u,v)II-.o

B'(xo,yo):X

xY -'Z

(u, v) -+ B(xo, v) + B(u, yo).

6. Let X, Y be two isomorphic Banach spaces and CQL (X, Y) the space of all continuous linear operators T from X onto Y, which are continuously invertible. Then the mapping

F : gL (X,Y) -r gc(Y,X)

F(t) = T-1 is differentiable in every point To and

F'(To)H = -To 1HTo 1.

Differential Calculus in Banach Spaces

23

Proof : If IIHII < IIT

1II-1'

then

IIF(To + H) - F(To) - F'(To)HII =II(T¢+H)-'-To1+To1HTo1I{

=II(I+To1H)-1.To1-Tot+To'HTo'II

=IIF, (-To1H)j.To1-Tot+To1HTo1II 00

j=o

= II F(-To 1H)j11.1IT6lII < IIHII2 j=2

IIIT6- 3

1-IITo HII

hence lim I IF(To +

H) - F(To) - To 'HTo 1I I

X-0o

=0.

I IHI I

0 Theorem 2.1 (Chain Rule) Let X, Y, Z be Banach spaces, U C X, V C Y open subsets.

Let F : U -- Y, G : V - Z be continuous mappings, such that F(U) C V, yo = F(xo) and let the Frdchet derivatives F'(xo) and G'(yo) exist. Then GoF is Frechet differentiable and (GoF)'(xo) = G'(yo)oF'(xo)

(The derivative of the composition is the composition of the derivative).

Proof : Let H = GoF, y = F(x) and cpF(x) = F(x) - F(xo) -- F'(xo)(x - xo) WG = G(y) - G(yo) -- G'(yo)(y

- Yo)

Nonlinear Functional Analysis

24

then w

= H(x) - H(xo) - G'(yo)(F(x) - F(xo)) = H(x) - H(xo) - G'(yo)[F'(xo)(x - xo) + coF(x)1 = H(x) - H(xo) - G'(yo)F'(xo)(x - xo) - G'(yo)coF(x)

hence

UGH = H(x) - H(xo) -

G'(yo)F'(xo)(x

- xo)

G'(yo)SPF(x) + pa(F(x)) Now it suffices to prove

lim

4PH (x)

I1x-x011,

Since G'(xo) is a bounded linear operator IIG'(yo)coF(x)II < IIG'(yo)1I.11cPF(x)I1

F is differentiable, hence

Ve>03b,VxEUIIx-xoll
be>0 3b2 VyEV IlY - yoll
Since F is differentiable, we have

3-y>0353>0 Vx EU IIx - xoll
`db2>0 3b4>0 bxEU 11x-xoIl!5 64 II F(x) - F(xo) 11 < b2.

Differential Calculus in Banach Spaces

25

Thus we obtain for x E U, llx - xoll < min(61, b8, b4) IIPH(x)Il < IIG'(yo)If.IIWF(x)II + Ilcov(F(x))II < IIG'(vo)II.E.IIx - xol) + E.IIF(x) - F(xo)II <-

IIG'(yo)II.E.IIx - xoll + E.'7.llx - roll

< e.(I IG'(vo)II +

xoll,

hence

lim II(PH(x)If = o IIx

- xoll

and

H'(xo) = G'(yo)oF'(xo).

0 Theorem 2.2 Let U C X be an open subset and F : U - Y be a homeomorphism from U onto the open subset V C Y, V = F(U). Let F be Frechet differentiable at xo E U, such that F'(xo) : X Y is a linear homeomorphism. Then the inverse

mapping G = F-1 : V -+ X is Frichet differentiable at yo F(xo) and G'(yo) = F''(xo)-'.

(The derivative of the inverse is the inverse of the derivative).

Proof : If F is differentiable at xo,

VE>o 3b>o VXEU

IIx-xoll<-b*

ItWF(x)II = IIF(x) - F(xo) - F'(xo)(x - xo)ll < E.llx - xoll

<

xo)II.

26

.

Nonlinear Functional Analysis

Let e' = e. I J F' (xo) -1 I I < 1. Then by the theorem of the local homeormorphism (Theorem 1.8(5°)) IIG(y) - G(yo) - F'(xo)-1(y - Yo)II :5

yolJ

with p(e') = 1"**.IIF'(xo)II2.IIF'(xo)-lII9. This shows, that G is differentiable at yo and

G'(yo) = F'(xo)-l. 13

Theorem 2.3 (Mean Value Theorem) Let X be a Banach space, U C X an open subset of X. Let xo, xo + h E U, such that the segment

(xo,xo+h]={xo+Th,O5 T<1}CU. Let F : U -- Y be continuous and Fr&Aet differentiable for all x E [xo, xo + h). Then IIF(xo + h) - F(xo)II <_ Jjhil.

sup zE (z0,zo+hJ

IIF'(x)II.

Proof : Let y' E Y' be a continuous linear functional. The real valued function r' 1of the real variable r

zli(T) = y'(F(xo +Th))

0
is differentiable by the chain rule and

0(1) - 0(0) = y'(F(xo + h) - F(xo)) _ 1'(Tm) = y'(F'(xo +Tmh)h),

0 < T,n < 1.

Differential Calculus in Banach Spaces

27

By the Hahn - Banach theorem we have

IIF(xo + h) - F(xo)II =

sup

y'(F(xo + h) - F(xo))

IIti II=1

y'EY'

<

sup IIF'(xo + Th)II.IIhII. 0
0

Remarks : 10

If dim X > 1, then in general there is no equality in the statement of the mean value theorem. Let

f:R --- C, f (t) = e", h = 27r, then

f (t + h) -f(t) = 0, but

f'(t + 2°

0.

In the situation of Theorem 2.3 we have for u E [xo, xo + h] : IIF(xo + h) - F(xo) - F'(u)hl1 <- IIhII IIF'(v) - F'(xo)II. This can be verified by the application of the mean value theorem to the function

G(x) = F(x) - F(u)x.

foru=xo. The statement 2° can be sharpened, if F(x) is uniformly bounded for all x E U. Let 30

k = sup IIF'(x)II < oo. xEU

Then for all x, y E U II F(x) - F(y)II <- kllx - yII This follows from the equations

co(t) = F(x + t(y - x))

Nonlinear Functional Analysis

28

cp'(t) =F'(x+t(y-x))(y--x) II11

IIF(x)

- F(y)II =

II '(1) - c(o)il =

i'(t)dtlI

< f IIW'(t)Ildt < klix - yli. If additionally F' is Lipschitz continuous in U, i.e. IIF'(y)

- F'(x)II <- mlly - xiI

then

IIF(y) - F(x) - F'(x)(y

- x)11:5 ¶iiv - xil2,

since

t'(t) = F(x + t(y - x)) - tF'(x)(y - x) 'b'(t) = F'(x + t(y - x)) (y - x) - F'(x)(y - x)

implies that

IIF(y) - F(x) - F'(x)(y - x)II = II

f1'(t)dtll

< mf1 dt.ll y - x112.

2.2

Higher Derivatives

Before going to higher derivatives, we will study multilinear mappings between Banach spaces. Let X1, ..., X,,, Y be Banach spaces, and uC(X1,..., X,,; Y) the vector space of all n-linear mappings

M: X1x...xXX-->Y i.e. the space of those mappings which are linear in each variable.

Differential Calculus in Banach Spaces

29

Proposition 2.1 A n - linear mapping M : X1 x ... x Xn -+ Y is continuous if there is a constant 7, such that for all

(x1,...,xn)EX1x...xXn IIM(x1,...,xn)II <- 7.IIxiII...IIxnII.

Proof : "

". Let n = 2; and (al, a2) E Xl x X2. To show that M

is continuous in (al, a2), we write M(x1, X2) - M(al, a2) = M(xi - a1, x2) + M(a1, x2 - a2) then

IIM(x1, X2) - M(al, a2)11:5 7IIx1 -

x211

II(xl,x2) - (a,, a2)11 = Ilx1- alll + 11x2 -- a211

0

implies the continuity of M. " =O.". Let M be continuous in (0, 0). Then there is a ball B = {(x1, x2) E X1 X

X2, I Ix11I + IIx211

< r}

such that for (xl, x2) E Xl x X2 we have I I M(x1, X2)11:5 1. Let

(x1, x2) 34 (0, 0), zi = 2" Then

,

i.e. (z1, z2) E B, I IM(zl, z2)I I S 1. 2

M(zl, Z2) = M

= 211x1 l I' 211x211

-M (X1, X2)411x1 11.11x211

Hence r2

<-1

30

Nonlinear Functional Analysis

implies IIM(x1,x2)II <_ ;724111-4211-

0 For M E 14C(Xl, ..., X,,; Y) we define 11M11= SUP

This defines a norm on AC(X1, ..., X,,; Y).

Proposition 2.2 Let X, Y, Z be Banach space. The space L(X, C(Y, Z)) and pC(X x Y; Z) are norm isomorphic, if X x Y is endowed with the norm II(x,y)II = max(IIxII , 1Iyl1).

Proof : Let MEµC(X xY;Z). For x E X we define M. E L(Y, Z) by M. (y) = M(x, y). The mapping TM : X -- L(Y, Z), defined by TMx = MM is linear and the linear mapping

M --'TM is an isomorphism of pC(X x Y; Z) onto L(X, L(Y, Z)) I10(M)I1= IITMII = sup IIMMII = sup sup II M(x,y)II = IIM1I IIxII51 IIvII<1

IIxII<1

If T E L (X, L (Y, Z) ), then for x E X, Tx E L (Y, Z),

i.e. M : (x, y) --> Ty is bilinear and II M(x, y)II = IITxyII <

Iy1I

and

IIMII =

sup

IIxII<1 IIyII<1

IITxyII =11TH.

Differential Calculus in Banach Spaces

31

Together with the above statement this means: 0 is bijective

0

and 11011 = 1.

Let U C X be an open subset of the Banach space X and F : U > Y. We say that F is continuously differentiable, if F: U -. £(X, Y) is continuous. We say that F is twice differentiable at xo E U, if F' is continuous and F' is differentiale at xo. Then we write F"(xo) and F"(xo) E G(X,G(X,Y)). Since G (X, G (X, Y)) can be identified with the space t&C(X x X ; Y) of

bilinear operators, we understand F"(xo) as a bilinear operator with (g, h) -+ F' (xo) (g, h).

Proposition 2.3 Let F be twice differentiable at xo E U, let g E X. then the map x - F'(x)g from U into Y is differentiable at xo, and its derivative is the linear operator

h -, F"(xo)(h,g)

Proof : The mapping x --+ F'(x)g is the composition of T -' Tg (from (G(X,Y) into Y) and x - F'(x) (from U into G(X,Y)). By the chain rule we obtain the result.

0

Theorem 2.4 Let F be twice differentiable at xo. Then the bilinear mapping (g, h)

-

F"(xo)(g, h) is symmetric, i.e.

F"(xo)(g,h) = F"(xo)(h,g)

Proof : Let cp : [0,11 -, Y be defined by

W(-r) = F(xo + rg + h) -- F(xo + rg)

32

Nonlinear F inctional Analysis

where we assume I IgI l < 6/2, the mean value theorem

l IhI I

- 6/2,

IIso(1) - V(0)II < sup IIc'(T)

B(xo, S) C U. By

- W(0)11

and by the chain rule

(F'(xo + Tg + h) - F'(xo +,r9))9 (F'(xo + Tg + h) - F'(xo))9 - (F'(xo +,r9) - F'(xo))g.

FZ

The differentiability of F'(x) implies, that for given e > 0 there exists 6* > 0, such that, if IIhII : 62/2, II91I < 6 *12 (F'(xo+Tg+h)-F'(xo))9-F"(xo)(T9+h,9)II S

II

II (F'(xo +T9) - F'(xo))9 - F"(xo)(T9,9)II <_ E.TII9II2.

Therefore we obtain II co'(r) - F"(xo)(h,9)II < 2.e(Il9II + IIhII). Now we obtain IIcp(1) - cp(0) - F"(xo)(h,9)II

IIco(1) - W(O) - V(O)II + IIw'(0) - F"(xo)(h,g)II < sup IIV (T)II + IIcQ'(OY - F"(xo)(h,9)II < sup IIco'(T) T

- F"(xo)(h,9)II + IIF"(xo)(h,9) - V(0)II

+IIV(0) - F"(xo)(h,9)II < 3.2.E(IIgII + IIhII)2.

The term cp(1)

- cp(0) = F(xo + g + h) - F(xo + h) - F(xo + g) + F(xo)

Differential Calculus in Banach Spaces

33

is symmetric in g and h, hence II F"(xo)(h,9)

- F"(xo)(9, h) 11

5 11w(1) - AP(O) - F"(xo)(h,g)II

+IIc(1) - W(o) - F'(xo)(g, h)II < 12.e(IIg11 + IIhII)',

whenever 119II < 6/2, IIhII < 6 *12. If we choose II(g,h)II = max(IIg11, I1h11), then we see II F"(xo)(h,g) - F"(xo)(9, h)II <- 48.EII(9, h)II',

this means, that the bilinear map (g, h) -+ F"(xo) (h, g) - F"(xo) (g, h) has norm zero; and this is the symmetry of the second derivative.

0 By induction, we can define k-times differentiable mappings:

Let F : U --+ Y, F is said to be k-times differentiable at zo, if the derivative of order k - 1 is continuous and differentiable at x0. We write Ftkl(zo).F(k)(xo) can be identified with *a k - linear

map, F(k)(zo) E µC(X x ... X X;Y).

Example Let f : R - R be a twice continuously differentiable function. Let F : C[O,1] -- C[O,1] defined by

F(x)(t) = f(x(t)). Then F"(xo)(g, h)(t) =

We conclude this chapter with the two versions of the theorem of Taylor.

Nonlinear Functional Analysis

34

Theorem 2.5 Let X, Y be Banach spaces, U C X open, x,

x + h E U, such that [x, x + h] C U; F : U -* Y

be

p-times differentiable and the derivatives up to the order p - 1 continuous. Then

IIF(x+h)-F(x)-F'(x)h-...1

(p-1)!

F(V-1)(x)(h,..., h)II

< Ilhlp up IIF(")(u)II PI (,s+hJ

Proof : Let y* E Y' be a continuous linear functional. The real valued function

O(r) = y*(F(x + Th)), T E [0,1] is p-times differentiable and ',p(k)(T)

= y*(F{k)(x+Th)(h,...,h)).

The Taylor formula for real functions gives ?-1 1

y*(F(x + h) - E j:i F(') (x) (h, ..., h) ) j=0

'

P-1 1

_ (1) _ E

(0)

J0

=

1 pf

(Tm), O < Tm < 1

= y*(- F(x+Tmh)(h,...,h)) Taking sup of y*, such that IIy*II <_ 1 and T,,,-E [0,1] the proof is complete.

Differential Calculus in Banach Spaces

Corollary 2.1 Let F

: U --' Y as in Theorem 2..5,r at tl ti i 44F()

(x) be continuous. Then for every e > 0 there is ,.

.,.

such that for all h E X, I IhII < b P

I I F(x + h) - E 4j F(j) (x) (h, ..., h) I I < E. I Ihl IP. 1=0

Proof : Let p = 2 and

G(x) = F(x) - F(y) - F'(y)x

-

F"(y)(x,x)

Then

G(x + h) = F(x + h) - F(y) - F'(y)(x + h) --ZF'(y)(x + h,x + h) G'(x)h = F'(x)h - F'(y)h -- F"(y)(x, h) and

G(x+h)-G(x)-G'(x)h = F(x+h)-F(x)--F'(x)h-1 F(y) (h, h). By Theorem 2.8 we have 2

IIG(x + h) - G(x) - G'(x)hII <

II

Il

2

uExup

[,+l

It is

G"(x)(h, h) = F"(x)(h, h) - F"(y)(h, h) thus

sup IIG'(u)II = sup u

and, letting y = x

u

IIF"(u)

- F,(y)II

IIG"(u)II

36

Nonlinear Functional Analysis

I IF(x + h) - F(x) - F'(x)h II 2112

-

F"(x)(h, h)I I

IIF"(u) - F"(x)II.

sup uE (x,x+hl

The continuity of F" guarantees the existence of 6 > 0, such that IIF"(u) -- F"(x)II S E if IIhII < S. In general, let G(x) = F(x) Q 7,FU)(y)(x,...x). Theorem 2.5, applied to C gives for y = x P

I IF(x + h) - E 1 FUN (x) (h, ..., h) I I

,o

<

hp I

P.

sup

1IF(P)(u) -- F(P)(x)I

uE (x,x+h]

The continuity of F(P) guarantees the existence of b > 0, such that IIF(P)(u) - F(P)(x)II < E, if IIu - xli < S and IIhII <_ S. D

2.3

Partial Derivatives

LetX=X1xX2,UuCXjopen, (al,a2) EU1xU2=U. Let F : U -- Y be differentiable. We define the partial maps

x1 - F(x1,a2),

x2 -- F(al,x2)

from Ui into Y and say, that F is partially differentiable in (al, a2) E U, if the maps x1 -i F(xl, a2) and x2 --, F(a2i x2)

are differentiable in aj. By D;F(al, a2) we denote the partial derivative of F with respect to the first (resp. second) coordinate.

Differential Calculus in Banach Spaces

37

Theorem 2.6 Let F : Ul x U2 --> Y be a continuous map from the open set U = Ui x U2 C X into Y. Then F is continuously differentiable in (al, a2) E U if and only if F is partially differentiable and the partial derivatives are continuous mappings (x1, x2) --' D1F(xi, X2)

U -' £(Xi,Y) and

(xi, x2) -' D2F(x,, x2)

U - G(X2,Y). The (total) derivative of F in (ai, a2) is given by F'(ai, a2)(hl, h2) = D1F(ai, a2)hi + D2F(ai, a2)h2

Proof : " =" : The mappings Gl : xl -- F(xl, a2) is the composition of F and ii' : xl -- (xi, a2). The derivative of the second map is the linear map it : hl - (h,, 0). By the chain rule we get

D1F(ai,a2)hi = G' (ai)hi = (Foi 2)'(ai)hi = F'(ii2(al))Oii(hi) = F'(al, a2)(hl, 0) and similarly,

D2F(al, a2)h2 = G2(al)h2 = (Foil' )'(a2)h2 = F' (i2' (a2))oi2(h2)

= F'(al, a2) (0, h2) Since it + i2 = id, we obtain F'(ai, a2)(hi, h2) = F'(ai, a2)(hi, 0) + F'(ai, a2)(0, h2) D1F(ai, a2)hi + D2F(al, a2)h2.

(*)

Nonlinear Functional Analysis

38

" 4--" : We will show that for given E > 0 there is a 6 > 0,

such that, if II(hi, h2)1I < 6, we have

0 =

I IF(ai + hl, a2 + h2) - F(ai, a2)

-D1F(ai, a2)hi - D2F(ai, a2)h211 < E.II(hl, h2)I I By the differentiability of the partial maps we obtain for 11h111<-b1

IIF(al + hl, a2) - F(al, a2) - D1F(al, a2)h1I I < E.IIh1I1,

by the mean value theorem we obtain

IIF(al + hl, a2 + h2) - F(ai + h1i a2) - D2F(ai + hl, a2)h211 < I Ih2I I

sup

I ID2F(ai + hl, a2 + z) - D2F(al + hl, a2) I I

IIz)ISIIh211

by the continuity of D2F, there is a 62 > 0, such that IIhiII <- b2, 11h211 < S2

I ID2F(ai + hi, a2) - D2F(ai, a2)11 < E and

sup I ID2F(ai + hl, a2 + z) - D2F(al + hi, a2)I I

<- E.

lIzII562

Thus we obtain for II(hi, h2)11 5 min (61, 62) A < 3E.11(hl, h2)II

The continuity of F follows from (*). O

Chapter 3 NEWTON'S METHOD Since it is important to know how fast the convergence of a sequence (xn) to the limit z really is, one usually introduces the order of convergence as a first asymptotic test.

