Nonlinear Functional Analysis
Rajendra Akerkar
Narosa Publishing House New Delhi
Madras Bombay Calcutta London
Raje...
13 downloads
242 Views
2MB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
Nonlinear Functional Analysis
Rajendra Akerkar
Narosa Publishing House New Delhi
Madras Bombay Calcutta London
Rajendra Akerkar Department of Computer Studies Chh. Shahu Central Institute of Business Education & Research University Road, Kolhapur, India
Copyright © 1999 Narosa Publishing House
NAROSA PUBLISHING HOUSE 6 Community Centre, Panchsheel Park, New Delhi 110 017 22 Daryaganj, Prakash Deep, Delhi Medical Association Road, New Delhi 110 002 3536 Greams Road, Thousand Lights, Madras 600 006 306 Shiv Centre, D.B.C. Sector 17, K.U. Bazar P.O., New Mumbai 400 705 2F2G Shivam Chambers, 53 Syed Amir Ali Avenue, Calcutta 700 019 3 Henrietta Street, Covent Garden, London WC2E 8LU, UK
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publishers. All export rights for this book vest exclusively with Narosa Publishing House. Unauthorised export is a violation of Copyright Law and is subject to legal action. ISBN
817319.2308
Published by N.K. Mehra for Narosa Publishing House, 6 Community Centre, Panchsheel Park, New Delhi 110 017 and printed at Replika Press Pvt. Ltd., Delhi 110 040 (India).
PREFACE This book presents the central ideas of applicable functional analysis in a vivid and straightforward fashion with a minimum of fuss and formality. The book was developed while teaching an upperdivision course
in nonlinear functional analysis. My intention was to give the background for the solution of nonlinear equations in Banach Spaces, and this is at least one intention of applicable functional analysis. This course is designed for a onesemester introduction at postgraduate
level. However, the material can easily be expanded to fill a two semestercourse. To clarify what I taught, I wrote down each delivered lecture. The
prerequisites for this text are basic theory on Analysis and Linear Functional Analysis. Any student with a certain amount of mathematical
maturity will be able to read the book. The material covered is more or less prerequisite for the students doing research in applicable mathematics. This text could thus be used for an M.Phil. course in the mathematics. The preparation of this manuscript was possible due to the excellent facilities available at the Technomathematics Research Foundation, Kolhapur. I thank my colleagues and friends for their comments and help. I specially thank Mrs. Achala Sabne for the excellent job of preparing the camera ready text. Most of all, I would like to express my deepest gratitude to Rupali, my wife, in whose space and time this book was written. R. AKERKAR
CONTENTS Preface
1. Contraction Banach's Fixed Point Theorem 1.2 The Resolvent Operator 1.3 The Theorem of the Local Homeomorphism 1.1
2. Differential Calculus in Banach Spaces
V
1 1
9 11
17
The Derivative
17
2.2 Higher Derivatives
28 36
2.1
2.3 Partial Derivatives
3. Newton's Method
39
4. The Implicit Function Theorem
47
5. Fixed Point Theorems
55
5.1 The Brouwer Fixed Point Theorem
5.2 The Schauder Fixed Point Theorem
6. Set Contractions and Darbo's Fixed Point Theorem Measures of Noncompactness
55 60
65
6.2 Condensing Maps
65 72
7. The Topological Degree
77
6.1
Axiomatic Definitions of the Brouwer Degree in R" 7.2 Applications of the Brouwer Degree 7.3 The LeraySchauder Degree 7.4 Borsuk's Antipodal Theorem 7.5 Compact Linear Operators 7.1
77 80 87 92 99
x
Contents
8. Bifurcation Theory 8.1
An Example
8.2 Local Bifurcation 8.3 Bifurcation and Stability 8.4 Global Bifurcation
9. Exercises and Hints References Index
105
105 110 116 123
129
153 155
Chapter 1 CONTRACTION 1.1
Banach's Fixed Point Theorem
Let (X, d), (Y, d) be metric spaces. A mapping F : X + Y is said to be Lipschitz continuous, if there exists a constant k > 0, such that for all x1, x3 E X d(F(xi), F(xs)) < k.d(xi, X2)
F is called a contraction, if for all x1i x2 E X, x1
,
x2
d(F(xi),F(x2)) < d(xi,xs). F is called a strict or a kcontraction, if F is Lipschitz continuous with a Lipschitz constant k < 1.
If X C Y, F : X  Y, then t E X is called a fixed point
of F, if flt) =.t. If an equation
H(x) = y
(1.1)
is to be solved, where H : U  X is a continuous mapping from a subset U of a normed space X into X, then this equation can be transformed in a fixed point problem : 1
Nonlinear Functional Analysis
2
Let T : X  X be an injective (linear) operator, then (1.1) is equivalent to
TH(x) = Ty x = x  TH(x) + Ty hence
x = F(x)
(1.2)
where F(x) = x  TH(x) + Ty. The (unique) fixed point x of (1.2) is a (the unique) solution of (1.1), since T Is injective. T can be chosen, such that some fixed point principles are applicable. Now we will start with the most important fixed point theorem.
Theorem 1.1 (Banach's Fixed Point Principle) Let X be a complete metric space. Let F : X + X be a kcontraction with 0 < k < 1, i.e. V x1, x2 E X d(F(xl), F(x2)) < k.d(x1, x2) Then the following hold 10 There exists a fixed point x of F. 20 30
40
This fixed point is unique.
If xo E X is arbitrarily chosen, then the sequence (xn), defined by xn = F(xn_1) converges to x.
For all n the error estimate is true d(xn, ±)
kj.d(xl, xo) j>n
Proof : This proof is analogous to the proof of Theorem 1.1.
d(xn+j+1, xn)
:5
d(xn+j+1, xn+j) +
+ d(xn+l, xn)
< d(Fn+jx1, F`+jxo) + ... + d(F`xl, Fnxo)
< (k,,+... + kn).d(x1, xo) Thus, (xn) is a Cauchy sequence. Let x = lim xn, then F(x) = lim F(xn) = lim xn±1 = i, i.e. 1 is fixed point; if 1 is a fixed point of F, so x is a fixed point for all Fn, hence
d(F`i, Fnx) < kn.d(2, x), implies x = i, since kn < 1 for almost all n and d(i, xn) < 1i m d(xn+j+1, xn) < E kj.d(xl, xo). j>n
C)
Contraction
5
If F is just a contraction, then F does not necessarily have a fixed point: Let X = 10, oo) and F : X + X be defined by
F(x) = x +
1
x+l.
F(x) = x + z+l+ # x, but
F(x)  F(y) =
i.e. 1 
(1+47
y) = 1
)2
(1 +
(x  Y)
< 1, thus, if x # y, I F(x)  F(y)I < I x  yl.
If we additionally assume that (X, d) is a compact metric space,
then we obtain the following result.
Theorem 1.3 Let X be a compact metric space, F : X > X a contraction. Then F has a unique fixed point and 1 with = 1imxn,
xn = F(xn1), x0 E X .
Proof : Since X is compact, the sequence (F(xn)) has a convergent subsequence (F(xn, )). Let
= joo lim F(xn, ), then F(2) = l lip F(xn,+1).
If i L F(x), there exist disjoint closed neighbourhoods U of x and V of F(i). The mapping P:UxV
R, P(x, y) =
F(y)) ,y) d(Fdx),
(
6
Nonlinear Functional Analysis
is continuous, and attains its maximum k < 1. Let p E N, such
that for j > p F(xn,) E U, F(xn,+l) E V. Then d(F(xn,+2), F(xn,+1)) < k.d(F(xn,+1), F(xn,)) and
d(F(xn), F(x }1 )) R, f (0) = 0, f (e, rl) _
+ 712).
77
Let x' = 0. Then (f (0 + ry)
 f(0)  x'(y)) =
T.T .r2(S2 +712)
hence lim 1 f (ry) = 0
and f (0) = 0 is the Gateaux derivative of f , but f is not continuous at 0. This definition can be extended to map between Banach spaces. Let X, Y be Banach spaces, U C X an open subset
and F:U'Yamap. Definition 2.2 1° A continuous linear map T : X  Y is the Gateaux derivative of F at x° E U if
+ ru)  F(xo)  rTuII = 0. Vu E X lim 1.IIF(xo r 20 A linear map T : X + Y is the Frechet derivative of F
atx°EU if I1lim
I ICI I
I I F(x°
+ u)  F(xo)  Tu11 = 0.
Differential Calculus in Banach Spaces 3°
19
T is the, weak Gateaux derivative of F at xo i ff
Vu E X Vx' E X' lim 1 x'(F(xo + Tu)  F(xo)  TTu) = 0. T
41
T is the weak Frechet derivative of F at xo if
Vx' E X'
lim
1
Hull 40 11uII
x'(F(xo + u)  F(xo)  Tu) = 0.
We denote by F'(xo) the derivative of F at xo. (Weakly) Frechet differentiable maps are (weakly) continuous, but there are weakly continuous maps, which are not continuous,
e.g. F. [0,11 +co
A(t)
0
F(t) = (fis(t)) F(O) = 0 _1 It '
linear and continuous elsewhere
(a) limn.., f (t) = 0, since fn (t) = 0, if n > 1 hence (fis(t)) E co. (b) I IF(0 + h)  F(O)II = maxis I fn (h) I =1, h =
,
hence
F is not continuous at 0.
(c) co = 4. Let x' = (Cn) E ll. Choose for e > 0 an integer no, such that E ISn I nO
then
1x`(F(h))I = I Ftnfn(h)1 np
The finite sum of continuous functions is continuous
hence, if IhI <  , then Ix'(F(h))I < e, and F is weakly continuous at 0.
