non-linear functional analysis

Nonlinear FunctionalAnalysis

Rajendra Akerkar

Narosa Publishing HouseNew Delhi Madras Bombay Calcutta

London

Rajendra AkerkarDepartment of Computer StudiesChh. Shahu Central Institute of Business Education & ResearchUniversity Road, Kolhapur, India

Copyright © 1999 Narosa Publishing House

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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 analysisin a vivid and straightforward fashion with a minimum of fuss andformality.

The book was developed while teaching an upper-division coursein non-linear functional analysis. My intention was to give thebackground for the solution of nonlinear equations in Banach Spaces,and this is at least one intention of applicable functional analysis. Thiscourse is designed for a one-semester introduction at post-graduatelevel. However, the material can easily be expanded to fill a twosemester-course.

To clarify what I taught, I wrote down each delivered lecture. Theprerequisites for this text are basic theory on Analysis and LinearFunctional Analysis. Any student with a certain amount of mathematicalmaturity will be able to read the book.

The material covered is more or less prerequisite for the studentsdoing research in applicable mathematics. This text could thus be usedfor an M.Phil. course in the mathematics.

The preparation of this manuscript was possible due to the excellentfacilities available at the Technomathematics Research Foundation,Kolhapur. I thank my colleagues and friends for their comments andhelp.

I specially thank Mrs. Achala Sabne for the excellent job of preparingthe 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 V

1. Contraction 1

1.1 Banach's Fixed Point Theorem 1

1.2 The Resolvent Operator 91.3 The Theorem of the Local Homeomorphism 11

2. Differential Calculus in Banach Spaces 17

2.1 The Derivative 17

2.2 Higher Derivatives 282.3 Partial Derivatives 36

3. Newton's Method 39

4. The Implicit Function Theorem 47

5. Fixed Point Theorems 55

5.1 The Brouwer Fixed Point Theorem 555.2 The Schauder Fixed Point Theorem 60

6. Set Contractions and Darbo's Fixed Point Theorem 65

6.1 Measures of Noncompactness 656.2 Condensing Maps 72

7. The Topological Degree 77

7.1 Axiomatic Definitions of the Brouwer Degree in R" 777.2 Applications of the Brouwer Degree 807.3 The Leray-Schauder Degree 87

7.4 Borsuk's Antipodal Theorem 927.5 Compact Linear Operators 99

x Contents

8. Bifurcation Theory

8.1 An Example8.2 Local Bifurcation8.3 Bifurcation and Stability8.4 Global Bifurcation

9. Exercises and Hints

ReferencesIndex

105

105

110

116

123

129

153155

Chapter 1

CONTRACTION

1.1 Banach's Fixed Point TheoremLet (X, d), (Y, d) be metric spaces. A mapping F : X -+ Yis said to be Lipschitz continuous, if there exists a constantk > 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 con-tinuous with a Lipschitz constant k < 1.

If X C Y, F : X - Y, then t E X is called a fixed pointof 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 canbe transformed in a fixed point problem :

1

2 Nonlinear Functional Analysis

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

TH(x) = Tyx = 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 somefixed 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 hold10 There exists a fixed point x of F.

20 This fixed point is unique.

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

40 For all n the error estimate is true

k knd(xn, ±) < d(xn, xn-1) < d(x1, xo).

1-k 1-k

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 Xis 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 esti-mates by

lim d(xn+j-1, xn)H0 d(i,xn) < 1k

k.d(xn,xn-1)

kn< 1 k.d(xl,xo)

13

This theorem meets all requirements for a useful mathemat-

ical statement: Existence, Uniqueness, Construction and Error

Estimate.

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 forallx,yEX

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

Ekn < 00-

Then 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, thenF(x) = lim F(xn) = lim xn±1 = i, i.e. 1 is fixed point; if 1is 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). C)

j>n

Contraction 5

If F is just a contraction, then F does not necessarily havea 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) = y) = 1- (1 + )2 (x - Y)

< 1, thus, if x # y,i.e. 1 - (1+47

I F(x) - F(y)I 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

= lim F(xn, ),j-oo

thenF(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 : U x V R, P(x, y) =d(Fdx), F(y))

( ,y)

is continuous, and attains its maximum k < 1. Let p E N, suchthat 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. Theuniqueness 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, ))

0The following example shows that there exist contractions

on compact spaces, which are not strict:2Let F : [-1, 01 -- [-1, 0] be defined by F(x) = x + x.

Contraction

Then

F(x)-F(y) = x+x2-y-y2=x-y+(x+y)(x-y)(1+x+y)(x-y).

7

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 wechoose (X, d) = (C[0,1], dry) with

dry(xi, x2) = max Ixi(t) - x2(t)I a-try.

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))Ila-2,max 1 t Ik(t,T,xi(T)) - k(t,T,x2(T))Idr.e-2'

tmaxf 'YI xl (T) - x2 (T) I e-27 dT.e-2

td-y(xi, x2).-y. max f e2''Tdr.e-27t

o<t<i o

1 - 2,yt 2'yt -d7(xi, x2)-7-- max a (e 1)

Id,(xl, x2)

By Banach's fixed point theorem we have the following result.

The Volterra integral equation (*) has for every continuousfunction 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 completemetric 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 itselfis a complete metric space. A sufficient condition for such asituation 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. Letxi 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 kk

+ 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 acontinuous mapping. Let V C X be a subset, such that for ally 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.

Theorem 1.5 Let X be a Banach space and F : X - X beLipschitz continuous with Lipschitz constant k < 1. Then theresolvent 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 definedby

Fo(x) = F(x) + y

is a k-contraction and has a fixed point x E X. So a mappingG : 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 andx? - 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 : 1k

k Ilyl - Y211

0The representation

x = y - R(y)

of the solution x of equation (1.1) shows that properties of Rdetermine the structure of the solution. We will illustrate thisfact by the following result.

Contraction 11

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

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). Letk : 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 dif-ferentiable functions. If further the resolvent operator R to Fexists, 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 LocalHomeomorphism

The strongest version of the solution of the equation

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

x = F-1(y)

The existence (and construction) of this inverse map in somecases is possible. In this section we will see, that F is (contin-uous) 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) aclosed ball in X with centre xo and radius r. Let F : U -+ Y bea mapping, and let T : X --+ Y a continuous linear homeomor-phism, such that there exist q with 0 < q < 1, such that for allu,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 ballV = B(F(xo), p) with centre F(xo) and radius pis 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 neighbourhoodof xo, and FIUo : Uo - V is a homeomorphism fromUo onto V.

50 For all yl, y2 E V we have the inequality

IIF-lyl - F-1y2 - T-1(yl - y2)11:5

plIT-1(yl - y2)II

with p =1 .IITI12 IIT-1II2.

Contraction 13

Proof :1° Let F(u) = F(v) for some u, v E U. Then (*) implies

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

and since q < 1, Tu = Tv and u = v by the injectivity of T.

2° Let y E V. Then F(x) = y is equivalent to the equations

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)III 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

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< qr IIT-'11-'+ (1 - q)r.IIT-ill-i

= 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 fixedpoint 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

G1(z) = z - F(T-'z)+ yi

02(z) = z -F(T-'z)

+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-x2111 1

For x; = F-i(y?) we obtain

iI1x1-x211 = 11T

i(zi-x2)11 < II1- gl.Ilyi-y2II

Differential Calculus in Banach Spaces

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.

15

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

IIwIlIIT-1II.IITwII < F(x2) - T(xi - x2)11

1IT-1II.q.IIT(xi - x2)11

X211

11T-1112.IITII.1 q

q

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

This proves 50.0

Chapter 2

DIFFERENTIALCALCULUS INBANACH 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 bean open subset of X, xo E U and f : U -+ R a real valuedfunction. 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 (f (X0 + u) - f(xo) - x*(u)) = 0.!lull-.o Null

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

Clearly, 20 = 10,

17

18 Nonlinear Fhnctional Analysis

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

Ve>O 36>0 dyEX IIyII R, f (0) = 0, f (e, rl) _

77

Let x' = 0. Then

(f (0 + ry) - f(0) - x'(y)) = T.T .r2(S2 +712)

hence

lim 1f (ry) = 0

and f (0) = 0 is the Gateaux derivative of f , but f is not con-tinuous at 0.

This definition can be extended to map between Banachspaces. Let X, Y be Banach spaces, U C X an open subsetand F:U-'Yamap.

Definition 2.2 1° A continuous linear map T : X - Y is theGateaux derivative of F at x° E U if

Vu E X lim 1.IIF(xo + ru) - F(xo) - rTuII = 0.r20 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 19

3° T is the, weak Gateaux derivative of F at xo i ff

Vu E X Vx' E X' lim 1x'(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 x'(F(xo + u) - F(xo) - Tu) = 0.Hull 40 11uII

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

F(t) = (fis(t))F(O) = 0

_1A(t) 0 It -'

linear and continuous elsewhere

(a) limn.., f (t) = 0, since fn (t) = 0, if n > 1hence (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 aninteger 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 isweakly continuous at 0.