Definition 3.1 Let (xn) be a sequence in the Banach space X with lim xn = i. The sequence (xn) is said to be convergent of order p > 1, if there exist positive meals Q, ry, such that for all

nEN Ilin - ill :5 If the sequence (xn) has the property, that

Ilxn - ill< -aqn, 0
Proposition 3.1 If I is the fixed point of F : U -' X and if there exist meals p > 1, fp > 0, such that for all x E U

IIF(x) - ill
-illp

(F(xn_1)) converges of order p. 39

Nonlinear Functional Analysis

40

Proof : Let r > 0, B(1, r) C U, such that a =

-1 ,Qp.rp-1

< 1,

7 = per > 0. Let x1 E B(1, r), then 1)1)

I lxl - ±11 < r =

a a

= ,0. exp

log a p

- 1 = /3. exp(-7p)

By induction we obtain from I Ix" - iI I < /3. exp(-'rp")

the estimate apIIxri - uII' -
Ilxn+l - ±11

= Q. exp(-7p"+1) < r.

0

Example : If f : [a, b] -, R is p - times continuously differentiable at t = f (t) E [a, b], such that f'(t) = 0, ..., f (r-1)(t) = 0, then t" = f (4_1) is convergent of order p, since by Taylor

If (t) - tl = If (t) - f(t)1= -,.It - ill. sup If(P)(T)I. If g : [a, b] -' R is twice continuously differentiable,

g(i) = 0,g'(i) # 0, then

f (t) = t t"+1 = tn

g(t) 91(t)

9(t") - 9'(t")

(3.1)

is convergent of order 2, since

fl(t)

= 1-

f'(t) = 0.

1(t)2-

g11(t)

g1(t)2

g (t)

Newton's Method

41

The equation (3.1) defines Newton's method for differentiable functions. Now we will consider Newton's method for solving the equation

F(x) = 0 in Banach spaces.

Theorem 3.1 Let X, Y be Banach spaces and F : B(xo,r) -Y continuously differentiable, such that (a) F'(xo)-1 E G(Y, X), IIF'(xo)-1.F(xo)II = a, IIF'(xo)-Ill =,3

(b) IIF'(u) - F'(v)II < k1lu - vll, u, v E B(xo, r)

(c) 2ka/ < 1, 2a < r are satisfied. Then F has a unique zero x in B(xo, 2a) and the Newton iterates xn+1 = xn - F'(xn)-1F(xn)

(3.2)

converge quadratically to i and satisfy

llxn - ill

q2n-1

2Q 1

=

q.2n-2n--y2")

(3.3)

with q = 2afk < 1, -y = -log q > 0.

Proof : At first we remark that for the real valued function w(t) _ y'(F(u + t(v - u))). We obtain (y* E Y*)

w'(t) = y`(F'(u + t(v - u))(v - u))

jw'(t)dt = w(1) - w(0), and therefore

F(v) - F(u) = f F'(u + t(v - u))(v - u)dt. 1

0

42

Nonlinear Functional Analysis

Suppose, that (xn) is defined and let IF'(xn)-'

an =IIxn+1 - xnl 1, Qn = I

I I, 7n

= kanjn

Then

an = <-

IIF'(xn)-'F(xn)II

IIF'(xn)-lI I IIF(xn) - (F(xn-1) + F'(xn-1)(xn - xn-l))II

< On f IIF'(xn-1 + t(xn - xn-1)) - F'(xn-1)Ildt.

< IIxn - x-1Qna-1

j

kIIt(xn - xn-1)Ildt

< Nnan-1.2. Since

F'(xn) = F'(xn-1) + F'(xn) - F'(xn-1) = F'(xn-1).[I +

F'(xn-1)-l(F'(xn)

- F'(xn-1))],

we also have On

= IIF'(xn)-lll =

II

F'(xn-1)-l(F'(xn) - F'(xn-1))II < Nn-1.kan-1 -'In-1 < 1

then

7n-l)-',

Nn

hence 7n-1)-1

an C

and

an-1 an 2 kan3n = 7n < 12' (1 1'n 1

Newton's Method

43

Since 'Yo = kao,Qo < kaf3 < 1, the inequalities (3.4) imply y,, <

and consequently an < ! an-1 for all n > 1. Hence, an < 2-na, and n

n

I Ixn+1 - X011 < E I lxj+l - xj ll <- E 2-1a < 2a < r. J=O

J=O

Thus it is obvious that (xn) is well defined, and is a Cauchy sequence with lim xn = I E "R(xo) 2a) C B(xo, r).

Clearly, F(±) = 0. Further

Ilxn+l - Ill = Ilxn - i - F'(xn)-1(F(xn) - F(I))Il < Qnll F'(Xn)(xn - 2) - F(xn) - F(I)lI < Qn f lIF'(xn - t(xn - 2)) F'(xn)lldt.Ilxn - ill

-

Qnl lxn - Ill f ktll xn - ±Ildt < 2anllxn -1112, hence

IIxn+1 - -+II <_ Cl lxn - 1112, C =

2

SUP On < 00

since Qn -- IIF'(x)-111

If z E 77(xo, 2a) is another zero, then Ill-Ill

<-

allF(-t)-F(I) --F'(xo)(i-I)ll

< Qkllx-Ill. f I 1l.4 +t(z-I)-xolldt < 2a,QkllI -- .II

and there i = I. Finally, to obtain equation (3.3), let an -yn(1

-

z

Jn = - I n hence

imply

-yn)-1. Then inequalities (3.4) and ryn <

- 7n)-1 < 2-fn < 6n-1,

44

Nonlinear Functional Analysis

and consequently bon-'

tan-lan-1 <

an

a n-1 <_ ... < 2-n.60

-1ao

< 2-n.602-' a, by (3.4) Therefore 00

I Ixn - xI I

< liM IIxn - xn+dI <_ E aj j=n 00

< E 2-f.g2J_1a j=n

<

2.2-n.g2n-1a

= 2a 1q2^-1 O

Newton's method is quadratically convergent, provided that it is possible to find xo and r > 0, such that I I F" (xo) -1 I I and IIF'(xo)-1F(xo)II are sufficiently small and F is Lipschitz continuous in B(xo, r). Certainly this is another difficult problem without a general recipe for the solution. A simplified Newton's type algorithm guarantees only linear convergence:

Theorem 3.2 Let X, Y be Banach spaces and F : B (xo, r) -Y continuously differentiable such that (a)

F'(xo)-1 E G(Y, X), I IF'(xo)-1I I = Q,

IIF'(xo)-'F(xo)II = a (b) IIF'(u) - F'(v)II < kIIu - vII, u, v E B(xo,r)

(c)ry=2kap<1,ro=2a.1-I
The Implicit Function Theorem

45

converge to x and satisfy n

I Ixn -- 1 1 1

-< 1 - q

'

°

l

1

with

q=1- 1-y. Proof : The mapping G : B(xo, ro) --+ B(xo, ro), defined by

G(x) = x

- F'(xo)-1F(x)

is a contraction. Let x, y E B(xo, ro)

G(x) - G(y)

= x - y - F'(xo)-1(F(x) - F(y)) = F'(xo)-1[F'(xo)(x - y) - F(x) + F(y)1 = F'(xo)-1 j1(F'(xo)

-F'(y + t(x - y)))(x - y)dt.

/3k.Ilxo-y-t(x-y)II.IIx-yIl

IIG(x) -G(y)II

< Qkrollx Since

ro =

- yIl

1- 1-'y < Qk

1

,0k

G is a q-contraction with q = /3kro = 1 show that G maps B(xo, ro) into itself. Let x E B (xo, ro). Then

JIG(x) -- roll = Ilx - xo 3.IIF'(xo)(x

< < p.

f

'

1 - 'y. It remains to

F'(xo)-'F(x)II

- xo) - F(x) + F(xo)II + a

IIF'(xo) - F'(xo + t(x - xo))IIdt.Ilx - xell + a

I8kro+a=ro. 0

Chapter 4 THE IMPLICIT FUNCTION THEOREM Let X, Y, Z be Banach spaces and U C X, V C Y neighbourhoods of x0 and yo respectively. Let F : U x V -' Z. In this chapter, we will consider the following problem.

If the equation F(x, y) = 0 has solution x E U, if y E V, provided that F(xo, yo) = 0. This is the generalization of the solution of G(x) - y = 0. The answer is given in the following theorems.

Theorem 4.1 Let X, Y, Z be Banach spaces, U C X, V C Y neighbourhoods of xo E U, Yo E V. Let F : U x V - Z be continuous and continuously differentiable with respect to y. Suppose that F(xo,yo) = 0 and Fy ' (xo, yo) E C(Z, Y). Then

there exist balls B(xo, r) C U, B(yo, s) C V and exactly one map G : B(xo, r) -* B(yo, s), such that G(xo) = yo and for all x E B (xo, r)

F(x, G(x)) = 0. This map G is continuous.

47

48

Nonlinear Functional Analysis

Proof : Without loss of generality xo = 0, yo = 0 since the general situation may be reduced to this one by a translation. Let T = Fy(0, 0) and I be the identity on Y.- Since F(x, y) _ 0 is equivalent to

y+T-1F(x,y)-y=0, we show that for x E U H(x, y) = T-1F(x, y)

-y

is a contraction on a properly chosen ball. Since H.' (0, 0) _ T- 1 F(0, 0) - I = 0 and Hy is continuous, we can fix k < 1 and

find s > 0, such that I IH'(x, y) I I < k for (Ix (I < s, I ly(I S s. Thus, IIH(x,yl) - H(x,y2)II < kIjy1 - Y2 11

by the mean value theorem. Furthermore, since H(0, 0) = 0, and H(., 0) is continuous,

there exist r < s, such that IIH(x,0)II < s(1 - k), if IIx(I < r. Then the mapping

Q : B(0, s) - Y

Q(y) _ -H(x,y) maps B(0, s) into itself and is a contraction: IIQ(y)II

II H(x, y) - H(x, 0)II + IIH(x, 0)(I

kIIyII + k(1 - s) < s.

Thus Q has a unique fixed pointy E B(0, s), i.e. if x E B(0, r), then there is a unique y = G(x), which is the fixed point of Q, hence

Q(y) = -H(x, G(x)) = y - T-1F(x, G(x)) = y

The Implicit unction Theorem

49

this shows

F(x, G(x)) = 0.

If x = 0, then y = 0, since 0 - H(0, 0) = 0 implies that H(0, y) has the fixed point y = 0, hence G(0) = O.G is continuous, since

G(xi) + H(xi, G(xi)) - G(x2) - H(x2, G(x2)) =

0

implies

IIG(xi) - G(x2)11 = IIH(xi,G(xi)) - H(x2,G(x2))II II H(xi, G(xi)) - H(xi, G(x2))II

+ I I H(xi, G(x2)) - H(x2, G(x2))I I

< kIIG(xi) - G(x2)II +

I H(xi, G(x2)) - H(x2, G(x2))II

I

hence

IIG(x,) - G(x2)1I

<- 1 1

IIH(xi, G(x2)) - H(x2, G(x2))II

k

which tends to zero, if xi - -T2-

0 Corollary 4.1 Let F : U -- Y be continuous, xo E U. If F'(xo) is continuously invertible, then there exist neighbourhoods Uo of xo and Vo of yo = F(xo), such that Flu,, = FO is a homeomorphism and Go = F6-1 is differentiable at yo GG(yo)

=

Fo(xo)-i

Proof : Let q5: Y x U -' y be defined by

50

Nonlinear Functional Analysis

Then 0z (yo, xo) = F'(xo) is invertible. By the implicit function theorem there exist Uo C U, Vo C Y and Go Vo - Uo, such that for y E Vo we have :

q5(y, Go(y)) = F(Go(y))

- y = 0.

F is invertible, this implies by theorem 2.2, that Go is differentiable at yo with G' (yo) = FF(xo)-1 O

Corollary 4.2 Under the assumptions of Theorem 4.1 and additionally, that F : U x V -+ Z is continuously differentiable (in both variables), then the map G is continuously differentiable and G'(x) = -F,(x, G(x))-11.(x, G(x))

Proof : Let x, x + s E B(0, r), t = G(x + s) - G(x), then

F(x + s, G(x) + t) = 0. Then the differentiability guarantees for E > 0 the existence of 6 > 0, such that IIs1I < b implies

II F(x + s, G(x) + t) - F(x, G(x)) - F(x, G(x))s - Fv(x, G(z))tlI < E(IIsII + IItII)

IIF2(x, G(x))x + FF(x, G(x))tII < E(IIsII + IItiD).

Since Fy(0, 0) is an isomorphism, there exist a neighbourhood, such that FF(x, G(x)) is an isomorphism, thus I It + F,', (x, G(x))'1F.(x, G(x))sII < E' (I Is II + IItH)

(*)

The Implicit Function Theorem

51

The definition of t = G(x + s) - G(x) shows, that G is differentiable with derivative

G'(-T) = -FF(x,G(x))-1Fz(x,G(x)),

if we show, that IItiI < i'IIsjj. But this follows from the continuity of G and the inequality (*).

0 The solution of the implicit problem

F(x, y) = 0 F(xo, yo) = 0

}

can be obtained by iterative methods of Banach or of Newton type.

Theorem 4.2 Let F : U x V - Z be as in Theorem 4.1. Let Fy be continuous in (xo, yo). Let

Yn+i(x) = yn(x) - F. (x,yo)-1F(x,yn(x))

Then (yn) converges to the solution G of equation (4.1), such that Ilyn(x) - G(x)II <- gnllyo(x) - G(x)II and n

Ilyn(x) - G(x)II

<-

lq

yo(x)II

n

<

g.1

1II ITIIT1II IIF(x,yo(x))II

1

where E, q are given by equations (4.2), (4.9) below.

52

Nonlinear Functional Analysis

Proof : Let T = FV(xo, yb) and let p(r, s) be reels, such that for x E R(xo, r), yi, y2 E R(xo, s) by the mean value theorem and

the continuity of F IIF(xi, yi)

- F(x, y2) - F,(xo, yo)(yi - y2)11:5 p(r, s)I Iyi - y211

with lim I IT-' I Ip(r, s) = 0. r,e-o If ro, so are sufficiently small, such that

f
IIT-111 1

- , 1 IT-l I I

IT-i I I-i

(4.2)

(p(ro, so) + E) < 1

IIFF(x,yo) - FF(xo,yo)II < E

IIF(x,yo)II <

(4.4)

EIIT-111) (1- q)so(1-i II II

then

H(x, y) = y - F,,(x,

yo)-1F(x,

(4.3)

y)

has the property

IIH(x,yi) - H(x,y2)II Ilyl - y2 - FF(x, yo)-'.fF(x, yi) - F(x, y2)1II F(x,y2) -F;,(x, yo)(yi - y2)II Since Fy (x, yo) = Fy (xo, yo) + F, (x, yo) - Fy (xo, yo)

= T(I +T-i(F,'(x, yo) - F(xo,yo)))

(4.5)

The Implicit Function Theorem

53

we obtain

1II

IIFF(x,yo)-'II < 1 IIIITT1 II

and

I I H(x, yl) - H(x, y2)II

IIT-111

< 1

<-

-EIIT-111

(P(ro, so) +

Y211

q.IIy1 - y211

Further

IIH(x,y) - yoII

<_

IIH(x,yo) - H(x,yo)II + IIH(x,yo) - yoII

<_

qII y -yoII + I I FF(x, yo)-1F(x, yo) I I

<

q.,90 + 1

II

T'

TI 1 1II , I

I F(x, yo) I I

II IIT-'I1

< q.so + 1- EIIT-1I1'

(1 - q)so(1- EIIT-1I1) IIT-111

< so. By Banach's fixed point theorem we obtain the convergence and the error estimates. O

If we use Newton's method, then we will obtain superlinear convergence.

Theorem 4.3 Let F be given as in Theorem 4.1. In a neighbourhood of (xo, yo) we assume the Hoelder condition II FF(x,yl) - Fy(x,y2)II <_'1'IIy1 - Y211°`

with 0 < a < 1. Let F and F' be continuous at (xo, yo). Then there exist ro, so, such that for x E B(xo, ro) yn+1 = yn(x) - FF(x,

yy(x))-1F(x,

yn(x))

Nonlinear Functional Analysis

54

exist and converge to the solution F of equation (4.1). Furthermore there exists a constant ko such that for any k with

k>

[IIT-11I7(1

+ a)-1]1/a = ko

we obtain

IIyn(x) - G(x)II <- 1 [kllyo(x) -

G(x)I11(1+a0^

Proof : We assume, that ro, so are so small, that FF(x, y) is continuously invertible whenever x E B(xo, ro), y E B(yo, so) and IlF;,(x,y)-11I <- (1 + e)IIT-111. Then 1IVn(x) - G(x)II

s IIyn-1(x) - G(x) - F,(x,

yn-1(x))-1

(F(x, yn-1(x)) - F(x, G(x))) I E)11T-111.11

(1 +

+T(G(x)

Jot F,1 (x, yn-1(x)) - Fv,(x, G(x)

-

G(x)II 11 o

(1 +

(1 +

I

(1- r)`dT.IIyn-1(x)

- G(x)lll+a

E)IIT-1IIy.IIyn-1(x) G(x)II1+«

,

1+a

hence (k.I1yn-1(x) - G(x)I1]1+'

ktIyn(x) - G(x)II <where k

__

(1 +

E)1IT-111 y 1/0

1+a

We remark, that klIG(x) - yoll may be arbitrarily small, if the initial value yo(x) is sufficiently close to G(x).

Chapter 5 FIXED POINT THEOREMS 5.1

The Brouwer Fixed Point Theorem

Brouwer's fixed point theorem is basic for many fixed point the-

orems. It states that a continuous map, which maps a convex bounded closed set in R" into itself, has a fixed point. Before we prove the Brouwer fixed point theorem, we observe

that the case of complex scalars is a consequence of the case of real scalars. This follows from the fact that the complex space C" is isometric with the natural space R', and the unit spheres in these spaces correspond in a natural way. Thus we restrict our attention to real Euclidean space. We need following lemma.

Lemma 5.1 Let f be an infinitely differentiable function of n + 1 variables (z0, ..., x") with values in R". Let D, denote the determinant whose colunmns are the n partial derivatives 55

56

Nonlinear Functional Analysis Then

(-1)`. i=0

a

axi

Di = 0.

(5.1)

Proof : For every pair i, j of unequal integers between 0 and n, let Cij denote the determinant whose first column is fx{zf and whose remaining columns are fxo, .., fx arranged in order of increasing indices, and where fx, and fx, are omitted from the enumeration.

Clearly Cij = Cji, and by the laws governing differentiation of determinants and interchange of columns in them we have a ax Hence

Di r E(-1)jC1, + E(-1)j-'Cij. i

j
a

j>i

n

(-1)i axi Di = i=0 where a(i,j) = 1 if j i. Thus E(-1)i+jCijo'(i,j),

n

n

i=0

Q=0

E(-1)i. a iDi = E (-1)i+1Cijcr(i,j) Interchanging the dummy indices i, j in this latter expression

and using the fact that o(i, j) _ -a (j, i), we see that n

E (-1)i+1Cijo'(i,j) _

i,j=0

(-1)j+iC43c(j,i) i,j=0

(-1)

(-1)i+3Cija(i,j)

Thus all the three equal quantities in this formula must be zero, and formula (5.1) is proved.

0

Fixed Point Theorems

57

Theorem 5.1 (8rouwer) If 0 is a continuous mapping of the' closed unit sphere B = {x E X, I xI < 1) of Euclidean n-space into itself, then

there is a point y in B such that b(y) = y.

Proof : We have remarked that if suffices to consider real Euclidean space. Further, the Weierstrass approximation theorem for continuous functions of n variables implies that every continuous map ¢ of B into itself is the uniform limit of a sequence (qk) of infinitely differentiable mappings of B into itself. Suppose that the theorem were proved for infinitely differentiable maps. Then, for each integer k there is a point yk E B such that ok(yk) = yk. Since B is compact, some subsequences (ykt) converge to a point y in B. Since limi--.,,o Or,, (x) = 4(x) uniformly on B,

O(y) = m Ok, (yk,) = li Yk, = Y1

This shows that it is sufficient to consider the case that 0 is infinitely differentiable.