20
Nonlinear Functional Analysis
Remark 2.1 Let F : U + Y be continuous and Frechet differentiable at xo, then F(xo) is a continuous linear map. Let > 0, b > 0 such that if llull < b < 1
IIF(xo + u)  F(xo)ll < 2 and
IIF(xo + u)  F(xo)  F'(xo)ul1 < 2.1Iu11 2. Then
IIF'(xo)ull < E if Iluli < 6.
This is the continuity of F'(xo) at the origin, hence everywhere.
Remark 2.2 If F is Frechet differentiable at xo, then there exist y > 0, 5 > 0, such that for all x E U
llx  xoll o 3b>o VXEU
IIxxoll 0 there is ,.
.,.
such that for all h E X, I IhII < b P
I I F(x + h)  E 4j F(j) (x) (h, ..., h) I I < E. I Ihl IP. 1=0
Proof : Let p = 2 and
G(x) = F(x)  F(y)  F'(y)x

F"(y)(x,x)
Then
G(x + h) = F(x + h)  F(y)  F'(y)(x + h) ZF'(y)(x + h,x + h) G'(x)h = F'(x)h  F'(y)h  F"(y)(x, h) and
G(x+h)G(x)G'(x)h = F(x+h)F(x)F'(x)h1 F(y) (h, h). By Theorem 2.8 we have 2
IIG(x + h)  G(x)  G'(x)hII
[IIT11I7(1
+ a)1]1/a = ko
we obtain
IIyn(x)  G(x)II i
n
(1)i axi Di = i=0 where a(i,j) = 1 if j i. Thus E(1)i+jCijo'(i,j),
n
n
i=0
Q=0
E(1)i. a iDi = E (1)i+1Cijcr(i,j) Interchanging the dummy indices i, j in this latter expression
and using the fact that o(i, j) _ a (j, i), we see that n
E (1)i+1Cijo'(i,j) _
i,j=0
(1)j+iC43c(j,i) i,j=0
(1)
(1)i+3Cija(i,j)
Thus all the three equal quantities in this formula must be zero, and formula (5.1) is proved.
0
Fixed Point Theorems
57
Theorem 5.1 (8rouwer) If 0 is a continuous mapping of the' closed unit sphere B = {x E X, I xI < 1) of Euclidean nspace into itself, then
there is a point y in B such that b(y) = y.
Proof : We have remarked that if suffices to consider real Euclidean space. Further, the Weierstrass approximation theorem for continuous functions of n variables implies that every continuous map ¢ of B into itself is the uniform limit of a sequence (qk) of infinitely differentiable mappings of B into itself. Suppose that the theorem were proved for infinitely differentiable maps. Then, for each integer k there is a point yk E B such that ok(yk) = yk. Since B is compact, some subsequences (ykt) converge to a point y in B. Since limi.,,o Or,, (x) = 4(x) uniformly on B,
O(y) = m Ok, (yk,) = li Yk, = Y1
This shows that it is sufficient to consider the case that 0 is infinitely differentiable.
We suppose that 0 is an infinitely differentiable map of B into itself and that O(x) x, x E B. Let a = a(x) be the larger root of the quadratic equation Ix + a(x  O(x))I2 = 1, so that
1 = (x + a(x  O(x)),x + a(x  O(x))) Ix12+2a(x,xO(x))+a21xO(x))I2.
By the quadratic formula a(x). I x  O(x) I2
= (x, O(x)  x) +{(x, x  O(x))2 +(1 Ix12)Ix O(x)12}

(5.2)
Nonlinear Functional Analysis
58
Since Ix  ¢(x)l # 0 for x E B, the discriminant (x, xO(x))2+(1 lxl2)IxO(x)i2 is positive when IxI # 1. Also if I x I = 1, then (x, x  0(x)) # 0, for otherwise (x, O(x)) = 1 and the inner product of two vectors with length at most 1 can be equal to 1 only when they are equal. Thus the discriminant is never zero for x in B. Since the function t1/2 is an infinitely differentiable function of t for t > 0, and since Ix  O(x) l , 0, x E B it follows from formula (5.2) that the function a(x) = 0 for fix( = 1 is an infinitely differentiable function of x E B. Moreover, it follows from formula (5.2), that a(x) = 0 for IxI = 1.
Now, for each real number t, put f (t; x) ; x + ta(x)(x 
0(x)). Then f is an infinitely differentiable function of the n + 1 variables t, x1, ..., x, with values in B. Since a(x) = 0 for IxI = 1, we have ft (t, x) = 0 for lxj = 1. Also f (0, x) = x , and from the definition of a we have If (1, x)l = 1 for all x E B. Denote the determinant whose columns are the vector fz, (t, x), fz,, (t, x) by Do (t, x) and consider the integral
1(t) = J Do(t, x)dx.
(5.3)
It is clear that 1(0) is the volume of B and hence 1(0) # 0. Since f(1, x) satisfies the functional dependence 1(1, x)+ = 1. It follows that the Jacobian determinant Do(1,x) is identically zero, hence I(1) = 0. The desired contradiction will be obtained
if we can show that 1(t) is a constant; i.e., that I'(t) = 0. To prove this, differentiate under the integral sign and employ (5.1) to conclude that I'(t) is a sum of integrals of the form
±f 
Di (t,
x)dx
where Di(t, x) is the determinant whose columns are the vectors 1.
f
IE,{_1(t1x),f.1j,4.3(t,x),...,fZ'(t'
).
Fixed Point Theorems
59
By the Gau$ Theorem we see that
fe t D1 (t, x)dx = f8B Di(t, x)77{dw. i9
ForxEBB,jxj =1,and fi(t,x) = a(x)(x  O(x)) implies ft (t, x) = 0 for x E 8B, hence D.(t, x) = 0. This implies
I'(t) = 0, 1 = constant, I(1) = 0 # 1(0). This contradiction shows, that there is an x E B such that O(x) = x.
0 A simple application of the Brouwer fixed point theorem is the following existence principle for systems of equations.
Proposition 5.1 Let B = B(0, r) C Rn be the closed ball with radius r and gj : B ' R be continuous mappings, j = 1, 2, ..., n. If for all x = (6, 6, ..., tn) E Rn, 11xI ( = r n
E ga (x)ei >_ 0
(5.4)
7=1
then the system of equations
g,(x) = 0, j = 1, 2, ..., n
(5.5)
has a solution i with IJuI` < r.
Proof : Let g(x) = (g1(x),...,gn(x)) and assume g(x) # 0 for all x E B. Define
1(x) 
rg(r) 119(x}11
60
Nonlinear Functional Analysis
f is a continuous map of the compact convex set B into itself. Therfore there exists a fixed point x, of f with I I±I I= I I f (x) II = r. Furthermore E9,i(x)
a=
1
=
rIjg(z)jt.ECj2
0, E 1 Ti = 1, then F,1 rjxi E A. The convex hull coB of B C X is the intersection of all convex sets A, which contain B. It is given by n
n
Tixi, xi E B, T i ! 0,
COB = {x
Ti=1,nE
Ar}.
i=1
i=1
Theorem 5.2 Let X be a Banach space, A C X a closed subset
and F : A + Y a continuous map from A into the Banach space Y. Then there exists a continuous extension F of F with
F:X*Y,FIA=F and F(x) C coF(a).
Fixed Point Theorems
61
Proof : (a) We construct a partition of unity. Let X E X \ A, Bx = B(x, rk) be an open ball with diameter 2rx less than the distance of A to Bx, e.g.
rx < 6 dist(x, A) Then X \ A = UXEx\ABx
A result of General Topology states, that there exists a locally finite open refinement {UA, A E A} of {B.,, x E X \ A}, i.e. there exists a covering of X \ A, such that
dxEX\A 3B: cp{A:UACBx} 0. Let cpa (x) =
dist(x, X \ UA) a(x).
cpa is continuous, 0 < cpa < 1, Ecp,, (x) = 1. {cpa, A E Al is a partition of unity for X \ A. (b) We now construct the extension.
VAEA3xEX\A U,\ c B,, hence dist(A, UA) > dist(A, Bx) > 2rx > 0.
For every A E A we choose as E A, such that dist(a,,, U,,) < 2dist(A, U,,).
62
Nonlinear Functional Analysis
Let
F(x) =
F(x)
,
Ecp,%F(aa)
,
if X E A if x E X \ A
Then F : x  Y is an extension of F, and F(x) C co(F(A)), since EWA(x) = 1, coa(x) > 1.
(c) F is continuous :
F is continuous for all x E 8A. Let xo ¢' 8A. Since F is continuous at xo,
VE>O 35>0 VxEA IjxxolI <S IIF(x)F(xo)1I <E. This implies
VxEX JIx  xol) < 6/4
)1F(x)F(xo)J1 <E
F(x)  F(xo) = Eco (x)(F(aa)  F(xo))
IIF'(x)  F(xo)Ij : EWA(x)jjF(aa)  F(xo)II If cp,,(x)34 0, then x E JA arid
lix  aAII < flxul(+IluaAII 0, and xl, ..., xn an e  net for K, i.e.
KCU{x,+eB}, or `dxEK 3x,llxx,ll <e. (B=B(0,1)={xEX:11xl1 0 there exists a finite covering by balls of diameter E.
3° M can be covered by M with diam M. 4°
Every cover of B2 is a cover of B1.
50
Let M1, ..., M,,, be a cover of N1, ..., N. a cover of B2, then all sets Mj + Nk form a cover of B1 + B2 B1,
and
diam(Mj + nk) < diam m; + diam Nk. 6°
Note that diam (AB) = ()Idiam B.
7°
From B C B follows X(B) < X(B). Conversely if B C UM,, then B C UMj with diam Mj = diamj, so X(B) < X(B).
8°
Let B = B1 U B2 and Q = max{X(Bi), X(B2)}. Then it follows from Bj C B that X(B) < X(B) and 0 < X(B). Conversely let for f > 0 given convergins Bj C Mjk with diam Mjk nj
M1j,nM2jznMn=o which contradicts
MnCM2CU;M2j and Mn n M1 j,
for all n.