Remark 2.1 Let F : U -+ Y be continuous and Frechet dif-ferentiable 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 thereexist -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 deri-vative 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.

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) isthe linear operator

F'(x) : C[O,1] - C[O,1]

jF'(x)u(t) = 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))

Then

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

F'(x)u(t) =i.e. the Frechet derivative of F is the multiplication oper-ator, 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.

IBI I = Sup 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)II-.o II(u,v)ll

Therefore B is Frechet differentiable and B'(xo, yo) is the

linear map

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 1HTo1.

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'II00

=IIF, (-To1H)j.To1-Tot+To1HTo1IIj=o

= II F(-To 1H)j11.1IT6lII < IIHII2IIIT6- 3

j=2 1-IITo HII

hence

lim I IF(To + H) - F(To) - To 'HTo 1I I

X-0o I IHI I=0.

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 thatF(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)

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 <b, *- I1(PF(x)ll

G is differentiable, hence

be>0 3b2 VyEV IlY - yoll <b2

IIWG(Y)II :5 e.Ily - yoII

Since F is differentiable, we have

3-y>0353>0 Vx EU IIx - xoll <b3

IIF(x) - F(xo)II < 7.I Ix - xoll

Since F is continuous

`db2>0 3b4>0 bxEU 11x-xoIl!5 64

II F(x) - F(xo) 11 < b2.

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

and

lim II(PH(x)If = oIIx - xoll

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

0

Theorem 2.2 Let U C X be an open subset and F : U - Ybe a homeomorphism from U onto the open subset V C Y, V =F(U). Let F be Frechet differentiable at xo E U, such thatF'(xo) : X Y is a linear homeomorphism. Then the inversemapping G = F-1 : V -+ X is Frichet differentiable at yoF(xo) and

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

(The derivative of the inverse is the inverse of the deriva-tive).

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. Letxo, 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 allx E [xo, xo + h). Then

IIF(xo + h) - F(xo)II <_ Jjhil. sup IIF'(x)II.zE (z0,zo+hJ

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 < T < 1

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.

By the Hahn - Banach theorem we have

IIF(xo + h) - F(xo)II = sup y'(F(xo + h) - F(xo))IIti II=1y'EY'

 1, then in general there is no equality inthe statement of the mean value theorem. Letf:R --- C,f (t) = e", h = 27r, then

f (t + h) -f(t) = 0, butf'(t + 0.

2° In the situation of Theorem 2.3 we have foru 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 meanvalue theorem to the function

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

The statement 2° can be sharpened, if F(x) is uniformlybounded for all x E U. Let

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))

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

IIF(x) - F(y)II = II '(1) - c(o)il =II11

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.lly - x112.

2.2 Higher Derivatives

Before going to higher derivatives, we will study multilinearmappings 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.

Proposition 2.1 A n - linear mapping M : X1 x ... x Xn -+ Yis 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" , i.e. (z1, z2) E B, I IM(zl, z2)I I S 1.Then

2

M(zl, Z2) = M211x1 l I' 211x211

=411x1 11.11x211

-M (X1, X2)-

Hence

<-1r2

implies

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

0For 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 spaceL(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), definedby 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 = IIM1IIIxII<1 IIxII51 IIvII<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 IITxyII =11TH.IIxII<1IIyII<1


Together with the above statement this means: 0 is bijectiveand 11011 = 1. 0

Let U C X be an open subset of the Banach space X andF : U > Y. We say that F is continuously differentiable, ifF: U -. £(X, Y) is continuous. We say that F is twice differ-entiable at xo E U, if F' is continuous and F' is differentiale atxo. Then we write F"(xo) and F"(xo) E G(X,G(X,Y)). SinceG (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 operatorwith

(g, h) -+ F' (xo) (g, h).

Proposition 2.3 Let F be twice differentiable at xo E U, letg 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, l IhI I - 6/2, B(xo, S) C U. Bythe mean value theorem

IIso(1) - V(0)II 0 there

exists 6* > 0, such that, if IIhII : 62/2, II91I < 6 *12

II (F'(xo+Tg+h)-F'(xo))9-F"(xo)(T9+h,9)II S

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

+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)

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.

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

Let F : U --+ Y, F is said to be k-times differentiable at zo, ifthe derivative of order k - 1 is continuous and differentiable atx0. 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).

ExampleLet f : R - R be a twice continuously differentiable func-

tion. 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.

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 - 1continuous. Then

IIF(x+h)-F(x)-F'(x)h-...-1 F(V-1)(x)(h,..., h)II(p-1)!

< Ilhlp up IIF(")(u)IIPI (,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

?-11

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

P-1 1_ (1) _ E (0)

J0

= 1pf (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 j4 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-1F(y) (h, h).

By Theorem 2.8 we have

2

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

2Il

uExupIIG"(u)II

[,+lIt is

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

thus

sup IIG'(u)II = sup IIF"(u) - F,(y)IIu u

and, letting y = x

I IF(x + h) - F(x) - F'(x)h - F"(x)(h, h)I I

II 2112 sup IIF"(u) - F"(x)II.uE (x,x+hl

The continuity of F" guarantees the existence of 6 > 0, suchthat 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

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

,ohp

 0, suchthat 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. LetF : 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) coordi-nate.


Theorem 2.6 Let F : Ul x U2 --> Y be a continuous map fromthe open set U = Ui x U2 C X into Y. Then F is continuouslydifferentiable in (al, a2) E U if and only if F is partially differ-

entiable 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. (*)

" 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

 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 asequence (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 Xwith lim xn = i. The sequence (xn) is said to be convergentof order p > 1, if there exist positive meals Q, ry, such that for all

nENIlin - ill :5

If the sequence (xn) has the property, that

Ilxn - ill< -aqn, 0 <q<1

then (xn) is said to be linearly convergent.

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

IIF(x) - ill <- -illp

then (xn ) (F(xn_1)) converges of order p.

39

Proof : -1Let r > 0, B(1, r) C U, such that a = ,Qp.rp-1 < 1,

7 = per1)1)

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

I lxl - ±11 < r =aa

= ,0. explog a

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

By induction we obtain from

I Ix" - iI I < /3. exp(-'rp")

the estimate

Ilxn+l - ±11 apIIxri - uII' -<NlN . exp(-rypn+1)

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

0Example :

If f : [a, b] -, R is p - times continuously differentiable att = f (t) E [a, b], such that f'(t) = 0, ..., f (r-1)(t) = 0, thent" = 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 - g(t)91(t)

t"+1 = tn9(t")

(3.1)- 9'(t")

is convergent of order 2, sinceg

gfl(t) = 1-g1(t)2

1(t)2- 11(t) (t)

f'(t) = 0.

Newton's Method 41

The equation (3.1) defines Newton's method for differentiablefunctions. Now we will consider Newton's method for solvingthe 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 theNewton iterates

xn+1 = xn - F'(xn)-1F(xn) (3.2)

converge quadratically to i and satisfy

llxn - ill 2Q 1q2n-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

Suppose, that (xn) is defined and let

an =IIxn+1 - xnl 1, Qn = IIF'(xn)-'

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.

j< IIxn - x-1Qna-1 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 =

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

then

hence

Nn 7n-l)-',

an C7n-1)-1

andan 2 1 an-1

< 1 1'nkan3n = 7n 2' (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 Cauchysequence 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)-111If 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 - -yn)-1. Then inequalities (3.4) and ryn <

zimply

Jn = - I n - 7n)-1 < 2-fn < 6n-1,

hence

and consequently

an tan-lan-1

Therefore

<bon-'

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

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

00

I Ixn - xI I < liM IIxn - xn+dI <_ E ajj=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 con-tinuous in B(xo, r). Certainly this is another difficult problemwithout 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 <r.Then F has a unique zero x E B (xo, ro) and the iterates of

the modified Newton method

The Implicit Function Theorem 45

converge to x and satisfyn

l 1I Ixn -- 1 1 1 -< 1 - q ' °

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.

IIG(x) -G(y)II /3k.Ilxo-y-t(x-y)II.IIx-yIl< Qkrollx - yIl

Since 1- 1-'y 1ro =

Qk<

,0k'

G is a q-contraction with q = /3kro = 1 - 1 - 'y. It remains toshow that G maps B(xo, ro) into itself.