We suppose that 0 is an infinitely differentiable map of B into itself and that O(x) x, x E B. Let a = a(x) be the larger root of the quadratic equation Ix + a(x - O(x))I2 = 1, so that

1 = (x + a(x - O(x)),x + a(x - O(x))) Ix12+2a(x,x-O(x))+a21x-O(x))I2.

By the quadratic formula a(x). I x - O(x) I2

= (x, O(x) - x) +{(x, x - O(x))2 +(1- Ix12)Ix O(x)12}

-

(5.2)

Nonlinear Functional Analysis

58

Since Ix - ¢(x)l # 0 for x E B, the discriminant (x, x-O(x))2+(1- lxl2)Ix-O(x)i2 is positive when IxI # 1. Also if I x I = 1, then (x, x - 0(x)) # 0, for otherwise (x, O(x)) = 1 and the inner product of two vectors with length at most 1 can be equal to 1 only when they are equal. Thus the discriminant is never zero for x in B. Since the function t1/2 is an infinitely differentiable function of t for t > 0, and since Ix - O(x) l , 0, x E B it follows from formula (5.2) that the function a(x) = 0 for fix( = 1 is an infinitely differentiable function of x E B. Moreover, it follows from formula (5.2), that a(x) = 0 for IxI = 1.

Now, for each real number t, put f (t; x) ; x + ta(x)(x -

0(x)). Then f is an infinitely differentiable function of the n + 1 variables t, x1, ..., x, with values in B. Since a(x) = 0 for IxI = 1, we have ft (t, x) = 0 for lxj = 1. Also f (0, x) = x , and from the definition of a we have If (1, x)l = 1 for all x E B. Denote the determinant whose columns are the vector fz, (t, x), fz,, (t, x) by Do (t, x) and consider the integral

1(t) = J Do(t, x)dx.

(5.3)

It is clear that 1(0) is the volume of B and hence 1(0) # 0. Since f(1, x) satisfies the functional dependence 1(1, x)+ = 1. It follows that the Jacobian determinant Do(1,x) is identically zero, hence I(1) = 0. The desired contradiction will be obtained

if we can show that 1(t) is a constant; i.e., that I'(t) = 0. To prove this, differentiate under the integral sign and employ (5.1) to conclude that I'(t) is a sum of integrals of the form

±f -

Di (t,

x)dx

where Di(t, x) is the determinant whose columns are the vectors 1.

f

IE,{_1(t1x),f.1j,4.3(t,x),...,fZ'(t'

).

Fixed Point Theorems

59

By the Gau$ Theorem we see that

fe -t D1 (t, x)dx = f8B Di(t, x)77{dw. i9

ForxEBB,jxj =1,and fi(t,x) = a(x)(x - O(x)) implies ft (t, x) = 0 for x E 8B, hence D.(t, x) = 0. This implies

I'(t) = 0, 1 = constant, I(1) = 0 # 1(0). This contradiction shows, that there is an x E B such that O(x) = x.

0 A simple application of the Brouwer fixed point theorem is the following existence principle for systems of equations.

Proposition 5.1 Let B = B(0, r) C Rn be the closed ball with radius r and gj : B -' R be continuous mappings, j = 1, 2, ..., n. If for all x = (6, 6, ..., tn) E Rn, 11xI ( = r n

E ga (x)ei >_ 0

(5.4)

7=1

then the system of equations

g,(x) = 0, j = 1, 2, ..., n

(5.5)

has a solution i with IJuI` < r.

Proof : Let g(x) = (g1(x),...,gn(x)) and assume g(x) # 0 for all x E B. Define

1(x) -

rg(r) 119(x}11

60

Nonlinear Functional Analysis

f is a continuous map of the compact convex set B into itself. Therfore there exists a fixed point x, of f with I I±I I= I I f (x) II = r. Furthermore E9,i(x)

a=

1

=

-rIjg(z)jt.ECj2<

0.

This contradicts formula (5.4), hence equation (5.5) has a zero.

0

5.2

The Schauder Fixed Point Theorem

We will begin this section with a theorem on the extension of continuous mappings. Let A C X be a subset of the normed space X.

A is convex, if for all x, y E A, T E [0,1], the segment T2 + (1 - T)y E A.

Then by induction, if x1, ..., x,,

E A,

Ti > 0, E 1 Ti = 1, then F,1 rjxi E A. The convex hull coB of B C X is the intersection of all convex sets A, which contain B. It is given by n

n

Tixi, xi E B, T i ! 0,

COB = {x

Ti=1,nE

Ar}.

i=1

i=1

Theorem 5.2 Let X be a Banach space, A C X a closed subset

and F : A -+ Y a continuous map from A into the Banach space Y. Then there exists a continuous extension F of F with

F:X-*Y,FIA=F and F(x) C coF(a).

Fixed Point Theorems

61

Proof : (a) We construct a partition of unity. Let X E X \ A, Bx = B(x, rk) be an open ball with diameter 2rx less than the distance of A to Bx, e.g.

rx < 6 dist(x, A) Then X \ A = UXEx\ABx

A result of General Topology states, that there exists a locally finite open refinement {UA, A E A} of {B.,, x E X \ A}, i.e. there exists a covering of X \ A, such that

dxEX\A 3B: cp{A:UACBx}<+oo.

Leta:X\A--*1 a(x) =

dist(x, X \ UU,). AEA

This sum is always finite, and x E X\ = a(x) > 0. Let cpa (x) =

dist(x, X \ UA) a(x).

cpa is continuous, 0 < cpa < 1, Ecp,, (x) = 1. {cpa, A E Al is a partition of unity for X \ A. (b) We now construct the extension.

VAEA3xEX\A U,\ c B,, hence dist(A, UA) > dist(A, Bx) > 2rx > 0.

For every A E A we choose as E A, such that dist(a,,, U,,) < 2dist(A, U,,).

62

Nonlinear Functional Analysis

Let

F(x) =

F(x)

,

Ecp,%F(aa)

,

if X E A if x E X \ A

Then F : x - Y is an extension of F, and F(x) C co(F(A)), since EWA(x) = 1, coa(x) > 1.

(c) F is continuous :

F is continuous for all x E 8A. Let xo ¢' 8A. Since F is continuous at xo,

VE>O 35>0 VxEA Ijx-xolI <S IIF(x)-F(xo)1I <E. This implies

VxEX JIx - xol) < 6/4

)1F(x)-F(xo)J1 <E

F(x) - F(xo) = Eco (x)(F(aa) - F(xo))

IIF'(x) - F(xo)Ij : EWA(x)jjF(aa) - F(xo)II If cp,,(x)34 0, then x E JA arid

lix - aAII < flx-ul(+Ilu-aAII
lixo - aall< ilxo - xUU+Ilx - aall< 411xo-x1l
lif(aa) - F(xo)II < Ecpa(x).E = E.

Therefore F : X -+ Y is continuous.

0

Fixed Point Theorems

63

Corollary 5.1 Let C C X be a closed convex subset of X. Then there exists a mapping R : X -- C, such that RI C = Id.

Proof : For F = Id1, apply the extension theorem. O

Corollary 5.2 (Brouwer's Fixed Point Theorem) Let C be a compact convex set in Rn, f : C - C a continuous mapping. Then f has a fixed point.

Proof : Let r > 0, such that B(0, r) D C, let f : IV -- Rn be an extension of f with f (Rn) C coC. Then

f (B(0,r)) C coC C C C 77(0,r).

f has a fixed point i in C, since f (±) = i E C. O

Theorem 5.3 (Schauder's Fixed Point Theorem)

Let K C X be a convex compact set, and F : K - K continuous. Then F has a fixed point.

Proof : Let e > 0, and xl, ..., xn an e - net for K, i.e.

KCU{x,+eB}, or `dxEK 3x,llx-x,ll <e. (B=B(0,1)={xEX:11xl1<1}). Let h' (x)

_ -

0,

e-

lIx-xjII,

if

llx-x,1l>e

if

1lx-xj11 <

E.

64

Nonlinear Functional Analysis

and

h,(x) = Eh?(x)xj Eh, (x)

h is continuous, if x E K, then Ilh(x) - -xl I =

Eh'(x)(xj 11

- x)

Eh; (x)

II

E.

Let KO = c6{xl,..., x,a}.Ko is compact, KO C K, and

hE o F

fixed point

xE E KO, hence

h,(F(x,)) = xE and

IIF(xE) - xEII = IIF(xE) -- h.(F(xE))II <- E.

The set {xE, E > 0} C K has a point of accumulation i E K, hence II F(x)

-III <_ IIF(I)

-

limh_,.,, xE,,, = I implies F(x) = I.

IIF(xE) - xIII + I Ix, - ill

Chapter 6 SET CONTRACTIONS AND DARBO'S FIXED POINT THEOREM 6.1

Measures of Noncompactness

Let B be the family of all bounded subsets of the Banach space X; recall that B C X is bounded if B is contained in some ball. B E B is relatively compact, if there exists for any e > 0 a finite covering of B by balls of radius e. If B E B is not relatively

compact then there exists an E > 0, such that B cannot be covered by a finite number of e - balls, and it is then impossible to cover B by finitely many sets of diameter < 6. Recall that

diam B=sup{IIx-yII,x,yEB} is called the diameter of B.

Definition 6.1 Let X be a Banach space and B its bounded sets. Then X : B --* R+, defined by X(B) = in f {6 > 0, B admits a finite cover by sets of diameter < 6} is called the (Kumtowski-) measure of noncom65

66

Nonlinear Functional Analysis

pactness and

/3 : B -+ R+, defined by Q(B) = inf {p > 0, B can be covered by finitely many balls of radius p } is called the ball measure of noncompactness.

Evidently, for B E B ,8(B):5 X(B) <_ 2Q(B),

but there exist B E B, such that the strict inequalities hold. The properties of the measures of noncompactness are collected in the following statements.

Proposition 6.1 Let X be a Banach space, B, B; E B. Then 1°

X(O) = 0

20

X(B) = 0 if B is relatively compact

3°

0 < X(B) < diamB

40

B1 C B2

51

X(Bi + B2) :5 X(Bi) + X(B2)

6°

X(AB) = I AI X(B) A E K

7°

x(B) = x(R)

8°

X(Bi U b2) = rnax(X(Bi), X(B2))

X(Bi) <_ X(B2)

Set Contractions and Darbo's Fixed Point Theorem

67

Proof : 1°

By definition diam 0 = 0.

20 B is relatively compact iff for every e > 0 there exists a finite covering by balls of diameter E.

3° M can be covered by M with diam M. 4°

Every cover of B2 is a cover of B1.

50

Let M1, ..., M,,, be a cover of N1, ..., N. a cover of B2, then all sets Mj + Nk form a cover of B1 + B2 B1,

and

diam(Mj + nk) < diam m; + diam Nk. 6°

Note that diam (AB) = ()Idiam B.

7°

From B C B follows X(B) < X(B). Conversely if B C UM,, then B C UMj with diam Mj = diamj, so X(B) < X(B).

8°

Let B = B1 U B2 and Q = max{X(Bi), X(B2)}. Then it follows from Bj C B that X(B) < X(B) and 0 < X(B). Conversely let for f > 0 given convergins Bj C Mjk with diam Mjk <_ X(B,) + e <_ Q + E. All of these

Mjk's together form a covering of B, so that X(B) :5,8 + E, i.e. X(B) <,8.

Proposition 6.2 Let B be a bounded set in X, then X(B) = X(coB) = X(coB).

Nonlinear Functional Analysis

68

Proof : Since B C coB, X(B) < X(coB), it remains to be shown that X(coB) < X(B). It is

diam B = diam coB, since for x, y E coB

x=Eajxj y=EAkyk

xjEB,Aj>O,Eaj=1 ykEB,ilk - O,EAk1

x - y = EAjxj - y = E)1j(xj - y) = EAj(xj -/ Eµkyk) = EAjEµk.(xj yk) Ilx - yll

EAj Eµk Ilxj - ykll < EAj Eµk diam B = diam B. <-

Now, let B C U711Mj with diam Mj < X(B) + E; and we can assume, that the Mj are convex. Let in

A = {(a1i ..., Am) E Rm, E Aj = 1, Aj j-1 and for A E A

in

A(A) _ F AiMp j=1

Then X(A(A)) <_ EAjX(Mj) <_ X(B) +,E.

Now we show that C = UAEAA(A) is convex.

0}l

Set Contractions and Darbo's Fixed Point Theorem

69

Let x, y E C, i.e. X E A(A), y E A(µ), and t E [0, 11. Then

x=EAjxj, AEA, xjEM, y=E/.ijyj, AEA, yjEMj

Let vj = tAj + (1 - t)µ3,

Pj=

1

0

if vj = O

tA3/vj

if vj#0

then 0
with v = (vi, ..., v,) E A. Finally we have to show the finiteness. Since

BCU? 1MjCUAEAA(A)=C, coB C C. Since A is compact, there exist AM, ..., A('') E A such that for A E A there is a A(k>, such that in

IA,-A'k)IsEpIIXII -<E. B

Let X E A(A), x = E7 Ajxj. Let y = Iix

yII

<EIAj-A(

Then k)I.11xjII <-E.

This means (U denotes the unit ball) C = UAEAA(A) C Uk_1A(A(k)) + EU

70

Nonlinear Functional Analysis

x(C) C maxx(A(a(k))) + E < X(B) + 2E. Thus X(coB) < X(C) < X(B) + 2E.

This implies the equality

X(coB) _ 6(B).

0 Proposition 6.3 Let X be a Banach space and (Mj) a decreasing sequence of nonempty closed bounded subsets of X, i.e.

M1_DM22...;? MnD_

---

such that M oX(Mn) = 0-

Y%-

Then

m. = nnENMn is nonempty and compact.

Proof : M,,. is compact, since X(Mo0) < X(M,n) - 0 and M,,. is closed. We have to show that M,,. $ 0. Let Mk, be chosen, such that Mk = Uj k1 Mk1 X( Mk) 3

Claim

<

X(Mk

1

)+ k

:

3j, VnEA( M31nMn#ci. If not, there is for each j an index n3, such that

m, nM=,=o)

Set Contractions and Darbo's Fixed Point Theorem then, for all ri

71

n3

MnnMi,=qb therefore

Mn = MnnMi = Mnn(UM1i) C MnnM1j =¢. This contradicts Mn #

Claim :

3j2 VnEN

m, j4

.

If not, then for each j there is an index nj, such that for n > nj

M1j,nM2jznMn=o which contradicts

MnCM2CU;M2j and Mn n M1 j,

for all n.

Finally by induction, we find that for every k there is an index jk such that for all n, m

Mnn(U,ti 1Mkjk)

(*)

Let Nm = Um k=lMkjk

...andM,nCMmjm then (*) implies, that for all m Nm 54 0

diam Nn < diam M,n < X(M,n) + m hence

72

Nonlinear Fbnctional Analysis

Choose a sequence (z,,,), such that z,,, E

Then (zn) is a

Cauchy sequence, which is convergent,

Cf1M,, =M,,,,. Therefore M,,,, is nonempty. 13

Condensing Maps

6.2

A continuous map F is called bounded, if it maps bounded sets onto bounded sets, and compact, if it maps bounded sets into compact sets.

Definition 6.2 Let X be a Banach space and U C X. An operator F : U -+ X is called a k-set contraction, 0 < k < 1, if F is continuous, maps bounded sets onto bounded sets, such that for all bounded sets B C U X(F(B))

kX(B)

F is said to be condensing, if for all bounded sets B with X(B)

0

X(F(B)) < X(B) Obviously, every k-set contraction is condensing. Every compact map is a 0-set contraction. The following example is an important one.

Example 6.4 Let X be a Banach space, U C X and K : U -+ X Lipschitz continuous with Lipschitz constant k < 1. C : U -+ X is compact.

Set Contractions and Darbo's Fixed Point Theorem

73

Then

,F=K+C is a k-set contraction.

Proof : Let B C U be a bounded set. Then

F(B) C K(B) + C(B) X(F(B)) < X(K(B)) + X(C(B)) < kX(B).

0 A continuous map is said to be proper, if the preimage of each compact set is compact. Lemma 6.1 Let B C X be closed and bounded and F : B -, X condensing. Then I - F is proper and I - F maps closed subsets of B onto closed sets.

Proof : Let K be compact and A = (I - F)-'K the preimage of K under I - F. Since I - F is continuous, A is closed. From K = (I - F)A we see, A C F(A) + K, hence X(A) :5 X(F(A)) + X(K) = X(F(A)) Since F is condensing, X(A) = 0, i.e. A is compact.

Now let A be closed and (xn) C A, yn = (I - F)x,,. Let y = lim yn. The set K = {y} U {yn, n E ,H} is compact, i.e. lim x,, = x exists and belongs to A, since A is closed. Thus X E A implies

y = (I - F)x E (I - F)A,

Nonlinear Fbnctional Analysis

74

and (I - F)A is closed. 13

We are now able to prove the fixed point theorem of Darbo (1955) and Sadovski (1967) which is a common generalization of Banach's and of Schauder's fixed point theorem.

Theorem 6.1 Let C C X be a nonempty,, closed, bounded and

convex subset of a Banach space X and let F : C -+ C be condensing. Then F has a fixed point.

Proof : Without loss of generality we may assume, that 0 E C.

Let F be a strict k-set contraction with k < 1. Define the decreasing sequence Co = C, C1 = MF(Co), Cn = oF(Cn_1). We have COQ

x(Cn) < kx(Cn-1) < ... < knX(Co), hence

lim x(Cn) = 0 and

Coo = nnENCn

is compact. Furthermore, C,,o is convex and

F(CC) C C,,.

Therefore, Schauder's fixed point theorem shows, that F has a fixed point in C. C C. Now let F be condensing and let (kn) be a. sequence with

0 < kn < 1, lim kn = 1. Then, since 0EC,Fn=k,,Fmaps C into C, and has a fixed point x,,,

The Topological Degree

75

thus

xn - F(xn) = xn - knF(xn) - (1 - kn)F(xn) implies

lim xn - F(xn) = 0.

The set K = {0} U {xn - F(xn)} is compact. Thus, since

I - F is proper by Lemma 6.5, (I - F)-1K is compact, i.e. {xn, n E .11(} has a point of accumulation z, i.e. there exists a subsequencewith lim xn,, =.f = F(x) = lim F(xn,, ). Therefore

F has a fixed point i in C. 11

Chapter 7 THE TOPOLOGICAL DEGREE 7.1

Axiomatic Definition of the Brouwer Degree in R'ti

The Brouwer degree is a tool, which allows an answer to the question, if a given equation

f(x)=y has a solution x E 0, where 12 C Rn is open and bounded, and f : -p Rn is continuous, and y does not belong to the image f (8Q) of the boundary 81 of 0. More precisely, for each admissible triple (f, ), y) we associate an integer d(f, S2, y) such that d(f, 0, y) 76 0 implies the existence of a solution x E S2 of this equation f (x) = V. This integer is uniquely defined by the following properties:

If f is the identity map, and y E Rn, then f (x) = y has a solution x E S2, iffy E 0, i.e.

