Advirory Editor C. S. Seshadri. Chennai Mathematical Inst., Chennai.

Managlng Edltor Rajendra Bhatia, Indian Statistical Inst., New Delhi. Nonlinear Functional Analysis

A First Course Edltors V. S. Borkiir. Tata Inst. oi Fundamental Rescarch, Mumhai. ProhoI Cliaudliuri, Iiidian Statistical Lnst., Kolkata. R. L. Karaiidikor, lndian Statistical Inst.. New Delhi. M. Ram Murty, Queeii's University, Kingeton. C. Miisili, Vignan School of Sciences, Hydrrabad. V. S. Sundcr. Inst. of Matliematical Sciences, Chennai. M. Vaiiiiinathan, TIFR Centre, Bangalore. T . N. Vciikatarumaiia, Tara Inst. oiFundamental Research. Mumhai.

Already Publirhed Volumes R. B. Bapat: Linear Algebra and Linear Models (Second Edition) Rajendra Bhatia: Fourier Series ( Second Edition) C. Musili: Representationi of Fiiiite Groups Institute of Mathematical Sciences H. Helson: Linear Algebra (Second Edition) D. Sarason: Notes ori Complrx Function Theory M. C. Nadkarni: Basic Ergodic Theory (Second Edition) H. Hclai>n: Hciriniinir Analysis (Second Edition) K , Cliiiiiilriirrkli;ii.iin: A Courie oii Iiitegration Theory K. C l i i i i i i l r i i i i i ~ l < l ~ ~ ~ r i ~ ~ ~ : A Ciiursc on Topological Gruups I{. 131i;iil~i (cil.): Aii;ilyais,

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ISRN 81.85931.46.1

Preface Nonlinear Functional Analysis studies the properties of (con-

l,iiiuous) inappings between iiorrried liiiex spaces and evolves meth- 011s to solve nonlinear equations involving such mappings. Two ~llajor approaches to the solution of riuiiliiiear equations could bc

vi

~)ilit theorems (in particular, Schauder's theorem) are proved and l L c:Ortaili 'Ia& of compactness' lias beeii built. Another instante applications are given. i,, l,lie theory of r - convergen,ce. This t,heory studies the conver-

gc!llce of the miiiima and miiiiiiiizers of a family of f~nct,ionals. Tlie foiirtli cliapter deals witli bifurcation tlieory. ~ h i ~ studies ARaiii, while t,he theory can be developed in t,he very general coll-

I.110 nt~biirc: cif the set of solut,iotis to equations dependent, on a pa- ~,(:xt of a topological space, alot of t,eclinical result,~ in Sobolev rn~ii(!l.(!r, iii tlir neighbourhood of a 't,rivial solution'. Science and ~llaces are needcd in order to preseiit reasonably interest,i% re- e!iigiiic(!riii~: are fui1 of iustatices of such problems. A variety of slllts. The applicat,ions of this t,heory me inyriad, rangi% from inolhcidti for the identificatioti of bifurcation point,~ - topological Iioiilinear elasticity to liomogeniz;it,ioii tlieory. Sueli topies, i11 rny i~iid v>~ri;~t,ional - are preseiited. (q)inioii, would be ideal for a seque1 t,o t,his volui~ie, meant for

;hdva,nced course on iionlinear analysis, specifically aimed at st,u- TIi(! c:ncluding chapter deals with t,he exist,ence alld inulti. rl(:nts working in applications of iiiat~1ieiiiatic.s.

1'1il:iby of criticai poiiits of fuiictioiials defined on Bana& spaces. Wiiiic iiiiiiimieatioti is one niethod, other critical points,'likc sad. The mat,erial presented here is classical and no claim is made cllc poitits are found by usiiig results like the mountain pass thee ~,()wal.ds oryinality of presentat,ioii (except for some of mY own icm, or, more generally, what are knowti as min - max theorems, work included in Chapter 4). My treat,ment of t,he subject has

Ijeeii greatly iiifluenced by the works of Cartan 141: Deimling [7], Nonlinear Analysis, today, has a bewildering a,rray of tools. In Kaviaii [ll], Nirenberg [19] aiid Rabinowitz [20].

~rlcrting thr nhove t,opics, a conscioiis choice has been made with I.iir hiinwiiip; ol~joc:t,ivr~ in mind:

~ h i s book grew out of the notes l~repared for courses that I Lo ~~t'ovi(l(! i i I;t!xt Ijook wliirh can he used for an introductory gwe on various ocasions to doetaoral students at the TIFR Cen-

l,re, Baiigalore, India (where I worked earlier), tlie Dipartimento Olll! - Hl!ill(!Hk!r i:oll~H(: i:ovcriiig clas~i(:al material; (1; Matematica G. Catelnuovo, Universiti degli Stiidi di Roma

I,() I)(! (ir iiil,(!rost 1.0 ;L gcn,er.al stiidcilt, of liiglicr mathematics, u ~ a sapienzan, Rorne, Italy and the Laboratoire MMAS, Univer- Sitk de Met,z, Met,z, Rance. I would like t,o take this oppurtunity

'i'lic! i!~aiii]>l(!~ aiid exercises tltat are found throiigllout the text t.O thank these itist itutions for their facilities and hospitality. linv(! I)ccii clios~ii to be in tune wit,h these objectives (t,hollgh, from

hitii(! time, my own bias towards differential equatioiis does &ow 1 1 1 ~ ) I like to thank the Insbitute of Matliematical Sciences,

(:llcnnai, India, for its exellent facilities and research eiiviron- Ib ror (ilii~ renson t,hat some of t,he t,ools developed more lli~:iit, wliich perrnitted me to bring out this book. I also t,hank

(!HIII~IY ~ I ~ L V I ! lit?

viii

V. S. Borkar, who egg;ed me on to give the lecture at ~~~~~l~~~ in the first lace anrl kept insisting that I publisli the notes, I alsO wish to thank one of niy summer students, Mr. hivanand Dwivedi, who. while learning the material from the lIianuscript,

did valuable proof reading. Finally, for numerous personal reasOns, I thank the members of my faniily and fondly dedicate this book to them.

Chennai. S. Kesavan October, 2003.

I

Contents

1 Differential Calculus on Normed Linear Spaces . . . . . . . . . . . . . . . . 1.1 The Fr.cliet Derivative

. . . . . . . . . . . . . . 1.2 Higher Order Derivatives . . . . . . . . . . . . . 1.3 Sonie Important Theorems

. . . . . . . . . 1.4 Extrema of Real Valued Functions

2 The Brouwer Degree . . . . . . . . . . . . . . . 2.1 Definition of the Degree . . . . . . . . . . . . . . . 2.2 Properties of the Degree

. . . . . . . . 2.3 Brouwer's Theorein and Applicatious 2.4 Borsuk's Theorem . . . . . . . . . . . . . . . . . . 2.5 The Genus . . . . . . . . . . . . . . . . . . . . . .

3 The Leray - Schauder Degree . . . . . . . . . . . . . . . . . . . . . 3.1 Prelimiiiaries

. . . . . . . . . . . . . . . 3.2 Definitiou of the Degree

. . . . . . . . . . . . . . . 3.3 Properties of tlie Degree . . . . . . . . . . . . . . . . :).4 Fixed Poirit Theorems

:3.5 The Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 An Application to Differential Eqiiations

4 Bifurcation Theory 4.1 Introduction. . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . 4.2 The Lyapunov - Sclmidt Rlethod . . . . . . . . . . . . . . . . . . . . 4,:l hforse's Lemnia

. . . . . . . . . . . . . . . 4.4 A Perturbation Method

4.5 Krasnoselsk'ii's Theorern . . . . . . . . . . . . . . . i 11 4.6 Rabinowitz' Theorclll . . . . . . . . . . . . . . . . 113 4.7 A Vari&t.ioiial Method . . . . . . . . . . . . . . . . 117

5 Criticai Points of Functionals 125 5.1 Miiiiniization of fiilctionals . . . . . . . . . . . . . 125 5.2 Sad(l1ePoiiits . . . . . . . . . . . . . . . . . . . . . 132 5.3 Tlie Palais - Sindc Condition . . . . . . . . . . . . 137 5.4 The Deformation Lemnia . . . . . . . . . . . . . . 144 5.5 The Mountain Pass Theorern . . . . . . . . . . . . 150 5 Multiplicity of Critica1 Poiiits . . . . . . . . . . . . 155 5.7 Critica1 Points with Constraints . . . . . . . . . . . 159

Bibliography 171

Index 175

Chapter 1

Differential Calculus on Normed Linear Spaces

1.1 The fichet Derivative

In this chapter we will review sorne of the important results of the differential calculus on norined linear spaces.