Let x E B (xo, ro). Then

JIG(x) -- roll = Ilx - xo - F'(xo)-'F(x)II

< 3.IIF'(xo)(x - xo) - F(x) + F(xo)II + a

< p. fIIF'(xo) - F'(xo + t(x - xo))IIdt.Ilx - xell + a

I8kro+a=ro.0

Chapter 4

THE IMPLICITFUNCTION THEOREM

Let X, Y, Z be Banach spaces and U C X, V C Y neighbour-hoods of x0 and yo respectively. Let F : U x V -' Z. In thischapter, 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 thesolution of G(x) - y = 0. The answer is given in the followingtheorems.

Theorem 4.1 Let X, Y, Z be Banach spaces, U C X, V C Yneighbourhoods of xo E U, Yo E V. Let F : U x V - Zbe continuous and continuously differentiable with respect to y.

Suppose that F(xo,yo) = 0 and Fy ' (xo, yo) E C(Z, Y). Thenthere exist balls B(xo, r) C U, B(yo, s) C V and exactly onemap G : B(xo, r) -* B(yo, s), such that G(xo) = yo and for allx E B (xo, r)

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

This map G is continuous.

47

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

hence

IIG(x,) - G(x2)1I <- 1 1 k IIH(xi, G(x2)) - H(x2, G(x2))II

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 homeomor-phism and Go = F6-1 is differentiable at yo

GG(yo) = Fo(xo)-i

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

Then 0z (yo, xo) = F'(xo) is invertible. By the implicit functiontheorem there exist Uo C U, Vo C Y and Go : Vo - Uo, suchthat 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 differen-

tiable at yo with G' (yo) = FF(xo)-1

O

Corollary 4.2 Under the assumptions of Theorem 4.1 andadditionally, 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 forE > 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 definition of t = G(x + s) - G(x) shows, that G isdifferentiable with derivative

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

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

0The solution of the implicit problem

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

can be obtained by iterative methods of Banach or of Newtontype.

Theorem 4.2 Let F : U x V - Z be as in Theorem 4.1. LetFy 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), suchthat

Ilyn(x) - G(x)II <- gnllyo(x) - G(x)II

and

n

Ilyn(x) - G(x)II <- lq yo(x)II

n 1II

<1

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

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

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 andthe 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 < I IT-i I I-i

IIT-1114 = 1 - , 1 IT-l I I

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

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

IIF(x,yo)II < (1- q)so(1-i EIIT-111)

II II

then

H(x, y) = y - F,,(x,yo)-1F(x,

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

(4.2)

(4.3)

(4.4)

(4.5)

Fy (x, yo) = Fy (xo, yo) + F, (x, yo) - Fy (xo, yo)

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

we obtain

and

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

II

I I H(x, yl) - H(x, y2)II < IIT-111-EIIT-111 (P(ro, so) + Y211

1

<- 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

T'q.,90 + 1 I I TI 1< 1II , I I F(x, yo) I I

II

< q.so +IIT-'I1 (1 - q)so(1- EIIT-1I1)

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 neigh-bourhood 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). Thenthere exist ro, so, such that for x E B(xo, ro)

yn+1 = yn(x) - FF(x,yy(x))-1F(x,

yn(x))

exist and converge to the solution F of equation (4.1). Further-more 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 contin-

uously 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 I

(1 +E)11T-111.11 Jot F,1 (x, yn-1(x)) - Fv,(x, G(x)

+T(G(x) - G(x)II

o(1 +11 (1- r)`dT.IIyn-1(x) - G(x)lll+a

(1 +E)IIT-1IIy.IIyn-1(x)

G(x)II1+«

1+ahence

,

ktIyn(x) - G(x)II <- (k.I1yn-1(x) - G(x)I1]1+'

where__ (1 + E)1IT-111 y 1/0

k 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 POINTTHEOREMS

5.1 The Brouwer Fixed PointTheorem

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

orems. It states that a continuous map, which maps a convexbounded 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 spaceC" is isometric with the natural space R', and the unit spheresin these spaces correspond in a natural way. Thus we restrictour attention to real Euclidean space. We need following lemma.

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

55

Then

aDi(-1)`. = 0. (5.1)

i=0 axi

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 ofdeterminants and interchange of columns in them we have

aax

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

Hencea n

(-1)i Di = E(-1)i+jCijo'(i,j),axi i=0

where a(i,j) = 1 if j <z,o,(i,j) = 0 if i = j, ando(i,j) = -1 ifj > i. Thus

n a n

E(-1)i. iDi = E (-1)i+1Cijcr(i,j)i=0 Q=0

Interchanging the dummy indices i, j in this latter expression

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

nE (-1)i+1Cijo'(i,j) _ (-1)j+iC43c(j,i)i,j=0 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 con-

tinuous functions of n variables implies that every continuousmap ¢ of B into itself is the uniform limit of a sequence (qk) of

infinitely differentiable mappings of B into itself. Suppose thatthe 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, = Y-1

This shows that it is sufficient to consider the case that 0 isinfinitely 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 largerroot 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 (5.2)

+(1- Ix12)Ix - O(x)12}


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. Alsoif I x I = 1, then (x, x - 0(x)) # 0, for otherwise (x, O(x)) = 1and the inner product of two vectors with length at most 1 canbe equal to 1 only when they are equal. Thus the discriminantis never zero for x in B. Since the function t1/2 is an infinitelydifferentiable 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. Toprove 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 vectors1. f IE,{_1(t1x),f.1j,4.3(t,x),...,fZ'(t' ).

By the Gau$ Theorem we see that

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

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 contradictionshows, that there is an x E B such that O(x) = x.

0A 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 withradius 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. Definerg(r)

1(x) -119(x}11

f is a continuous map of the compact convex set B into itself.Therfore there exists a fixed point x, of f withI 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 PointTheorem

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

A is convex, if for all x, y E A, T E [0,1], the segmentT2 + (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 coBof B C X is the intersection of all convex sets A, which contain

B. It is given by

n n

COB = {x Tixi, xi E B, T i ! 0, T i = 1 , n EAr}.

i=1 i=1

Theorem 5.2 Let X be a Banach space, A C X a closed subsetand F : A -+ Y a continuous map from A into the Banachspace Y. Then there exists a continuous extension F of F withF:X-*Y,FIA=F and

F(x) C coF(a).

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 locallyfinite open refinement {UA, A E A} of {B.,, x E X \ A}, i.e. thereexists 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

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

cpa is continuous, 0 < cpa < 1, Ecp,, (x) = 1. {cpa, A E Al is apartition 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,,).

LetF(x) = F(x) , if X E A

Ecp,%F(aa) , 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 iscontinuous at xo,

VE>O 35>0 VxEA Ijx-xolI <SIIF(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<diam U,,+(fu-aall< 2rr + dist(aa, UA)

< dist(A, B=) + 2 dist(A, UA)

< 3 dist(A, BT) < 311xo - x1i

and

lixo - aall< ilxo - xUU+Ilx - aall< 411xo-x1l <b

together with JIF(a,\) - F(xo)f l < E implies

lif(aa) - F(xo)II < Ecpa(x).E = E.

Therefore F : X -+ Y is continuous.0

Corollary 5.1 Let C C X be a closed convex subset of X. Thenthere 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

0, if llx-x,1l>eh' (x)

_- e- lIx-xjII, if 1lx-xj11 < E.

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

Eh, (x)

h is continuous, if x E K, then

Ilh(x) - -xl I = 11Eh'(x)(xj - x)

I I E.Eh; (x)

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

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

Chapter 6

SET CONTRACTIONSAND DARBO'S FIXEDPOINT 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 afinite 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 becovered 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 boundedsets. 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 noncom-

65

pactness and

/3 : B -+ R+, defined by

Q(B) = inf {p > 0, B can be covered by finitely many ballsof 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 col-

lected 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 X(Bi) <_ X(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))

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 thereexists 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 B1, N1, ..., N. a coverof B2, then all sets Mj + Nk form a cover of B1 + B2and

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 ifB 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)}. Thenit follows from Bj C B that X(B) < X(B) and0 < X(B).Conversely let for f > 0 given convergins Bj C Mjkwith diam Mjk <_ X(B,) + e <_ Q + E. All of theseMjk's together form a covering of B, so thatX(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).

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 xjEB,Aj>O,Eaj=1

y=EAkyk 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 canassume, that the Mj are convex. Let

in lA = {(a1i ..., Am) E Rm, E Aj = 1, Aj 0}j-1

and for A E A

Then

inA(A) _ F AiMp

j=1

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

Now we show that

C = UAEAA(A) is convex.

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,

1 0 if vj = OPj= tA3/vj if vj#0

then 0<pj<1,zj=pjxj+(1--pj)yjEMj and

tx + (1 - t)y = EtAjxj + (1 - t)µjyj

= Evjpjxj + (vj - tAj)yj

= Evjpjxj + vj(1 --- pj)yj

Evvz3 E A(v) C C

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

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

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

Iix yII <EIAj-A( k)I.11xjII <-E.