Finally by induction, we find that for every k there is an index jk such that for all n, m
Mnn(U,ti 1Mkjk)
(*)
Let Nm = Um k=lMkjk
...andM,nCMmjm then (*) implies, that for all m Nm 54 0
diam Nn < diam M,n < X(M,n) + m hence
72
Nonlinear Fbnctional Analysis
Choose a sequence (z,,,), such that z,,, E
Then (zn) is a
Cauchy sequence, which is convergent,
Cf1M,, =M,,,,. Therefore M,,,, is nonempty. 13
Condensing Maps
6.2
A continuous map F is called bounded, if it maps bounded sets onto bounded sets, and compact, if it maps bounded sets into compact sets.
Definition 6.2 Let X be a Banach space and U C X. An operator F : U + X is called a kset contraction, 0 < k < 1, if F is continuous, maps bounded sets onto bounded sets, such that for all bounded sets B C U X(F(B))
kX(B)
F is said to be condensing, if for all bounded sets B with X(B)
0
X(F(B)) < X(B) Obviously, every kset contraction is condensing. Every compact map is a 0set contraction. The following example is an important one.
Example 6.4 Let X be a Banach space, U C X and K : U + X Lipschitz continuous with Lipschitz constant k < 1. C : U + X is compact.
Set Contractions and Darbo's Fixed Point Theorem
73
Then
,F=K+C is a kset contraction.
Proof : Let B C U be a bounded set. Then
F(B) C K(B) + C(B) X(F(B)) < X(K(B)) + X(C(B)) < kX(B).
0 A continuous map is said to be proper, if the preimage of each compact set is compact. Lemma 6.1 Let B C X be closed and bounded and F : B , X condensing. Then I  F is proper and I  F maps closed subsets of B onto closed sets.
Proof : Let K be compact and A = (I  F)'K the preimage of K under I  F. Since I  F is continuous, A is closed. From K = (I  F)A we see, A C F(A) + K, hence X(A) :5 X(F(A)) + X(K) = X(F(A)) Since F is condensing, X(A) = 0, i.e. A is compact.
Now let A be closed and (xn) C A, yn = (I  F)x,,. Let y = lim yn. The set K = {y} U {yn, n E ,H} is compact, i.e. lim x,, = x exists and belongs to A, since A is closed. Thus X E A implies
y = (I  F)x E (I  F)A,
Nonlinear Fbnctional Analysis
74
and (I  F)A is closed. 13
We are now able to prove the fixed point theorem of Darbo (1955) and Sadovski (1967) which is a common generalization of Banach's and of Schauder's fixed point theorem.
Theorem 6.1 Let C C X be a nonempty,, closed, bounded and
convex subset of a Banach space X and let F : C + C be condensing. Then F has a fixed point.
Proof : Without loss of generality we may assume, that 0 E C.
Let F be a strict kset contraction with k < 1. Define the decreasing sequence Co = C, C1 = MF(Co), Cn = oF(Cn_1). We have COQ
x(Cn) < kx(Cn1) < ... < knX(Co), hence
lim x(Cn) = 0 and
Coo = nnENCn
is compact. Furthermore, C,,o is convex and
F(CC) C C,,.
Therefore, Schauder's fixed point theorem shows, that F has a fixed point in C. C C. Now let F be condensing and let (kn) be a. sequence with
0 < kn < 1, lim kn = 1. Then, since 0EC,Fn=k,,Fmaps C into C, and has a fixed point x,,,
The Topological Degree
75
thus
xn  F(xn) = xn  knF(xn)  (1  kn)F(xn) implies
lim xn  F(xn) = 0.
The set K = {0} U {xn  F(xn)} is compact. Thus, since
I  F is proper by Lemma 6.5, (I  F)1K is compact, i.e. {xn, n E .11(} has a point of accumulation z, i.e. there exists a subsequencewith lim xn,, =.f = F(x) = lim F(xn,, ). Therefore
F has a fixed point i in C. 11
Chapter 7 THE TOPOLOGICAL DEGREE 7.1
Axiomatic Definition of the Brouwer Degree in R'ti
The Brouwer degree is a tool, which allows an answer to the question, if a given equation
f(x)=y has a solution x E 0, where 12 C Rn is open and bounded, and f : p Rn is continuous, and y does not belong to the image f (8Q) of the boundary 81 of 0. More precisely, for each admissible triple (f, ), y) we associate an integer d(f, S2, y) such that d(f, 0, y) 76 0 implies the existence of a solution x E S2 of this equation f (x) = V. This integer is uniquely defined by the following properties:
If f is the identity map, and y E Rn, then f (x) = y has a solution x E S2, iffy E 0, i.e.
(dl)
d(id, 0, y) = 77
1
for y E 0
0 for y V U
Nonlinear Functional Analysis
78
The second condition is a natural formulation of the desire that d should yield information on the location of solutions. Suppose that III and 112 are disjoint open subsets of SI and suppose, that f (x) = y has finitely many solutions in SI1 U 02i but no solution in \ (SIA U S22). Then the number of solutions in c is the sum of the numbers of solutions in 01 and SZ2, and this suggests that d should be additive in its argument c, that is
d(f, c, y) = d(f, 01, y) + d(f, ci2, y) whenever cl and SI2 are disjoint open subsets of 11, such that (d2)
Uc12))
The third and the last condition reflects the desire, that for complicated f the number d(f, St, y) can be calculated by d(g,11, y) with simpler g, atleast if f can be continuously deformed into g such that at no stage of the deformation we get solutions on the boundary. This leads to d(h(t,.), 11, y(t)) is independent of t E [0, 1] whenever h : [0,1] x SI  R" and y : [0,1] + R" are continuous (d3)
and y(t) V h(t, 80) for all t E [0, 1). In principle, it is inessential how to introduce degree theory, since there is only one IIvalued 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 11valued mapping d, which associates every admissible triple (f , 0, y), where c C R", f : R" is continuous and y E R" \ f (8Sl) an integer d(f, St, y), 0 with the following properties :
The Topological Degree
79
1. If d(f , ill y) # 0, then there exist x E N such that f (x) = Y.
2. d(id, 0, y) = 1 if y E ), d(id,1, y) = 0, if y ¢ 1. [NORMALIZATION]
S. Let h : [0,1] x  Rn be continuous and y 0 h(t, Oil) for t E [0, 1], then d(h(t,.), ii, y) is independent from t. [HOMOTOPY INVARIANCE]
4. If g
IV is continuous and I If  g I I < dist(y, f (aQ) ),
then
d(f,11,y) = d(g,il,y) 5. If z E Rn, IIy
 zII < dist(y, f(911)), then d(f,1, y) = d(f, 0, z) 
6. If U{ `_1it C ill u1 Um 1 f (Mi), then
C S7, fl is open, disjoint, y m
d(f, fl, y) _ > d(f, clt, y) t=1
[ADDITIVITY] 7. If g : S tt
Rn is continuous and f 1,9n = glen then
d(f,il,y) =d(g,Q,y) 8. If A is a closed subset of SZ, A #
and y ¢ f (A), then
d(f, il, y) = d(f, SZ \ A, y).
[EXCISION PROPERTY]
9. d(f, 0, y) = d(f (.)  y, Q, 0)
80
Nonlinear Functional Analysis
10. Let m < n, cl C R" 'be open and bounded and f continuous, y E R' \ (I  f)(8SZ). Then
d(I  f, 0, y) = d(I  f J .
m, fl
R."
n Rm, y). [REDUCTION]
7.2
Applications of the Brouwer Degree
Theorem 7.1 (Brouwer) Let D C R" be a nonempty compact convex set and let f : D + D be continuous. Then f has a fixed point. The same is true if D is only homeomorphic to a compact convex set.
Proof : Suppose first that D = Br (0). We may assume that f (x) # x
on aD. Let h(t, x) = x  t f (x). This defines a continuous h : [0, 1] x D + Rn such that 0 ¢ h([0,1] x 8D), since by assumption
Ih(t,x)l > Ixltlf(x)I > (1t)r > Din [0,1)xBDand f(x) # x for Ixj = r. Therefore d(id  f, D, 0) = d(id, B,.(0), 0) = 1, and this proves existence of an x E B,.(0) such that x  f (x) = 0. Next, let D be a general compact and convex set. By Theorem 5.4 we have a continuous extension f : R" of f, and if we look at the defining formula in the proof of this result, we see that f (R") C conv f (D) C D since m
m
wiWi(x)f(a')
is defined for m = m(x) being sufficiently large, and belongs to cony f (D). Now, we choose a ball B,.(0) 3 D, and find a fixed
The Topological Degree
81
point x off in $,.(0), by the first step. But j (x) E D and therefore x = AX) = f (x). Finally, assume that D = h(Do) with Do compact convex and h a homeomorphism. Then h` 1 f h : Do  Do has a fixed point x by the second step and therefore f (h(x)) = h(x) E D. 13
Let us illustrate this important theorem by some examples. Example 7.1 (Perron  Frobenius) Let A = (a j) be an n x n matrix such that aid > 0 for all i, j. Then there exist A > 0 and x 0 such that xi > 0 for every i and Ax = Ax. In other words, A has a nonnegative eigenvector corresponding to a nonnegative eigenvalue. To prove this result, let m
D= xER":xi>0 for all i and Exi=1 i=1
If Ax = 0 for some x E D, then there is no need to prove this result, with A = 0. If Ax j4 0 in D, then E 1(Ax)i > a in D for some a > 0. Therefore, f : x  Ax/ E 1(Ax)i is continuous in D, and f (D) C D since ail > 0 for all i, j. By Theorem 7.2 we have a fixed point of f , i.e. an xo E D such that Axo = Axo with A = Es 1(Ax0 )i
Example 7.2 Lets consider the system of ordinary differential equations u' = f (t, u), where u' = di and f : R x R" + R" is w  periodic solutions. Suppose, for simplicity, that f is continuous and that there is a ball B,.(0) such that the initial value problems
u' = f (t, u), u(0) = x E B,.(0) have a unique solution u(t; x) on [0, oo).