Given a function f : R + R we know what is meant by its derivative (if it esists) at a point a E R. It is a iiuinber denoted by f l (n) (or D f (a) or %(a)) such tliat

lim f (a + h) - f (a) = f l ( a ) (1.1.1) h i O h,

or, equivalently,

I f (a + h,) - f (a) - f1(a)h.l = ~ ( h ) (1.1.2) where, by the syrnbol o(h,) we understand that the right-hand side is equal to a function ~(11) such that

i O as h,l i 0. (1.1.3) Ihl

l i we wish to generalize this notion of the derivative to a function ilitfined in an open set of R? or, rnore generally, to a fuiiction

, tli!iiiied in an open set of a ilornled linear space E and taking vrrliicis in another normed linear space F , it will be convenient

2 1.1 The Bch,et Derivative CH.l DIFFERENTIAL CALCULUS

to regard f l ( a ) h as tlie result of a linear operation on h. Thus, that point. . f ' ( a ) is now considered as a bounded linear operator on R which

~~~~~l~ 1.1.1 Let E = R2 arid F = R. Define satisfies (1.1.2). We now define the notion of differentiability for functiorix defined on a normrd linear space.

if ( z , y) # (0,O) Let E and F be normed linear spaces (over R).We denote by f ( x , y ) = if ( T , y) = (0,O). L ( E , F ) thr space of bounded linear transformations of E into F. L

Deflnition 1.1.1 Let U C E be a n open set and let f : U -i F be a gitien function. The ftinction f is said to be differentiable nt ( I E U if there ezists a bounded linear transforrnation f l (a) E L ( E , F ) svch that

I l f (a+ h ) - ) ( a ) - f1(a)hIl = o(Ilhl1). (1.1.4) Equil~alently, we can write

f ( a + h ) - f ( a ) - f l ( a ) h =&(h) (1.1.5) where $j _t 0 as l(h/l -t O.. R e m a r k 1.1.1 The following facts are simple consequences of t,lic: i~l>e)vo de:fiiiit,ioii: (i)if f is differentiable at a E U , then f is (:oi~I;iiit~oti~ i t t t h t poilit; (ii)if f is differentiable at a E U , then t.lio ilorivat,ivi: f t ( ( l ) f ( E , F ) is uniqiiely definrd. It is for the iiiii

4 CH. 1 DIFFERENTL~L CALCVLUS 1.1 Th,e Frch,et Derivalalie 5

where (.,.I staiids for tlie iiiiier-product iIi E. ~h~~~ I.CRUlt, < ' ~ ~ t i l l ~ o ~ s from L'($)) into itself (since, $2 beiiig bounded, 1 Lm(n) is contained in L2(R)). Then, N is also differentiable and f (X + h) - J(z) = a(x, h) + -a(h, h) - (b,h). 2 i i h E L Z ( ~ ) > then the fuiictioii ~ ' ( u ) h E L ~ ( R ) is given by

Sincc a(., .) is contiuuous. a f N1(u)(h)(x) = %(xl u ( x ) ) h ( ~ ) . !o(h,h)l 5 Mllhl12. TO see this, notice that by the U(X) at + Q(~)h(x) )h(x) .

1 Hciice, deiioting the norm in L 2 ( 0 ) by i/.\\. we get Example 1.1.4 (The Nemylskii Operator) Let R C RN he a doriiain and let f : R x R + R be a givcn fuiictioii such tliat the riiapping x H J(z: t) is ineltsural~le for a11 fixed t E R and the iiiapping t H f (x, t) is continuous for alinost a11 x E R. Such a fiinc:tion is called a Carath,odory Juractiola. Let W be a vector sl>fi(:(: OS roa1 - valiicd functioiis ou fl. The Nem.ytskii operator ~I.HHO(:~RI.I!(I to f is a iioiiliiicar mapping defiiicd o11 W by

h ii:iiiarkal~l(: t,lieoi.(:rii, diic t.o Krasiioselsk'ii [14] (see also Joshi aiid Bose 1101 for a proof), is tliat if ( l l p ) + ( l l q ) = l1 where 1 5 ~ , q 5 w, aiid if IV maps LP(C1) into Lq(R), then this inapping is c:~iitiniious aiid bounded, P.e. it maps bounded sets int bouiided HI!I;H. A typical conditioii on f would be a growtli coiiditiun of the I,yl>I f (x. t)i < a ( z ) + btlpIq wli(:rc! (1. in :L iioii-iiegative function in Lq(R) and b is a positive (!O~Int~l~ilI~.

L(!O 6 2 i>(: ib 1)oiiiidcd domain and let p = q = 2. Assume that, iii >~iltlil;ioii, j i$ i i ~ $1 x IR and that %(,c: .(,c)) is in Lm(R) if 2 1 E L ) Tli i i~, I,li(: iiiapl)iiig- o H %(., u ( . ) ) isl by Kras~ioselsk'ii's

I 13." t,hn coiit,iiluiry of the Nemytskii operstor associa,tcd to %, it follows that t~he terni on the right teiids to zero as h -r O in L ( Q ) aiid tliis proves our clnim. . Exarnple 1.1.5 Lct fl C RN be a hounded domaiii and let f : 62 x R -+ R b r functioii as iii the preceding exaniple. Let H; (fl) he t,he usual Soholev spa

6 CH.1 DIFFERENTIAL CALCUL US 1 .1 The Frch,et Der'ivatiue

We claim that T is differentiable and that T i ( u ) h = Z. Indeed, if differentiable at the origin. v = T(.u + h ) , theii, C = 11 - ,w - z vanishes on 3a and satisfies

~h~ derivative follows the usual rules of the calciilus. F~~ instante, if f and g are two functions which are differentiable at a point a and if we define f + g and Af (for A E R) b~

in a. By standard estimates, we know that J I C J J L ~ ( n ) is bourided by the norm in ~ ~ ( a ) of the expression in the right-hand side, whicli, iii turn, is of the order 0 ( l l h l l ~ 2 ( ~ ) ) as seen in the precediiig example. This establishes our claim. 1

( f + g) ' (a ) = f l ( a ) + gl(a) , ( A f ) '(a) = Af ' (a) Exercise 1.1.1 Let a C IRn, be a bounded domain, where 5 3. as can he easily yee~i. Another important rule relates to the deriva- Then it is known that (cf. Kesavan [13]) H, '(a) c P ( a ) , with tive of the coinposition of two differentiable functions. c0lltiilu0us inclusion, if 1 _< p 5 6. Let f E L 2 ( a ) be given, Show that the functional J defined for v E H; (a ) by Propoition 1.1.1 Let E , F and G be norm.ed linear spaces, U

an open set in E an,d V an open set in F . Let f : U 4 F,.q : V G such th,at for a giuen point a E U , we haue f ( a ) = b E V . On the open set U' = f - ' (V), wh,ich contains a , define

is differentiable and that h = g o f : U 1 - i G .

~f f is diflerentirjble at a and g at b, thr71. h i s diflcrentiahlr at a whcre < ., . > denotes the duality bracket between H,'(Q) and its h l ( a ) = g ' ( f ( a ) ) o f l ( a ) . dual (dciiotcd by H - ] (a)) . .

Proof: We have Exercise 1.1.2 Let M ( n , R) denote the space of aii 7~ x 7~ matrices with real entries. Let G L ( n , R) be the set of a11 invertible iriatrices f ( z ) = f ( a ) + f f ( a ) ( z - a ) + 4% - a ) , in M ( n , R ) . g ( y ) = g ( b ) + g l ( b ) ( y - b ) + r ) ( : ~ - b ) , (i) Show that G L ( n , R) is an open set in M ( n , R ) (provided with

wticre E ( X - a ) = o((Ix - a ( ( ) and q ( y - b) = ~ ( ( I Y - h( \ ) . Now, tlir tisiia1 topology of R"'). (ii) Sliow that the mappiiig f : G L ( n , R) i M ( n , R) defiiied by h ( x ) - h ( a ) = r ( f ( z ) ) - ~ ( f (a ) ) f (A) = A-' is differentiable and that = g l ( f ( a ) ) ( f ( z ) - f ( n ) ) + r / ( f ( x ) - f ( a ) ) .

~ ' ( A ) H = -A-'HA-'.. Exercise 1.1.3 Let E be a nornied linear space. Show that the map f : E i R defined by f ( z ) = Jlz(l for a11 z E E is never

8 CH.l DIFFERENTIAL CALCULUS 1.1 The Bchet Derivative

But Proof: If f is differeiitiahle, so is f,, since it is the compositiou i d f and a cont,inuous linear map (which is always differentiahle;

l l d ( f ( a ) ) ~ ( z - a ) l l 5 l1g1(f ( a ) ) ~ i ~ ~ ~ ( x - n ) ~ ~ = o ( ~ ~ x - a ~ ~ ) . (1.1.9) 1 1 f1(u)lI , then, for ) x -ali sinal1 enough, f:(a) = Pi f l ( a ) .