This means (U denotes the unit ball)

C = UAEAA(A) C Uk_1A(A(k)) + EU

in

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 decreas-ing 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

Mk) <X(Mk 1

X( 3 ) + k

Claim :

3j, VnEA( M31nMn#ci.

If not, there is for each j an index n3, such that

m, nM=,=o)

then, for all ri 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 anindex jk such that for all n, m

Mnn(U,ti 1Mkjk) (*)

Let

Nm = Umk=lMkjk

then ...andM,nCMmjm(*) 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 aCauchy sequence, which is convergent,

Cf1M,, =M,,,,.

Therefore M,,,, is nonempty.

13

6.2 Condensing Maps

A continuous map F is called bounded, if it maps bounded sets

onto bounded sets, and compact, if it maps bounded sets intocompact sets.

Definition 6.2 Let X be a Banach space and U C X. Anoperator F : U -+ X is called a k-set contraction, 0 < k < 1,if F is continuous, maps bounded sets onto bounded sets, suchthat 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 com-

pact map is a 0-set contraction. The following example is an

important one.

Example 6.4Let X be a Banach space, U C X and K : U -+ X Lip-

schitz continuous with Lipschitz constant k < 1. C : U -+ X iscompact.

Then

,F=K+Cis 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).

0A 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 -, Xcondensing. Then I - F is proper and I - F maps closed subsetsof 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. FromK = (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,,. Lety = 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. ThusX E A implies

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

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 andconvex subset of a Banach space X and let F : C -+ C becondensing. 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 thedecreasing 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 Cinto 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, sinceI - 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 asubsequencewith lim xn,, =.f = F(x) = lim F(xn,, ). ThereforeF has a fixed point i in C.

11

Chapter 7

THE TOPOLOGICALDEGREE

7.1 Axiomatic Definition of theBrouwer 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 theimage 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 ofthis 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 asolution x E S2, iffy E 0, i.e.

(dl) d(id, 0, y)1 for y E 0

= 0 for y V U

77


The second condition is a natural formulation of the desirethat d should yield information on the location of solutions.Suppose that III and 112 are disjoint open subsets of SI andsuppose, that f (x) = y has finitely many solutions in SI1 U 02ibut 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, andthis suggests that d should be additive in its argument c, thatis

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

Uc12))The third and the last condition reflects the desire, that

for complicated f the number d(f, St, y) can be calculated byd(g,11, y) with simpler g, atleast if f can be continuously de-formed into g such that at no stage of the deformation we get

solutions on the boundary. This leads to

(d3) 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

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),

with the following properties :0

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) fort 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 C S7, fl is open, disjoint,Um1 f (Mi), then

m

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

y

[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)

10. Let m < n, cl C R" 'be open and bounded and f R."

continuous, y E R' \ (I - f)(8SZ). Then

d(I - f, 0, y) = d(I - f J . m, fl n Rm, y).

[REDUCTION]

7.2 Applications of the BrouwerDegree

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 sameis 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) # xfor Ixj = r. Therefore d(id - f, D, 0) = d(id, B,.(0), 0) = 1, andthis 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 The-orem 5.4 we have a continuous extension f : R" of f, and if welook at the defining formula in the proof of this result, we see

that f (R") C conv f (D) C D since

m mw-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


point x off in $,.(0), by the first step. But j (x) E D andtherefore x = AX) = f (x).

Finally, assume that D = h(Do) with Do compact convexand h a homeomorphism. Then h` 1 f h : Do - Do has a fixedpoint 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 iand Ax = Ax. In other words, A has a nonnegative eigenvector

corresponding to a nonnegative eigenvalue.

To prove this result, let

D= xER":xi>0 for all i andm

Exi=1i=1

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

with A = Es 1(Ax0 )i

Example 7.2Lets consider the system of ordinary differential equations

u' = f (t, u), where u' = di and f : R x R" --+ R" is w - periodicsolutions. 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) (7.1)

have a unique solution u(t; x) on [0, oo).

Now, let ptx = u(t; x) and suppose also that f satisfies theboundary 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 everyt E R+, since

dtIu(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) attime t. Furthermore, Pt is continuous, as follows easily fromour assumption that equation (7.1) has only one solution. Thuswe find P, has a fixed point x,,, E Br(0), i.e. u' = f (t, u) has asolution such that u(0; x,,) = x, = u(w, x,,). Now, we may easilycheck 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 ifx is a fixed point of P.Example 7.3

It is impossible to retract the whole unit ball continuouslyonto its boundary such that the boundary remains pointwisefixed, i.e. there is no continuous f : B1(0) -' 0B1(0) such thatf (x) = x for all x E 8B1(0).

Otherwise g = -f would have a fixed point x0, by Theorem7.2, 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 pointwhere 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.


Since t(x) is continuous, g would be such a retraction which does

not exist by assumption.

Surjective MapsIn this section we shall show that a certain growth condition

of f E C(R't) implies f (R") = R". Let us first consider thatfo(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 - 00asIxI -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 TheoremUp 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 letf : 8f -- R" \ {0} be continuous. Suppose also that the space


dimension n is odd. Then there exist x E 8f2 and A 0 suchthat f (x) = Ax.

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. Ifd(f, SZ, 0) # -1, then h(t, x) = (1 - t) f (x) - tx must have azero (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 toh(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. Incase Cl = B1 (0) the theorem tells us that there is at least onenormal such that f changes at most its orientation. In otherwords: 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'.

However, f (x) = (x2, -x1, ..., x2,,,, 1) is a nonvanishing

tangent vector field on S C R1.The proof of existence and uniqueness of the Brouwer de-

gree and its construction is based on the fact, that f (Sf (0)) isof 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 ap-proximations 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.

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 andthat M C Rn has measure zero (i.e. µn(M) = 0) iff to everye > 0 there exist at most countably many intervals Js such thatMCU;JJarid <e.

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

Since an open set 11 in Rn may be written as a count-able 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 uniformcontinuity of f' on Q, given e > 0, we then find m E .fU suchthat I f'(x)- f'(Y)I < E for all x,Y E Q with Ix-zI < 6 = f.,and therefore

jI f (x) -- f (y) - f'()(x - )I < I f'(Y

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

for any such x, T. So let us decompose Q into r cubes Qk ofdiameter 6. Since b/ f is the lateral length of Qk, we haver = 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; letA = f'(T) and g(y) = f (T + Y) - F(T) for y E = Qk _X.

Then we have

9(y) = Ay + R(y)

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 Rnwith 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

ng(y) =>(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 < cfor some c, in particular IAI < c. Therefore, f (Sj(Q)) C Uk=1Jk

withr

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 letf o : T7 - Rn be continuous and twice continuously differentiable

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 atwice 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 boundedsubset of X, F : 3I -+ X compact and y V (I - F) (Oil). Onthese 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 ofSl 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

Banach space, SZ C X open and bounded, and F : N -. Xcontinuous. Let {x1, ..., x"} be a basis in X, {e1, ..., e"} the unitvector 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 thatthis definition is independent of the choice of the basis of X.

In the second step we will assume, that X is an arbitraryreal Banach space, SZ an open and bounded subset of X andF : --> X a compact map. By .F(?7) we denote the compact

mappings F : St -+ X such that F(SZ) is contained in a finitedimensional 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))IIMEM

inf (IIy-(x-F(x)II-IIF,(x)-F(x)II)>0.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

This number is independent of the choice of Fi and X1. LetF2 E such that I IF - F21I < R and X2 C X, such thatdimX2 <ooandyEX2,F2(3'I)CX2,02 =SlnX2960. Nowlet Xo = span{Xi, X2} be the finite dimensional subspace of Xgenerated by X1 and X2, flo = Xo fl Cl, then by the reductionproperty (1) of the Brouwer degree for j = 1, 2

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 thaty 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.

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

Theorem 7.4 Besides (DI), (DQ), (DS), the Leray - Schauderdegree 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 - Xcompact [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), wheneverGlen = Flon.

(D8) D(I - F, S2, y) = D(I - F, SZ., y) for every opensubset Q. of 0 such that y ¢ (I - F)(3I \ S2.).

(D9) Let Y be a closed subspace of X, S2 C X be open andbounded, S2 fl Y # 0. Let F Y compact,y $ Y \ (I - F) (09% ThenD(I - F, 0, y) = D((I - F)InnY, 12 f1 Y, V).

Proof :(D4) Let Fn E .F(U2) such that IIF FnhI < nr. Choose

Xn, S2n according to the Definition 7.7. Then

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) = yforanxE7,xit 8S2,i.e. xES2.