(dl)

d(id, 0, y) = 77

1

for y E 0

0 for y V U

Nonlinear Functional Analysis

78

The second condition is a natural formulation of the desire that d should yield information on the location of solutions. Suppose that III and 112 are disjoint open subsets of SI and suppose, that f (x) = y has finitely many solutions in SI1 U 02i but no solution in \ (SIA U S22). Then the number of solutions in c is the sum of the numbers of solutions in 01 and SZ2, and this suggests that d should be additive in its argument c, that is

d(f, c, y) = d(f, 01, y) + d(f, ci2, y) whenever cl and SI2 are disjoint open subsets of 11, such that (d2)

Uc12))

The third and the last condition reflects the desire, that for complicated f the number d(f, St, y) can be calculated by d(g,11, y) with simpler g, atleast if f can be continuously deformed into g such that at no stage of the deformation we get solutions on the boundary. This leads to d(h(t,.), 11, y(t)) is independent of t E [0, 1] whenever h : [0,1] x SI --- R" and y : [0,1] -+ R" are continuous (d3)

and y(t) V h(t, 80) for all t E [0, 1). In principle, it is inessential how to introduce degree theory, since there is only one II-valued function d satisfying (dl), (d2), and (d3). Therefore we refer to an excellent book by K. DEIMLING, "Nonlinear Functional Analysis", and start with the following theorem, which collects the properties of the Brouwer degree. Definition 7.1 There exists a unique 11-valued mapping d, which associates every admissible triple (f , 0, y), where c C R", f : R" is continuous and y E R" \ f (8Sl) an integer d(f, St, y), 0 with the following properties :

The Topological Degree

79

1. If d(f , ill y) # 0, then there exist x E N such that f (x) = Y.

2. d(id, 0, y) = 1 if y E ), d(id,1, y) = 0, if y ¢ 1. [NORMALIZATION]

S. Let h : [0,1] x - Rn be continuous and y 0 h(t, Oil) for t E [0, 1], then d(h(t,.), ii, y) is independent from t. [HOMOTOPY INVARIANCE]

4. If g

IV is continuous and I If - g I I < dist(y, f (aQ) ),

then

d(f,11,y) = d(g,il,y) 5. If z E Rn, IIy

- zII < dist(y, f(911)), then d(f,1, y) = d(f, 0, z) -

6. If U{ `_1it C ill u1 Um 1 f (Mi), then

C S7, fl is open, disjoint, y m

d(f, fl, y) _ > d(f, clt, y) t=1

[ADDITIVITY] 7. If g : S tt

Rn is continuous and f 1,9n = glen then

d(f,il,y) =d(g,Q,y) 8. If A is a closed subset of SZ, A #

and y ¢ f (A), then

d(f, il, y) = d(f, SZ \ A, y).

[EXCISION PROPERTY]

9. d(f, 0, y) = d(f (.) - y, Q, 0)

80

Nonlinear Functional Analysis

10. Let m < n, cl C R" 'be open and bounded and f continuous, y E R' \ (I - f)(8SZ). Then

d(I - f, 0, y) = d(I - f J .

m, fl

R."

n Rm, y). [REDUCTION]

7.2

Applications of the Brouwer Degree

Theorem 7.1 (Brouwer) Let D C R" be a nonempty compact convex set and let f : D --+ D be continuous. Then f has a fixed point. The same is true if D is only homeomorphic to a compact convex set.

Proof : Suppose first that D = Br (0). We may assume that f (x) # x

on aD. Let h(t, x) = x - t f (x). This defines a continuous h : [0, 1] x D -+ Rn such that 0 ¢ h([0,1] x 8D), since by assumption

Ih(t,x)l > Ixl-tlf(x)I > (1-t)r > Din [0,1)xBDand f(x) # x for Ixj = r. Therefore d(id - f, D, 0) = d(id, B,.(0), 0) = 1, and this proves existence of an x E B,.(0) such that x - f (x) = 0. Next, let D be a general compact and convex set. By Theorem 5.4 we have a continuous extension f : R" of f, and if we look at the defining formula in the proof of this result, we see that f (R") C conv f (D) C D since m

m

w-iWi(x)f(a')

is defined for m = m(x) being sufficiently large, and belongs to cony f (D). Now, we choose a ball B,.(0) 3 D, and find a fixed

The Topological Degree

81

point x off in $,.(0), by the first step. But j (x) E D and therefore x = AX) = f (x). Finally, assume that D = h(Do) with Do compact convex and h a homeomorphism. Then h` 1 f h : Do - Do has a fixed point x by the second step and therefore f (h(x)) = h(x) E D. 13

Let us illustrate this important theorem by some examples. Example 7.1 (Perron - Frobenius) Let A = (a j) be an n x n matrix such that aid > 0 for all i, j. Then there exist A > 0 and x 0 such that xi > 0 for every i and Ax = Ax. In other words, A has a nonnegative eigenvector corresponding to a nonnegative eigenvalue. To prove this result, let m

D= xER":xi>0 for all i and Exi=1 i=1

If Ax = 0 for some x E D, then there is no need to prove this result, with A = 0. If Ax j4 0 in D, then E 1(Ax)i > a in D for some a > 0. Therefore, f : x -- Ax/ E 1(Ax)i is continuous in D, and f (D) C D since ail > 0 for all i, j. By Theorem 7.2 we have a fixed point of f , i.e. an xo E D such that Axo = Axo with A = Es 1(Ax0 )i

Example 7.2 Lets consider the system of ordinary differential equations u' = f (t, u), where u' = di and f : R x R" --+ R" is w - periodic solutions. Suppose, for simplicity, that f is continuous and that there is a ball B,.(0) such that the initial value problems

u' = f (t, u), u(0) = x E B,.(0) have a unique solution u(t; x) on [0, oo).

(7.1)

82

Nonlinear Functional Analysis

Now, let ptx = u(t; x) and suppose also that f satisfies the boundary condition (f (t, x), x) = E 1 f{(t, x)x{ < 0 for t E [0,w] and 1 x) = r. Then, we have Pt : R,.(0) - B,.(0) for every t E R+, since dt

Iu(t)12 = 2(u'(t), u(t)) = 2(f (t, u(t)), u(t)) < 0

if the solution u of equation (7.1) takes a value in 8B,. (0) at time t. Furthermore, Pt is continuous, as follows easily from our assumption that equation (7.1) has only one solution. Thus we find P, has a fixed point x,,, E Br(0), i.e. u' = f (t, u) has a solution such that u(0; x,,) = x, = u(w, x,,). Now, we may easily check that v : [0, oo) - R'&, defined by v(t) = u(t - kw, xk,) on [kw, (k + 1)w], is an w - periodic solution of equation (7.1). The map P,, is usually called the Poincare operator of u' = f (t, u), and it is now evident that u(.; x) is an w - periodic solution if x is a fixed point of P.

Example 7.3 It is impossible to retract the whole unit ball continuously onto its boundary such that the boundary remains pointwise fixed, i.e. there is no continuous f : B1(0) -' 0B1(0) such that f (x) = x for all x E 8B1(0). Otherwise g = -f would have a fixed point x0, by Theorem 7.2, but this implies Ixol = 1 and therefore x0 = -,f (xo) = -xo, which is nonsense. This result is in fact equivalent to Brouwer's theorem for the ball. To see this, suppose f : B1(0) -+ Bi (0) is continuous and has no fixed point. Let g(x) be the point where the line segment from f (x) to x hits 8B1(0), i.e. g(x) _ f (x) + t(x)(x - f (x)), where t(x) is the positive root of t2lx

- f(x)12 + 2t(f (x),x - f(x)) + If(x)12 =1.

The Topological Degree

83

Since t(x) is continuous, g would be such a retraction which does not exist by assumption.

Surjective Maps In this section we shall show that a certain growth condition

of f E C(R't) implies f (R") = R". Let us first consider that fo(x) = Ax with a positive definite matrix A. Since det A 54 0, fo is surjective. We also have (fo(x), x) > cIx12 for some c > 0 and

every x E R", and therefore (fo(x), x)/I xl - oo as fix) -> oo. This condition is sufficient for surjectivity in the nonlinear case too, since we can prove the following theorem.

Theorem 7.2 Let f E C(R") be such that (fo(x), x)/Ixl - 00

asIxI -goo. Then f(R")=R". Proof : Given y E R", let h(t, x) = tx + (1- t) f (x) - y. At JxI = r we have

(h(t, x), x) > r[tr + (1 - t)(f (x), x)/lxl - I yI ] > 0 for t E [0, 11 and r > IyI being sufficiently large. Therefore, d(f, B,.(0), y) = 1 for such an r, i.e. f (x) = y has a solution. 13

Hedgehog Theorem Up to now we have applied the homotopy invariance of the degree as it stands. However, it is also useful to use the converse namely: if two maps f and g have different degree then a certain h that connects f and g cannot be a homotopy. Along these lines we shall prove the following theorem.

Theorem 7.3 Let SZ C R" be open bounded with 0 E Q and let

f : 8f -- R" \ {0} be continuous. Suppose also that the space

Nonlinear Functional Analysis

84

dimension n is odd. Then there exist x E 8f2 and A that f (x) = Ax.

0 such

Proof : Without loss of generality we may assume f E C(), by Proposition 5.1. Since n is odd, we have d(-id, Cl, 0) = -1. If d(f, SZ, 0) # -1, then h(t, x) = (1 - t) f (x) - tx must have a zero (to, xo) E (0, 1) x 8f1. Therefore, f (xo) = to(1- to)-1xo. If, however, d(f, Cl, 0) = -1 then we apply the same argument to h(t, x) _ (1 - t) f (x) + tx. 13

Since the dimension is odd in this theorem, it does not apply in Cn. In fact, the rotation by 2 of the unit sphere in C = R2, i.e. f (x1, x2) = (-x2, x1), is a simple counter example. In

case Cl = B1 (0) the theorem tells us that there is at least one normal such that f changes at most its orientation. In other words: there is no continuous nonvanishing tangent vector field

on S = 8B1(0), i.e. an f : S -> Rn such that f (x) # 0 and (f (x), x) = 0 on S. In particular, if n = 3 this means, that `a hedgehog cannot be combed without leaving tufts or whorls'. 1) is a nonvanishing However, f (x) = (x2, -x1, ..., x2,,,,

tangent vector field on S C R1. The proof of existence and uniqueness of the Brouwer degree and its construction is based on the fact, that f (Sf (0)) is of measure zero, where S f(fl) is the set of critical points of f, i.e. Sf(ci) = {x E 11, Jf(x) = det f'(x) = 0} (Sard's lemma), and approximations of continuous functions by differentiable functions.

Proposition 7.1 Let Cl C Rn be open and f E C' (Q). Then (Sf)) = 0, where An denotes the n - dimensional Lebesgue measure.

The Topological Degree

85

Proof : We need to know here about Lebesgue measure /pn is that

µn(J) = rj 1(bi - a{) for the interval J = [a, b] C RI and that M C Rn has measure zero (i.e. µn(M) = 0) iff to every e > 0 there exist at most countably many intervals Js such that

MCU;JJarid

<e.

Then it is easy to see that at most countable union of sets of measure zero also has measure zero.

Since an open set 11 in Rn may be written as a countable union of cubes, say SZ = U1Q1, it is therefore sufficient to show An (f (Sf(Q))) = 0 for a cube Q C 11, since f (Sf(f )) = U; f (Sf(Q1)). Let p be the lateral length of Q. By the uniform continuity of f' on Q, given e > 0, we then find m E .fU such

that I f'(x)- f'(Y)I < E for all x,Y E Q with Ix-zI < 6 = f., and therefore

I f (x) -- f (y) - f'()(x -

)I

<

j

I f'(Y

-f'(T)1Ix - YIdt < EIx - TI

for any such x, T. So let us decompose Q into r cubes Qk of diameter 6. Since b/ f is the lateral length of Qk, we have r = mn and f (x) = f (T) + f'(T)(x - 7) + R(x, 7) with I R(x, T)1 < eb for x, r E Qk.

Now, assuming that Qk n s f # 0, choose r E Qk n S f; let = Qk _X. A = f'(T) and g(y) = f (T + Y) - F(T) for y E Then we have

9(y) = Ay + R(y)

Nonlinear Functional Analysis

86

with IR(y) I = IR(T + y, 7) I <_ eb on Qk.

Since det A = 0, we know that A(Qk) is contained in a (n - 1) - dimensional subspace of Rn. Hence, there exists bl E Rn with Ib1I = 1 and (x, b1) = 1 xib2 = 0 for all x E A(Qk). Extending bl to an orthonormal base {b1, ..., bn} of IV, we have g(y) =

n

>(9(y), b`)b$ i=1

with I(9(y),b1)I <- IR(y)IIb1I < eb and

I9(y), bi)I <_ IAIIyl + I R(y)I <- IAIS + e6 for i = 2,..., n, where IAI = I(aij)I =

(E1 aj)'). Thus, f (Qk) _ .f (m)+9(Qk)

is contained in an interval Jk around f (T) satisfying /in(Jk) = [2(IAIS +

E8)]n-1.2e6

= 2n(IAI + )n-1.6n.

Since f is bounded on the large cube Q, we have If'(x) I < c for some c, in particular IAI < c. Therefore, f (Sj(Q)) C Uk=1Jk with r

µn(Jk) :5 r.2n(c + E)"-1.6n = 2n(C + E)n-1(/ p)nE, k=1

i.e. f (Sf(Q)) has measure zero, since E > 0 is arbitrary.

0 Having this result we give the following definition.

Definition 7.2 (a) Let 1Z C Rn be open and bounded, and let f o : T7 - Rn be continuous and twice continuously differentiable

The Topological Degree

87

in Q. Let yo E R" \ fo(O U Sfo(Sl)). Then

d(fo, IZ, yo) _ E sgnJfo(x) xE fa 1 {y}

(b) Let f R" be continuous and y 0 f (80). Choose a twice continuously differentiable function fo : 3i -- R", and yo 0 fo(eci U Sfo(Sl)), such that 11f -foil < r, Ily - yoll < r, where

r = dist(y, f (80)). Then by (a) we set d(f,11, y) = d(fo, cl, yo).

7.3

The Leray-Schauder Degree

Let X be a real Banach space, Sl C X an open and bounded subset of X, F : 3I -+ X compact and y V (I - F) (Oil). On these admissible triplets we want to define a II- valued function D that satisfies the three basic conditions corresponding to (d 1), (d2) and (d3) of the Brouwer degree, namely, (D1)

D(I, Sl, y) = 1 for y E Sl

(D2)

D(I-F,52,y)=D(I-F,521iy)+D(I-F,fl2,y) whenever S21 and 112 are disjoint open subsets of

Sl such that y 0 (I -- F)(? \ (Sli U D2)) (D3)

D(I - H(t,.), Sl, y(t)) is independent oft E [0, 11, whenever H : [0, 1] x N - X is compact, y : [0,1] -+ is continuous and y(t) ¢ (I - H(t, .))(8SZ) on [0, 11.

In the first step we will extend the Brouwer degree to finite

dimensional Banach spaces. Let X be a n - dimensional real

88

Nonlinear Functional Analysis

Banach space, SZ C X open and bounded, and F : N

-. X

continuous. Let {x1, ..., x"} be a basis in X, {e1, ..., e"} the unit vector basis of R", and h : X - R" the linear homeomorphism, defined by

h(xi) = ej,

j = 1, 2,..., n.

Then h((2) is open and bounded in R", f = hFh-1 : h(O) --+ R" is continuous and h(y) 0 f (ah(Sf)). Then we define d: (F, (1, y) = d(hFh-1, h(11), h(y)).

It is easy to check by the properties of the Brouwer degree that this definition is independent of the choice of the basis of X.

In the second step we will assume, that X is an arbitrary real Banach space, SZ an open and bounded subset of X and F : --> X a compact map. By .F(?7) we denote the compact mappings F : St -+ X such that F(SZ) is contained in a finite dimensional subspace of X. Let G = I - F and y g X \ G(&I). By Lemma 6.5 G(&) is closed, and r = dist(y, G(8 )) > 0. Now let F1 E .F(SI) be an approximation of F, such that supZEn I I F(x) - Fi (x) I I < r. Then we have

y¢(I-F1)(afZ) since

dist(y, (I - F1)(af2)) = inf IIy - (.T- F1(x))II MEM (IIy-(x-F(x)II-IIF,(x)-F(x)II)>0. inf XEM

Now we choose a finite dimensional subspace X1 of X, such that y E X1, Sl fl Xi 0,0 and Fi (3'7) C X1. Then Q, = Sl fl X1 is open and bounded in X1, and by the first step we have the existence

of

The Topological Degree

89

This number is independent of the choice of Fi and X1. Let F2 E such that I IF - F21I < R and X2 C X, such that

dimX2
dz((I - Fj)I?ji,III, y) = dz((I - Fj)I?1o,flo,y) The homotopy H : [0,1] x No - Xo with H(t, x) = t(x - F1(x)) + (1 - t)(x - F2(x))

connects I - Fi with I - F2 and it is IIx - F(x) - H(t,x)II = IIF(x) - tF1(x) - (1 - t)F2(x)II < tIIF(x) - Fl(x)II + (1-- t)IIF(x)

- F2(x)II < r,

hence y ¢ H([0,1] x 8S?,o). Therefore by the homotopy invariance

principle

dx((I -

slo, y) = dx((I - Fa)I?jo, slo, y)

Thus it is legitimate to give the following definition.

Definition 7.3 (Leray - Schauder Degree) Let X be a real Banach space, Cl C X a bounded and open subset of X, F : 37 --+ X a compact map, y E X such that y 0 (I - F) (8S1). Let D(I -- F, il, y) = dz((I - Fi)lci,, Sli, y), where Fi E .F(!') is chosen, such that with a finite dimensional subspace X1 of X

IIF- Fill < dist(y, (I - F)(8)) F1(?6)

cX1,yEX1,C11 =onXi 96o,

this number D(I - F, Cl, y) is called the Leray-Schauder degree.

90

Nonlinear Functional Analysis

Now it is rather clear that we obtain nearly all properties of the Brouwer degree for the Leray - Schauder degree, too.

Theorem 7.4 Besides (DI), (DQ), (DS), the Leray - Schauder degree has the following properties:

(D4) D(I - F, S2, y) 00 implies (I - F)'1{y} 00. (D5) D(1- F, S2, y) = D(I - G,12, y) for G : 3'E - X compact [IF - GII < r = dist(y, (I - F)(8S2)).

(D6) D(I - F,S2,y) = D(I - F,S2,z) forz E X, Ily - zII < r = dist(y, (I - F)(812)).

(D7) D(I - F, S2, y) = D(I - G, S2, y), whenever Glen = Flon.

(D8) D(I - F, S2, y) = D(I - F, SZ., y) for every open subset Q. of 0 such that y ¢ (I (D9)

- F)(3I \ S2.).

Let Y be a closed subspace of X, S2 C X be open and bounded, S2 fl Y # 0. Let F Y compact,

y $ Y \ (I - F) (09% Then D(I - F, 0, y) = D((I F)I nnY, 12 f1 Y, V).

-

Proof : (D4) Let Fn E .F(U2) such that IIF

FnhI < Xn, S2n according to the Definition 7.7. Then

nr.

Choose

dd((I - Fn)Inn,On,y) 0 0 i.e. 3xn E On such that xn - Fn(xn) = y. Thus

Ilxn-F(xn)-yII <- Ilxn-Fn(xn)-yll+llF(xn)-Fn(xn)il <-

nr --+ 0.

Since (I - F) (?I) is closed, y E (I - F)( ?7), hence x - F(x) = y

foranxE7,xit 8S2,i.e. xES2.

The Topological Degree

91

(D5) Choose GI E F(?t) such that JIG - G111 < min{dist(y, (I - G)(90)),

dist(y, (I - F)((952)) - IIF - GII}. Then

D(I - G,Sl,y) = d=((I - G1)In,,ni?y)

From JIG,

- FII <_ IIG - FII + IIG - G1II < dist(y, (I - F)(852))

follows D(I - F, 52, y) = d,, (I - Gi) In,, 521, y).

(D6) is similar to (D5).

(D7) Let H(t, z) = tF(x) + (1 - t)G(x). Then y ¢ H([O,1] x 052) if Flan = GIan.

Thus D(I-F,52,y)=D(I-G,0,y).

-

(D8) (I F) (S2 \ St.) is closed. Let r = dist (y, (I - F) (3 \ 52.)) > 0 and F1 E ,F(?) such that IIF - Fi I I < L. Then

dist(y,(I-Fi)(

\Q.))?r-IIF-FiII>

2

and for properly chosen X1 C X, 521 = SZ fl Xi

D(F,0,y) = dx((I -Fi)In1, !ill y) dx((I - Fi)In1, Sl, fl Xi, y)

D(F,Sl.,y) (D9) Since F(52) C Y for r = dist(y, (I - F)(81)) there exists F1 E F(? I), I I F - FiII < r and F, (?7) C Y. Let X1 C X,

92

Nonlinear Functional Analysis

dim Xi < 00,y E Xi, st n-Y n Xi # ¢. Then

D(I - F, i, y) = dx(I - Filnnx1, fZ n Xi, y) = dx(I - Filnnynx,, SZ n Y n Xi, y)

= D(I-F( ,y,11nY,y) by reduction property (1) of Theorem 7.1.