I l f ( x ) - f (a) l l I MIIz - all. li, conversely, fi is differentiahle for each i, we get, from (1.1.11), aiid so l I f ( 2 ) - f (a)ll + O as Ilx - a11 i O. Tlius rn

11~1( f (x ) - f (a ) ) l l < l / 7 ) ( f ( z ) - f (a))ll + O as I l z - i I l f ( x ) - f (a)l l llz - all

and again, as ui is a linear map, it is differentiable and (1.1.12) wliicli proves that Cllows.This coiripletes the proof. .

I I v ( f ( x ) - f (a))ll = ~ ( l l ~ - ali). Let us now consider the case where E is the product of nortned linear spaces. Let E = Ei x ... x E, and Li C E an open set and

The relations (1.1.8)-(1.1.10) prove (1.1.7). . f : U i F a given map. Given a = (ai, ..., a,) E E , we define We look at some special situations where E and F are product

spaces. Let 11s assuine that F = Fl x ... x F,,, the product of iiornied linear spaces. For 1 5 i 5 m, definc tlic projection Ai(xi) = ( a l , . .ai-l,xi,a.i.+~,..a~).

p , : F + Fi, Proposition 1.1.3 If f is differentiable at a E U: then for each i , f o Ai is diflerentiable at ai. Further, i ~ i i ~ l 11.t I r , : Ri + F 11re tlir iiijectioii defined by

n

7~;(:zi) = ( O , ..,O, % i , O, ..,O), f f ( a ) ( h l , ... h,) = x(f o hi) '(ai)hi i=l (witli O everywliere except. iii the i-tli place). Then

fr>r u,1,y 1' = (h,, ..., hn) t El X ... X En = E . Proof: If t ~ i is the inject,iori of Ei into E as defined previously, we

(wlic!rc I E denotes the ideritity map in a norined linear space E). Xi(zi) = a + ui(x i - ai) . Proposition 1.1.2 Lct U C E be an open set and f : U + F Tlieri (cf. Example 1.1.2), I I I I T/?,I?TL f is di$eren,tiable nt a E U ;if, an,d only z:f, I ' , , -- - 11; o J' : LI + fi; is diflcrentiab1.e at a for each i, 1 5 i < n ~ . In X:(xi) = oi, for all z.; E Ei. / / l , i ~

10 CH.l DIFFERENTIAL CALCULUS

Once again we have

whirh eives

which is jiist a reforrnulation of (1.1.14). i Definition 1.1.3 The deriuadiue of f o Ai ot ai is ealled the i - th par t ia l derivative of f a,t a and is denoted by %(a) or by a i f ( a ) . i Example 1.1.6 Let U C R" be an open set arid let f : U i R be a given function differentiable a t a point a t U . Then the partial derivat.ives of f at a are the usual ones we kiiow from the calculus of functioiis of severa1 variables. Fiirther the relation (1.1.14) implies that f l (a) can be represented as follows (since L(Rn,n.IW) Rn):

(wliich is also deiioted by V f (a)). It can also be seen that g ( a ) is the Gteau derivative of f along ei , the i-tli standard basis vec- tor of R1'..

Example 1.1.7 Let U be as i11 the previous example aiid let f : U + Rm be differentiable at a t U. Then f'ial t LIR". Rml

. \ , - \ , , wliich can be represented by an m x n matrix. Iiidred, if f (x) = (.fi (:c), ..., f,,,(x)),theii, by (1.1.12) and (1.1.14), we deduce that f l(a) is giveii by the usiial Jacobian niatrix,

" ( a ) dri . . "(a) 1 1 .. .. . . . .

Rernark 1.1.3 As shown by Exa~nple 1.1.1, the converse of Propo- iribion 1.1.3 is false; a11 partial derivatives of f may exist at a point I~iit f could fail to be differentiable there. However, we will prove (c:f. Proposition 1.1.4) the differentiability of f given the existence i~ f partial derivatives under some additional hypotheses. i

We now discuss ali important result of differential calculus, i the mean value theorem. Oiie of the earliest forms of this I.luiorem states that if f : R i R is differentiable in an iiiterval i:oiitaining [a, a + h], then there exists a 0 t (0 , l ) such that

f (a + h) - f (a) = f l ( a t 0h)h. (1.1.15)

11. is clear that (1.1.15) caiinot be true in more general situations for arbitrary normed linear spaces E and F. Indeed, even if we I,i~ke E = R and F = R2 and set

f (t) = (COS t, sin t ) ,

I.lic:n it follows from Proposition 1.1.2 that

f l ( t ) = ( - sint, cos t). 'rhus, while f (O) = f (2n), we can never have f l ( t ) = O for any 1 E (O, 2n). However, there are other versions of this result which ILU! trne.

Indeed, from (1.1.15) we deduce that if f is a real valued func- I.lon of a real variable which is differentiable in an interval con- t,i~iiiing [a, a + h], theii

12 CH.l DIFFERENTIAL CALCULUS 1.1 The Frchet Deriuative

Definition 1.1.4 Let E be a normed linear space and let a , b E E . Then by the interual [a, b] loe mean the set

l l f ( c ) - f ( a ) l I 5 ( k + c ) ( c - a ) + . { x E E lx = ( 1 - t ) a + tb,O 5 t 5 I } . (Ioiiibining the two inequalities ahove,we deduce that for c < x 5

The open interual ( a , b) is sim,ilarly defined using O < t < 1.. I. t q , x 6 V which contradicts the definition of c. This shows that V = 0 and hence proves (1.1.17).

Theorem 1.1.1 (Mean Value Theorem)Let E and F be normed linear spaces and U an open set in E . Let f : U i F be differen- Sl,ep 2. Defiiie, for O

14 CH.l DIFFERENTIAL CALCULUS

k = O. Thus f is locally constant. Then for any b E F, f- '(b) is an open set and since F is Hausdorff, it is also closed. Thus, by the connectedness of U, f - ' (h) = 0 or U. If we choose b = f (x,) for some x , E U. we get that f ( x ) = b for a11 x E U..

The mean value theorem has nunlerous applications. We present below one such resiilt, the converse of Proposition 1.1.3, a s proniised earlier (cf. Remark 1.1.3). Some other iiiiportant applications will be discussed in Section 1.3.

Proposition 1.1.4 Let E = El X ... x E,, the product of normed linear spaces and let F be a normed linear space. Let U c E be an open set and let f : LI i F be gi,uen. If the partia1 derituatives &x) exist at eoery point of U und i f the maps r H g(x) are continuous at a E U for euch 1 5 i < n, then f is differentiable at a.

Proof: If f were differentiable, then we know what f ' ( a ) should be, i.e. we need to show that

Consider

f O be any arbitrarily sinal1 positive quantity. Let

Then

1.1 The Frchet Deriuative 15

But g is differentiable and

Since is continuous, there exists 17 > O such that llgl([i)ll < E whenever llxi - ai ( ( < q for each 1 < i _< n.Hence,by the mean value theorem,

Ilg(x1) -g(a1)lI 5 Ell. -ali. We have similar estimates for the ternis in each row of (1.1.21) which in turn proves that

This shows that f is differentiable at a and that its derivative. at a is given by (1.1.20). .

Exercise 1.1.4 If f : U C E i $ then, show that, under the hypotheses of Thoerem 1.1.1, there exists a point c E ( a , b) such that

Exercise 1.1.5 We say that f : U C E i F is Gateau differen- tiable at a point a E U if there exists a continunus linear operator df ( a ) : E -t F such that

lim t -t O

(a + th) - ( a ) = df ( a ) h , for a11 h E E. t

If F = R, and f is Gteau differentiable at a11 points of U, show that we can still have the same conclusion a s in the preceding exrecise. If the mapping x tt d f ( x ) is continuous at a E U, show that, in that case, f is Frchet differentiable at a and that f l ( a ) = df (a)..