The Topological Degree

(D5) Choose GI E F(?t) such that

JIG - G111 < min{dist(y, (I - G)(90)),

91

Then

From

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

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

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). Theny ¢ 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)) thereexists F1 E F(? I), I I F - FiII < r and F, (?7) C Y. Let X1 C X,

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 thatf (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 forodd maps Borsuk's theorem states that d (f , 0, 0) is odd, hence

different from zero. We say that SZ C X is symmetric withrespect to the origin, if S2 = -0, and f : fZ -' X is said to beodd 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


can be chosen arbitrarily small, hence

d(f, S2, 0) = d(f, 1Z, 0).

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

d(f, S2, 0) = d(g, 12, 0) = sgnJ9(0) + E 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 ofv!'IA if 0 is a regular value of by IA,

since for x E A,v

(x) = y, thus hy(x) = 0 implies by the quotient

rule (v )'(x) _ ' h,(x), because

u'(x) v'(x)u(x) .[u'(x) - v`(x).uU

(x)]V v(x) v(x) V(X) v

= v(x) (u'(x) - v,(x)y),

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

94 Nonlinear Fltnctional Analysis

Now we define odd functions hl, ..., h" E G1(1l) withl f - hk I I < E if x E ? and 0 is regular value of hk Ink, where

Hk = {x = S") E R", Sk = 0}

cik = cl\(H1n...nHk).

Let h1 (x) = f y1i where y1 E R", such that 0 is a regularvalue 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) andhk+1(x) = hk(x), for x E 1k n Hk+1, by induction we gethk+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", then0 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 and0 E Sl. Let f R" be continuous such that for all A > 1

andxE8Sl

f (x) 0, f (-x) - Af (x) 0.

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

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)

Let SZ C R" be as before, f : c 9Q R' be continuous andm < n. Then there exists an x E 8Q such that

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 symmetricwith 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

Let x E A then f3 (x) = 1, f j (-x) = -1 and x E -A; impliesf,(x) = -1, fj (-x) = 1. Let f = (f'), then f (-x) does notpoint in the same direction as f (x), thus f (-x) # A f (x) on ffor 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 infinitedimensional Banach space X. Then the measure of noncompact-

ness X(B) = 2.

ProofLet S = aB be the boundary of B and

S=U 1MM

be a finite covering of S by closed sets M; C S such thatdiam(M3) < 2. Let X be an n - dimensional subspace of X,then S n Xn = Ul 1(M; n is the boundary of the unit ball in

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 showsX(B) = 2.

11

If X is infinite dimensional, the Borsuk theorem follows im-

mediately from the finite dimensional case:

Theorem 7.7 Let 1 C X be open, bounded and symmetric withrespect to 0 E 0, F : fZ --> X be compact, G = I - F and 0 0G(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 isodd.

Proof :

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

x - H(t, x) - 1 +t

t1(1 + t)x - t F(x) + F(-x))t .(1(x - F(x)) - (-x - F(-x))l+t t

1- t.(1 G(x) - G(-x)) jA 0.t

Let Fo(x) = z (F(x) - F(-x)), then Fo is odd and thusD(I - F, ft, 0) = D(I - Fo, ft, 0). Choose F1 E .F() withIIFo - 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 symmetricwith 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. Thenthere exists a fixed point xo E aft of F.

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), whichcontains 0, there exists an x E ), such that x - F(x) = y, i.e.

C0 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 symmetricwith respect to 0 E 1, and F X compact. Let (I-F)(17) C

CY # X. Then there is an x E 8SZ such that

x - F(x) = -x - F(-x).

Proof :Let G(x) = z(F(x) - F(-x)). By Corollary 7.4 there is an

x E 81 with G(x) = x, thus I (x - F(x)) +z(x + F(-x)) = 0

or x - F(x) = -x - F(-x).11

Corollary 7.6 Let F : X -+ X be compact, and L : X -+ Xcompact and linear. If there exists a A E R, such that for allxEX,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),


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 isthe 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 - Schauderdegree: 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 linearmaps 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.


Proposition 7.2 Let S, T : X - X be compact linear opera-tors, such that 1 is not an eigenvalue of S or of T. Then forr>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 - Tand (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 compactlinear 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).


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 reductionproperty (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 continuouslyinvertible } be the spectrum of T. Then

1° u(T)C {AEC:IAI <IITII}o,(T) is countable, the only point of accumulation ispossibly 0 .

2° If A ¢ o(T ), T - AI is an isomorphism in X .

30 If A E o,(T) \ {0}, there exists a minimal numberk(A) e N, such that forR(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 isdim ker(T - AI).

Theorem 7.11 Let X be a real Banach space, T : X ---- Xbe a compact linear operator, A 0 and A` not aneigenvalue of T . Let 0 C X be open and bounded and0 E SZ . Then D(I - AT, 0, 0) = (-1)m() where m(A) isthe 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. Bytheorem (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 > IA-1I, say

µi, ...,1u7,. Let

V = ®;=1N(µ1), W = lj=1R(1_ij)


We show that X = V ® W. First of all v n w = ¢, sincex 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)

henceP

xl = x - E xj E R(Al) n N(jul) = {0}j=2

and similarly we obtain x2 = ... = x, = 0. Now, any x E Xmay be written as x = xj + yj with xj E n(µj), yj E R(µj) bytheorem (7.10, (3°)) again, we have

Px-E=x-xk-rxj -yk-1: xj ER(/Aj)j=1 j#kk jOk

hence

and

P

x - L xj E W = nR(pk)j=1

X=v wBy Theorem 7.9 we have

But

D(S, Sl, 0) = D(SIv, o n V, 0).D(SI w,11 n W, 0)

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 haveP

D(SI v, 9 n V, 0) = jj d(SIN(µi), st n N(µj), 0).j=1


Since h(t, x) = (2t -1)x - tATx is an admissible homotopy fromSI N(µ;) to -II N(iz,) (this is true because (2t - 1)x - tATx = 0

and I Ix I I = 1 implies 1a =2t-71

)hence t = 2_aµ, >1 )thus

D(SI N(j,i), Q n N(/j), 0) = D(-IINC,,,), Sl n N(µ), 0)

and thereforeD(S, Q, 0) _ 1)m(''),

where m(A) _ IAI,>i n(µ3).If there are no such µ at all, then X

D(S, Q, 0) = 1 = (-1)°.

W and

0

Chapter 8

BIFURCATIONTHEORY

8.1 An Example

Let X be a Banach space, 1 C X an open bounded subset, andF : SZ -+ X compact. We consider problems of the following

type. Assume 0 E S2 and F(0) = 0. Then for every real A theequation

Ax = F(x) (*)

has trivial solution. The following example shows that thereexist real \o that

0 < IIx,,II <E, Pa - Aol < E

Ax,, = F(xa),

i.e. a branch of nontrivial solutions of (*) starts in the point(Ao, 0). The point (A - 0, 0) E R x ) is called a bifurcationpoint. We will solve the following integral equation

Ax(s) = njr

sin s sin t + b sin 2s sin 2t][x(t) + x3(t)]dt(8.1)

105

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 befunctions of the real parameter A). Substituting in equation(8.1), we have

A[A sins + B sin 2s] = 2 T. + b sin 2s sin 2t]

. [A sin t + B sin 2t + (A sin t + B sin 2t)3]dt

= 2a. sin s[A f sin2 tdt + A3 f sin4 tdtn o 0

ine t. sine 2tdt]+3AB2 fo s

+ 2 b. sin 28 [BJ

if

sin2 tdt + 3A2B TO sin2 2t. sin2 tdt

+B3 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:

1x

sin t. sin 2tdt =To

sin3 t. sin 2tdt = T sin t. sin3 2tdt = 0.0 0

Equating coefficients of sin s and sin 2s, we obtain a pair of non-

linear simultaneous algebraic equations:

AA = aA + 3aA3 + 3aAB24 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. Wecancel A # 0 to obtain

A =a+3aA2

whence

A=f a/ -1a .

The corresponding solution of equation (8.1) is

= f 3xl (s, A) A/a - 1 sins,

defined and real for A > a.

3. A = 0, B # 0; only the second equation is nontrivial. Wecancel B # 0 to obtain

A = b +3bB2

whence

B=±23 A/b-1.fThe 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 inequation (8.2).

We obtain two ellipses:

4A2+2B2=--1(8.3)

Solutions of equation (8.3) are given by intersections of theseellipses. Solving, we get

A24.[2a-b,_,] B22b-a\-19ab 9' ab

so that we have the following solutions of equation (8.1):

X3(s,

A) = f2 . 2a - b,\

- 1 sins ±2 2b - a.

- 1 sin 2s.3 ab 3 ab

(8.4)

Clearly 2a - b > 0 since we assumed that b < a. Hence thequestion 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>2We have the following cases:

Case I : a < 2; x3(s, A) is real for no real A.