(7.1)
82
Nonlinear Functional Analysis
Now, let ptx = u(t; x) and suppose also that f satisfies the boundary condition (f (t, x), x) = E 1 f{(t, x)x{ < 0 for t E [0,w] and 1 x) = r. Then, we have Pt : R,.(0)  B,.(0) for every t E R+, since dt
Iu(t)12 = 2(u'(t), u(t)) = 2(f (t, u(t)), u(t)) < 0
if the solution u of equation (7.1) takes a value in 8B,. (0) at time t. Furthermore, Pt is continuous, as follows easily from our assumption that equation (7.1) has only one solution. Thus we find P, has a fixed point x,,, E Br(0), i.e. u' = f (t, u) has a solution such that u(0; x,,) = x, = u(w, x,,). Now, we may easily check that v : [0, oo)  R'&, defined by v(t) = u(t  kw, xk,) on [kw, (k + 1)w], is an w  periodic solution of equation (7.1). The map P,, is usually called the Poincare operator of u' = f (t, u), and it is now evident that u(.; x) is an w  periodic solution if x is a fixed point of P.
Example 7.3 It is impossible to retract the whole unit ball continuously onto its boundary such that the boundary remains pointwise fixed, i.e. there is no continuous f : B1(0) ' 0B1(0) such that f (x) = x for all x E 8B1(0). Otherwise g = f would have a fixed point x0, by Theorem 7.2, but this implies Ixol = 1 and therefore x0 = ,f (xo) = xo, which is nonsense. This result is in fact equivalent to Brouwer's theorem for the ball. To see this, suppose f : B1(0) + Bi (0) is continuous and has no fixed point. Let g(x) be the point where the line segment from f (x) to x hits 8B1(0), i.e. g(x) _ f (x) + t(x)(x  f (x)), where t(x) is the positive root of t2lx
 f(x)12 + 2t(f (x),x  f(x)) + If(x)12 =1.
The Topological Degree
83
Since t(x) is continuous, g would be such a retraction which does not exist by assumption.
Surjective Maps In this section we shall show that a certain growth condition
of f E C(R't) implies f (R") = R". Let us first consider that fo(x) = Ax with a positive definite matrix A. Since det A 54 0, fo is surjective. We also have (fo(x), x) > cIx12 for some c > 0 and
every x E R", and therefore (fo(x), x)/I xl  oo as fix) > oo. This condition is sufficient for surjectivity in the nonlinear case too, since we can prove the following theorem.
Theorem 7.2 Let f E C(R") be such that (fo(x), x)/Ixl  00
asIxI goo. Then f(R")=R". Proof : Given y E R", let h(t, x) = tx + (1 t) f (x)  y. At JxI = r we have
(h(t, x), x) > r[tr + (1  t)(f (x), x)/lxl  I yI ] > 0 for t E [0, 11 and r > IyI being sufficiently large. Therefore, d(f, B,.(0), y) = 1 for such an r, i.e. f (x) = y has a solution. 13
Hedgehog Theorem Up to now we have applied the homotopy invariance of the degree as it stands. However, it is also useful to use the converse namely: if two maps f and g have different degree then a certain h that connects f and g cannot be a homotopy. Along these lines we shall prove the following theorem.
Theorem 7.3 Let SZ C R" be open bounded with 0 E Q and let
f : 8f  R" \ {0} be continuous. Suppose also that the space
Nonlinear Functional Analysis
84
dimension n is odd. Then there exist x E 8f2 and A that f (x) = Ax.
0 such
Proof : Without loss of generality we may assume f E C(), by Proposition 5.1. Since n is odd, we have d(id, Cl, 0) = 1. If d(f, SZ, 0) # 1, then h(t, x) = (1  t) f (x)  tx must have a zero (to, xo) E (0, 1) x 8f1. Therefore, f (xo) = to(1 to)1xo. If, however, d(f, Cl, 0) = 1 then we apply the same argument to h(t, x) _ (1  t) f (x) + tx. 13
Since the dimension is odd in this theorem, it does not apply in Cn. In fact, the rotation by 2 of the unit sphere in C = R2, i.e. f (x1, x2) = (x2, x1), is a simple counter example. In
case Cl = B1 (0) the theorem tells us that there is at least one normal such that f changes at most its orientation. In other words: there is no continuous nonvanishing tangent vector field
on S = 8B1(0), i.e. an f : S > Rn such that f (x) # 0 and (f (x), x) = 0 on S. In particular, if n = 3 this means, that `a hedgehog cannot be combed without leaving tufts or whorls'. 1) is a nonvanishing However, f (x) = (x2, x1, ..., x2,,,,
tangent vector field on S C R1. The proof of existence and uniqueness of the Brouwer degree and its construction is based on the fact, that f (Sf (0)) is of measure zero, where S f(fl) is the set of critical points of f, i.e. Sf(ci) = {x E 11, Jf(x) = det f'(x) = 0} (Sard's lemma), and approximations of continuous functions by differentiable functions.
Proposition 7.1 Let Cl C Rn be open and f E C' (Q). Then (Sf)) = 0, where An denotes the n  dimensional Lebesgue measure.
The Topological Degree
85
Proof : We need to know here about Lebesgue measure /pn is that
µn(J) = rj 1(bi  a{) for the interval J = [a, b] C RI and that M C Rn has measure zero (i.e. µn(M) = 0) iff to every e > 0 there exist at most countably many intervals Js such that
MCU;JJarid
<e.
Then it is easy to see that at most countable union of sets of measure zero also has measure zero.
Since an open set 11 in Rn may be written as a countable union of cubes, say SZ = U1Q1, it is therefore sufficient to show An (f (Sf(Q))) = 0 for a cube Q C 11, since f (Sf(f )) = U; f (Sf(Q1)). Let p be the lateral length of Q. By the uniform continuity of f' on Q, given e > 0, we then find m E .fU such
that I f'(x) f'(Y)I < E for all x,Y E Q with IxzI < 6 = f., and therefore
I f (x)  f (y)  f'()(x 
)I
X a compact map. By .F(?7) we denote the compact mappings F : St + X such that F(SZ) is contained in a finite dimensional subspace of X. Let G = I  F and y g X \ G(&I). By Lemma 6.5 G(&) is closed, and r = dist(y, G(8 )) > 0. Now let F1 E .F(SI) be an approximation of F, such that supZEn I I F(x)  Fi (x) I I < r. Then we have
y¢(IF1)(afZ) since
dist(y, (I  F1)(af2)) = inf IIy  (.T F1(x))II MEM (IIy(xF(x)IIIIF,(x)F(x)II)>0. inf XEM
Now we choose a finite dimensional subspace X1 of X, such that y E X1, Sl fl Xi 0,0 and Fi (3'7) C X1. Then Q, = Sl fl X1 is open and bounded in X1, and by the first step we have the existence
of
The Topological Degree
89
This number is independent of the choice of Fi and X1. Let F2 E such that I IF  F21I < R and X2 C X, such that
dimX2 0, hence 0 ¢ H([0,1] x (911). By homotopy invariance of the degree
D(I  F, B(0, r), 0) = D(I  AL, b(0, r), 0) is odd, since linear opeators are odd.
0
7.5
Compact Linear Operators
One of the most useful applications of the antipodal theorem is the fact, that the degree is different from zero. We will continue these considerations and specialize to linear compact operators.
Theorem 7.8 The product formula for the Leray  Schauder degree: Let fZ C X be open bounded, F0 : N + X compact,
Co : X  X compact, F = I  Fo, G = I  Go, y ¢ GF(8) and KA, A E A the connected components of X \ F(e). Then
D(G,F,ll,y) =
D(F,S2,KA)D(G,KA,y) AE A
where only finitely many terms are non zero and
D(F, St, KA) = D(F,1l, z) for any z E KA. This product formula in the sequel we need only for linear maps and y = 0, thus we will omit the proof of Theorem 7.8. Unfortunately, for the simplest proof of the product formula we need the approximation property of X, but in spite of this loss of generalization we will prove the following results.
Nonlinear Functional Analysis
100
Proposition 7.2 Let S, T : X  X be compact linear operators, such that 1 is not an eigenvalue of S or of T. Then for
r>0 D((I  S)(I  T), B(0, r), 0) = D(I  S, B(0, r), 0).D(I  T, B(0, r), 0).
Proof : Let B = B(0, r).. Since 0 V (I  S)(cB) U (I  T) (8B), all degrees are defined, and the mappings I  S, I  T and (I  S) (I  T) are isomorphisms, hence ker(I  S) , ker(I  T ), ker(I  S) (I  T) consist only of the zero element. Let us now assume that X has the approximation property, i.e. every compact linear map is the uniform limit of compact linear maps with finite rank, we find finite rank operators So, To such that I I S SoII and IIT ToII are sufficiently small. Then


D((I  S)(I  T), B, 0) = D((I  So)(I  To)Ieo, Bo, 0) = sgn det(I  So)(I  To)IB0 = sgn det(I  So)IB0.det(I  To)IBo
= D(I  So, B, 0).D(I  To, B, 0) 11
The following result is a special case of Proposition 7.2.
Theorem 7.9 Let X = X1 ® X2 be a topological composition, and T : X + X a compact linear operator with TXj C XX, j= 1, 2. Let I  T be an isomorphism of X . Then for each open
ball B=B(0,r) inX D(I T,B,0) = D(I T1 1,BnX1i0),D(I TI2fBnX2,0).