There are severa1 applications of the inean value theorem which interested readers can find in Cartan [4]. We will use it again in

r Section 1.3 to prove some important results.

16 CH. 1 DIFPERENTI.4L CALCUL TIS 1.2 Higher Order DeriT~atiTies

1.2 Higher Order Derivatives i,e, the biliiiear form is sgmmet,ric. We onlit the proof. Let. E aiid F be nornied linear spaces aiid let, 24 C E be an opeli E ~ ~ ~ ~ ) ~ 1.2.1 ~~t us consider the function defined in Example set.. Let f : U + F be a given mapping which is differeiitiable in 1.1.3. ~~t E be a Hilbert space and let a(., .) be a symmet,ric U. Consider the map I,ilinear form on E and b t E. We have

x H f l(x) E L(E, F). This is a mapping from U into L(E, F) and one could again inves- tigate its differentiability. Its derivative at a poiiit a, if it exists, is

f1(z,)h = a(xo, h) - (bi h). a linear map of E iiiro L(E, F) and thus belongs to t (E , L(E, F ) ) . Tliis is called tlie secoiid derivative of f at (L and is denoted by It is now easy to see that f"(a) (or D2f(a)) .

fu(x0)(h, k) = a(h, k).. Remark 1.2.1 To define the second derivative at a point a E U ,

~~t us llow assume that E = Ei X ... x E,, the product of we need not assume that it is rlifferrntiable at every point of U.

normed linear spaces and that U C E. Assume that f is twice It suffices to assume that it is differentiable in a neighboiirhood V differentiable at a E U . Theii f ' exists at each point in a neigh- of a, aiid in that neighbourhood, the inapping

bourhood of a. Now x c+ f '(x)

is differcntiable at a,.

Deflnition 1.2.1 The finction f is said to be twice differentiable In the same way, in U zf ftl(a) erists at euery point a of U. If the map x H f(Z) 2s continuoos from U into L(E, C(E, F)), ,we say that f i8 of class C2..

Recall that t,he space L(E, C(E, F)) is isomorphic to &(E, F), the space of continuoiis biliiiear forms from E x E into F. Thus fl'(a) can be thouglit of as a bilinear forin and if (h, k) E E x E,

fl'(a)(h, k ) = (fl'(u)h)k. ~~t~ that %(a) L(Ei ,C(E,F)) aiid so (1.2.3) defines an Note tliat f1'(o)h L(E;F) and so (1.2.1) makes sense.

As ali application of the ~nean valur theorem, oiie can prove ment of F. Now, tliat (cf.Cartan [$I)

f "(a) (h, A:) = fl'(a) (k, Ir),

18 CH. 1 DIFFERENTIAL C A L CUL 1.2 High,er Order Deriuatives 19

We by &(a) ( 0rby L?>f ( a ) ) the term & ( % ( a ) ) which differentiable at a e if it is (n - 1)-times differentiable in a is an element of L ( E i , f ( E j , F ) ) , i.e. a bilinear form on E, x neighbourhood V C U of a and the niaP with values in F . Thus we can rewrite (1.2.3) as

x H f ( n - i ) ( x ) E L,-i(E, F ) 7L a2f ( f V ( a ) k ) h = (- ( a ) ki) h j . (1.2.4) is differentiable ai a. The derivative f ! n ) ( a ) belongs to & ( E , F),the

i , j=i i)xii3xj space of n-linear forms on E with values in F.lf the ma^ x But ( f " ( a ) k ) h = ( f "(a)h)k, which gives f ( n ) ( x ) is ~ontinuous, we say that f is of class C"'.

O . ' . 2

~ ~ f i ~ i t i ~ ~ 1.2.2 p'e say that 'f is of class C ?f zt rs continu,ous. ) k ) h = Cn. ,,,,=i ( -(a)hi)kj tve 6ay that j is o j class C* if it is of class Cn for ~osi t iue (1.2.5)

= ~ t ~ , ~ ( & ( r ~ ) h ~ ) k ~ . integers n.. ~~~~~k 1.2.2 in the case of the second derivative, thc: n-th In particular, if E = Rnli.e. E, = B for 1 5 i 5 n, then we

can identify P) with F aiid so L ( R , L (R , F ) ) ~ i t h F again, derivative is a syrnmetric n-linear fornl. ThuS, if 0 is a permuta- Tl~en the bilinear nlap &(a) : R x !R + F is just giveil by tion of the synlbols { I , 2, ..., n.1: then,

j ( " ) (a ) (h i , ..., h,) = f ("')(a)(h,(i), ...,h C(,)).. (AI P) &jA&

por completeuess, we state helow the various generalizations where &j = &( F be we can define the n-th derivative of f at a point a , denoted by (n + l).timrs diflerentiable in U and assume that f (,)(a) Or D n f ( a ) , 1. induction: if the concept of the first (n - 1 ) derivatives already been defined, then f is said to be n-times l l f ( 7 1 + 1 ) ( ~ ) i j 5 M for euery 5 t U.

20 GH. 1 DIFFERENTIA L GA L GUL US 1.2 High,er Order Deri'Uati'Ues

Then Finally, we describe the Taylor formula with the "integral for111 of the remainder" .

If F is a Bana& space and f : [a, b] C R + F is a conti~uous ma*, then we can define the integral

Remark 1.2.3 The relation (1.2.8) is stronger thaii (1.2.7). While (1.2.7) is asyinptotic in the sense that it just tells us what hap- Pens as llh,li -f 0,(1.2.8) gives us an estimate of the erros in the n-th order expansion. It is sometinies called lhe Taylor's formula as a vector i11 P. This can be done by defining it as the ,as with Lagrange form of the reniainder. Note however that the hy. + m, of appropriate R'iemann sums. Using the continuity of f , potheses 0f Tlieorem 1.2.2 are also stronger than those of Theorem we can show that the Riemann sums form a Cauchy sequence and 1.2.1. . hence converge, siiice F is complete. The integral als0 turns oiit to

be the unique vector y E F (unique, by virtue of the Hahn-Banach Example 1.2.2 Let E = F = R and consider the function Theorem) such that for any (o E F* (the dual of F),

It is easy to check that f is n,ot twice differentiable at the origin. However, where the integral on the right-hand side is now that 0f a real

f(.) = o(11~11~) valued continuous function on [a, b]. and so it possesses a "Taylor expansion" of arder 2 at tlie origin,

Theorem 1.2.3 Let F be n Ban,ach space and E a norrned linear f (z) = ao + ais + n2z% +(1(ill2) space and U C E an open set. Let f : U + F be of class C(n+').

Let [a,a + h,] C U . Then, with ao = ai = a2 = 0..

FLemark 1.2.4 As the above example shows, a function f : u c f (a + h,) = f (a) + f1(o,)h + ... + & ~ ( ~ ) ( a ) ( h , ) ~ (l.2,10) R + lnay possess an n-th order Taylor expansion at a point in the form

Remark 1.2.5 Observe now that we need even StrOIlger hypothe- a2 2 f (a + h) = ao + aih + -h + ... + -h?' + o(lh,ln) (1.2.9) on f .We need tliat the derivative of f of order (n + 1) be 2! n!

continuous and that F be complete. H but fail to be n-times differeiitiable at a. However, if it is n-times

~f f is of class C', a particular case of (1.2.10) gives differentiable at a and admits a Taylor expansion of the form given in (1.2.9), then necessarily,

ao = f (a) , ai = f(")(a) , 1 5 i 5 .n.H

..

23 22 CH.l DIFFERENTIAL CALCULUS 1.3 Some Important Theorems

1.3 Some Important Theorems Proof: Step 1. Define y : fl + Ez by

In this section we wiil present sonie very irnportant results in y ( z l , x 2 ) = 2 2 - A-lf ( x i , ~ ) . (1.3.4) Anal~sis which will also be fieqiiently used iri the sequel. It will I

Lookiilg for solutioiis of (1.3.1) is the samc as findiiig fixed point,s also be seen tliat the mean value theorem will play an importarit z2 of g for given 21. Now, role in the proof of these results.