Case II : a >2;

x3(s, A) is real for A > max(2ab b, zaba

).Since a > b, this means x3(s, A) is real when A > 2aa

In Case I above, i.e. when a < 1, the only real solutions ofequation (8.1) are the trivial solution x(s, A) - 0, and the two

main branches:

xl(s, A) = f73

. a - 1 sin s

X2(5, ,1) = f733

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)


In Case Ii, i.e. when > 2, we again have trivial solutionx(s, A) = 0, and the two main branches

xi (s, A) = f2

. a - 1 sin s735

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,namely that in equation (8.4). Note that as A -- A > ab

2b-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 . Thusas \ --,

2bba, we see that x3(8, A) --> X3 18, 26ba] = x1 [s, 26 baTherefore at A = 2aa' the sub-branch (twig)

x3 (s, A)=3 /2b_aA_ 1sinSf3 abaA-lsin2s

joins the main branch, i.e. x3 [s, 2bba] = xl [s, 2b. J while thesub-branch (twig)

x3(s,A)2 2b-aA-1sinsf 2 [2b- aA -1 sin2s3 ab 3 ab

joins the negative part of the main branch, i.e.,ab abx3[3'2b-a] -x1[s'2b-a

We have in Case II, when a > 2, the phenomena of "sec-ondary bifurcation", or the forming of sub-branches or twigswhich bifurcate from the main branches. The main branchesbifurcate from the trivial solution at the eigenvalues of the

linearization of equation (8.5), while the twigs bifurcate fromthe main branches.

Thus solutions of the nonlinear equation (8.1) exist as contin-

uous 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, andits 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 forthe main branches. If a > 2, i.e. Case II, two sub-branches or

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 inthis simple problem is a "condition preventing secondary bifur-

cation".

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 E1, 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 andA 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.

Proposition 8.1 In the situation of Definition 8.1 let

(a) F, Fz are continuous in a neighbourhood of the .bifur-

cation point (Ao, 0). Then FF(ao, 0) is not a homeo-morphism.

(b) If X = Y, F(A, x) = x -- ATx + G(A, x) with a contin-

uous C : (a, b) x 1-+ X, such that

sup IIG(A,x)II 0AE(a,b) lixll

if IlxII -- 0, then ) 1 belongs to the spectrum of T.

Proof :

(a) If F(Ao, 0) is a homeomorphism then the implicitfunction Theorem 4.1 tells us, that F has a uniquesolution, 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


In this case (ao, 0) is called a vertical bifurcation point.

ExampleLet 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)ERxXis not a bifurcation point.

If we represent F in the form of Proposition 8.1 (b), thenF(A,x) = x-Ax+G(A,x) with G(,\, x) = (713,-713), thus Ao = 1

is of multiplicity two.

ExampleLet X = Y = R2 and

F(A, x) =( I ) - a

1 1 )()+A(;3).77 1 77

A0 = 1 has geometric multiplicity one, algebraic multiplicitytwo, but F(A, x) = 0 implies

t-arl=071-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 (*)

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 bifurcationpoint (µ, 0) with µl < µ < µ2.

In the first case (µ, 0) is the only solution of (*), hence homo-

topy 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 alge-

braic multiplicity.

Theorem 8.1 Let X be a real Banach space, K : X --+ X becompact 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 withlim,.,o co(r) = 0 and 8 > 0, such that for all (A, x)(A, x) E S2

with is - AoI< 6,Ilxll <r,IIYII <r

IIG(A,x) - G(7,T)II <- W(r).IIx - YII.

Then (Ao, 0) is a bifurcation point for

F(A, x) = x - AKx + G(A, x) = 0.

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 The-

orem 7.2 X = N(Ao) ® R(A0), and both subspaces are invariant

under K.T is a homeomorphism from R(o) onto itself and ) 1is the only eigenvalue of KI N(ao). Let P : X - N(0) be thepro-

jection 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) = 0equivalent to

x=v+za=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, thenG is a contraction in the space of R(Ao) - valued continuousfunctions. 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

IIz(p,v)11 CI {vI 1,supIIZ(p,v)11 < p}.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 _ I l (p,v+z)-G(p,0)11

< cp(r)I Iv + zII < cp(r) (1 + c) I IvI1 < W(r) (1 + c)r.

Bifurcation Theory

The operator

z -+ (I - ASK) -1S(I - P)V(p, v + z)

115

maps M into itself and is a contraction, if r (and thus cp(r)) ischosen 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 acompact 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)

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) _ (8.4)

where +µj) is the sum of the algebraic multiplicities ofthe 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)

where K : X -' X is linear and compact, A01 is a simple eigen-value of K, G : R x 0 is continuous differentiable and SZ is anopen 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(<A01}.Let xo E X, I Ixol I = 1 be an eigenvector of K to the eigen-

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

-oaao + 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

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 continuouslydifferentiable 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 G1Rx)xR-+X,such that for all µ,aER,xES1witha=<x,xo>

G(A, ax) = a2G1(A, x, a).

Equation (8.4) then has the form

-o

+ a < G1(µ, x(a, µ), a), xo >= 0. (8.6)

The function

f (a, µ) = - o + a < x(a, A), a), xo >

has the property

By the implicit function theorem there exists a continuouslydifferentiable function µ : (-a°, a°) - It, such that for all awith 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 additionalsolution 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 negativeeigenvalues if A < A0, since

(I - AK)x = vx implies (1 - v)x = AKx.

If v < 0, then 1-' > a > -L would be an eigenvalue of K. Then,A AO

since G(A, 0) = 0, we have with ) = B(0, e)

i(I - F, 12) = i(AK + G(A, .), )) = i(AK, SZ)

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},

theni(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 thefollowing:

Since µ'(a) # 0 for small a # 0, there is an f > 0 such thatfor all x with 0 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 equicontin-

uous, hence G(A0, xn) - G(An, xn) --s 0. From this follows

lim (I - AOK)xn + G(ao, xn) = 0.

Since F(Ao, .) is proper, there is an ±, 1 1

By equation (8.7)= 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. IfA 76 A0, the solutions are isolated, therefore d(A) is the sum ofindices 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 trivialsolution has index - 1, hence the bothnontrivial solutions for a < 0 and a > 0have the index +1.

Case 2° : Since µ'(a) # 0, we have nontrivial solutionsfor A < A° and A > A0, and the index of thenontrivial solution of absolute value 1, wehave d(A) = 1 + index of nontrivial solution, ifA < Ao d(A) = -1 + index of nontrivial solution,if A > a° therefore, if A < AO, then the indexof the nontrivial and the index of thenontrivial solution is +1, if A > A0,and d(A) = 0.

Case 30 : d(A) = -1, since the trivial solution has theindex -1, if A > A0, thus both nontrivialsolutions 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

and f (ao, x) = Ax + R(x), where A is an x x n - matrixand Rx = 0(I IxI I) . Then the trivial solution x = 0 ofx' = f (ao, x), x(0) = xo is said to be stable, if for every e > 0there exists 8 > 0, such that the solution exists on R+ andsatisfies I I x(t) I I < e whenever I Ixo I I < 6. The trivial solution ofx' = Ax is stable if Re o < 0 for all p E o(A) and every p, suchthat Re p = 0 has algebraic multiplicity equal to its geometricone. If Re p < 0 for all p E Q(A) then x = 0 is stable as asolution of x' = f (x). If Re p > 0 for some u E Q(A) then x = 0is 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 simpleeigenvalue 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 <I],IA-AoI <Il implies

o,(I - AK + G'(A, x)) C UCEQ(I-AOK)B((, 2)

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, Ais real; and A 0 for small solutions x = x(a.p(a)), sincep'(cx) # 0. Therefore the stability depends upon A > 0 or A < 0.

Since Ao is not a vertical bifurcation point, for the nontrivialsolution (A, x) we have A 36 Ao.

For A > A0 the index of the nontrivial solution is +1, andthis is true if A > 0, these solutions are stable. If A < A0, theindex 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 provemore about the global behaviour of components of the solutionset 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 thatG(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 beexplained 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 # 0of (Al,0) E C for another characteristic value Al # Ao of K. Incase 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 andG are compact. Suppose next that we are able to find an open

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 thatthe intersection of ?o and real line is given by J = [A0 - 8, Ao +

8] with 8 > 0 so small that no other characteristic value ofK satisfies IA - Aof < 28. By the homotopy invariance of theLeray - 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 de-grees 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. In-deed, 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 homotopyinvariance 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 withUa = {(A, x) E SZ : dist ((A, x), C) < S}. Evidently, Ua n M is

compact and c fl 8U6 = 0. Note that U6 fl A? is not connectedunless 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 thendiet (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 M2and M1 fl M2 = 0.

Proof :To use a good substitute for possibly missing pathwise con-

nectedness, 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 thereare 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. LetAE = {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 sinceBE(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 anand 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

En - chains Mn joining ao and bo, for every n > 1. Consider thelimit 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 EC2 are En - chainable. Hence, MO is connected. Consequently,

Mo C A since as E MO n A and A is maximal connected, andtherefore bo E A n B, a contradiction.