The Topological Degree
101
Proof : Let Bj = B fl Xj, Pj : X  Xj be the linear projection onto Xi and Aj = (I  T)Pj. Then x = P1x + P2x. Let
Si=(IT)Pi+P2, S2=P1+(IT)P2. Sj is injective, since Six = 0 implies P2x = 0 and (I T)Pix = 0 but I T is injective, hence P1x = 0. Observe that Sj = I TPj,
and 1 is not an eigenvlaue of T, thus Sj is an isomorphism. I  Sj = TPj is compact, (I  Sj)(X) C Xj, by the reduction property (9) of the Leray  Schauder degree we obtain D(Sj, B1 + B2i 0)
= D(SjI (B1+B2)nx;, (Bi + B2) fl Xj, 0)
= D(Sj, Bj, 0).
S1S2 = (I TP1)(I TP2) = I T(Pi +P2)+TP1TP2 = I T since
TPiTP2 = TP1P2T = 0 implies by Proposition 7.2
D(I  T, Bi + B2, 0) = D(S1S2, Bi + B2, 0)
= D((I  TP1)lx,, B1, 0).D(I  TP2)`x2, B2, 0) = D(I T1x,,Bi,0).D(I T1x2,B2,0). 0 Theorem 7.10 Let X be a Banach space, T : X + X be linear and compact, let o (T) = {A E C : T  AI is not continuously invertible } be the spectrum of T. Then
102
Nonlinear Functional Analysis 1°
u(T)C {AEC:IAI 1, i.e. sgn p = sgn A and IµI > µi, ...,1u7,. Let
V = ®;=1N(µ1), W = lj=1R(1_ij)
IA1I, say
The Topological Degree
103
We show that X = V ® W. First of all v n w = ¢, since x E V n W implies P
x = I: xj, xj E n(µj) and x E R(µj) j=1
for j = 1, 2, ..., p. By theorem (7.10, (30)) we have
x2+x3+...+x+pER(p1) hence P
xl = x  E xj E R(Al) n N(jul) = {0} j=2
and similarly we obtain x2 = ... = x, = 0. Now, any x E X may be written as x = xj + yj with xj E n(µj), yj E R(µj) by theorem (7.10, (3°)) again, we have P
xE=xxkrxj yk1: xj ER(/Aj) j=1
j#kk
jOk
hence P
x  L xj E W = nR(pk) j=1 and
X=v w By Theorem 7.9 we have D(S, Sl, 0) = D(SIv, o n V, 0).D(SI w,11 n W, 0)
But D(SIw, S2 n w, 0) = 1 since T I w has no eigenvalue It with pA > 1 and x  tATx defines an admissible homotopy from I  AT to I. By the same theorem we have P
D(SI v, 9 n V, 0) = jj d(SIN(µi), st n N(µj), 0). j=1
Nonlinear Functional Analysis
104
Since h(t, x) = (2t 1)x  tATx is an admissible homotopy from
to II N(iz,) (this is true because (2t  1)x  tATx = 0 2t71 hence t = 2_aµ, >1 )thus and I Ix I I = 1 implies 1a = SI N(µ;)
)
D(SI N(j,i), Q n N(/j), 0) =
D(IINC,,,), Sl n N(µ), 0)
and therefore D(S, Q, 0) _
1)m(''),
where m(A) _ IAI,>i n(µ3).
If there are no such D(S, Q, 0) = 1 = (1)°.
µ
at all, then X
W and
0
Chapter 8 BIFURCATION THEORY 8.1
An Example
Let X be a Banach space, 1 C X an open bounded subset, and F : SZ + X compact. We consider problems of the following type. Assume 0 E S2 and F(0) = 0. Then for every real A the equation
Ax = F(x)
(*)
has trivial solution. The following example shows that there exist real \o that 0 < IIx,,II <E,
Pa  Aol < E
Ax,, = F(xa),
a branch of nontrivial solutions of (*) starts in the point (Ao, 0). The point (A  0, 0) E R x ) is called a bifurcation
i.e.
point. We will solve the following integral equation
Ax(s) = n
jr
sin s sin t + b sin 2s sin 2t][x(t) + x3(t)]dt (8.1) 105
Nonlinear Functional Analysis
106
which has a secondrank kernel. We suppose that 0 < b < a. Because of the form of the kernel, any solution of equation (8.1) is necessarily of the form x(s) = A sin s + B sin 2s
with undetermined constants A, B (which will turn out to be functions of the real parameter A). Substituting in equation (8.1), we have A[A sins + B sin 2s] = 2 .
+ b sin 2s sin 2t]
T.
[A sin t + B sin 2t + (A sin t + B sin 2t)3]dt
= 2a. sin s[A f sin 2 tdt + A3 f sin4 tdt 0 n o +3AB2 fo s ine t. sine 2tdt] if
+ 2 b. sin 28 [B
+B3
J
sin2 tdt + 3A2B TO sin2 2t. sin2 tdt
sin4 2tdt] 0
2a.sinsj7rA+
8A3+34AB2]
+2b. sin 2s[2B +4A2B + 8 B], where use has been made of the following values of integrals: x
10
sin3 t. sin 2tdt = T sin t. sin3 2tdt = 0.
sin t. sin 2tdt =
0
To
Equating coefficients of sin s and sin 2s, we obtain a pair of nonlinear simultaneous algebraic equations:
AA = aA + 3 aA3 + 3 aAB2 4
2
aB = bB + 2 bA2B + 3bB3. There are four kinds of solutions of equations:
1. A = B = 0; this gives the trivial solution of equation (8.1).
Bifurcation Theory
107
2. A # 0, B = 0; only the first equation is nontrivial. We cancel A # 0 to obtain A
=a+3aA2
whence
A=f
a/a 1 .
The corresponding solution of equation (8.1) is xl (s, A)
=f3
A/a  1 sins,
defined and real for A > a.
3. A = 0, B # 0; only the second equation is nontrivial. We cancel B # 0 to obtain A = b +3bB2
whence
f
B=±23 A/b1. The corresponding solution of equation (8.1) is x2 (s, 1) =
f0 A/b  1 sin 2s,
defined and real for A > b, where we recall that b < a.
4. A # 0, B # 0; here both A and B may be cancelled in equation (8.2). We obtain two ellipses:
4A2+2B2=1 (8.3)
Nonlinear Functional Analysis
108
Solutions of equation (8.3) are given by intersections of these ellipses. Solving, we get
A24.[2ab,_,]
B22ba\1
9ab
9'
ab
so that we have the following solutions of equation (8.1): (s, X3
A) = f
2a  b,\
2.
ab
3
 1 sins ±
2
2b  a.
3
ab
 1 sin 2s. (8.4)
Clearly 2a  b > 0 since we assumed that b < a. Hence the question of whether or not solutions of the form of Equation (8.4) can be real hinges upon whether or not 2b  a > 0, or
a>2 We have the following cases:
Case I : a < 2; x3(s, A) is real for no real A. x3(s, A) is real for A > max(2ab b, za ba ).
Case II : a >
Since a > b, this2; means x3(s, A) is real when A > 2aa In Case I above, i.e. when a < 1, the only real solutions of equation (8.1) are the trivial solution x(s, A)  0, and the two main branches:
xl(s, A) = f X2(5, ,1) = f
73
733
.
a  1 sin s
b  1 sin 2s
The solutions x1 and x2 branch away from the trivial solution x  0 at the eigenvalues a, b of the linearization of equation (8.1) at the origin:
Ah(s) = 7r
0
{a sin s sin t + bsin2ssin2t]h(t)dt.
(8.5)
Bifurcation Theory
109
In Case Ii, i.e. when
> 2, we again have trivial solution
x(s, A) = 0, and the two main branches
xi (s, A) = f
2 .
735
a  1 sin s
x2 (s, A) = f 2 , b  1 sin 2s which bifurcate from x
0 at the primary bifurcation points,
which are the eigenvalues a, b of linearized equation (8.5). More
over, for A > 26a > a, a third type of solution branch appears, ab namely that in equation (8.4). Note that as A  2ba' A > 2ba' the coefficients V a A  1 > 0 and s A  1 > _ . On the other hand note that a  I  26a as A  za ba . Thus
as \ ,
2b
ba, we see that x3(8, A) > X3 18, 26ba] = x1 [s, 26 ba
Therefore at A
= 2a a' the subbranch (twig)
x3 (s, A)=3
/2b_aA_ 1sinSf3
joins the main branch, i.e. x3 [s, 2bba] = xl [s, 2b. subbranch (twig)
abaAlsin2s
J while the
2 [2b aA 1 sin2s x3(s,A)23 2baA1sinsf ab 3 ab joins the negative part of the main branch, i.e., ab
ab
x3[3'2ba] x1[s'2ba We have in Case II, when a > 2, the phenomena of "secondary bifurcation", or the forming of subbranches or twigs which bifurcate from the main branches. The main branches bifurcate from the trivial solution at the eigenvalues of the linearization of equation (8.5), while the twigs bifurcate from the main branches.
Nonlinear Functional Analysis
110
Thus solutions of the nonlinear equation (8.1) exist as continuous loci in (A, sins, sin 2s) space. There are two main branches:
x1(s, A) splits off from the trivial solution x  0 at A = a, and its two parts x+, xi differ only in sign; x2(s, A) joins the trivial solution at A = b, and its two parts xz , x2 differ only in sign. a and b on the A axis are the primary bifurcation points for the main branches. If a > i.e. Case II, two subbranches or 2,
twigs split away from x1(s, A) at A = sb ba , which is known as a secondary bifurcation point. The question of whether or not secondary bifurcation of the eigensolutions of equation (8.1) takes place therefore hinges on
whether we have a > Z, or a < z. The condition a < z in this simple problem is a "condition preventing secondary bifurcation".
8.2
Local Bifurcation
We will use the Leray  Schauder degree to study bifurcation points.