The first result we will examine concerris the solution set of a f a g the equation - ( z l , x 2 ) = I - A ' - 8x2

("1,.2) 8x2 f h V) = O (1.3.1)

where f : E x F + G is a continuous function, E, F and G and, by (ii) above, 2 is continuous on n. Further, being nornied linear spaces. Of course, the very general nature of the problem prevents us froin having a precise general result

-(a,, a g b) = O , y(a, b) = b. (1.3.5) on the global structiire of the set of solutions to (1.3.1). However, 8x2 given a solutiori, say, (a, b), of the equation, iinder some reasonable Hence, by the continuity of the partia1 derivative, there exist conditioris, we can describe locally the set of solutioiis of (1.3.1). neighbourlioods Ui of o and V of b such that In fact, we will show that t1e solutioiis "close" to ( a , b) lie on a "curve". More precisely, we have the followingresult. 1

-(x1.x2)I 5 li (1.3.6) Theorem 1.3.1 ( Im~l ic i t Function Theorem) Let E l , Ez and F be nonri.ed linear spaces an,d assume that Ez is complete. Let for every ( z L , x 2 ) 6 Ul x V. Without loss of generality, we may r2 C E1 x E, be an open set and let f : fl -t F be a function, assume that V = B ( b ; r ) , the closed ball with centre at b arid such thut radius r > O. By the coiitinuity of y, we deduce the existente of a (i) f i8 con,tinuous;

neighbourhood U C Ui of a such that (ii)for every ( x I , x ~ ) E fl, g ( x 1 , z 2 ) exists and is continuous on (1.3.7) R; I (g ( z l ,b ) -g (a ,b ) ( l 5 r / 2 , for every si E U.

(iii)f ( a , b) = O and A = %(a, b) is inuertible ,uith continuous inuerse. Step 2. Let xl E U be fixed. Define g,, : V + Ez by Then, there exist neighbourhoods U of a and V of b and a contin- Uous function

24 CH.1 DIFFERENTIAL CALCULUS I.!! Some ImportanA Theorems

by virtue of (1.3.6) and the inean value theorem alid hy wliore ? ( h ) -t O as IJhll -t 0. But (1.3.7). Thus, g ~ , maps V i1it0 ilself a ~ i d , again by (1.3.6) and the mean vdue thmrem, qz, is a contraction (with Lipschitz constanl equal to 1/21 and so there exists a unique fixed poiiit x2 = in V . Thus the only solutions to (1.3.1) in U x V are giverl h,y (21, < P ( z I ) ) . Cleuly, p(a) = b.

Siiice p is contiiinous, J J k J j -t O as IjhJJ -t O alld 30 ~ ( h j k ) -t O Step 3. (Cont~inuity of p) Let xp E U. Then ILs 1JhlJ + O. Thus there exists r, > O such that if (Ihll I T o , then

P ) - = ~ ~ ~ ( x I , P ( : ~ I ) ) - g(x?, ~ ( 4 ) - g(z:>v(x:))Il . llkil 5 (20 + l)iihll Thus,

wlfich (1.3.8) and completes the proof of the theorem. 8 Ilv(x1) - v(x7)Il 5 2 I l Y ( x l > ~ ( x ? ) ) -Y(x7,p(z:))Il

Remark 1.3.1 1f f is a C2 functio~l, theii the right-hand side of and the result follows from the contiiluity of g. (1.3.3) is C1 aild so p will be C'. By induction, if f is of class c'p>

I,hen p will also be of ciass CP.m Step 4. (Differentiability of p) Assume now thal f is differentiable at ( a , b). Let h E E , small eiioiigl~ such that a + h E U. Set nemark 1.3.2 We can derive (1.3.3) heuristically by implicit dif-

k = p(a + h ) - p(a) . krentiation. We have

Now, f (.i> p(z1) ) = 0. o = f ( a + h, d a + h ) ) - f (a , v ( a ) ) Ilifferentiating tliis w.r.t x l , we get

= H ( a , b ) h B r i + z ( a . b ) k + e(h,k)( j lh( l + Ilkll) wllere &(hl k ) -t O when IlhJl -t O aiid IlkJ( -t O. Tben,

wliich yields (1.3.3). n The folloa.i~ig consequence of the implicit fu~iction theorem

t,i!lls us when a mapping is a local Iiomeomorphism. To prove (1.3.3), it sufices to show that

Theorem 1.3.2 (Inuerse finction Theonm) Let 8 and F be Ba- (IIhlI + I l ~ l I ) l I A 1 ~ ( h , ~ ) l I = q(h)lIhJI * rl,nc:h spaces and f : 0 c E -t F be a CP- map, for some i, > 1.Let

26 CH.1 DIFFERENTIAL CALCULUS 1.3 Some Important Theorems 27

a E C2 with f ( a ) = b and let f l ( a ) : E -t F be an isomorphisrn. i~iid Then there ezists a neighbourhood V of b in F and a unique C P f ( x ) f 0 for any x E R'.. function g : V -t E such that

We conclude with oiie more classical result. = g(b)

{ f ( ; Y ) ) = Y (1.3.9) Theorem 1.3.4 (Sard's Th,eorem) Let Q C E' be an open set nnd let f : R -t R" be a C' function. Let

for e v e y g E V . S = { x E R ( J f ( x ) = O } ,

Proof: Define $ : C2 x F -+ F by ,where J f ( z ) = d e t ( f l ( z ) ) . Then f ( S ) is of measure zero in R".

$ ( x , Y ) = f ( x ) - Y. Proof: Step 1. Let C be a cube of side a coiitained in Q. We

The result now follows immediately on applying the implicit f ~ ~ i c - ilivide it into kn sub-cubes each of side a / k . Since f is of class tioii theorem to $.. C'; the map z H f l ( x ) is uniformly continuous on C. Thus, for

A stronger version of Theorem 1.3.2 exists, wlierein sufficient E > 0, there exists 6 > 0 such that conditions are given for f to be a global homeomorphism. We state it without proof (cf. Schwartz [23]). IIx - Y I I 6 * Ilf l(z) - f l (y) l l < E . (1.3.10) Theorem 1.3.3 Let f : E -t F be a C' map such that for e v e y We choose k large eiiough so that &a/k i 6, i.e. the diameter z E E, ~ u e have that f l ( z ) : E -t F is an isomorphism of E onto of each sub-cube is less than 6. E'. Assume further thot there ezists a constant K > O S U C ~ that Since f ' is bounded on C, the function f is Lipschitz continu- 1/(f1(z))-'11 I K for e v e y 3: E E. Then f is a homeomorphism 011s on C by the mean value theorem. Thus, of E onto F..

l ( f ( x 1 ) - f ( x a ) ) ( 5 L l z 1 - ~ 2 1 1 , for every ~ 1 ~ x 2 E C , (1.3.11) Example 1.3.1 The above result is not trile if ( f l ( z ) ) - I is not uniformly bouiided in L(F, E).Consider E = F = and where

L = sup IIf1(x)ll. f ( 2 ) = (eZ1 sinxi, eZ1 cos 2 2 ) , for :c = ( z l I 4. X E C Then Step 2. Let x E C n S . Then there exists a sub-cube 6 such that

eZi sin 2 2 eZ1 cos za I :i: E C. Tlwii for every y E 6, f f ( x ) = [ 9' coszz - e"L sinx2 I l f (3 : ) - f ( ~ ) h i L&a/k; (1.3.12)

and d e t ( f l ( x ) ) = - eZz1 # O and so f ' ( x ) is invertible for each x E R' but the norm of its inverse is ulibounded. The mapping f Iiy (1.3.11). On the other hand, is neither one-one nor oiito for

.1

f ( ~ 1 ~ x 2 + 2nv) = f ( 2 1 , $ 2 ) for all n f(7 , ) - f ( z ) - f r ( z ) ( y - X ) = 1 O ( f t b + t ( Y - $ ) I - f l ( x ) ) ( y - x)dt

-

28 CH.1 DIFFERENTIAL CALCULUS 1.4 Extrema of Real Valued Functions

which, by (1.3.10), yields Remark 1.3.4 A more general versioii of Sard's theorem states t,llat if f : 0 C Rn + W is a mappiiig of class C"-m+' and if

I.f(l/) - f ( ~ ) - f l ( X ) ( y - 5 1 1 ~ - ~ 1 1 5 c&/k. (1.3.13) 2 m, then f (S) is of iiieasiire zero, where now S is the set of ali points z E C1 such that t,lie rarik of f l (z ) is < m (cf. Sard [22]).

Step 3. NOW f l ( z ) is singular and so f ' ( z ) ( ~ ~ ) = H is a subspace She result is not trile iii general if f is orily of class (cf. of dimensioil I n - 1. Hence, by (1.3.13), we liave