0The 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 aneighbourhood of (,\o, 0), G : St -+ X be completely continuous

and G(A,x) = 0(IJx`J) as x -+ 0, uniformly in A. Let k belinear 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 multi-

plicity.

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, abounded neighbourhood S1o of C satisfying Mn8S2o = 0 containsonly a finite number of points (Ak, 0) with Ak 1 E u(K), sayAl < ... < 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),-) forj = 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) - D(F(Akl, ), Bp(0), 0)] = 0-k=1

This evidently implies that we have an even number of jumpsin the degree. Since we have one at A0, and since the jumpsoccur only at characteristic values of odd algebraic multiplicity,

the theorem is proved.O

Chapter 9

EXERCISES ANDHINTS

Every exercise is not answered here. Readers will learnbest if they make a serious attempt to find their ownanswers before peeking at these hints.1. Let X be a Banach space. A : X -+ X is linear andcontinuous, b E X. For every A in the spectrum of A holdsJAI < 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 - bF2 (x) = A2x - Ab - b

F3(x) = A3x - Alb - b

F'(x) = A"x-A"-lb-...-Ab-b

129

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 = IAInIIxIISo it does not converges in IAI > 1.)

2 Let (X, d) be a complete metric space, V' : [0, oo) -+ [0, oo)

continuous, O(t) < t, if t > 0. Let F : X -> X, such that forx,yEX

d(F(x), F(y)) 0, Cn is decreasing sequence.

2limCn=C>0

C<V(C)=:t,. C=0.

Next, show (Fnx) is a Cauchy sequence.

If not z 2E > 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, F'nkx) < d(F'kx, Fmk+lx)

+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, andF(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)) <d(X,X)

= d(X, X) = 0 X = X Uniqueness)3 Let P be a metric space. For each p let Fp : X - X be ak-contraction. 0 < k < 1. Let for fixed po E P and for all x E X

Then, for each p E P, Fp has a unique fixed point xp, andxp = xpo.

4.tt tt

Define f : R -' R2 by f (t)tt= (t, t2) and g > R2 - R by9 (Si, S2) = 1 if t2 = ti and 9(e1, b2) = 0 if 1 Y-' Q2 . Show thatthe Fr chet derivative. f'(t) exists for all t E R, the Gateaux-derivative Dg(0, 0) exists, but (gof)'(0) # Dg(0, 0) f'(0) (andhence the chain rule fails for Gateaux - derivatives).

(hint : f : R - R2 is defined by f (t) = (t, t2),

9 : R2 --+ R by e2) =If 2 = 1

0 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

T)lim

9(TS1, Te - 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

(b) e2 0 9(Te1,Te2) = 0 for ITI < fa

So the limit exists i.e. Dg(0, 0) exists.

Now (9of)(t) = 9(t,t2) = t .Therefore (gof)'(0) = 1. But Dg (0, 0), f'(0) = 0. )

5. Suppose that X is a real Banach space and f : X - R isGateaux - differentiable on X. If f reaches a relative minimumat 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 ifand 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 <limf(xo+th)- f(xo)

t+

f'(xo)h = lim f `x0 + th) - f (xo) < 0t-+o- t

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) =x0h(r+t-rt) = 0T+t-rt=0

-t t

1-t t 1

f (xo) = f (,r(xo + h) + (1- r)(xo + th))

t 1 1 f (xo + th)i.e., f (xo) < t f(xo + th) - t 1

tf(xo) -f(xo) > f(xo+h)- 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

Px= jb x(r)

-f(T, s)dsdro J

134 Nonlinear Thnctional Analysis

a continuously differentiable functional on C[a, b] is defined. Inaddition If (t, s)I < c(1 + IsIvA) on [a, b] x R, with c > O,p > 0and n + a = 1, then co is continuously differentiable on Lp[a, b].

7. DefinitionA set A is called a Banach algebra, if

A is a Banach apace, and if

- there is defined an associative distributive continuous multi-

plication of elements of A, i.e.

if X, y E A, then x.y E A and I I < IIxI.I IyIILet 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 differen-tiable. 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

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. I

Ilf(x + k) - f(x) - f'(x)kll :5Elllkli

gF - diff *363>0 s.t. (Ikll <-63

= Ilg(x + k) - g(x) - 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

135

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

Since E J Ilx-'hlln is convergent and A is complete,00

E (x-' h)n (E - x-' h)n=0

6= `((X-lh)n - (x-1h)n+l)

l ln=O

k

= lim 1:((x-1h)n - (x-lh)n-1)n--O

= k (E - (x-lh)k+1) = E

x - h is invertible.So, B(x, 1771 ) < gA 9A is open.

Again, f (x) = x-'; take IihIl < 21 1 ,. So,

II(x +h)-' - x-1 + x-'hx-1I1

= II(x(E +n-'h))-1 - x-' + x-'hx-'II

= I I (E + x-'h)-1- (e + x-'h)x-11100

=I I(> (-x-lh)n - (e + x-1h)x-1IIn=000

= IIE(-x-1h)n.x-'II

n=200

= lI(-x-'h)2 E(-x-1h)n-21I

n=o00

< IIx-11131 IhI12 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' 211xE'113

,

Il.f(x + h) - f(x) + x-lhx-11I <_ IIh1I

Now

Exercises

8. Consider the eigenvalue problem

Ax = ABx,

137

where A and B are real (n x n) matrices and where the normcondition < x, x > = 1 is satisfied. An important trick isto 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).

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) =21

7' 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)

sin r(x(T) + x3(r))dr= 2a sin t Toit o

x+. b sin 2t

Jsin 2T(x(r) + x3 (r) )dr.

If x is a solution then x should be linear combination asx(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)37r 0

T2

_ (2asint)(I,) + (2

b sin 2t) (12)Ir 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)AB = bB(1 + 3-4B2 + 2A2)

(i)

(ii)

Case 1 A = B = 0 x(t) = 0 is a solution for every A E R.

Case 2 A = 0, B j40.from (ii) A = b(1 + 4B2)

B731Vb- ERe*a>b

= the solution

X2 (t) = f fFb- sin 2t for A - b.

Case 3 A # 0, B = 0 = equation (ii) fulfilled everywhere. Then

from (i)


A 32 b - 1 sin t.x4 a (t) = f

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)) - 2J

k(t,T)xtaudr

- 2 J x(t,,r)x3(T)dT7r 0

(A, x) exists and cont. in xH,'

HH(A, x)h(t) _ Ah(t) - f k(t,T)h(T)dT

j-3.k(t,T)x2(T)h(T)dT7r

0

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

Letsin t sin 2te1=TI ' e2

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 - 0is the unique solution in U,,.Case 4

A#0,B34 0 A=a(1+3A2+3B2)

A=b(1+3B2+2A2)

=f3 baA )-1

B=f3Since 0<b<a=(b-Q)>O. But we need

tab b <a(>0)

andA(a-1)> 1sinceA>O*2-1>Os a<2bwehaveA>aba>a.

Ifa<2bandA> bathen

xs>7,s,9('\)(t) = f3 t(b -a ) - 1 sint

3

'\(ab) - 1 sin 2t.)

10.: Let X be a complete normed space, and x : [0,1] -- X becontinuous. Show that the Riemann integral

i

Jx(t)dt

is well defined, linear and that

11Jo' x(t)dtll j'IIx(t)(Jdt.

11. : Let F : Rn -, 1Zn be continuously differentiable and letF'(x) # 0 on Rn. Then F is a homeomorphism onto Rn if andonly 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 iflimlI.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) -- Y-

3(x,,,,,) = F(lim F(x) = y E F(R").Since F(Rn) is open (from (b) and also F(Rn) is closedF(Rn) = 0 or Rn, here it is not empty. So F(Rn) = IV =:,. F

is surjective.

(d) F is injective. )

12. : Let (afl) be a real (n x n) matrix with aj > 0 for alli, 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 possessesa positive eigenvalue.

(PERRON - FROBENIUS).(hint : a+j E Mat(n, n, R), ai f > 0 V i, j then A possessesnon-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 Aand X is a eigenvector.

Case 2 : Ax # 0 d x E k E{ 1(Ax) j# 0 we define

f (x) Ax),' f : k --- k is continuous ),=1(

13. : BROUWER's fixed point theorem is equivalent to thefollowing 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 continuousf : 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 continu-

ous without fixed point (U is also a compact).

(b) U = (0, 1), f : U --+ U by f (X) = 1X f is continuouswithout fixed point.