Definition 8.1 Let X, Y be Banach spaces, Q C X open, 0 E 1, F : (a, b) x ci  Y continuous, such that for all A E (a, b) we have F(A, 0) = 0. The point (Ao, 0) E (a, b) x St is said to be a bifurcation point, if for all E > 0 there exist x,\ E St and A E (a, b) with 1.\  Ao I < E, 0 < I I x,\ I I < E, such that
E(A, xa) = 0.
If we assume that F can be linearized near (ao, 0), then it is easy to give necessary conditions for bifurcation in terms of the linearization.
Bifurcation Theory
111
Proposition 8.1 In the situation of Definition 8.1 let (a) F, Fz are continuous in a neighbourhood of the .bifurcation point (Ao, 0). Then FF(ao, 0) is not a homeomorphism.
(b) If X = Y, F(A, x) = x  ATx + G(A, x) with a continuous C : (a, b) x 1+ X, such that sup IIG(A,x)II AE(a,b)
0
lixll
if IlxII  0, then ) 1 belongs to the spectrum of T.
Proof : (a) If F(Ao, 0) is a homeomorphism then the implicit function Theorem 4.1 tells us, that F has a unique solution, i.e. only the trivial solution near (Ao, 0).
(b) If )b 1 ¢ o(T), then I  AoT is an isomorphism for all A close to AO and
x = (I 
AT)1G(A,
x)
0
contradicts
1 < II(I 
AT)11IIIG(,\,x)ll
+0.
11 4
0
If C = 0, i.e. if F is linear, and
is an eigenvalue of T with an eigenvector xo, then for all a E 1 the pair A(a) = Ao, x(a) = axo solves
Nonlinear Fbnctional Analysis
112
In this case (ao, 0) is called a vertical bifurcation point.
Example Let
X = Y = R2,x =
77),F(7,x) = (1  x)(
71) +
(7j3, t3) then Fx(1, 0) = 0 but F(A, x) = 0 implies
(1A)e+713=(1A)rl.3=0 and(1A)C71=714=1;4,hence C=71= 0, i.e. (1,0)ERxX is not a bifurcation point. If we represent F in the form of Proposition 8.1 (b), then F(A,x) = xAx+G(A,x) with G(,\, x) = (713,713), thus Ao = 1 is of multiplicity two.
Example Let X = Y = R2 and F(A, x) =
1
(I)a 77
1
1 )()+A(;3). 77
A0 = 1 has geometric multiplicity one, algebraic multiplicity two, but F(A, x) = 0 implies
tarl=0 71A(C+71)+A 3=0 hence
71(1 A2A+\4712) =0, therefore rl = 0 or r12 = _A4(l  A \2) , but the second solution is not close to the trivial solution, hence (1, 0) is not a bifurcation point. The following theorem will be based of the following degree
jump principle: Let
X  H(µ, x) = 0
(*)
Bifurcation Theory
113
µ E R, x E X and assume for 1 C R x X bounded and open (JP1) H : )  X is compact and H(µ, 0) = 0 for all µ (JP2) For µl < µ2 we have D(I  H(p1i .), x n SZ, 0)
D(IO  H(µ2i .), x n Q, 0).
Now if (JP1) is satisfied and (µ0i 0) is not a bifurcation point of (*), then D(I  H(µ, .), X nil, 0) is constant in a neighbourhood of µo. If (JP1) and (JP2) are satisfied, then (*) has a bifurcation point (µ, 0) with µl < µ < µ2. In the first case (µ, 0) is the only solution of (*), hence homotopy invariance yields the constancy of the degree; in the second case the jump of the degree yields the existence of a nontrivial solution (µ, x) # (µ, 0). We will now assume that Aj1 is an eigenvalue of odd algebraic multiplicity.
Theorem 8.1 Let X be a real Banach space, K : X + X be compact and linear, S2 C R x X a neighbourhood of (Ao, 0), G : 0 + X be compact and G(A, 0) = 0. Suppose also that (a) is an eigenvalue of K of odd algebraic multiplicity.
(b) There exists a continuous function cp : R  R with lim,.,o co(r) = 0 and 8 > 0, such that for all (A, x)(A, x) E S2 with is  AoI< 6,Ilxll 1, or 0, if there are no such eigenvalues. Since A5 1 is the only eigenvalue of KI N(Ao) and either A01(Ao + pi) > 1 and A 1(Ao +.U2) < 1 or vice versa, one of the degrees in equation (8.4) is + 1 while the other one is 1. Hence Go must have a zero in [Al, u2] x 8B (0, p) since Go would be an admissible homotopy otherwise. If (pa, vp) is such a zero, then (ao + A p, xp) with x,, = vo + z(tp, vp) is a nontrivial zero of F. Since µj and p may be chosen arbitrarily *close to zero, we have shown, that (\o, 0) is a bifurcation point of F(A, x) = 0. O
8.3
Bifurcation and Stability
In this section we will study the situation of Theorem 8.3 more carefully. Again let us assume, that
F(A, x) = (I  AK)x + G(A, x)
(8.1)
Bifurcation Theory
117
where K : X ' X is linear and compact, A01 is a simple eigenvalue of K, G : R x 0 is continuous differentiable and SZ is an open neighbourhood of 0, such that for all A E R (a) G(A, 0) = 0 (b) Gz(A, 0) = 0 (c) G(,\,.) is compact (d) {G(., x), x E Q} is equicontinuous
(e)o(K)\{a01}C {(EC:Re( xo. Again we have the decomposition of X = R(Ao) ® N(Ao) and equation (8.1) decomposes into
(I  AK)Px + PC(A, x) = 0
(8.2)
(I  AK) (P)x + (I  P)G(A, x) = 0.
(8.3)
If we again denote by T = I  \oK,
A= Ao+A, MA, X) =G(ao+µ,x), then by definition of P equation (8.2) reduces to

o
aao + PP(µ, x) = 0
with a = < x, xo >. If we again denote by S = [(I  AoK) IR(Ao) I]1,
the equation (8.3) becomes
(I  P)x  ASK (I  P)Z7(,u, x) = 0
Nonlinear Functional Analysis
118
and with
H((a, µ), x) = x  pSK(I  P)x + S(I  P)?7(µ, x)  aao H((a, µ), x) = 0
(8.5)
where H : R2 x SZ  X is continuously differentiable. Since Hz (0, 0) = 1, by the implicit function theorem there exist neighbourhoods of 0 in R2 and in Sl and a continuously differentiable function
(aI µ) ' x(a, µ) such that locally H((a, µ), x(a, µ)) = 0, x(0, 0) = 0. Now we additionally assume (f) There exists a continuously differentiable mapping G1
Rx)xR+X,such that for all µ,aER,xES1with
a=<x,xo> G(A, ax) = a2G1(A, x, a).
Equation (8.4) then has the form

+ a < G1(µ, x(a, µ), a), xo >= 0. o
The function
f (a, µ) =  o + a < has the property
x(a, A), a), xo >
(8.6)
Bifurcation Theory
119
By the implicit function theorem there exists a continuously differentiable function µ : (a°, a°)  It, such that for all a with Ia I < a° 0 F(a, µ(a)) = 0, µ(0) = 0
and so we obtain a solution x(a, p(a)) of equation (8.1). Now in the sequel we will assume, that for sufficiently small
al
0 we have µ'(a)
0. In this case we have three possibilities
for the behaviour of x(a, µ(a)) : 1°
µ'(a) N. In Case 30 µ(a) < Po fora54 0. If µ'(a) 0, a 0, then ao is not a vertical bifurcation point, since pc(a) j4 µ(0) = 0. Let (AO, x°) be an isolated zero of F{, and ) an open neigh
bourhood of x°, such that 1 does not contain an additional solution of F(A, x) = 0. Then D(F, Sl, 0) = i(I  F, SZ)
is called the fixed point index. We now will determine the index of the trivial solution: The linear operator I  AK does not have negative eigenvalues if A < A0, since
(I  AK)x = vx implies (1  v)x = AKx.
If v < 0, then 1' > a > LAOwould be an eigenvalue of K. Then, A since G(A, 0) = 0, we have with ) = B(0, e)
i(I  F, 12) = i(AK + G(A, .), )) = i(AK, SZ)
Nonlinear Functional Analysis
120
D(I  .1K, f, 0) = (1)m(\) = 1 since m(a) = 0 (no eigenvalues > A 1). If we choose A > A0 such
that 41 is the only eigenvalue of K in the set { E C, Re(> .1}, then i(I  F, St) = i(AK,1l) = (1)m(.\) _ 1, since m(A) = 1(71 is a simple eigenvalue). Thus the trivial solution has the index
i(AK,SZ)=1 if A< \
i(AK,S1)=1 if A A0, and the index of the nontrivial solution of absolute value 1, we have d(A) = 1 + index of nontrivial solution, if A < Ao d(A) = 1 + index of nontrivial solution, if A > a° therefore, if A < AO, then the index of the nontrivial and the index of the nontrivial solution is +1, if A > A0, and d(A) = 0.
Case 30 : d(A) = 1, since the trivial solution has the index 1, if A > A0, thus both nontrivial solutions have index 1. When we have an evolution problem governed, for example, by the differential equation
x' = f(A,x) in an appropriate space, then the results tell us something about the existence and number of equilibria, i.e. timeindependent solutions. Consider
x'(t) = f (A, x(t)) with f (A, 0) = 0
Nonlinear Functional Analysis
122
and f (ao, x) = Ax + R(x), where A is an x x n  matrix
and Rx =
Then the trivial solution x = 0 of x' = f (ao, x), x(0) = xo is said to be stable, if for every e > 0 there exists 8 > 0, such that the solution exists on R+ and 0(I IxI I)
.
satisfies I I x(t) I I < e whenever I Ixo I I < 6. The trivial solution of
x' = Ax is stable if Re o < 0 for all p E o(A) and every p, such that Re p = 0 has algebraic multiplicity equal to its geometric one. If Re p < 0 for all p E Q(A) then x = 0 is stable as a
solution of x' = f (x). If Re p > 0 for some u E Q(A) then x = 0 is an unstable solution of x' = f (x).