~ ( f (.Y), f ( ! E ) + H ) 5 ~&a/k:, for every ;y E 6 (1.3.14) Definition 1.3.1 If f : Q C R" + Rn' and I: E Q is such that fliz) is of rank < m, then, z is said to be a critica1 point of f . wliere p(a, .y) denotes the distance of a point a fr0111 a set X. com-

bining (1.3.12) and (1.3.14), we deduce tliat f (6) is coiitained in If ~ ~ o t , 5 is a regular point. A uector y E Rm is ralled a critical a cylindrical block of radius L&a/k: and height 2&&a/k:. l'hus "alue i f there ezists a critical point x E Q such that f (x) = :y.

Otheruiise, it s cnlled a regular value. .

Thus, Sard's tiieorem states that if f : 0 C R'' 4 R" is of (:lass C1, tlieii aliiiost every value f takes is regular, i.e. tlie set of w h e r ~ A is a constaiit depending ouly ori n and rn(X) is the

(Lebesgue) measure of a set ,Y. Thiis

30 CH. I DIFFERENTIAL C A L CUL US 1.4 Extrema of Reol Valued Functions

Theorem 1.4.1 Let f : U C E -t !R adrnit a relatiue eztrem,um, where V : U C E -t F is a given mapping. We t h ~ i i look 01 a ai u E U . If f i.5 differentiable at u , then relative extremum of f : 24 C E -t R in K. Notice that K is not

ali open set. In fact, i v is continuoiis, then h' is closed. Thus f l( .) = O. the prcvious theorem caiinot be applied.

Proof: Let v E E be an arbitrary vector. Since U is open, we cai] Exercise 1.4.1 Let E be a vector space aiid let g and gi , 1 < i 5 find an open iiiterval J C !R containing the origin such that fOr all m be linear functionals on E. Assume that t E J , u + t v E U . Define

n z l K e r ( g i ) C Ker(g). p ( t ) = f ( u + t v ) .

Show that there exist scalars A;, 1 5 i < m such that Theri p is differeiitiahle at the orjgin and m

pl(0) = f l ( u ) u . i= 1

Assnnie that f attaiiis a relative minimum at u,. Then Theorem 1.4.2 Assum.e that E is a, Banach space and that p E

K = {v E E / ~ ( u ) =O}. Thus, ~ ' ( u ) P I = O and since u was arbitrary, (1.4.1) ollows. The Assum,e, further, that for a11 u E h', we have pt (u) # 0. I f f E case o a relative maximum is siitiilar. C1(E; R) and if u E K is such that

Remark 1.4.1 The relation (1.4.1) is called Euler's equation cor- f (.) = .UEK inf f ( v ) , responding to the extrema1 problem.

th,en there ezists A E R 8uch that f t ( u ) = Apl(U). Remark 1.4.2 I E = !Rn, then f'(u) = O is equivalent to the Proofi Since u E K, by hypothesis, pl(u,) # O. Hence, there system o equations

exists tu0 E E such that ((tua(( = 1 and < pt(u),wo >= 1 where < .;. > denotes the duality bracket between E and its dual. Set

aif ( ~ i , u z , ..., u,) = O for a11 1 5 i < n E. = Ker(vt(u,)) . Then, clearly, E = E0 @ R{wo}. Define : where u = ( ~ l . i , ~ a ; ..., ?L,,,) and this is the well knowii iiecessary E. x R + R by conditiou for the existence o a relative extremuni. H

@(Pl,t) = ~ ( u + V + ~ ' u ) o ) . We iiow consider the case o extrema under constraints. Let E Then @ ( O , O ) = O. Further, we also liave that

aiid F be iioriiied linear spaces and let U C E he an open siibset. Lr t, & @ ( O , O ) = ~ ' ( U , ) ( E ~ = O

K = {V E U I ~ ( v ) = O} a t @ ( O , O ) =

32 CH. 1 DIFFERENTIAL CAL CUL US 1.4 E~trama of Ren.1 Valued Functions

Thus, by the iinplicit function theorem, there exists a iieighboiir- x ~ , 1 5 ,i 5 m, we solve the system of n + m ecl1iations hood V of O ir1 E0 and a C1 function C : V + R such tliat C ( O ) = r(O) = O and if 0 = u + L', then tlie only points in f 2 n K are of the forin ,w = u + v + C(v)luci with v E L'. N ~ w , setting f ( v ) = f ( u + v + C(v).uo). for v E V, we see that f attairis its miniiiium over the open set V at O. Tlius, < ? ( O ) , v >= O for every v E Eo. Since . Thiis, it follows that Ker(pl(u)) C Ker ( f l (u ) ) and tlie conclusioii follows from the pre- cediiig exercise.

which is exactly the well known niethod of Lagrange mul t i~ l ie~s Remark 1.4.3 In the saine way, it can be slic,wn that if vi, 1 5 i11 the calciilus of severa1 variables. . i 5 ni are in C1 (E; R) and if we define K by Exercise 1.4.2 Let A be a syminetric n X n mat rk with real

K = { v E E / pi(v) = O , 1 5 i 5 m); entries aiid let B be a syrnmetric and positive definite matrix. Characterize the relative extrema of the functinal

tlien, if f attains a relative extremiiin at u E K and if pi(u) , 1 5 i 5 m are liiiearly independent, there exist scalars 1 5 i 5 m such that

on the set K giveii by m

f ' ( 7 4 = 1 ADp:(i~).. i=1

where (., .) denotes the usual iniier prodiict in R".. Example 1.4.1 Let f : U C R" -t R. Let p; : U i R, 1 5 i we will now take into account the second order derivatives o* m, 1 _< m < n be given functions. Let f to characterize extrema1 points.

Theorem 1.4.3 Let f : U C E -t R be deffere7itzable in and K = {v E U pi(v) = O , 15 i 5 m). tupjce d,j,ffeTentiable ot u E L.(. ~f f admits o ~elatioe minim'.um ot

v, tlirn; for a11 v E E. Thus, if at a point u E U , tlie m vectors pt(u), 1 5 i 5 m f f l ( u ) ( v , v ) 2 O. are liiiearly independent, a necessary coiidition that f att,aiiis a relative extremum at u ; Ly the preceding tlieorein, is the existente proofi ~~t v # O \if: an arbitrary vector in E. Then, there exist's of A . E R, 1 I i 5 m such that f l ( u ) - C(=, Aip$( i~) = O. Let an interval J C R

34 CH.l DIFFERENTIAL CALCULUS 1.4 Eztrema of Real Valued Flinctions

f ( u+ tu) 2 f (u). BY Taylor's formula (cf. Theoreiii 1.2.1) and the fact that f l ( u ) = 0; we get

t2 ~~~~~i~~ 1.4.3 ~ e t A be a symmetric n x n. ~ n a t r i i with real 0 5 f ( . + tu ) - f(.) = Z ( f l l ( u ) ( u > u ) + ( t ) ) entries. Define

where ~ ( t ) i O as t 9 0, from which (1.4.3) follows.

Remark 1.4.4 If f admits a relative maximiim at 11 E U , then, where b E Rn is a given vector. under the above conditions, the inequality in (1.4.3) will be re- (i) show that f admits a strict minimum in Rn if, and only if, A versed. . is positive definite.

(ii) show that f attains its miniinum if, and only if, A is positive Theorem 1.4.4 (Suficient conditions) Let f : C E 9 R be semi.definite and the set, s = { w E R* 1 A w = b) is non-empty. differentPable in U and let u E U be such that fr(u) = 0. (iii) l f the matrix A is psi t ive semi-definite and the set S is empty, ( i ) If f twice differentiable at u and i f there ezists a > 0 such that inf f ( u ) = -m.

f f 1 ( ~ 1 ) ( u , ~ ) 2 alluli2 .UER~ for a11 u E E, then f admits a s t ~ c t relatiue minimum at 21. (iv) ~f the infimum of f over W is a real iiumber, show that the (i) If f i s tuiise differentiable in U and fhere ensts a bnll B(u; r ) C

matrix A is positive semi-defiriite and that the set S is non-enlPtY. U such that f U ( u ) ( w , w ) 2 O

for ali u E B(u;T) and a11 u1 E E, then f admits a relatiue mini- mum at u .

Proof: (i) For a11 w with siifficiently small norm, we have, by Taylor's formula,

where E ( W ) -f O as I/w/I -f O. Thus there exists r > O soch that as soon as /lulll < T we have e ( w ) < a. Then f ( u + U J ) > f (u) for a11 u +w E B(u; r ) and so f admits a strict relative minimiiin at 1'. (ii) Since f is real valued, there exists a u in the open interval (u, u + UJ) C B (u; i) such that