0

7.4

Borsuk's Antipodal Theorem

Whenever we want to show by means of degree theory that f (x) = y has a solution in 0, we have to verify d(f,11, y) # 0. For symmetric domains fl (with respect to the origin) and for odd maps Borsuk's theorem states that d (f , 0, 0) is odd, hence different from zero. We say that SZ C X is symmetric with respect to the origin, if S2 = -0, and f : fZ -' X is said to be

odd if f(-x)=-f(x)forxEIl. We start with X=R. Theorem 7.5 Let S2 C R" be open, bounded, symmetric. with 0 E 9. Let f : St -- IV be odd and 0 ¢ f (80). Then d(f, 0, 0) is odd.

Proof : 1. We may assume that f is continuously differentiable in Sl and jf (0) 0. To see this, approximate f by a differentiable gi and consider 92 with g2(x) = 1(91(x) - gl(-x)), choose S which is not an eigenvalue of g2(0). Then AX) = g2(x) - S - x is continuously differentiable, odd, If(x) # 0, and

Ilf -- Ill = supI2(f(x) -gi(W)) - 2(f (-x) -gl(x)) -Sxl < Iif -gill + 6 diam 0

The Topological Degree

93

can be chosen arbitrarily small, hence d(f, S2, 0) = d(f, 1Z, 0).

2. Let f be continuously differentiable and Jj (0) # 0. Now to prove the theorem it suffices to show that there is an odd g : 0 - R", differentiable, close to f such that 0 5t 9(S9(S2)), since then

E

d(f, S2, 0) = d(g, 12, 0) = sgnJ9(0) +

sgnJ9(x)

x 0 x E g-101

where the sum is even, since g(x) = 0 if g(-x) = 0 and J9(x) is even, since g(x) = -g(-x) implies g'(x) = g'(-x). 3. We begin with the following observation. Let

u:12lR", v:Sl-+R be continuously differentiable, v(x) # 0 if x belongs to an open subset A of Q. For y E R" define hy(x) = u(x) - v(x).y,

then y is a regular value of !'IA if 0 is a regular value of by IA, since for x E A, (x) = y, thusv hy(x) = 0 implies by the quotient

rule (v )'(x) _ 'vh,(x), because u'(x) v'(x)u(x) V

v(x)

v(x)

= v(x) (u'(x)

.[u'(x) V(X)

v`(x).uU

v

(x)]

- v,(x)y),

thus (by Sard's lemma) 0 is a regular value of NIA for almost all y E R". (*)

Nonlinear Fltnctional Analysis

94

Now we define odd functions hl, ..., h" E G1(1l) with l f - hk I I < E if x E ? and 0 is regular value of hk Ink, where S") E R", Sk = 0} Hk = {x = cik = cl\(H1n...nHk).

Let h1 (x) = f

y1i where y1 E R", such that 0 is a regular

value of h1 In, and ly1 I sufficiently small. If hk is defined, we define hk+1 by hk+1(x) =

with a sufficiently small yk+1, such that 0 is a regular value of hk+1In\Hk+i. If x E Sl n Hk+1 we have hk+1(x) = hk(x) and

hk+1(x) = hk(x), for x E 1k n Hk+1, by induction we get hk+1(x) = 0, then hk+1(x) is regular. Since

(IlknHk+1)u(Il\Hk+1)

= (Sl\h1n...nHk)nHk+lu(n\Hk+1) = Sl\Hl n...nHk+1 =clk+1 we have that 0 is a regular value of hk+1 Ink+,. Let g = h", then 0 is a regular value of h"In\{o}, but by definition of g = h" we obtain g'(0) = f'(0), 0 is a regular point of g.

0 Corollary 7.1 Let Q C R" be open, bounded, symmetric and R" be continuous such that for all A > 1 0 E Sl. Let f

andxE8Sl f (x) Then d(f, i, 0) is odd.

0,

f (-x) - Af (x)

0.

The Topological Degree

95

Proof : The mapping h(t, x) = f (x) - t(-x) defines a homotopy in R" \ {0} between f and the odd function g(x) = f (x) - f (-x). 11

Corollary 7.2 (Borsuk - Ulam) R' be continuous and m < n. Then there exists an x E 8Q such that Let SZ C R" be as before, f :

c 9Q

f (x) = f

(-x).

Proof :

If not, g(x) = f (x) - f (-x) # 0 for all x E M. Again by g we denote a continuous extension to of these boundary values. Then d(g, ), y) = d(g, SZ, 0) for all y in a properly chosen neighbourhood U of 0 (by (5) of definition (7.1)). Thus U C g (?I) C Rm, which is not possible.

0 Theorem 7.6 Let SZ C 1Z" be open, bounded and symmetric with respect to 0 E 0 and {A1, ..., Ap} be a covering of 8St by closed sets A,, C 8fl such that A; n (-A5) = 0 for j = 1, 2,...) p. Then p ? n + I.

Proof : Suppose that p < n, let f j (x) = 1 on A,, and f, (x) _ -1 on -A; for j = 1, 2,..., p-1 and f9 (x) = 1 on SZ for j = p, p+l, ..., n. Extend fl,..., fp_ 1 continuously to f. Let x E Ap, then

-x ¢ Ap, and therefore -x E A, for some j < p - 1, i.e. x E -A,, . Hence

Nonlinear Functional Analysis

96

Let x E A then f3 (x) = 1, f j (-x) = -1 and x E -A; implies

f,(x) = -1, fj (-x) = 1. Let f = (f'), then f (-x) does not point in the same direction as f (x), thus f (-x) # A f (x) on f for A > 0, by Corollary 7.1 d(f, S2, 0) is odd, i.e. there exists an x E f such that f (x) = 0, which contradicts ,,(x) = 1. 13

No:, we are able to compute the measure of noncompactness of the unit ball.

Corollary 7.3 Let B = B(0,1) be the unit ball in an infinite dimensional Banach space X. Then the measure of noncompactness X(B) = 2.

Proof Let S = aB be the boundary of B and

S=U 1MM

be a finite covering of S by closed sets M; C S such that diam(M3) < 2. Let X be an n - dimensional subspace of X, is the boundary of the unit ball in then S n Xn = Ul 1(M; n Xn, and therefore one of the Mj n X contains a pair of antipodal points x and -x by theorem 7.6. Hence diam M,, > IIx - (-x)II = 2IIxIf = 2. This contradiction along with the fact X(B) < 2 shows X(B) = 2. 11

If X is infinite dimensional, the Borsuk theorem follows immediately from the finite dimensional case:

The Topological Degree

97

Theorem 7.7 Let 1 C X be open, bounded and symmetric with respect to 0 E 0, F : fZ --> X be compact, G = I - F and 0 0 G(afZ). If for all A > 1 and for all x E aft, G(-x) - AG(x) 34 0, then D(I - F, f2, 0) is odd. In particular, this is true if Flan is odd.

Proof : Let H(t, x) = then H is compact and x # H(t, x) for all x E afl, t E 10,1] since

x - H(t, x)

-

t 1+

t1(1 + t)x - t F(x) + F(-x))

t .(1(x - F(x)) - (-x - F(-x))

l+t

t

1- t.(1t G(x) - G(-x)) jA 0. Let Fo(x) = z (F(x) - F(-x)), then Fo is odd and thus D(I - F, ft, 0) = D(I - Fo, ft, 0). Choose F1 E .F() with IIFo - Fill < dist(0, (I - Fo)(afi)) and F2(x) = 2(Fi(x) Fi(-x)), then F2 is odd and IIFo - F2I1 <- IIFo - Fill,

-

therefore

D (I - Fo, c, 0) = d ((1- F2) I n, 112, 0) is odd.

We will conclude this section with the following corollaries.

Corollary 7.4 Let fZ C X be open, bounded and symmetric with respect to 0 E Il, and F : S7 -+ X compact and odd. Let Y C X be a proper linear subspace and (I - F) (3I) C Y. Then there exists a fixed point xo E aft of F.

Nonlinear Functional Analysis

98

Proof : If 0 0 (I - F)(t9 Z), then D(I - F, Q, 0) is odd, i.e. for all y E A, then connected component of X \ (I - F) (8cl), which contains 0, there exists an x E ), such that x - F(x) = y, i.e. C 0 C (I - F) (Q) C Y # X. This contradicts the openness of 0.

0 Corollary 7.5 Let fZ C X be open, bounded and symmetric X compact. Let (I-F)(17) C with respect to 0 E 1, and F C Y # X. Then there is an x E 8SZ such that

x - F(x) = -x - F(-x). Proof : z(F(x) - F(-x)). By Corollary 7.4 there is an Let G(x) = x E 81 with G(x) = x, thus I (x - F(x)) + (x + F(-x)) = 0

or x - F(x) = -x - F(-x).

z 11

Corollary 7.6 Let F : X -+ X be compact, and L : X -+ X compact and linear. If there exists a A E R, such that for all xEX,IIxII =r we have I I F(x) - ALxl I < I Ix - ALxl I ,

then D(I - F, B (0, r), 0) is odd.

Proof : Let H : [0,11 x B (0, r) -' X be defined by

H(t, x) = x - (1 - t)ALx - tF(x),

The Topological Degree

99

then

IIH(t,x)II ? Iix - ALxII - tiIF(x) - ALxII > 0, hence 0 ¢ H([0,1] x (911). By homotopy invariance of the degree

D(I - F, B(0, r), 0) = D(I - AL, b(0, r), 0) is odd, since linear opeators are odd.

0

7.5

Compact Linear Operators

One of the most useful applications of the antipodal theorem is the fact, that the degree is different from zero. We will continue these considerations and specialize to linear compact operators.

Theorem 7.8 The product formula for the Leray - Schauder degree: Let fZ C X be open bounded, F0 : N -+ X compact,

Co : X -- X compact, F = I - Fo, G = I - Go, y ¢ GF(8) and KA, A E A the connected components of X \ F(e). Then

D(G,F,ll,y) =

D(F,S2,KA)D(G,KA,y) AE A

where only finitely many terms are non zero and

D(F, St, KA) = D(F,1l, z) for any z E KA. This product formula in the sequel we need only for linear maps and y = 0, thus we will omit the proof of Theorem 7.8. Unfortunately, for the simplest proof of the product formula we need the approximation property of X, but in spite of this loss of generalization we will prove the following results.

Nonlinear Functional Analysis

100

Proposition 7.2 Let S, T : X - X be compact linear operators, such that 1 is not an eigenvalue of S or of T. Then for

r>0 D((I - S)(I - T), B(0, r), 0) = D(I - S, B(0, r), 0).D(I - T, B(0, r), 0).

Proof : Let B = B(0, r).. Since 0 V (I - S)(cB) U (I - T) (8B), all degrees are defined, and the mappings I - S, I - T and (I - S) (I - T) are isomorphisms, hence ker(I - S) , ker(I - T ), ker(I - S) (I - T) consist only of the zero element. Let us now assume that X has the approximation property, i.e. every compact linear map is the uniform limit of compact linear maps with finite rank, we find finite rank operators So, To such that I I S SoII and IIT ToII are sufficiently small. Then

-

-

D((I - S)(I - T), B, 0) = D((I - So)(I - To)Ieo, Bo, 0) = sgn det(I - So)(I - To)IB0 = sgn det(I - So)IB0.det(I - To)IBo

= D(I - So, B, 0).D(I - To, B, 0) 11

The following result is a special case of Proposition 7.2.

Theorem 7.9 Let X = X1 ® X2 be a topological composition, and T : X --+ X a compact linear operator with TXj C XX, j= 1, 2. Let I - T be an isomorphism of X . Then for each open

ball B=B(0,r) inX D(I -T,B,0) = D(I -T1 1,BnX1i0),D(I -TI-2fBnX2,0).

The Topological Degree

101

Proof : Let Bj = B fl Xj, Pj : X - Xj be the linear projection onto Xi and Aj = (I - T)Pj. Then x = P1x + P2x. Let

Si=(I-T)Pi+P2, S2=P1+(I-T)P2. Sj is injective, since Six = 0 implies P2x = 0 and (I -T)Pix = 0 but I -T is injective, hence P1x = 0. Observe that Sj = I -TPj,

and 1 is not an eigenvlaue of T, thus Sj is an isomorphism. I - Sj = TPj is compact, (I - Sj)(X) C Xj, by the reduction property (9) of the Leray - Schauder degree we obtain D(Sj, B1 + B2i 0)

= D(SjI (B1+B2)nx;, (Bi + B2) fl Xj, 0)

= D(Sj, Bj, 0).

S1S2 = (I -TP1)(I -TP2) = I -T(Pi +P2)+TP1TP2 = I -T since

TPiTP2 = TP1P2T = 0 implies by Proposition 7.2

D(I - T, Bi + B2, 0) = D(S1S2, Bi + B2, 0)

= D((I - TP1)lx,, B1, 0).D(I - TP2)`x2, B2, 0) = D(I -T1x,,Bi,0).D(I -T1x2,B2,0). 0 Theorem 7.10 Let X be a Banach space, T : X --+ X be linear and compact, let o (T) = {A E C : T - AI is not continuously invertible } be the spectrum of T. Then

102

Nonlinear Functional Analysis 1°

u(T)C {AEC:IAI
2° 30

.

If A ¢ o(T ), T - AI is an isomorphism in X

.

If A E o,(T) \ {0}, there exists a minimal number

k(A) e N, such that for R(A) = (AI - T)k(A)(X), N(A) = ker(AI - T)k(')

(a) X =R(A)®N(A) (b) TR(A) C R(A),TN(A) C N(A).

The number n(A) = dim N(A) is called the algebraic mul-

tiplicity of the eigenvalue A, the geometric multiplicity is dim ker(T - AI).

Theorem 7.11 Let X be a real Banach space, T : X ---- X be a compact linear operator, A 0 and A` not an eigenvalue of T . Let 0 C X be open and bounded and 0 E SZ Then D(I - AT, 0, 0) = (-1)m() where m(A) is .

the sum of the algebraic multiplicites of the eigenvalues µ of T; satisfying pA > 1, and m(A) = 0 if T has no eigenvalue of this kind.

Proof : Let S = I - AT = -A(T - A-'I) is a homeomorphism onto X. Hence it is sufficient to consider 0 = B, the unit ball. By theorem (7.10, (1°)) there are at most finitely many eigenvalues

µ E a(T) with Ap > 1, i.e. sgn p = sgn A and IµI > µi, ...,1u7,. Let

V = ®;=1N(µ1), W = lj=1R(1_ij)

IA-1I, say

The Topological Degree

103

We show that X = V ® W. First of all v n w = ¢, since x E V n W implies P

x = I: xj, xj E n(µj) and x E R(µj) j=1

for j = 1, 2, ..., p. By theorem (7.10, (30)) we have

x2+x3+...+x+pER(p1) hence P

xl = x - E xj E R(Al) n N(jul) = {0} j=2

and similarly we obtain x2 = ... = x, = 0. Now, any x E X may be written as x = xj + yj with xj E n(µj), yj E R(µj) by theorem (7.10, (3°)) again, we have P

x-E=x-xk-rxj -yk-1: xj ER(/Aj) j=1

j#kk

jOk

hence P

x - L xj E W = nR(pk) j=1 and

X=v w By Theorem 7.9 we have D(S, Sl, 0) = D(SIv, o n V, 0).D(SI w,11 n W, 0)

But D(SIw, S2 n w, 0) = 1 since T I w has no eigenvalue It with pA > 1 and x - tATx defines an admissible homotopy from I - AT to I. By the same theorem we have P

D(SI v, 9 n V, 0) = jj d(SIN(µi), st n N(µj), 0). j=1

Nonlinear Functional Analysis

104

Since h(t, x) = (2t -1)x - tATx is an admissible homotopy from

to -II N(iz,) (this is true because (2t - 1)x - tATx = 0 2t-71 hence t = 2_aµ, >1 )thus and I Ix I I = 1 implies 1a = SI N(µ;)

)

D(SI N(j,i), Q n N(/j), 0) =

D(-IINC,,,), Sl n N(µ), 0)

and therefore D(S, Q, 0) _

1)m(''),

where m(A) _ IAI,>i n(µ3).

If there are no such D(S, Q, 0) = 1 = (-1)°.

µ

at all, then X

W and

0

Chapter 8 BIFURCATION THEORY 8.1

An Example

Let X be a Banach space, 1 C X an open bounded subset, and F : SZ -+ X compact. We consider problems of the following type. Assume 0 E S2 and F(0) = 0. Then for every real A the equation

Ax = F(x)

(*)

has trivial solution. The following example shows that there exist real \o that 0 < IIx,,II <E,

Pa - Aol < E

Ax,, = F(xa),

a branch of nontrivial solutions of (*) starts in the point (Ao, 0). The point (A - 0, 0) E R x ) is called a bifurcation

i.e.

point. We will solve the following integral equation

Ax(s) = n

jr

sin s sin t + b sin 2s sin 2t][x(t) + x3(t)]dt (8.1) 105

Nonlinear Functional Analysis

106

which has a second-rank kernel. We suppose that 0 < b < a. Because of the form of the kernel, any solution of equation (8.1) is necessarily of the form x(s) = A sin s + B sin 2s

with undetermined constants A, B (which will turn out to be functions of the real parameter A). Substituting in equation (8.1), we have A[A sins + B sin 2s] = 2 .

+ b sin 2s sin 2t]

T.

[A sin t + B sin 2t + (A sin t + B sin 2t)3]dt

= 2a. sin s[A f sin 2 tdt + A3 f sin4 tdt 0 n o +3AB2 fo s ine t. sine 2tdt] if

+ 2 b. sin 28 [B

+B3

J

sin2 tdt + 3A2B TO sin2 2t. sin2 tdt

sin4 2tdt] 0

2a.sinsj7rA+

8A3+34AB2]

+2b. sin 2s[2B +4A2B + 8 B], where use has been made of the following values of integrals: x

10

sin3 t. sin 2tdt = T sin t. sin3 2tdt = 0.

sin t. sin 2tdt =

0

To

Equating coefficients of sin s and sin 2s, we obtain a pair of nonlinear simultaneous algebraic equations:

AA = aA + 3 aA3 + 3 aAB2 4

2

aB = bB + 2 bA2B + 3bB3. There are four kinds of solutions of equations:

1. A = B = 0; this gives the trivial solution of equation (8.1).

Bifurcation Theory

107

2. A # 0, B = 0; only the first equation is nontrivial. We cancel A # 0 to obtain A

=a+3aA2

whence

A=f

a/a -1 .

The corresponding solution of equation (8.1) is xl (s, A)

=f3

A/a - 1 sins,

defined and real for A > a.

3. A = 0, B # 0; only the second equation is nontrivial. We cancel B # 0 to obtain A = b +3bB2

whence

f

B=±23 A/b-1. The corresponding solution of equation (8.1) is x2 (s, 1) =

f0 A/b - 1 sin 2s,

defined and real for A > b, where we recall that b < a.

4. A # 0, B # 0; here both A and B may be cancelled in equation (8.2). We obtain two ellipses:

4A2+2B2=--1 (8.3)

Nonlinear Functional Analysis

108

Solutions of equation (8.3) are given by intersections of these ellipses. Solving, we get

A24.[2a-b,_,]

B22b-a\-1

9ab

9'

ab

so that we have the following solutions of equation (8.1): (s, X3

A) = f

2a - b,\

2.

ab

3

- 1 sins ±

2

2b - a.

3

ab

- 1 sin 2s. (8.4)

Clearly 2a - b > 0 since we assumed that b < a. Hence the question of whether or not solutions of the form of Equation (8.4) can be real hinges upon whether or not 2b - a > 0, or

a>2 We have the following cases:

Case I : a < 2; x3(s, A) is real for no real A. x3(s, A) is real for A > max(2ab b, za ba ).

Case II : a >

Since a > b, this2; means x3(s, A) is real when A > 2aa In Case I above, i.e. when a < 1, the only real solutions of equation (8.1) are the trivial solution x(s, A) - 0, and the two main branches:

xl(s, A) = f X2(5, ,1) = f

73

733

.

a - 1 sin s

b - 1 sin 2s

The solutions x1 and x2 branch away from the trivial solution x - 0 at the eigenvalues a, b of the linearization of equation (8.1) at the origin:

Ah(s) = 7r

0

{a sin s sin t + bsin2ssin2t]h(t)dt.