In our formulation the operator A is given by (I  AK), and the condition (e) in section 8.3 states that 0 is a simple eigenvalue of A and all other eigenvalues of A have negative real
part. Since 0 is an isolated point of the specturm, we find p > 0, such that
((7(I)0K)\{0})nB(0,4p) = 0 and we also find rl > 0 such that for all x E X, A E C with
IIx II 0, uniformly in A. Typical global
results about zeros of F(A, x) = x  AKx + G(A, x) will be explained f o r compact K and G. So, let A 1 be an eigenvalue of odd algebraic multiplicity of the compact K. Let
M={(A,x)EfZ:F(A,x)=0 and x#0} and C be the connected component of M containing (Ao, 0). Remember that components are closed and (A0) E M since (A0, 0) is a bifurcation point. We want to prove that C n t9 # 0 of (Al,0) E C for another characteristic value Al # Ao of K. In case S2 = R x X, C n 80 # 0 means that C is unbounded. So, let us first sketch how we arrive at a contradiction if we assume
Cn81=0, Cn(Rx{0})={(A0,0)}.
(8.1)
First of all C n 8SZ = 0 implies that C is compact, since K and G are compact. Suppose next that we are able to find an open
Nonlinear Functional Analysis
124
bounded go such that C C go C Uo C SZ and M n ago = 0. By the second part of equation (8.1) we may then assume that the intersection of ?o and real line is given by J = [A0  8, Ao + 8] with 8 > 0 so small that no other characteristic value of K satisfies IA  Aof < 28. By the homotopy invariance of the Leray  Schauder degree M n 8S1o = 0 then implies that D(A) _ D(F(A, .), SZo(.A), 0) is constant in j; remember that SZo(a) _ {x : (A, X) E SZo}. To see this let a = D(F(Ao,.), Qo(Ao), 0) and let A
= inf{A, IA  Aol < 6, D(F(A,.), Q(A), 0) 34 al
then there is an ± E 80(A), such that (A,±) EM n ago. Like in the proof of Theorem 8.1, we want to exploit the jump in the degree when A crosses ao. Hence, choose \1 and 1\2 such that 1\o  S < 1\1 < 1\0 < 1\2 < ho + b and note that
D( ) = D(F(as, ),11o(Aj) \ Bp(0), 0)  D(F(A1, ), Bp(0), 0) (8.2)
for i = 1, 2 with p > 0 sufficiently small. Since the D(F(A,.), Bp(0), 0) differ by a factor 1 and D(.\1) = D(A2), the first degrees on the righthand side of equation (8.2) must also be differ
ent. But it is easy to see that they are in fact equal to zero. Indeed, consider for example A3 > A2 so large that SZa(.\3) = 0 and
p > 0 so small that F(A, x) # 0 on Bp(0)\{0} for A E [1\2,,\o+261
and bo(a) n Bp(0) = 0 for A > \o + 26. Then the homotopy invariance for SZo \ ([A2, A3] x B p(0)) implies
D(F(A2, ), 00(A2) \ Be(0), 0) = D(F(A3i .), SZ(A3), 0) = 0.
Thus, the only problem is to find such a bounded nieghbourhood
go of C. Let us start with Ua = {(A, x) E SZ : dist ((A, x), C) < S}. Evidently, Ua n M is
Bifurcation Theory
125
compact and c fl 8U6 = 0. Note that U6 fl A? is not connected unless it equals C, since C is already a maximal connected subset of M. Of course we choose 11o = U6 if U6 n "M = C. If not then one may guess that, due to the disconnectedness of U6flM, there exist compact C1 D C and C2 D M fl 8U6 such that C1 fl C2 = 0
andU6flM= C1UC2. If this is true then diet (Cl, C2) = Q > 0 and we may choose the intersection of U6 and the ,Q/2  neighbourhood of C1 for f o.
Lemma 8.1 Let (M, d) be a compact metric space. A C M be a component and B C M closed such that A fl B = 0. Then there exist compact M1 D A and M2 J B such that M = M1 U M2 and M1 fl M2 = 0.
Proof : To use a good substitute for possibly missing pathwise connectedness, namely E  chains, let us recall that, given f > 0, two
points a E M and b E M are said to bee  chainable if there are finitely many points xl, ..., xn E M such that xl = a, xn = b and d(x;+1i xi) < E for i = 1, ..., n  1. In this case xl, ..., xn is an E  chain joining a and b. Let AE = {x E M : there exists a E A such that x and a are e chainable 1. Clearly A C AE and A, is both open and closed in M since BE(z)fl (M\AE) = 0 for z E AE, BE(z)flAE = 0 for z E M\AE.
It is therefore enough to show B fl AE = 0 for some e > 0, since then M1 = AE and M2 = M \ AE have all properties we are looking for. Suppose, on the contrary, that B fl A. # 0 for all E > 0. Consider en > 0, (an) C A and (bn) C B such that an and b are En  chainable. Since A and B are compact, we may assume an  ao E A and bn + bo E B, and therefore we have
126
Nonlinear Functional Analysis
En  chains Mn joining ao and bo, for every n > 1. Consider the limit set
Mo={xEM:x= limXnk with xnkEMnk}. Evidently, MO is compact and ao, bo E Mo. Suppose that MO is not connected. Then MO = Cl U C2 with Ci compact and dist (Up (C1), Up(C2)) > p for sufficiently small p > 0. For En < p this contradicts the obvious fact that any two cl E C1 and c2 E C2 are En  chainable. Hence, MO is connected. Consequently, Mo C A since as E MO n A and A is maximal connected, and therefore bo E A n B, a contradiction.
0 The reasoning given so far leads to a further result, which we are going to prove next.
Theorem 8.2 Let X be a real Banach space, SZ C R. x X a neighbourhood of (,\o, 0), G : St + X be completely continuous
and G(A,x) = 0(IJx`J) as x + 0, uniformly in A. Let k be linear and compact and AO a characteristic value of odd algebraic
multiplicity F(A, x) = x  AKx + G(A, x) and
M={(A,x)E1l:F(A,x)=0 and x740}. Then the component C of M, containing (ao, 0), has at least one of the following properties: (a) C n aQj4 0; (b) C contains an odd number of trivial zeros (At, 0) # (ao, 0), where \1 is a characteristic value of K of odd algebraic multiplicity.
Exercises
127
Proof : Suppose that C n CQ = 0. Then we already know that C is compact and contains another (A, 0) with A A0. Clearly, a bounded neighbourhood S1o of C satisfying Mn8S2o = 0 contains only a finite number of points (Ak, 0) with Ak 1 E u(K), say Al < ... < A2_1 < AO < Ai+1 < ... < A,,. We may assume that ?Ion (Rx {0}) = Uk=l[AkS, Ak+b] with b > 0 sufficiently small. Choosing Akl and Ak2 such that Ak6 < Akl < Ak < Ak2 < Ak+b, we have D(F(A,.), S2o(A), 0) = m on [A1  6, Ap + 6] for some m E fl.
m = D(F(Ak;, ),1lo(Akj), 0) = dki + D(F(Aki, ), Bp (0),) for j = 1, 2.and p > 0 sufficiently small, where
dk; = D(F(Ak,, .), 1l (Aki) \ Bp(0), 0). Furthermore d11 = 0 = dp2 and dk2 = dk+l,l. Hence p1
p
E dk+1,1 + L D(F(Akl, ), Bp(0), 0) k=1
k=1 P1
p
_ >2dk2 + k=1
k=1
and therefore P
E[D(F(Ak2, ), Bp(0), 0) k=1
 D(F(Akl, ), Bp(0), 0)] = 0
This evidently implies that we have an even number of jumps in the degree. Since we have one at A0, and since the jumps occur only at characteristic values of odd algebraic multiplicity, the theorem is proved. O
Chapter 9 EXERCISES AND HINTS Every exercise is not answered here. Readers will learn
best if they make a serious attempt to find their own answers before peeking at these hints. Let X be a Banach space. A : X + X is linear and continuous, b E X. For every A in the spectrum of A holds 1.
JAI < 1. Let F : X s X be defined by
F(x) = Ax  b.
Then there exists a unique x E X, F(i) = i . 1 = lim x,,, x = F(xi1),xo E X. If A has an eigenvalue A with JAI > 1, then
does not converge for every xo E X.
(hint : (a) Proof :
F(x) = Ax  b F2 (x) = A2x  Ab  b F3(x) = A3x  Alb  b
F'(x) = A"xA"lb...Abb 129
Nonlinear Functional Analysis
130
IIFn(x)  Fn(y)Il = IIAnx  AnyII 0, Cn is decreasing sequence.
2limCn=C>0 C 0 and 2(77k), (mk) s.t. 1. mk+1 > nk+1 > mk > 7tk
Exercises
131
2. d(F'"k+lx, F"kx) < e
.
3. d(Fkx,F"kx)>E. 4. d(F ,t1x, F"kx) < e
.
So,
e < d(Fmkx, F"kx)
< d(Fm x, Fmklx) +d(F'"klx,
F"kx)
G Cmk1 + e
limd(F'"kx,F"kx) = E. < d(F'kx, Fmk+lx)
d(F'"kx, F'nkx)
+d(F"k+lx, F"k+lx)
+d(F"k+lx, F"kx) < Cmk + iP(d(Fmkx, r `kx)) + C"k.