1 f ( 2 ~ + w ) = f (u) + S f " ( u ) ( w , w ) .

I 2.1 Definitio~, of th,e Degree 37

Chapter 2

The Brouwer Degree

2.1 Definition of the Degree

The topological degree is a useful to01 in tlie study of existence of solutions to nonlinear equations. In this chapter, we will study the finite dimensioiial version of the degree, known as the Brouwer degree.

Let R C Rn be a bounded open set. By Ck(9 ; Wn), we denote the space of functions f : R -t R'' wliich are k times differentiable in R such that these functions and a11 their derivatives upto order k can be extended continuously to n. We denote tlie bouudary of R by 80.

Let f E C 1 ( n ; P ) . R,ecall that f '(x) E L(R7',R7') and hence f l (x) can be represented by an n x n niatrix. Let S be the set of critica1 points of f (cf. Definition 1.3.1). Definition 2.1.1 Let f : R -t 1" be a function in c l (n ;RrL) and let b @ f (S) U f (8R). Then we define the degree of f zn R with respect to b as

o, if f-'(b) = 0, d(f, R, h) =

sgn(Jf (x)), otherwise. i (2.1.1)

The functiori sgn denotes the sign (= +1 if positive and = - 1 if negative) and Jf (2) denotes the determiiiant of fl(x).

Remark 2.1.1 We will now verify tliat the ahove definition makes sense. Sincc b @ f (S) U f (8R), we know that f l (x) is well defined for 3: E f-'(b') and that Jf (x) # O. Thus, Jf (z) has a defiiiite sign aiid, hy the inverse function theorem, f is invertible in a neigli- l>ourhood of x. Consequently, since D is compact, the set f- '(b) is finite and so (2.1.1) makes sense. 1

Example 2.1.1 Let I : Rn -t RTL be the identity map and set f = Iln for f2 c Rn. Then

More geiierally, if T : Rn -t Rn is a nonsingular linear operator, and if f = TIn, we have

Notice that sgn(detT) = ( - I ) ~ , where P is the sum of the (alge- braic) multiplicities of the negative eigenvalues of T..

Example 2.1.2 Let 0 = (-1,1) and define

f (x) = x2 - E2, for E < I

Then, f '(2) = 2% and f-'(0) = { + E , - E). Thus,

Remark 2.1.2 We defiiied the degree to be zero if the value h were iiot attaiiied by f . Tlie converse, as seen 1>y the above examplel is false. Nowever, if d(f, Q l b) # O> the the solutiou set to the cquation f (z) = b is indeed iioii-einpty. W

We wish to extend the drfiiiition of tlie degree to functions wliicli are rnerely coiitiriiioiis oii n. To do tliis we need soirie pi.eliiiiiiiary result,~. We start wit,li aiiother formula for the deg-ree.

38 CH2. THE BROUWER DEGREE 2.1 Definition, of the Degree

Proposi t ion 2.1.1 Let f E C1(a;R7') and let b 6' f (S) U f (80) . by an obvious cliaiige of variahle in each of the sets Wi and the Then there exists E, wch that, for a11 O < s < E,, right-hand side is exactly d(f, Q, b) thanks to (2.1.3).8

We use t,he formula (2.1.2) to prove the robustness of t,he de- gree iii the sc:nse that it remains stable when b ar f iu slightly

ulhere ip, : JW" + R i8 a Cm function whose snpport i8 contained perturbed. In order to do this, we need a teclinical result. in the ball B(O;E) with centre at O und radius E and wch that

Lemma 2.1.1 Let g E C 2 ( a ; Rn-' ). Set Bi = det(8lg, ..., ai-~ig, 8i+ig, ..., 8ng).

Proofi If f-'(b) = 0, theii we choose s, < p(b, f (fi)) (where p(x,A) derioteu the distance of the point x from the se t .A). If 1 ( - 1 ) ~ 8 i ~ i = O. ip, is a s above, we then have ip,(f (3:) - 6) = O and so (2.1.2) is i=l trivially true. Proofi Let 1 5 i 5 n. Set Cii = O. If j < i., define Let us now assume that f-'(h) = {xl ,xz, ..., s,). For eacli 1 5 i I m, we have Jf (xi) # O, aiid so, by the inverse functiori theorem, there exists a iieighbourhood Ui of xi and a neighbour- liood Vi of b such that the Ui are a11 pairwise disjoint and and, if j > i. set

C, - det(&g, ... ; 8i-lg,ai,~ig, ..., a,-1g,&,ig,8j+ig ,..., an,g). f lu, : ui + vi is a honieomorphism. Further, by shrinking the neiglibourhoods if Then, clearly, &Bi = C:=, C,,, by the riile for differentiating necessary, we can also ensure that Jf lui has a constant sign. Now deterininants. Thus the left liand side of (2.1.4) eqiials choose > O s u d ~ that n

B(b; E,) C nzlvi. Set Wi = f -'(B(b; E , ) ) nu,. Then the W, are a11 pairwise diujoiiit Since g is C', 8ijg = 8jig and so, by t,he property of deter~iiinants arid Jf is of constant sign in each of them. Hence, if O < E < relating to transposition of coluninsl it is easy to see that as ~ , ( f (2) - 6) = O outside the sets Wi, we liave

rn

vZ(f - b ) J f ( ~ ) d ~ = 1 ipdf ($1 - b)Jf(x)dx and tlie lenima follows easily.. ,.=i '

m Lemma 2.1.2 Let ,f E C2(a;RT1). Let Ay(z) denote the cofnctor = s g n ( ~ f ( ~ i 1 ) 1 ipc(f (2) - b)lJf(~) ldx of the en,try 8;fj(z) i71 Jf (2). Then for all 1 5 j 5 n,

1=1 . w, rn

i=l

40 CH2. THE BROUWER DEGREE 8.1 Definition of the Degree

Proof: Recall that Aij is given hy Aij = (-1) "' d e t ( 8 ~ f k ) k f ~ , ~ + .

For fixed j, we apply tlie preceding lenima to Now, if y E f ( a R ) ,

g = ( f i , . . .>f j -~rf j+ir . , . > f n ) 19 - (1 - t ) b ~ - tbal = I(y - b) + (1 - t ) ( b - b l ) + t (b - b2)J

to get the desired result.. > PO - (1 - ~ ) ( P o - 6 ) - t (p0 - 6) = & > E .

Remark 2.1.3 The ahove lemma is essentially a consequence of the fact that the order of derivation is imiiiaterial for C2 functiotis Since the support of (o, is contaiiied in B(O;E) , it follows that (and this was used ir1 the proof of Lemma 2.1.1). For iiistance, if w ( y ) = O for y E f ( a f l ) . Now, for 1 5 i 5 n, define n = 2, then

AI^ = az f2 , AZI = -al f2 ~ ~ ~ = - a z f ~ , ~ ~ ~ = a ~ f ~ ,

and we readily verify that, if f is C', By the preceding consideratio~is, .ui = O on an. Now

alAll + &A21 = %A12 t &A22 = 0.. Cy,k=1 z(f(~))%(~)~i.j(~) Proposition 2.1.2 Let f E C2(G; P) an,d let b 6 f ( 8 0 ) . Let po = p(b, f (an)) > O. Let bi E B(b;p,) for i = 1,2. If bi 6 f ( S ) , + C,"=l w j ( f ( x ) ) & A i j ( ~ ) . we have

d ( f , % b i ) = d ( f bz). Proof: Clearly, by choice, bi 6 f ( a R ) . Thus, hy hypotliesis, the C $ = i z(f (2)) g ( ~ ) ~ i j ( ~ ) ) degree d ( f , R , bi) is well-defined for i = 1,2. Let + C,"=i q(f ( x ) ) (cL &A&))

6 < p,-(b-b,l, i = l , 2 . BJ Leiiima 2.1.2, the second term on the right-hand side vanishes. Then tliere exists E < 6 such that Notice that, by the definition of the AV,

d ( f , fl.bi) = a( f ( I ) - bi )J l (x )dx . i = 1,2. where (o, is as in Proposition 2.1.1. Then

(oE(y - h) - p E ( y - bi) = $(o.(y ;h + t(bi - bz))dt = ( b ~ - 62). JO V p C ( y - bi + t (bi - b2))dt = div(w(y))

42 CH.9. THE BROUWER DEGREE 2.1 Definition of the Degree

Hence it follows that Proof: Case 1. Let b 6 f (R). Tlieii 7 = p(b, f (n)) > O. Set i.- = 7/2llg1, (where l l . l l m denotes the norm in C ( l ; R n ) ) . If d ( f , 0 ,b2) - d ( f , = .f,div(,w(f ( x ) ) J j ( z ) d x