(8.5)

Bifurcation Theory

109

In Case Ii, i.e. when

> 2, we again have trivial solution

x(s, A) = 0, and the two main branches

xi (s, A) = f

2 .

735

a - 1 sin s

x2 (s, A) = f 2 , b - 1 sin 2s which bifurcate from x

0 at the primary bifurcation points,

which are the eigenvalues a, b of linearized equation (8.5). More-

over, for A > 26a > a, a third type of solution branch appears, ab namely that in equation (8.4). Note that as A -- 2b-a' A > 2b-a' the coefficients V -a A - 1 -> 0 and s A - 1 -> _ . On the other hand note that a - I - 26-a as A - za ba . Thus

as \ --,

2b

ba, we see that x3(8, A) --> X3 18, 26ba] = x1 [s, 26 ba

Therefore at A

= 2a a' the sub-branch (twig)

x3 (s, A)=3

/2b_aA_ 1sinSf3

joins the main branch, i.e. x3 [s, 2bba] = xl [s, 2b. sub-branch (twig)

abaA-lsin2s

J while the

2 [2b- aA -1 sin2s x3(s,A)23 2b-aA-1sinsf ab 3 ab joins the negative part of the main branch, i.e., ab

ab

x3[3'2b-a] -x1[s'2b-a We have in Case II, when a > 2, the phenomena of "secondary bifurcation", or the forming of sub-branches or twigs which bifurcate from the main branches. The main branches bifurcate from the trivial solution at the eigenvalues of the linearization of equation (8.5), while the twigs bifurcate from the main branches.

Nonlinear Functional Analysis

110

Thus solutions of the nonlinear equation (8.1) exist as continuous loci in (A, sins, sin 2s) space. There are two main branches:

x1(s, A) splits off from the trivial solution x - 0 at A = a, and its two parts x+, xi differ only in sign; x2(s, A) joins the trivial solution at A = b, and its two parts xz , x2 differ only in sign. a and b on the A axis are the primary bifurcation points for the main branches. If a > i.e. Case II, two sub-branches or 2,

twigs split away from x1(s, A) at A = sb ba , which is known as a secondary bifurcation point. The question of whether or not secondary bifurcation of the eigensolutions of equation (8.1) takes place therefore hinges on

whether we have a > Z, or a < z. The condition a < z in this simple problem is a "condition preventing secondary bifurcation".

8.2

Local Bifurcation

We will use the Leray - Schauder degree to study bifurcation points.

Definition 8.1 Let X, Y be Banach spaces, Q C X open, 0 E 1, F : (a, b) x ci --- Y continuous, such that for all A E (a, b) we have F(A, 0) = 0. The point (Ao, 0) E (a, b) x St is said to be a bifurcation point, if for all E > 0 there exist x,\ E St and A E (a, b) with 1.\ - Ao I < E, 0 < I I x,\ I I < E, such that

E(A, xa) = 0.

If we assume that F can be linearized near (ao, 0), then it is easy to give necessary conditions for bifurcation in terms of the linearization.

Bifurcation Theory

111

Proposition 8.1 In the situation of Definition 8.1 let (a) F, Fz are continuous in a neighbourhood of the .bifurcation point (Ao, 0). Then FF(ao, 0) is not a homeomorphism.

(b) If X = Y, F(A, x) = x -- ATx + G(A, x) with a continuous C : (a, b) x 1-+ X, such that sup IIG(A,x)II AE(a,b)

0

lixll

if IlxII -- 0, then ) 1 belongs to the spectrum of T.

Proof : (a) If F(Ao, 0) is a homeomorphism then the implicit function Theorem 4.1 tells us, that F has a unique solution, i.e. only the trivial solution near (Ao, 0).

(b) If )b 1 ¢ o(T), then I - AoT is an isomorphism for all A close to AO and

x = -(I -

AT)-1G(A,

x)

0

contradicts

1 < II(I -

AT)-11I-IIG(,\,x)ll

-+0.

11 4

0

If C = 0, i.e. if F is linear, and

is an eigenvalue of T with an eigenvector xo, then for all a E 1 the pair A(a) = Ao, x(a) = axo solves

Nonlinear Fbnctional Analysis

112

In this case (ao, 0) is called a vertical bifurcation point.

Example Let

X = Y = R2,x =

77),F(7,x) = (1 - x)(

71) +

(7j3, -t3) then Fx(1, 0) = 0 but F(A, x) = 0 implies

(1-A)e+713=(1-A)rl-.3=0 and(1-A)C71=-714=1;4,hence C=71= 0, i.e. (1,0)ERxX is not a bifurcation point. If we represent F in the form of Proposition 8.1 (b), then F(A,x) = x-Ax+G(A,x) with G(,\, x) = (713,-713), thus Ao = 1 is of multiplicity two.

Example Let X = Y = R2 and F(A, x) =

1

(I)-a 77

1

1 )()+A(;3). 77

A0 = 1 has geometric multiplicity one, algebraic multiplicity two, but F(A, x) = 0 implies

t-arl=0 71-A(C+71)+A 3=0 hence

71(1 -A2-A+\4712) =0, therefore rl = 0 or r12 = _A-4(l - A \2) , but the second solution is not close to the trivial solution, hence (1, 0) is not a bifurcation point. The following theorem will be based of the following degree

jump principle: Let

X - H(µ, x) = 0

(*)

Bifurcation Theory

113

µ E R, x E X and assume for 1 C R x X bounded and open (JP1) H : ) -- X is compact and H(µ, 0) = 0 for all µ (JP2) For µl < µ2 we have D(I - H(p1i .), x n SZ, 0)

D(IO - H(µ2i .), x n Q, 0).

Now if (JP1) is satisfied and (µ0i 0) is not a bifurcation point of (*), then D(I - H(µ, .), X nil, 0) is constant in a neighbourhood of µo. If (JP1) and (JP2) are satisfied, then (*) has a bifurcation point (µ, 0) with µl < µ < µ2. In the first case (µ, 0) is the only solution of (*), hence homotopy invariance yields the constancy of the degree; in the second case the jump of the degree yields the existence of a nontrivial solution (µ, x) # (µ, 0). We will now assume that Aj1 is an eigenvalue of odd algebraic multiplicity.

Theorem 8.1 Let X be a real Banach space, K : X --+ X be compact and linear, S2 C R x X a neighbourhood of (Ao, 0), G : 0 --+ X be compact and G(A, 0) = 0. Suppose also that (a) is an eigenvalue of K of odd algebraic multiplicity.

(b) There exists a continuous function cp : R - R with lim,.,o co(r) = 0 and 8 > 0, such that for all (A, x)(A, x) E S2 with is - AoI< 6,Ilxll
Then (Ao, 0) is a bifurcation point for

F(A, x) = x - AKx + G(A, x) = 0.

114

Nonlinear Functional Analysis

Proof : 1. Let T = I - A0K, and k = k(A) be the smallest number, such that N(Ao) = ker T k = ker T k+1 Then, according to Theorem 7.2 X = N(Ao) ® R(A0), and both subspaces are invariant under K.T is a homeomorphism from R(o) onto itself and ) 1 is the only eigenvalue of KI N(ao). Let P : X - N(0) be the projection defined by the direct sum. Then Range (I - P) = R(Ao)

and writing v = Px, z = (I - P)x, we have F(o + p, x) = 0 equivalent to

x=v+z a=AO +ti Tv = pKv - PG(p, v + z)

(8.1)

Tz = pKz - (I - P)U(p,v + z)

(8.2)

where G(p, x) = G(AO + p, x). Let S = (T I R(Ao)) -1. Then the second equation becomes

z = pSKz - S(I - P)G(p, v + z).

(8.3)

2. We will solve equation (8.3) by applying Banach's fixed point theorem. If ii. < 1?, 1vI < r are sufficiently small, then G is a contraction in the space of R(Ao) - valued continuous functions. Let

J=

[-rl, r1J, B = R(0, r) fl N(Ao), and for p, c > 0

M = {z : J x b - R(AO) continuous with CI {vI 1,supIIZ(p,v)11 < p}.

IIz(p,v)11

µ,V

Then for z, z1i z2 E M, p E J I I G(p, v + zl)

- G(p, v + z2) I I eqW(r) I l z1- z21I

IiG(p,v+z)lI _ < cp(r)I Iv + zII

Il

(p,v+z)-G(p,0)11

< cp(r) (1 + c) I IvI1 < W(r) (1 + c)r.

Bifurcation Theory

115

The operator

z -+ (I - ASK) -1S(I - P)V(p, v + z) maps M into itself and is a contraction, if r (and thus cp(r)) is chosen sufficiently small.

Thus there exist a unique z(A, v), which solves equation (8.3), and w(p, v) = I I vII -1z(A, v) satisfies I IwI I, < c. Hence equation (8.2) implies by IIw(p, v)II _<

II(I-ASK)-'S(I-P)II II?°(,u, v+Il vll w(,u,v))II.Il vll -1

uniformly in A E J lim I Iw(A, V) I I = 0.

IIt'Ik o

Thus, all possible zeros of F in a small neighbourhood of (Ao, 0)

are contained in { (A, X) : A = Ao + it, X = v + mz(A, v), IpI <- ii, I IvI I <- r}.

3. Now, let us insert z = z(A, v) into the bifurcation equation (8.1) on N(Ao). Let Go : J x B --+ n(Ao) be defined by Go(p, v) = (I - ()+o + p)K)v + P-O(p, v + z(p, v))

Clearly, Go is continuous, and is of the form I - Fo with a compact map Fo. Now we will apply degree theory. Choose Al, p2 such that -tl <- Al < 0 < A2:5 ,i and p E (0, r] such that for j = 1, 2 we have in N(Ao) \ {0}

Go(µ1,.) is homotopic to I - (Ao + p1)KIN(Ao) Go(p2i .) is homotopic to I - (Ao + µ2)KI N(ao)

116

Nonlinear Functional Analysis

in N(ao) \ {O}, e.g.

H(t,,u, v) = (I + (Ao + µ)K)v + tPZ7(µ, v + z) is an admissible homotopy, since H(t, µ, v) = 0 for O # 0 implies

v = -t(I + (7o + µ)K)-'PP(p,v + z) 1

< tll (I + (Ao + µ)K)-'PII.

(µ, v +

0

Then D(Go(µj, ), B(0, p), 0) = D(I - (Ao + pj)K, B(0, p), 0). By Theorem 7.11

D(I -- ('o +,uj)K, B(0, p), 0) _ where

(8.4)

+µj) is the sum of the algebraic multiplicities of

the eigenvalues A of KIN(AD) satisfying A(Ao + µj) > 1, or 0, if there are no such eigenvalues. Since A5 1 is the only eigenvalue of KI N(Ao) and either A01(Ao + pi) > 1 and A 1(Ao +.U2) < 1 or vice versa, one of the degrees in equation (8.4) is + 1 while the other one is -1. Hence Go must have a zero in [Al, u2] x 8B (0, p) since Go would be an admissible homotopy otherwise. If (pa, vp) is such a zero, then (ao + A p, xp) with x,, = vo + z(tp, vp) is a nontrivial zero of F. Since µj and p may be chosen arbitrarily *close to zero, we have shown, that (\o, 0) is a bifurcation point of F(A, x) = 0. O

8.3

Bifurcation and Stability

In this section we will study the situation of Theorem 8.3 more carefully. Again let us assume, that

F(A, x) = (I - AK)x + G(A, x)

(8.1)

Bifurcation Theory

117

where K : X -' X is linear and compact, A01 is a simple eigenvalue of K, G : R x 0 is continuous differentiable and SZ is an open neighbourhood of 0, such that for all A E R (a) G(A, 0) = 0 (b) Gz(A, 0) = 0 (c) G(,\,.) is compact (d) {G(., x), x E Q} is equicontinuous

(e)o(K)\{a01}C {(EC:Re(
value A01, i.e. (I - \oK)xo = 0 and let xO* E X` be chosen, such that projection P : X --- n(Ao) is given by Px =< x, xo > xo. Again we have the decomposition of X = R(Ao) ® N(Ao) and equation (8.1) decomposes into

(I - AK)Px + PC(A, x) = 0

(8.2)

(I - AK) (-P)x + (I - P)G(A, x) = 0.

(8.3)

If we again denote by T = I - \oK,

A= Ao+A, MA, X) =G(ao+µ,x), then by definition of P equation (8.2) reduces to

-

o

aao + PP(µ, x) = 0

with a = < x, xo >. If we again denote by S = [(I - AoK) IR(Ao) I]-1,

the equation (8.3) becomes

(I - P)x - ASK (I - P)Z7(,u, x) = 0

Nonlinear Functional Analysis

118

and with

H((a, µ), x) = x - pSK(I - P)x + S(I -- P)?7(µ, x) - aao H((a, µ), x) = 0

(8.5)

where H : R2 x SZ - X is continuously differentiable. Since Hz (0, 0) = 1, by the implicit function theorem there exist neighbourhoods of 0 in R2 and in Sl and a continuously differentiable function

(aI µ) -' x(a, µ) such that locally H((a, µ), x(a, µ)) = 0, x(0, 0) = 0. Now we additionally assume (f) There exists a continuously differentiable mapping G1

Rx)xR-+X,such that for all µ,aER,xES1with

a=<x,xo> G(A, ax) = a2G1(A, x, a).

Equation (8.4) then has the form

-

+ a < G1(µ, x(a, µ), a), xo >= 0. o

The function

f (a, µ) = - o + a < has the property

x(a, A), a), xo >

(8.6)

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119

By the implicit function theorem there exists a continuously differentiable function µ : (-a°, a°) - It, such that for all a with Ia I < a° 0 F(a, µ(a)) = 0, µ(0) = 0

and so we obtain a solution x(a, p(a)) of equation (8.1). Now in the sequel we will assume, that for sufficiently small

al

0 we have µ'(a)

0. In this case we have three possibilities

for the behaviour of x(a, µ(a)) : 1°

µ'(a) <0fora <0andp'(a)>0fora>0.

2°

p'(a) does not change its sign for small jal # 0

3°

µ'(a)>0fora<0andµ'(a)<0fora>0.

.

In Case 1° for all a 36 0, µ(a) > N. In Case 30 µ(a) < Po fora54 0. If µ'(a) 0, a 0, then ao is not a vertical bifurcation point, since pc(a) j4 µ(0) = 0. Let (AO, x°) be an isolated zero of F{, and ) an open neigh-

bourhood of x°, such that 1 does not contain an additional solution of F(A, x) = 0. Then D(F, Sl, 0) = i(I - F, SZ)

is called the fixed point index. We now will determine the index of the trivial solution: The linear operator I - AK does not have negative eigenvalues if A < A0, since

(I - AK)x = vx implies (1 - v)x = AKx.

If v < 0, then 1' > a > -LAOwould be an eigenvalue of K. Then, A since G(A, 0) = 0, we have with ) = B(0, e)

i(I - F, 12) = i(AK + G(A, .), )) = i(AK, SZ)

Nonlinear Functional Analysis

120

D(I - .1K, f, 0) = (-1)m(\) = 1 since m(a) = 0 (no eigenvalues > A 1). If we choose A > A0 such

that 41 is the only eigenvalue of K in the set { E C, Re(> .1}, then i(I - F, St) = i(AK,1l) = (-1)m(.\) _ -1, since m(A) = 1(71 is a simple eigenvalue). Thus the trivial solution has the index

i(AK,SZ)=1 if A< \

i(AK,S1)=-1 if A<\0. The index of the nontrivial solutions can be obtained by the following:

Since µ'(a) # 0 for small a # 0, there is an f > 0 such that for all x with 0< I Ix I f < E

(I - \oK)x + G(A0, x) 76 0. Then there is 6 > 0 such that for all A with Ia - Aol < 6, f Ixf I = E we have

(I - AK)x + G(A, x) 710

(8.7)

otherwise then we would have sequences (An), (xn), such that

1A-)10I: n,llxnII =E and 0 = F(An, xn) = ((I - AnK)xn + G(An, xn)

(I - A0K)x,, + G(A0, xn) = (I - A, K)xn + G(An, xn) +(A. - \o)Kxn + (G(A0i xn) - G(an, xn)) {Kxn, n E .l f} is bounded, (A1n - .\o)Kxn --* 0, G is equicontinuous, hence G(A0, xn) - G(An, xn) --s 0. From this follows

lim (I - AOK)xn + G(ao, xn) = 0.

Bifurcation Theory

121

Since F(Ao, .) is proper, there is an ±, By equation (8.7)

11

= E and F(Ao, x) = 0.

H(t, x) = F(A0 + 6(2t - 1), x) 4 0 for 0 < t < 1, JRxj I = E thus D(H(A,.), B(0, e), 0) =: d(,1) is constant if IA - Aol < 8. If A 76 A0, the solutions are isolated, therefore d(A) is the sum of indices of the trivial and the nontrivial solutions.

Case 1°:

For A < A0, the trivial solution has index 1, hence d(A) = 1, if A > A0 the trivial solution has index - 1, hence the both

nontrivial solutions for a < 0 and a > 0 have the index +1.

Case 2° : Since µ'(a) # 0, we have nontrivial solutions for A < A° and A > A0, and the index of the nontrivial solution of absolute value 1, we have d(A) = 1 + index of nontrivial solution, if A < Ao d(A) = -1 + index of nontrivial solution, if A > a° therefore, if A < AO, then the index of the nontrivial and the index of the nontrivial solution is +1, if A > A0, and d(A) = 0.

Case 30 : d(A) = -1, since the trivial solution has the index -1, if A > A0, thus both nontrivial solutions have index -1. When we have an evolution problem governed, for example, by the differential equation

x' = f(A,x) in an appropriate space, then the results tell us something about the existence and number of equilibria, i.e. time-independent solutions. Consider

x'(t) = f (A, x(t)) with f (A, 0) = 0

Nonlinear Functional Analysis

122

and f (ao, x) = Ax + R(x), where A is an x x n - matrix

and Rx =

Then the trivial solution x = 0 of x' = f (ao, x), x(0) = xo is said to be stable, if for every e > 0 there exists 8 > 0, such that the solution exists on R+ and 0(I IxI I)

.

satisfies I I x(t) I I < e whenever I Ixo I I < 6. The trivial solution of

x' = Ax is stable if Re o < 0 for all p E o(A) and every p, such that Re p = 0 has algebraic multiplicity equal to its geometric one. If Re p < 0 for all p E Q(A) then x = 0 is stable as a

solution of x' = f (x). If Re p > 0 for some u E Q(A) then x = 0 is an unstable solution of x' = f (x).

In our formulation the operator A is given by -(I - AK), and the condition (e) in section 8.3 states that 0 is a simple eigenvalue of A and all other eigenvalues of A have negative real

part. Since 0 is an isolated point of the specturm, we find p > 0, such that

((7(I-)0K)\{0})nB(0,4p) = 0 and we also find rl > 0 such that for all x E X, A E C with

IIx II
and {( E C : ICI < 7) n {u(I - \K + C' (A, x))} _ {A}, i.e. the A - neighbourhood of the simple eigenvalue 0 of I- AoK contains only the simple eigenvalue A of I - AK + Gz(A, x). Since the spectral values are real or conjugate complex, A

is real; and A

0 for small solutions x = x(a.p(a)), since

p'(cx) # 0. Therefore the stability depends upon A > 0 or A < 0. Since Ao is not a vertical bifurcation point, for the nontrivial solution (A, x) we have A 36 Ao.

Bifurcation Theory

123

For A > A0 the index of the nontrivial solution is +1, and this is true if A > 0, these solutions are stable. If A < A0, the index is -1, which requires A < 0, these solutions are unstable.

8.4

Global bifurcation

So far we have only considered local results, i.e. existence of solu-

tions in small neighbourhoods of a bifurcation point. However,

essentially under the same hypothesis, it is possible to prove more about the global behaviour of components of the solution set containing these points, as we are going to indicate in this section.