(c) f : 10, 1] -+ [0, 1]

f(X) =0 for 1<X<11 for 0<X<1f 2

* f is continuous without fixed point. )15. : Let X be a separable infinite dimensional Hilbert spacewith a complete orthogonal system {x0, x1, x2, ..., }. For

00

x = >2ajxjJ=O

defineCO

F(x) = 1- IIX112.xo + E aj-1xj-j=1

Show that F maps the closed unit ball into itself continuouslywith no fixed points. Thus, compactness of F is a necessaryhypothesis in the Schauder's fixed point theorem.

(hint : X is a separable infinite dimensional Hilbert space,

{xo, x1) ...., ..., } complete orthogonal system.

=100f(x) = 1- I lxl l2.0 + E aj-lxj, 1141j=1

to show that F : 77(0,1) --' R(0,1) is continuous and has nofixed point

00IIF(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

00_ E ajxj = E aj-1xjj=1 j=1

do = 0 by induction 6j = OVj E Ni =0.Finally, F is continuous

IIF(x) - F(y)112

II(V1IIx112 - "j-11y112)xo

00

+E((x,xj-1) - (y,xj-1)112j=1

= ( 1--11x1!2- 1-11y1I2)200

+ E(x - y,xj-1)2 < E2j=1

VUlx--y11<-6.

Since 1 - I lx 112 is continuous F is continuous. )16. : Let X be a partially ordered real Banach space, i.e. thereexists a " <", which is compatible with the linear and the topo-

logical structure in X, i.e.1. x<y= x+z<y+z2. AER,A>O,x<y=AX <Ay3.

4. limyri=0.Example : X = C[0,1].

X < y a b'0 < t < 1x(t) < y(t).

F is called monotone increasing iff x < yF(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

relatively compact, then there exists a fixed point i of Fsuch 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<...<yi5 yo

To show [xo, yol is convex : z1, z2 E [x0, yo], t E [0,1]

xo < z1 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 wecan 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<yn:5 yn-1F(xn-1) F(xn) <_ F(±) < F(yn) < F(yn-1)

= xn ! xn+l 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 in [0, 1]}

B2 {x E B1 :0 < x(t) < 2 in [0, 2] and

2 < x(t) < 1 in [2, 1] }

B3 = {x E.B1 : 0 < x(t) < 3 in[O, 2] and

3 < x(t) < 1 in [2,1]}.

Then f3(B3) = Z for j = 1, 2,3; and X(Bi) = 1, X(B2) = 2, and

X(B3) = 3(Fact : X(T) = 2,13(U) = 1 for the unit ball U in an infinitedimensional space. )

(hint : Since B2 C B3 C B1 then it is enough to show thatf3(B2) = 2. We want to show : /3(B2) > 2.

Assume that Ry1i..., y,n E C[0, 1], B2 C U!'_' 1B(y{, r),

148

r=z-E<r. Define

Nonlinear Functional Analysis

0 if tE[0,2--n]

Z+2(t-Z) if tE[2-n,2+

1 if tE [I+n,1]

1]

n

Let i be fixed,

Case I yi (I) >z

. Since yi is continuous, then

38 > 0 'tE [0,1] : It- 2I 

X is not in ball B(yi, r).On the otherhand

1 1CaselI yi (2) <

Ilxn -yell ? Ilxn(2 + y i ( +n2 _)II> 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,

2]: 0 < x(t) < 2, -2 < -y(t) < 0

-2 < x(t) - y(t) <_ 2(ii) If

t E [2,1] : 2< x(t) < 1 1t -1 < -y(t) < -2 -

2< x(t) - y(t) <

2

From (i) and (ii)

= Ix(t) - y(t) 1 <

2

Vt E [0, 1]

=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)<C.

Define : if 0<t<6X (t)

_1 if 6 t . 1

=xEB1.

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) < 3X(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.

Definea s3 'b

0 < t < b- -

x(t) 3 b<t<2

1+3t 2<t<1

xEb3

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" ontoits boundary, such that the boundary remains pointwise fixed,

i.e. there is no continuous f : 7(0,1) -- 8B(0,1), such thatf (x) = x for all x E 0- (0, 1). (This result is equivalent to

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 itis continuous. So, by Brower's fixed point theorem 3 an openball 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 xfor x E R(O,1), y = x + a(x)(x - f (x)), a(x) > 0. In our fixedpoint theorem, it was shown that

a(x) (x, x - f (x)) + (X, x - f (x)2 + (1 - 11 x112)(x - f (x))2

(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 thereexist a continuously differentiable function t -+ (p(t),x(t)), suchthat p : R -i R is strictly monotone increasing and for all t wehave 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 ofbread, ham and cheese. Can this be divided equitably among two

people with one slice of a knife, such that each person receivesan identical share of bread, ham and cheese?

In McDonalds dream Dr X and Dr Y appeared and general-

ized this problem to n dimensions.

If Bl,..., B are bounded, measurable subsets of R" with


n > 1. Then there is an (n --1) - dimensional hyperplane which

divides all the Bf in half.

References

References

DEIMLING, K.: " Nonlinear Functional Analy-sis", Springer, Heidelberg 1985.

DEIMLING, K.: "Nichtlineare Gleichungen andAbbildungsgrade", Springer, Heidelberg 1974.

EISENACK, G.; FENSKE, C.C.: "Fixpunkt-theorie", BI, Mannheim 1978.

KRASNOSELSKI, M.A.: "Topological Methodsin the Theory of Nonlinear Integral Equations",Pergamon, Oxford 1964.

KRASNOSELSKI, M.A. et al. : "ApproximateSolution 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 nichtlin-

eare Operatorengleichungen", Teubner, Leipzig

1976.

ZEIDLER, E. : "Nonlinear Functional Analysis

and its Applications I : Fixed - Point Theorems",

Springer, Heidelberg 1986.

153

INDEX

A

Algorithm 44Antipodal theorem

92, 99Antipodal points 96Asymptotic 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 point2, 7, 8, 53, 74,114

Bifurcation 105, 109,110, 112, 113, 115,123

Bilinear map 22, 30,31

Bijective 30Bilinear operator 31Bounded 24, 70, 72Brouwer's fixed point

55, 57, 59, 63, 64,82

Brouwer's degree 77,78,80,84,87,89,90

Borsuk 92, 96Borsji-Ulam 95

C

Contraction 1, 6, 9,48

Complete metric-space2,4, 7, 8

Cauchy sequence 3,4,43

Construction 3Compact metric space

5

Convergent 5, 39, 67Closed ball 8, 12, 14Continuous 19, 21,

26, 29, 34, 36Chain rule 23, 26,

32, 37Complex space 55Convex 60, 63, 74,

80Convex hull 60Compact operator 99

D

Derivative 17Dual space 17Diameter 65Darbo 74Deimling 78

E

Error estimate 3Enumeration 56

156

Euclidean space 57Eigen vector 81, 111,

117Eigenvalue 81, 109,

111,113, 119,120

F

Fixed point 1, 2, 6,14,55,64,75,80

Fixed point index 119Frechet derivative 17,

18, 19, 22, 23Frechet differentiable

20, 21, 23, 25, 26

G

GauB theorem 59Gateaux derivative

17, 18, 19General Topology 61

H

Halm-Banach theo-rem 57

Hedgehog theorem 83Hoelder condition 53Homeomorphism 12,

25, 49, 80, 81, 88,102, 111, 114

Homotopy 83, 89, 98,103, 116, 124

I

linear Functional Analysis

Inequality 12, 43Injective 2, 12Initial value problem

8Invertible 49, 50Implicit problem 51Isomorphic 22Isomorphism 30, 51

K

k-contraction 1, 2,10, 74

Kuratowski 65

L

Lebesgue measure 84Leray-Schauderdegree 87, 89, 90, 99,

101, 110Linear map 18, 20,

30Lipschitz continuous

1, 8, 9, 10, 28,44, 73

Lipschitz condition7

Lipschitz constant 9,10, 73

Local Homeomorphism11, 26

M

Maping 1Mean value theorem

26, 27, 31, 48, 52

Index

Measure of non-compactness65, 66, 96

Metric spaces 1Multilinear mapping

28

157

Real-valued function34, 41

Resolvant operator9, 10, 11

S

N

Neighourhood 5, 12,47, 49, 51, 53, 113,1151119t 11771241125, 126, 127

Newton iterates 42Newton's method 39,

41, 44, 53Nonlinear 7Non-linear functional

analysis 78Norm 30Normed space 1

Sadovski 74Sard's lemma 84,

94Schauder's fixed point

60, 63,74,75Sequence 9, 39Strict 1Surjective maps 83

T

Taylor formula 34Topological compo-

sition 100

UO

Ordinary DifferentialEquation 81Operator 8, 21

P

Partial derivatives 36Partial differentiable

36Picard - Lindelof 8

Proper 73

R

Unique fixed point48

Uniqueness 3

V

Volterra integral equa-tion 7, 8

W

Weakly continuous19

Weierstrass approx-imation theorem57

non-linear functional analysis

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