Taking limit, we get
E
=0
a contradiction
therefore (F"x) is Cauchy sequence. Since X is complete, 3X = lim F"x, and
F(X) = lim F'+1(X) = X X is a fixed point. Now it remains to show X is unique. Let X, X be two fixed points. Then
d(X,X) = d(F(X),F(X))
< ii(d(X,X)) O,p > 0 and n + a = 1, then co is continuously differentiable on Lp[a, b].
7. Definition A set A is called a Banach algebra, if A is a Banach apace, and if  there is defined an associative distributive continuous multiplication of elements of A, i.e.
if X, y E A, then x.y E A and I I < IIxI.I IyII Let A be a Banach algebra, and U C A an open subset. Let f, g : U + A be Frechet differentiable.
(a) Then h : U  A, h(x) = f (x).g(x) is Ftechet differentiable. Determine h'(x). (b) Let (jA = {x E A, x1 exists, x1 E A}. Determine the first and second derivative of f'(x) = x1. Show
that A is open in A (hint : : A Banach algebra, U 0 s.t. =
Ilg(x + k)  g(x)
(Ikll = I I (E + 00
n=0 00
= II
E(x1h)n.x'II
n=2 00
=
E(x1h)n21I
lI(x'h)2
n=o
< IIx11131 IhI12
00
0(IIx'IIIIhII)n
n=0
= IIx'1131IhI121
 11x1'Illlhll `
211x'11311h112
E > 0 given, we can choose 6 > 0 s.t.
b = min f
1
211x 'II'

Il.f(x + h)  f(x) + Now
,
211xE'113
xlhx11I = 1 is satisfied. An important trick is to change' this linear problem into a nonlinear one of the form T z = 0, where
z = (X,,\) and T z = (Ax  \Bx, < x, x > 1) . Show that the F  derivatives are T'((X, \)) (y,,u) = (Ay  pBx  \By, 2 < x, y >).
(hint :
Ax = ABx, where A, B E Mat(n, n; R), < x, x >= 1,x E W. And T (X, A) _ (Ax  ABx, < x, x > 1), (X, A) E R" X R_ = X
Let T : X , X. So, F  derivatives are
T'((X, A))(y, µ) = (Ay  pBx  ABy, 2 < x, y >)
T(x,A) _ (0,1)+(Ax,0)+(ABx, < x,x >), TO
T1
T2(x,A)
where To = constant and T1 = linear operator, where T2 (tx, tA) _
t2T2(X, A) = T2 is a homopol on X  into X = T2 is generated by a symmetric bilinear map M, i.e. by M E ML(X, X ; X) symmetric. So, M((xl, A1), (X2,,\2)) = 2 (A1Bx2  A2Bx2i 2 < xi, x2 >) Also,
M((x, A), (x, A)) = T2(x, A).
Nonlinear Functional Analysis
138
So we have To (constant) differentiable Ti (linear operator) is differentiable and T2 is differentiable. Therefore
T'(x,A)(y,µ) = To+Tl +TZ = 0 + (Ay, O) + 2M((x, x), (y,,u))
_ (Ay, 0) + (,By  µBx, 2 < x, y >) = (Ay  ABy  ,uBx, 2 < x, y > ) Also,
T"(x, A)(yi, 111)(Y24 12) = 0 + 2M((yi, al), (y2, 1\2))
since (Ay, 0) is continuous with respect to x = (µ, Bye µ2By1, 2 < yl, y2 >) this is continuous with respect to x
=T(") (x,\) =0 for n> 2.) 9. : Let F : C[0,,7r]  C[O, 7r] be defined by
F(x)(t) =
217' k(t, r) (x(r) + x3(r))dT
where, k (t, r) = a sin t sin r + b sin 2t sin 2T, 0 < b < a. Discuss the set of all solutions of
H(a, x) = Ax  F(x) = 0, where A is a real parameter.
(hint :
Ax(t) = F(x)(t) = 2 a sin t To sin r(x(T) + x3(r))dr it
o
+. b sin 2t
x
sin 2T(x(r) + x3 (r) )dr.
J If x is a solution then x should be linear combination as x(t) = A sin t + B sin 2t. AA sin t + AB sin 2t
Exercises
139
= 2a sin t
sinr(Asinr+Bsin 2r) TO
+(Asinr + Bsin2T)3)dr
+2bsin2tJ sin 2T(Asinr+Bsin 2T)+(Asin T+Bsin2T)3 0
7r
T2
_ (2asint)(I,) + ( Ir
2
b sin 2t) (12)
7r
I1=2A+ $A3+34AB2 I2 = 2 B + 8 B3 + 4 A2B. Since sin t and sin 2t are linearly independent in C(0, 7r], we can equate the coefficients of sin t and sin 2t to get,
AA=aA(1+3!A2+ZB2)
(i)
AB = bB(1 + 34B2 + 2A2)
(ii)
Case 1 Case 2
A = B = 0 x(t) = 0 is a solution for every A E R. A = 0, B j40. from (ii) A = b(1 + 4B2)
B
731
Vb ERe*a>b
= the solution
X2 (t) = f f Fb
sin 2t for A
 b.
Case 3 A # 0, B = 0 = equation (ii) fulfilled everywhere. Then from (i)
Nonlinear Functional Analysis
140
A
3
x4 a (t)
2
=f
b  1 sin t.
From theorem of local homeomorphism F : X + Y cont. diff. F'(xo) is invertible. 3 nbd u(xo) s.t. F is a homeomorphism on u(xo). We have
H(A, x)(t) _ (Ax  F(x))(t)
= A(x(t))  2
2J 7r
J
k(t,T)xtaudr
x(t,,r)x3(T)dT
0
H,' (A, x) exists and cont. in x
HH(A, x)h(t) _ Ah(t) 
f k(t,T)h(T)dT
j
0 3.k(t,T)x2(T)h(T)dT 7r
H,' (A, 0)h(t) = Ah(t)  2 J k(t,T)h(T)dr.
Since k(t,T) = asintsinr +bsin2tsin2T, sin t, sin 2t E L2 [0, 7r], I I sin tI 12 = I ( sin 2t 112 = 2
Let
sin t e1=TI
' e2
sin 2t
V2
2 [ k(t, T)h(T)dT = a(ei, h)el + b(e2, h)e2 a and b are the eigenvalue of this integral operator
Exercises
141
H(,\, 0) is invertible « a 3& A # b. So, if a # A # b then from local homeomorphism theorem = 3u,, a neighbourhood of 0 in C[O, 7r] * H(A, 0) is a homeomorphism on ua since H(A, 0) = 0  0 is the unique solution in U,,.
Case 4
A#0,B34 0
A=a(1+3A2+3B2)
A=b(1+3 B2+2A2)
=f3 ba )1 A
B=f3 Since 0 1sinceA>O*21>Os aaba>a. Ifa bathen xs>7,s,9('\)(t) = f3
'\(a 3
t(b
a )  1 sint
b)  1 sin 2t.)
10.: Let X be a complete normed space, and x : [0,1]  X be continuous. Show that the Riemann integral i
x(t)dt J is well defined, linear and that 11
Jo'
x(t)dtll
j'IIx(t)(Jdt.
142
11.
Nonlinear Functional Analysis
Let F : Rn , 1Zn be continuously differentiable and let F'(x) # 0 on Rn. Then F is a homeomorphism onto Rn if and only if :
lim IIF(x)II = oo.
11T100
(hint : F : Rn  R" continuously differentiable and det (F'(x)) j4 0 on R". Then F is a homeomorphism onto Rn if limlI.T11
.
IIF(x)II = oo.
" ," Assume 2(xn) in Rn with IIxnII  oo and IIF(xn)II 74 M. = 3(xnk) with I IF(xnk) I I
oo, which contradicts to F(xn)  Y3(x,,,,,) = F(lim F(x) = y E F(R").
Since F(Rn) is open (from (b) and also F(Rn) is closed F(Rn) = 0 or Rn, here it is not empty. So F(Rn) = IV =:,. F is surjective.
(d) F is injective. )
Let (afl) be a real (n x n) matrix with aj > 0 for all 12. i, j. then A possesses a nonnegative eigenvalue. The associated :
eigenvector can be chosen, such that all coordinates are non
Exercises
143
negative. If additionally E 1 a{ f > 0 for all j, then A possesses a positive eigenvalue.
(PERRON  FROBENIUS).
(hint : a+j E Mat(n, n, R), ai f > 0 V i, j then A possesses nonnegative eigenvalue, Ax = Ax, x # 0,
x > 0. If in addition E
1
at1 > OVj then .A > 0.
Proof: K=Ix ER":x>0,E.1x{=1}. Then k is bounded, closed
k is compact. Case 1 : Ax = 0 for some x E k 0 is an eigenvalue of A and X is a eigenvector.
Case 2 : Ax # 0 d x E k f (x)
E{ 1(Ax) j# 0 we define
Ax),' f : k  k is continuous )
,=1(
13. : BROUWER's fixed point theorem is equivalent to the following theorem of POINCARE (1886) and BOHL (1904): Let f : R" * Rn be a continuous mapping and suppose
3r>0 VA>0 VxER" (I IxI I = r =f(x) +Ax 54 0). Then there exists a point xo, IIxoIi < r, such that f (xo) = 0. 14. Construction of Counter Examples to the Brouwer's fixed point theorem. (a) Construct U C R, which is compact, and a continuous f : U + U without fixed points.
(b) Construct a convex bounded U C R, f : U  U
,
is
continuous, without fixed points. (c) Find f : [0, 1] + [0,1] without fixed points.
(hint : (a) Compact U = {0,1 }, f (0) = 1, f (1) = 0 = f is continuous without fixed point (U is also a compact).
144
Nonlinear Fbnctional Analysis
(b) U = (0, 1), f : U + U by f (X) = 1X without fixed point.
f is continuous
(c) f : 10, 1] + [0, 1]
f(X) =
0
f
1
for 1<X0:`dtE [0,1] n;(t)