Let X be a real Banach space, SZ C R x X a neighbourhood of (A0, 0), K be linear, G : -' X be continuous and such that G(A,x) = 0(IIxIl) as x -> 0, uniformly in A. Typical global

results about zeros of F(A, x) = x - AKx + G(A, x) will be explained f o r compact K and G. So, let A 1 be an eigenvalue of odd algebraic multiplicity of the compact K. Let

M={(A,x)EfZ:F(A,x)=0 and x#0} and C be the connected component of M containing (Ao, 0). Remember that components are closed and (A0) E M since (A0, 0) is a bifurcation point. We want to prove that C n t9 # 0 of (Al,0) E C for another characteristic value Al # Ao of K. In case S2 = R x X, C n 80 # 0 means that C is unbounded. So, let us first sketch how we arrive at a contradiction if we assume

Cn81=0, Cn(Rx{0})={(A0,0)}.

(8.1)

First of all C n 8SZ = 0 implies that C is compact, since K and G are compact. Suppose next that we are able to find an open

Nonlinear Functional Analysis

124

bounded go such that C C go C Uo C SZ and M n ago = 0. By the second part of equation (8.1) we may then assume that the intersection of ?o and real line is given by J = [A0 - 8, Ao + 8] with 8 > 0 so small that no other characteristic value of K satisfies IA - Aof < 28. By the homotopy invariance of the Leray - Schauder degree M n 8S1o = 0 then implies that D(A) _ D(F(A, .), SZo(.A), 0) is constant in j; remember that SZo(a) _ {x : (A, X) E SZo}. To see this let a = D(F(Ao,.), Qo(Ao), 0) and let A

= inf{A, IA - Aol < 6, D(F(A,.), Q(A), 0) 34 al

then there is an ± E 80(A), such that (A,±) EM n ago. Like in the proof of Theorem 8.1, we want to exploit the jump in the degree when A crosses ao. Hence, choose \1 and 1\2 such that 1\o - S < 1\1 < 1\0 < 1\2 < ho + b and note that

D( ) = D(F(as, ),11o(Aj) \ Bp(0), 0) - D(F(A1, ), Bp(0), 0) (8.2)

for i = 1, 2 with p > 0 sufficiently small. Since the D(F(A,.), Bp(0), 0) differ by a factor -1 and D(.\1) = D(A2), the first degrees on the right-hand side of equation (8.2) must also be differ-

ent. But it is easy to see that they are in fact equal to zero. Indeed, consider for example A3 > A2 so large that SZa(.\3) = 0 and

p > 0 so small that F(A, x) # 0 on Bp(0)\{0} for A E [1\2,,\o+261

and bo(a) n Bp(0) = 0 for A > \o + 26. Then the homotopy invariance for SZo \ ([A2, A3] x B p(0)) implies

D(F(A2, ), 00(A2) \ Be(0), 0) = D(F(A3i .), SZ(A3), 0) = 0.

Thus, the only problem is to find such a bounded nieghbourhood

go of C. Let us start with Ua = {(A, x) E SZ : dist ((A, x), C) < S}. Evidently, Ua n M is

Bifurcation Theory

125

compact and c fl 8U6 = 0. Note that U6 fl A? is not connected unless it equals C, since C is already a maximal connected subset of M. Of course we choose 11o = U6 if U6 n "M = C. If not then one may guess that, due to the disconnectedness of U6flM, there exist compact C1 D C and C2 D M fl 8U6 such that C1 fl C2 = 0

andU6flM= C1UC2. If this is true then diet (Cl, C2) = Q > 0 and we may choose the intersection of U6 and the ,Q/2 - neighbourhood of C1 for f o.

Lemma 8.1 Let (M, d) be a compact metric space. A C M be a component and B C M closed such that A fl B = 0. Then there exist compact M1 D A and M2 J B such that M = M1 U M2 and M1 fl M2 = 0.

Proof : To use a good substitute for possibly missing pathwise connectedness, namely E - chains, let us recall that, given f > 0, two

points a E M and b E M are said to bee - chainable if there are finitely many points xl, ..., xn E M such that xl = a, xn = b and d(x;+1i xi) < E for i = 1, ..., n - 1. In this case xl, ..., xn is an E - chain joining a and b. Let AE = {x E M : there exists a E A such that x and a are e chainable 1. Clearly A C AE and A, is both open and closed in M since BE(z)fl (M\AE) = 0 for z E AE, BE(z)flAE = 0 for z E M\AE.

It is therefore enough to show B fl AE = 0 for some e > 0, since then M1 = AE and M2 = M \ AE have all properties we are looking for. Suppose, on the contrary, that B fl A. # 0 for all E > 0. Consider en -> 0, (an) C A and (bn) C B such that an and b are En - chainable. Since A and B are compact, we may assume an - ao E A and bn -+ bo E B, and therefore we have

126

Nonlinear Functional Analysis

En - chains Mn joining ao and bo, for every n > 1. Consider the limit set

Mo={xEM:x= limXnk with xnkEMnk}. Evidently, MO is compact and ao, bo E Mo. Suppose that MO is not connected. Then MO = Cl U C2 with Ci compact and dist (Up (C1), Up(C2)) > p for sufficiently small p > 0. For En < p this contradicts the obvious fact that any two cl E C1 and c2 E C2 are En - chainable. Hence, MO is connected. Consequently, Mo C A since as E MO n A and A is maximal connected, and therefore bo E A n B, a contradiction.

0 The reasoning given so far leads to a further result, which we are going to prove next.

Theorem 8.2 Let X be a real Banach space, SZ C R. x X a neighbourhood of (,\o, 0), G : St -+ X be completely continuous

and G(A,x) = 0(IJx`J) as x -+ 0, uniformly in A. Let k be linear and compact and AO a characteristic value of odd algebraic

multiplicity F(A, x) = x - AKx + G(A, x) and

M={(A,x)E1l:F(A,x)=0 and x740}. Then the component C of M, containing (ao, 0), has at least one of the following properties: (a) C n aQj4 0; (b) C contains an odd number of trivial zeros (At, 0) # (ao, 0), where \1 is a characteristic value of K of odd algebraic multiplicity.

Exercises

127

Proof : Suppose that C n CQ = 0. Then we already know that C is compact and contains another (A, 0) with A A0. Clearly, a bounded neighbourhood S1o of C satisfying Mn8S2o = 0 contains only a finite number of points (Ak, 0) with Ak 1 E u(K), say Al < ... < A2_1 < AO < Ai+1 < ... < A,,. We may assume that ?Ion (Rx {0}) = Uk=l[Ak-S, Ak+b] with b > 0 sufficiently small. Choosing Akl and Ak2 such that Ak-6 < Akl < Ak < Ak2 < Ak+b, we have D(F(A,.), S2o(A), 0) = m on [A1 - 6, Ap + 6] for some m E fl.

m = D(F(Ak;, ),1lo(Akj), 0) = dki + D(F(Aki, -), Bp (0),-) for j = 1, 2.and p > 0 sufficiently small, where

dk; = D(F(Ak,, .), 1l (Aki) \ Bp(0), 0). Furthermore d11 = 0 = dp2 and dk2 = dk+l,l. Hence p-1

p

E dk+1,1 + L D(F(Akl, ), Bp(0), 0) k=1

k=1 P-1

p

_ >2dk2 + k=1

k=1

and therefore P

E[D(F(Ak2, ), Bp(0), 0) k=1

- D(F(Akl, ), Bp(0), 0)] = 0-

This evidently implies that we have an even number of jumps in the degree. Since we have one at A0, and since the jumps occur only at characteristic values of odd algebraic multiplicity, the theorem is proved. O

Chapter 9 EXERCISES AND HINTS Every exercise is not answered here. Readers will learn

best if they make a serious attempt to find their own answers before peeking at these hints. Let X be a Banach space. A : X -+ X is linear and continuous, b E X. For every A in the spectrum of A holds 1.

JAI < 1. Let F : X -s X be defined by

F(x) = Ax - b.

Then there exists a unique x E X, F(i) = i . 1 = lim x,,, x = F(xi-1),xo E X. If A has an eigenvalue A with JAI > 1, then

does not converge for every xo E X.

(hint : (a) Proof :

F(x) = Ax - b F2 (x) = A2x - Ab - b F3(x) = A3x - Alb - b

F'(x) = A"x-A"-lb-...-Ab-b 129

Nonlinear Functional Analysis

130

IIFn(x) - Fn(y)Il = IIAnx - AnyII <_ IIAnII IIx - yll. K-n

Now it is clear that if EKn < oo by the Weisinger theorem (Theorem - 1.2). It is sufficient to show that

lim sup n IIAnII = lim sup " Kn < 1 (Spectrum is a compact subset of A.) (b) If A is a eigenvalue with IAI > 1, then 30 # x E X s.t. Ax = Ax, IIFnO - FnxII = IIAn0 - AnxII = IAInIIxII So it does not converges in IAI > 1.) Let (X, d) be a complete metric space, V' : [0, oo) -+ [0, oo) 2

continuous, O(t) < t, if t > 0. Let F : X -> X, such that for

x,yEX d(F(x), F(y)) < i (d(x, y)).

Then T has a unique fixed point x, and x = lim F(xn), xn = F(xn-1),x0 E X.

(hint : Proof : Cn =

d(F(n+1)(x),

< tb(d(F"(x),

Fn(x)) Fn-1(x))

= 7/1(Cn - 1) < Cn-1Since Cn > 0, Cn is decreasing sequence.

2limCn=C>0 C 0 and 2(77k), (mk) s.t. 1. mk+1 > nk+1 > mk > 7tk

Exercises

131

2. d(F'"k+lx, F"kx) < e

.

3. d(Fkx,F"kx)>E. 4. d(F ,t-1x, F"kx) < e

.

So,

e < d(Fmkx, F"kx)

< d(Fm x, Fmk-lx) +d(F'"k-lx,

F"kx)

G Cmk-1 + e

limd(F'"kx,F"kx) = E. < d(F'kx, Fmk+lx)

d(F'"kx, F'nkx)

+d(F"k+lx, F"k+lx)

+d(F"k+lx, F"kx) < Cmk + iP(d(Fmkx, r `kx)) + C"k.

Taking limit, we get

E

=0

a contradiction

therefore (F"x) is Cauchy sequence. Since X is complete, 3X = lim F"x, and

F(X) = lim F'+1(X) = X X is a fixed point. Now it remains to show X is unique. Let X, X be two fixed points. Then

d(X,X) = d(F(X),F(X))

< ii(d(X,X))
132

Nonlinear Functional Analysis

Then, for each p E P, Fp has a unique fixed point xp, and xp = xpo.

4.

Define f : R -' R2 by f (t)tt= (t, t2) and g > R2 - R by tt tt 9 (Si, S2) = 1 if t2 = ti and 9(e1, b2) = 0 if 1 Y-' Q2 . Show that the Fr chet derivative. f'(t) exists for all t E R, the Gateauxderivative Dg(0, 0) exists, but (gof)'(0) # Dg(0, 0) f'(0) (and hence the chain rule fails for Gateaux - derivatives). (hint : f : R - R2 is defined by f (t) = (t, t2), 9 : R2 --+ R by

e2) =

If 0

2=1 elsewhere.

To show that f(t) exists Vt E R, Dg(0, 0) exists, but (gof)'(0) 71 Dg(0, 0) f'(0).

We can easily show f(t) = (1, 2t). So it exists Vt E R. Now 9(TS1, TeT)

lim

- 9(0, 0) = 0 = D9(0, 0)

(TS1)2 = Tel b

(1) e = 0 = 9(r6, 712) = 0 (ii) SI # 0

(a) 6 = 0

9(TC1, TS2) = 0

9(Te1,Te2) = 0 for ITI < fa So the limit exists i.e. Dg(0, 0) exists. (b) e2

0

Now (9of)(t) = 9(t,t2) = t .

Therefore (gof)'(0) = 1. But Dg (0, 0), f'(0) = 0. ) Suppose that X is a real Banach space and f : X - R is 5. Gateaux - differentiable on X. If f reaches a relative minimum

at xo E X (i.e., there is an r > 0 such that f (x) > f (xo) whenever Ix - xoi < r) show that f'(xo) = 0. If, in addition, f

Exercises

133

is convex, show that f reaches a relative minimum at xo E X if and only if f'(xo) = 0. f is convex

to`dx,yEXVrE[0,1] f (rx + (1 - T)y) < rf (x) + (1 - T )f (y)

(hint

:

(a) To show f'(xo) = 0 0


+

t

th) - f (xo) <0 f'(xo)h = lim f `x0 +

t

t-+o-

for thi < r since f (x) > f (xo) for Ix - xoi < r = f'(xo) = 0. (b) Suppose f'(--o) = 0 3rE (0,1)

:

T(xo+h)+(1-T)(xo+th) =x0

h(r+t-rt) = 0

T+t-rt=0 -t 1-t t

t

1

f (xo) = f (,r(xo + h) + (1- r)(xo + th)) i.e., f (xo) < t t

1

f(xo + th) - t

tf(xo) -f(xo) >

f(xo+h)-

1 1

f (xo + th)

f(xo+th)

f (xo + th) - f (xo) < f(xo + h) - f (xo)

t

f'(xo) = 0 < f (xo + h) - f (xo) min at xo.)

6.:

Let f : [a, b] x R - R be a continuous function. Then

by b

Px= j

x(r) -f(T, s)dsdr

oJ

Nonlinear Thnctional Analysis

134

a continuously differentiable functional on C[a, b] is defined. In addition If (t, s)I < c(1 + IsIvA) on [a, b] x R, with c > O,p > 0 and n + a = 1, then co is continuously differentiable on Lp[a, b].

7. Definition A set A is called a Banach algebra, if A is a Banach apace, and if - there is defined an associative distributive continuous multiplication of elements of A, i.e.

if X, y E A, then x.y E A and I I < IIxI.I IyII Let A be a Banach algebra, and U C A an open subset. Let f, g : U -+ A be Frechet differentiable.

(a) Then h : U - A, h(x) = f (x).g(x) is Ftechet differentiable. Determine h'(x). (b) Let (jA = {x E A, x-1 exists, x-1 E A}. Determine the first and second derivative of f'(x) = x-1. Show

that A is open in A (hint : : A Banach algebra, U <,A is Frechet differentiable open subset.

(a) h : U - A, h(x) = f (x).g(x)

II h(x + k) - h(x) - f'(x)k.g(x) - f (x

f (x)g(x)

-f'(x)k.g(x) - .f (x + k)g(x + k) - g(x) - g'(x)kI I

+11(f (x + k) - f (x)

- f'(x)k)g(x)II

+IIf(x + k) - f(x)g'(x)kfI IIf(x + k)IIIIg(x + k) - g(x) - g'(x)kII +IIf (x + k) - f (x) - f'(x)kII .I Ig(x)II +11f (x + k) - f (x)IIIIg(x)IIIIkII.

Exercises

135

E > 0 choose E/2 E/2

+El < E/3

f count * 361 > 0 s.t.

IIkII < 6

I If (x + k) - f(x) I IEl

F - di f f 362 > 0 rrns.t. Ilf(x + k) - f(x) - f'(x)kll :5Elllkli I

gF - diff *363>0 s.t. =

Ilg(x + k) - g(x)

(Ikll <-63

- 9'(x)kll :5 E3IIkII.

Also, Ilkl1 < 6 = min{61, 62i 63}

= I I h(x + k) - h(x)

- f'(x)kg(x) - f (x).g'(x)kl I

:5 (Ilf (x)II + E1)E31IkII + Ell lkI I.I Ig(x)II

+Elllg(x)IIIIkll : EIIkII

* h'(x)k = f'(x)k.g(x) + f (x).g'(x)k. (b) gA = {x E A, x-1 exists, x-1 E A}. gA is open. f : gA --+ gA by f(x) = x-1

Let X E gA = x-1 exists and x-1 E A= IIx-'II L 0. Consider IIhII <

We shall show (x - h)-1 exists.

So, x - h = x(e - x-1h). Since IIx-1hll < 1 * e - x-1h E gA = (e -

x-'h)-1

exists

Nonlinear Functional Analysis

136

Since E J Ilx-'hlln is convergent and A is complete, 00

- x-' h)

E (x-' h)n (E n=0

6

=

`((X-lh)n ll

- (x-1h)n+l)

n=O

- (x-lh)n-1)

k

= lim

1:((x-1h)n

n--O

= k (E x - h is invertible. So, B(x, 1771 ) < gA 9A is open. Again, f (x) = x-'; take IihIl <

21 1

h)-'

II(x +

= II(x(E +

(x-lh)k+1) = E

,. So,

- x-1 + x-'hx-1I1 n-'h))-1 - x-' + x-'hx-'II

x-'h)-1- (e + x-'h)x-111 (-x-lh)n - (e + x-1h)x-1II =I I(> = I I (E + 00

n=0 00

= II

E(-x-1h)n.x-'II

n=2 00

=

E(-x-1h)n-21I

lI(-x-'h)2

n=o

< IIx-11131 IhI12

00

0(IIx-'IIIIhII)n

n=0

= IIx-'1131IhI121

- 11x1'Illlhll `

211x-'11311h112

E > 0 given, we can choose 6 > 0 s.t.

b = min f

1

211x 'II'

-

Il.f(x + h) - f(x) + Now

,

211xE'113

x-lhx-11I <_ IIh1I

Exercises

137

8. Consider the eigenvalue problem

Ax = ABx,

where A and B are real (n x n) matrices and where the norm condition < x, x > = 1 is satisfied. An important trick is to change' this linear problem into a nonlinear one of the form T z = 0, where

z = (X,,\) and T z = (Ax - \Bx, < x, x > -1) . Show that the F - derivatives are T'((X, \)) (y,,u) = (Ay - pBx - \By, 2 < x, y >).

(hint :

Ax = ABx, where A, B E Mat(n, n; R), < x, x >= 1,-x E W. And T (X, A) _ (Ax - ABx, < x, x > -1), (X, A) E R" X R_ = X

Let T : X --, X. So, F - derivatives are

T'((X, A))(y, µ) = (Ay - pBx - ABy, 2 < x, y >)

T(x,A) _ (0,-1)+(Ax,0)+(-ABx, < x,x >), TO

T1

T2(x,A)

where To = constant and T1 = linear operator, where T2 (tx, tA) _

t2T2(X, A) = T2 is a homopol on X - into X = T2 is generated by a symmetric bilinear map M, i.e. by M E ML(X, X ; X) symmetric. So, M((xl, A1), (X2,,\2)) = 2 (-A1Bx2 - A2Bx2i 2 < xi, x2 >) Also,

M((x, A), (x, A)) = T2(x, A).

Nonlinear Functional Analysis

138

So we have To (constant) differentiable Ti (linear operator) is differentiable and T2 is differentiable. Therefore

T'(x,A)(y,µ) = To+Tl +TZ = 0 + (Ay, O) + 2M((x, x), (y,,u))

_ (Ay, 0) + (-,By - µBx, 2 < x, y >) = (Ay - ABy -- ,uBx, 2 < x, y > ) Also,

T"(x, A)(yi, 111)(Y24 12) = 0 + 2M((yi, al), (y2, 1\2))

since (Ay, 0) is continuous with respect to x = (-µ, Bye µ2By1, 2 < yl, y2 >) this is continuous with respect to x

=T(") (x,\) =0 for n> 2.) 9. : Let F : C[0,,7r] - C[O, 7r] be defined by

F(x)(t) =

217' k(t, r) (x(r) + x3(r))dT

where, k (t, r) = a sin t sin r + b sin 2t sin 2T, 0 < b < a. Discuss the set of all solutions of

H(a, x) = Ax - F(x) = 0, where A is a real parameter.

(hint :

Ax(t) = F(x)(t) = 2 a sin t To sin r(x(T) + x3(r))dr it

o

+. b sin 2t

x

sin 2T(x(r) + x3 (r) )dr.

J If x is a solution then x should be linear combination as x(t) = A sin t + B sin 2t. AA sin t + AB sin 2t

Exercises

139

= 2a sin t

sinr(Asinr+Bsin 2r) TO

+(Asinr + Bsin2T)3)dr

+2bsin2tJ sin 2T(Asinr+Bsin 2T)+(Asin T+Bsin2T)3 0

7r

T2

_ (2asint)(I,) + ( Ir

2

b sin 2t) (12)

7r

I1=2A+ $A3+34AB2 I2 = 2 B + 8 B3 + 4 A2B. Since sin t and sin 2t are linearly independent in C(0, 7r], we can equate the coefficients of sin t and sin 2t to get,

AA=aA(1+3-!A2+ZB2)

(i)

AB = bB(1 + 3-4B2 + 2A2)

(ii)

Case 1 Case 2

A = B = 0 x(t) = 0 is a solution for every A E R. A = 0, B j40. from (ii) A = b(1 + 4B2)

B

731

Vb- ERe*a>b

= the solution

X2 (t) = f f Fb-

sin 2t for A

- b.

Case 3 A # 0, B = 0 = equation (ii) fulfilled everywhere. Then from (i)

Nonlinear Functional Analysis

140

A

3

x4 a (t)

2

=f

b - 1 sin t.

From theorem of local homeomorphism F : X -+ Y cont. diff. F'(xo) is invertible. 3 nbd u(xo) s.t. F is a homeomorphism on u(xo). We have

H(A, x)(t) _ (Ax - F(x))(t)

= A(x(t)) - 2

-2J 7r

J

k(t,T)xtaudr

x(t,,r)x3(T)dT

0

H,' (A, x) exists and cont. in x

HH(A, x)h(t) _ Ah(t) -

f k(t,T)h(T)dT

j

0 -3.k(t,T)x2(T)h(T)dT 7r

H,' (A, 0)h(t) = Ah(t) - 2 J k(t,T)h(T)dr.

Since k(t,T) = asintsinr +bsin2tsin2T, sin t, sin 2t E L2 [0, 7r], I I sin tI 12 = I ( sin 2t 112 = 2

Let

sin t e1=TI

' e2

sin 2t

V2

2 [ k(t, T)h(T)dT = a(ei, h)el + b(e2, h)e2 a and b are the eigenvalue of this integral operator

Exercises

141

H(,\, 0) is invertible « a 3& A # b. So, if a # A # b then from local homeomorphism theorem = 3u,, a neighbourhood of 0 in C[O, 7r] * H(A, 0) is a homeomorphism on ua since H(A, 0) = 0 - 0 is the unique solution in U,,.

Case 4

A#0,B34 0

A=a(1+3A2+3B2)

A=b(1+3 B2+2A2)

=f3 ba )-1 A

B=f3 Since 0O. But we need

tab b 0)

andA(a-1)> 1sinceA>O*2-1>Os a<2bwehave A>aba>a. Ifa<2bandA> bathen xs>7,s,9('\)(t) = f3

'\(a 3

t(b

-a ) - 1 sint

b) - 1 sin 2t.)

10.: Let X be a complete normed space, and x : [0,1] -- X be continuous. Show that the Riemann integral i

x(t)dt J is well defined, linear and that 11

Jo'

x(t)dtll

j'IIx(t)(Jdt.

142

11.

Nonlinear Functional Analysis

Let F : Rn -, 1Zn be continuously differentiable and let F'(x) # 0 on Rn. Then F is a homeomorphism onto Rn if and only if :

lim IIF(x)II = oo.

11-T1-00

(hint : F : Rn - R" continuously differentiable and det (F'(x)) j4 0 on R". Then F is a homeomorphism onto Rn if limlI.T11

.

IIF(x)II = oo.

" ," Assume 2(xn) in Rn with IIxnII - oo and IIF(xn)II 74 M. = 3(xnk) with I IF(xnk) I I

<_ k

F(xnk) E B(0, k) since B is compact. F-' (B(0, k)) is also compact. (xnk) is bounded. Which is contradiction to IIxII --' 00. " (a) F is continuous. (b) F'(x) is invertible since F'(x) # 0. We can apply local homeomorphism.

=o- 3anbadU(x):F:U(x)=F(U(x))= F is open. (c) Suppose F(xn) - y E Rn = B(xnk) is bounded. Since if the opposite is true then I Ix" I I -* oo.

IIF(xn)II -> oo, which contradicts to F(xn) -- Y3(x,,,,,) = F(lim F(x) = y E F(R").

Since F(Rn) is open (from (b) and also F(Rn) is closed F(Rn) = 0 or Rn, here it is not empty. So F(Rn) = IV =:,. F is surjective.

(d) F is injective. )

Let (afl) be a real (n x n) matrix with aj > 0 for all 12. i, j. then A possesses a non-negative eigenvalue. The associated :

eigenvector can be chosen, such that all coordinates are non-

Exercises

143

negative. If additionally E 1 a{ f > 0 for all j, then A possesses a positive eigenvalue.

(PERRON - FROBENIUS).

(hint : a+j E Mat(n, n, R), ai f > 0 V i, j then A possesses non-negative eigenvalue, Ax = Ax, x # 0,

x > 0. If in addition E

1

at1 > OVj then .A > 0.

Proof: K=Ix ER":x>0,E.1x{=1}. Then k is bounded, closed

k is compact. Case 1 : Ax = 0 for some x E k 0 is an eigenvalue of A and X is a eigenvector.

Case 2 : Ax # 0 d x E k f (x)

E{ 1(Ax) j# 0 we define

Ax),' f : k --- k is continuous )

,=1(

13. : BROUWER's fixed point theorem is equivalent to the following theorem of POINCARE (1886) and BOHL (1904): Let f : R" --* Rn be a continuous mapping and suppose

3r>0 VA>0 VxER" (I IxI I = r =f(x) +Ax 54 0). Then there exists a point xo, IIxoIi < r, such that f (xo) = 0. 14. Construction of Counter Examples to the Brouwer's fixed point theorem. (a) Construct U C R, which is compact, and a continuous f : U --+ U without fixed points.

(b) Construct a convex bounded U C R, f : U -- U

,

is

continuous, without fixed points. (c) Find f : [0, 1] -+ [0,1] without fixed points.

(hint : (a) Compact U = {0,1 }, f (0) = 1, f (1) = 0 = f is continuous without fixed point (U is also a compact).

144

Nonlinear Fbnctional Analysis

(b) U = (0, 1), f : U --+ U by f (X) = 1X without fixed point.

f is continuous

(c) f : 10, 1] -+ [0, 1]

f(X) =

0

f

1

for 1<X<1 for 0<X<12

* f is continuous without fixed point. ) 15. Let X be a separable infinite dimensional Hilbert space with a complete orthogonal system {x0, x1, x2, ..., }. For :

00

x = >2ajxj J=O

define

CO

F(x) = 1- IIX112.xo + E aj-1xjj=1

Show that F maps the closed unit ball into itself continuously with no fixed points. Thus, compactness of F is a necessary hypothesis in the Schauder's fixed point theorem. (hint : X is a separable infinite dimensional Hilbert space, {xo, x1) ...., ..., } complete orthogonal system.

f(x) = 1- I lxl l2.0 + E00aj-lxj, 1141 =1 j=1

to show that F : 77(0,1) --' R(0,1) is continuous and has no fixed point 00

IIF(x)112 -1- IIxII21a.2f-1 =1, j=1

since

[X = Eajxj, IIxI12 =

Eaj] * F : B(0,1) - $(0,1).

Exercises

145

Next, if i = F(i) = (Jill =1 .

So,

00

_ E ajxj = E00 aj-1xj j=1

do = 0

j=1

by induction 6j = OVj E N

i =0.

Finally, F is continuous

IIF(x)

-

F(y)112

II(

V1IIx112

- "j-

11y112)xo

00

+E((x,xj-1) - (y,xj-1)112

j=1

= ( 1--11x1!2-

1-11y1I2)2

00

+ E(x - y,xj-1)2 < E2 j=1

VUlx--y11<-6.

F is continuous. ) 16. : Let X be a partially ordered real Banach space, i.e. there exists a " <", which is compatible with the linear and the topological structure in X, i.e. Since

1 - I lx 112 is continuous

1. xO,x
limyri=0.

4.

Example : X = C[0,1]. X < y a b'0 < t < 1x(t) < y(t).

F is called monotone increasing iff x < y F(x) <_ F(y). Let [xo, yo] _ {x = txo+(1-t)yo, 0 < t < 1}. If there exist xo, yo E X, xo _< yo, xo <_ F(xo), yo ? F(yo) and F[xo, yo] is

Nonlinear Functional Analysis

146

relatively compact, then there exists a fixed point i of F such that for all n E H and for xn = F(xn_1), y,, = F(yn_1)

x0 <x1<...<Xn<...5 i<...5

y15 y.

(hint : A partially ordered Banach space

[xo,yo]={zEx:no is continuous and monotonically increasing, xo < yo, xo < F(xo), yo >- F(yo), F[xo, yb]

is relatively compact. Xn := F(xn-1), yn = F(yn-1)

=:,. BxE[xo,yo]:F(i)=i and

xo5 x15 ...<x<...
txo < tz1

xo < z2, (1 - t)xo < (1- t)z2

= xo < tz1 + (1 - t)z2 < yo [x0, Yo] is closed

z E [xo, yo]

F(xo) < F(z) < F(y) xo < F(xo) < F(z) < F(yo) < yo F[xo, yo] < [xo, yol

F[x, y] is relatively compact. CoF[xo, yo] 9 [xo, yo]

Exercises

147

K = CoF[xo, yo] C [xo, yo], k is compact

F(k) C F[xo, yo] < k

F is a continuous map on convex compact set. Then we can apply Schauder's fixed point theorem = 3x E k : F(x) = i Finally, xo < x1 < ± < y1 < yb since I E k C [xo, yo]. Assume

xn-1<2n5 2
= xn ! xn+l

F(xn) <_ F(±) < F(yn) < F(yn-1) x -< yn+1 : yn

By induction, the assumption is true. ) 17. Consider the subsets B2 C B3 C b1 C C[O, 1], defined by

B1 =

{x:x(0)=0,x(1)=1, 0<x(t)<1

B2

{x E B1 :0 < x(t) < 2 in [0, 2] and

in [0, 1]}

2 < x(t) < 1 in [2, 1] }

B3 = {x E.B1 : 0 < x(t) < 3 3 < x(t) < 1 in

in[O,

2]

and

[2,1]}.

Then f3(B3) = Z for j = 1, 2,3; and X(Bi) = 1, X(B2) = 2, and

=3 (Fact : X(T) = 2,13(U) = 1 for the unit ball U in an infinite

X(B3)

dimensional space. )

Since B2 C B3 C B1 then it is enough to show that f3(B2) = 2. We want to show : /3(B2) > 2.

(hint :

Assume that Ry1i..., y,n E C[0, 1], B2 C U!'_' 1B(y{, r),

Nonlinear Functional Analysis

148

r=z-E
0 Z+2(t-Z)

if tE[2-n,2+ n

1]

if tE [I+n,1]

1

Let i be fixed,

Case I yi (I) >

Since yi is continuous, then

.

z

38 > 0 'tE [0,1] : It- 2I
- yi(I < E

Take ni such that

-

Iyilt)I :5 Iyi(2)I +

<6 2

.2

=0

Iyi(1

- 1)-yi(2)+yi(2

Iyi(2)I - Iyi(2 - n) - yi(2)I >

X is not in ball B(yi, r). On the otherhand 1

CaselI yi (2) < Ilxn -yell

1

? Ilxn(2 +

n2 _)II + yi(

> 1-(Iy(2)I+E) -

1-2-E=2-E=r

Exercises

149 xn

B(yi, r)

4. no := maxni =o- for n > no : xn ¢ UB(yi,r /3(B1) = 2 for j = 1, 213. We know /3(B) < x(B) < diam(B).

(*)

x,yEB2: (i) t E [0,

: 0 < x(t) < 2, -2 < -y(t) < 0

2] -2 < x(t)

- y(t) <_ 2

(ii) If

t

t E [2,1] : 2< x(t) < 1

- 2 < x(t) - y(t) < 1

-1 < -y(t) < -2

2

From (i) and (ii)

= Ix(t) - y(t) 1 <

Vt E [0, 1] 2

=IIx-yI1 <=0- diamB2<2 from (*) 2 < x(B2) < 2 = x(B2) = 2. Also we know X (B1) < 1. We assume 3Mi with diam(Mi) < 1 - e < 1(i = 1, ..., 7/6)B1 < U;"-_1Mi without loss of generality. Assume b1 n M{ 5 0,0. Choose xi E B1 n Mi

Mi C P(X1, r)

b1 C Us `_1B(xi, r) 2 UMs.

Since

X1EB1:Xi(0)=0

36>0:`dtE [0,1] n;(t)
Define X (t)

=xEB1.

_

: if 0
if 6

t

.

1

Nonlinear Functional Analysis

150

But IIx

- xill?

Ix(6)-xi(6)I>1-E=r

x V U7(xi,r)

x V UMi

x(B1)=1. If we define the function similarly as the case for B, we can show

diam(B3) < 3

X(B3) <- 3

Assume B3 C Ui_` 1Mi, diam(Mi) < s - E = r < 3. Choose xi E B3 fl Mi, 36 < 2tibt E [0, 1], t < 6 = xi(t) < E. Define

as

x(t)

3 'b

0< t
3

b
1+3t 2
Ilx - xill >

Ix(b)-xi(b)I > 3 -E=r

* x O Um 1Mi.)

18. It is impossible to retract the whole unit ball in R" onto its boundary, such that the boundary remains pointwise fixed, :

there is no continuous f : 7(0,1) -- 8B(0,1), such that f (x) = x for all x E 0- (0, 1). (This result is equivalent to

i.e.

Brouwer's fixed point theorem for the ball.) (hint : Assume that Bf : 7(0,1) --p 67(0,1) be continous s.t.

f (x) = x for all x E 87(0,1) then g(x) :_ -f (x), g :7(0,1) -87(0,1) is continuous for x E 87(0,1) we have g(x) _ - f (x) _

-xix.

Exercises

151

= g does not have a fixed point on the boundary. But it is continuous. So, by Brower's fixed point theorem 3 an open ball i.e. x E B(0,1) : g(x) = x. So, 1 = I If(x)I I = I jg(x)I I _ I IxiI < 1. Assume 3f : B(O,1) --> B(O,1) continuous f (x) j4 x for x E R(O,1), y = x + a(x)(x - f (x)), a(x) > 0. In our fixed point theorem, it was shown that a(x)

(x, x - f (x)) + (x - f (x))2

(X, x - f (x)2 + (1 - 11 x112)(x - f (x))2

(x - f (x))2

a : B(O,1) --* R+ is continuous. g(x) := x + a(x)(x -- f (x))

then g : 'B(0,1) and g(x) = x for x E aB(0,1) [ if x E OB = a(x) = 0]. ) 19. Let F : R" x R" be continuously differentiable. Let there exist a continuously differentiable function t -+ (p(t),x(t)), such that p : R -i R is strictly monotone increasing and for all t we

have F(p(t), x(t)) = 0. Let d(t) = det FF(p(t), x(t)). (a) If d(ti)d(t2) < 0 and tj < t2, then F(p, x) = 0 has a bifurcation point (p(t), x(t)) with ti < t < t2.

(*)

(b) If d changes sign at to, then (p(to), x(to)) is a bifurcation point of (*). 20. (McDonalds problem) : Given any sandwich, Hamburger,

Big Mac, Double Burger, Super Big Mac etc., consisting of bread, ham and cheese. Can this be divided equitably among two people with one slice of a knife, such that each person receives an identical share of bread, ham and cheese?

In McDonalds dream Dr X and Dr Y appeared and generalized this problem to n dimensions.

If Bl,..., B are bounded, measurable subsets of R" with

152

Nonlinear Functional Analysis

n > 1. Then there is an (n --1) - dimensional hyperplane which divides all the Bf in half.

References

153

References DEIMLING, K.: " Nonlinear Functional Analysis", Springer, Heidelberg 1985.

DEIMLING, K.: "Nichtlineare Gleichungen and Abbildungsgrade", Springer, Heidelberg 1974.

EISENACK, G.; FENSKE, C.C.: "Fixpunkttheorie", BI, Mannheim 1978.

KRASNOSELSKI, M.A.: "Topological Methods in the Theory of Nonlinear Integral Equations", Pergamon, Oxford 1964.

KRASNOSELSKI, M.A. et al.

"Approximate Solution of Operator Equations", Wolters Noordhoff, Groningen 1972. :

PIMBLEY, G.H.: " Eigenfunction

Branches

of Nonlinear Operators and their Bifurcation", Springer, Heidelberg 1969.

RIEDRICH, Th.

" Vorlesungen fiber nichtlineare Operatorengleichungen", Teubner, Leipzig :

1976.

ZEIDLER, E.

"Nonlinear Functional Analysis and its Applications I : Fixed - Point Theorems", Springer, Heidelberg 1986. :

INDEX C

A Algorithm 44 Antipodal theorem 92, 99 Antipodal points 96 Asymptotic test 39

B Banach space 9, 10, 11, 12,14,18, 22, 23, 26, 28, 30, 33, 39, 44, 47, 60, 65, 66, 70, 73, 87, 88, 89, 101, 102, 105, 110, 113, 123

Banach fixed point

2, 7, 8, 53, 74, 114

Bifurcation 105, 109, 110, 112, 113, 115,

Contraction 1, 6, 9, 48

Complete metric-space

2,4, 7, 8 Cauchy sequence 3,4, 43

Construction 3 Compact metric space 5

Convergent 5, 39, 67 Closed ball 8, 12, 14

Continuous 19, 21, 26, 29, 34, 36

Chain rule 23, 26, 32, 37 Complex space 55 Convex 60, 63, 74, 80 Convex hull 60 Compact operator 99

123

Bilinear map 22, 30, 31

Bijective 30 Bilinear operator 31 Bounded 24, 70, 72 Brouwer's fixed point 55, 57, 59, 63, 64, 82

Brouwer's degree 77,

78,80,84,87,89,

D Derivative 17 Dual space 17 Diameter 65 Darbo 74 Deimling 78

E

90

Borsuk 92, 96 Borsji-Ulam 95

Error estimate 3 Enumeration 56

linear Functional Analysis

156

Euclidean space 57 Eigen vector 81, 111, 117

Eigenvalue 81, 109, 111,113, 119,120

F

Inequality 12, 43 Injective 2, 12 Initial value problem 8

Invertible 49, 50 Implicit problem 51 Isomorphic 22 Isomorphism 30, 51

Fixed point 1, 2, 6,

K

14,55,64,75,80 Fixed point index 119 Frechet derivative 17, 18, 19, 22, 23 Frechet differentiable 20, 21, 23, 25, 26

G

GauB theorem 59 Gateaux derivative 17, 18, 19 General Topology 61

k-contraction 1, 2, 10, 74

Kuratowski 65 L

Lebesgue measure 84 Leray-Schauder degree 87, 89, 90, 99, 101, 110

Linear map 18, 20, 30

Lipschitz continuous

H

1, 8, 9, 10, 28, 44, 73 Lipschitz condition

Halm-Banach theorem 57 Hedgehog theorem 83 Hoelder condition 53 Homeomorphism 12,

Lipschitz constant 9, 10, 73 Local Homeomorphism

25, 49, 80, 81, 88, 102, 111, 114 Homotopy 83, 89, 98,

M

103, 116, 124

I

7

11, 26

Maping 1 Mean value theorem 26, 27, 31, 48, 52

Index

157

Measure of non-compactness 65, 66, 96 Metric spaces 1 Multilinear mapping

Real-valued function 34, 41

Resolvant operator 9, 10, 11

28

S

N

Sadovski 74

Sard's lemma Neighourhood 5, 12, 47, 49, 51, 53, 113, 1151119t 11771241

125, 126, 127

Newton iterates 42 Newton's method 39, 41, 44, 53 Nonlinear 7 Non-linear functional analysis 78 Norm 30 Normed space 1

84,

94

Schauder's fixed point 60, 63,74,75 Sequence 9, 39 Strict 1 Surjective maps 83

T Taylor formula 34 Topological composition 100

U O

Ordinary Differential Equation 81 Operator 8, 21

P

Unique fixed point 48

Uniqueness 3

V Volterra integral equa-

tion 7, 8

W

Partial derivatives 36 Partial differentiable

Weakly continuous

36 Picard - Lindelof 8

19

Weierstrass approximation theorem

Proper 73

R

57