= .fn div(.u(x))dx = O Itl < E , then ~ ( b , ( f + t y ) ( R ) ) > P/2 > O and thus b 6 ( f + t g ) ( R ) . IIence (2.1.7) is trivially true as both sides vanish.

since ,u vanishes on 80.. Case 2. Let b 6 f ( S ) and let f-'(6) = { x l , ..., x,} so tliat

Let f E C 2 ( n ; Rn) and b 6 f ( 8 0 ) . Let p, be as in the previous .7,(:ci) # O for 1 5 i 5 m. Define proposition. Since, by Sard's theorem (cf. Theorem 1.3.4), the

h ( t , x ) = f ( x ) + t g ( x ) - b. singular values of f are of measure zero, there exist regular values in the ball B(b; p,) aiid the degrees of all such poiiits are tlie same Then, for 1 5 i m, by the previous proposition. \Ve are thus led to tlie following definition. h(0 ,x i ) = O Definition 2.1.2 Let f E G 2 ( 1 ; R n ) and b 6 f ( 8 0 ) . Set p, = 8,h(O,x;) = f l ( z i ) p(b, f ( 8 2 ) ) . The degree o,f f in 0 ,with respect to b is defivled as and f l ( x i ) is invert,ible, by assu~nption. Hence, by the implicit

d ( f , R,'.)) = d ( f , 0 , b') function theorem, there exist neighbourhoods ( - E ; , E < ) of O in W and pairwise disjoint neighbourhoods Ui of xi in 0 and fuiictioiis

rrih,ere H is any regular ualue i, B(b,p,).. 9 ; : ( - E < , & ; ) + U.i s11ch that the only solutions of h ( t , x ) = O in (-E;,&;) x Ui are of tlie form ( t , 9 ; ( t ) ) . Further, by shrinking t,lie

Exercise 2.1.1 Let f , b a~ id p, be as above. If b i - bl < p0/2, neighbourhoods if necessary, we can ensure that sgn(JJ+ty ( x ) ) =

show that ng~a(J,(x,)) in each Ui. Now set

d ( f , 0 , b l ) = d( f ,R.b) . . Exercise 2.1.2 Let C denote the co~nplex plane and let 0 C C E = l 0 , b ) . plane and f ( z ) = z".. Proposition 2.1.3 Let f , y E C 2 ( t ; Rn,) 0.d let b 6 f ( 8 0 ) . Then Now choose E < I ~ ~ I I { E , , po/311gilm}. Clearly, b ( f + t y ) ( 8 0 ) for there ezists E = ~ ( f , g , a ) , such that for O < t l < E ; Itl < E . In fact, p(b, ( f + t g ) ( 8 0 ) ) 2 2p0/3 while

d ( f + tg ,%b) = d ( f , R , b ) .

44 CH2. T H E B R O U W E R D E G R E E

Consequently (cf. Exercise 2.1.1),

and the proof is complete. We are now in a position to define the degree for a11 con-

tinuou~ functio~is. Let f E C(2;Rn) and let b e f (80) . Let po = p(b, f ( a f l ) ) . We can always find g E C2(n;1") such that Ilg- f 1 1 , < po/2. Then clearl~, b $i g(aCl) and the degree d ( g , f l , b) is well defined. If gl and g2 are two such functions, set 5 = g , -92. Then, for O < t < 1 , we have ( 1 f - ( g z + t a ) I , < p, and, by P rope sition 2.1.3, the function

is locaily constant, aiid hence, by the connectedness of [ O , l j , is coiistant oii this intervai. Thus

This paves the way for the following definitiori.

Definition 2.1.3 Let f , b and p, be as aboue. T h e n the degree of f in Cl with respect to b i s given by

for any g E C2(2;Rn) such that 1 1 f - gll, < po/2.. Remark 2.1.4 Compare this with the result of Exercise 2.1.1. .

Proposition 2.1.4 Let f E C(2; IWn) and let b f ( a f l ) . T h e n

I and so, by definition,

d ( f , f l , b ) = d ( g , R , b ) aiid d ( f - b , f l ,O) = d ( g - b, f l ,O) If b is a singular value of g , then we can find a regulas valite bl of g such tliat

Ib - bil < p ( b , d a f l ) ) / 2 so that

d(g - b l , f l , O ) = d ( g - b , f l ,O) and d ( g , Cl, b,) = d ( g , f l , b) . Since bl is a regular value of g , it is trivial to see that

and the proof is complete.

2.2 Properties of the Degree

In this section, we prove the basic properties of the Brouwer degree and look at some of their simple consequences.

Theorem 2.2.1 (i) (Cont inui ty with respect to the function) Let f E C ( 2 ; P ) and let b $i f (8Cl). There ezists a neighbourhood U of f i n C (2 ; R") such that for euery g E U ,

(ii) (Inuoriance ,under homotopy) Let H E C ( 2 x [O, 11; R") such thot b $2 H ( a R x [ O , 11). T h e n d ( H ( . , t ) , Cl, b) i s independent of t. (i i i) The degree is constant, with respect to b, i n each connected con~ponent of Rn\ f ( 8 f l ) . ( i t i ) (AdditiuitylLet f l l n Clz = 0 a t ~ d b $i f (8f l l ) U f (a%), where f E C ( 2 ; R n ) , f l = fli U f12 . T h e n

46 CH2. THE BROUWER DEGREE 2.2 Properties of the Degree 47

ProoR (i) Define and the result follows. . u = { S E Rn) I l f - gll, < po;4) Exercise 2.2.1 Let f : [a, b] i IR he continuous and such that

where Po = ~ ( b , f (ao)). If g E U , then p(b,g(aQ)) _> 3p0/4. ~h~~ f ( a ) f (b ) # O. Show that b # g ( a R ) and the degree is well-defined. Let h E Cy(n; rp) s,ich that I i f - hllm < po/R. Tl~en 1 d( f , ( a , h) , O ) = - [ s g n ( f ( h ) ) - s g n ( f (a))] . .

3 1 2 119 - h / / < jjpo 5 Z ~ ( b , g ( B R ) ) .

proposi t ion 2.2.1 f f E C@; Rn) and b 6 f (n), then d ( f , b) = Hence, by definition, 0 . r;quivcrlentlll, if d ( f ; 0 , b) # O , then there ezists x E 0. such that

4 9 , n , b ) = d ( h , f i , b) = d ( f , R , h ) . f ( s ) = b.

(ii) BY preceding step, d(H( . , t ) , R, b ) is locally constant and proofi Let p, = p(b, f (R) ) . If g is C' such that 1 1 f - gll, < p0/2, lieilce continuous and therefore coiistaiit on [O, 11 hg conllected- ~l~~~~ b ,y(n). T l ~ ~ i s , as b is now a regular 'valile' of .q: we have ness. that d(g , n, b) = O and the result follows. .

(iii) BY virtueof (2.1.9), d ( f , n , b ) = d ( f -h,ll ,O) and soif I b b , / coro i ia ry 2.2.1 If d ( f , ll, h) # 0; then f (n) is a neighbourhood is sma'l, d(f - *,a, 0) = d ( f - b ~ , Q , O ) . Thiis, tbe degree is 10- of h. c a l l ~ constant and thus continiious and so constaiit. 01, coi,liected compotients. Proof: I,nt Cb he the conriected component of Rn\f (an) contain-

ing b. Tlien, for a11 c E C*, we have d ( f , n , c ) # O and $0, by t l e (iv) Lct PO bc as in Step (i) and let g be a C2 fuil~1i~11 s ~ C I I 1liat prerrding proposition, Ch C f ( O ) and the conclusioii follows as I l f - g/Im < ~,,/2.Thrri, it is clear that Cb is open. .

4 % R, b) = d ( f , R, h) Exercise 2.2.2 If f ( O ) is contained in a proper subspace of Rn, d (g , f ) i , h) = d ( f , n i , h), i = l ,2 . show that, for a11 b 6 f ( a R ) , d ( f , n , b ) = O..

*ow B = B ( h ; ~ " 1 2 ) is connected and is cont,ained in RrL\g(afi) as ~ e l l as in Rn\9(a%) for i = 1,2 and heiice iii one connected Exercise 2.22. Let f ( z ) = zn + alzn-I + aizn-2 + ... + a, be a COmPonent of each of these sets. By Sard's tlieorem, there exists polyIiomial with complex coefficients in the coniplex variable 2. c E B such that it is a regular value of g and 80 (i) Assume that lal(+lail+ ...+I a,l < 1. Then iising thepro~ert ies

of the degree (cf. Exercise 2.1.2), show that f has a root in the d ( g , n 3 c ) = d ( g , n , b ) and d(g , n i , c ) = d ( g , n , , b) unit disc D C C.

for i = 1,2. Since g is c2 and c is regular, it red i ly follows fron, (ii) Using the change of variable z = c tu where C > 0, show that the definition of the degree that we can reduce the search for a root of a general polyliomial f to

the preceding case and thiis prove t,he fimdamental theoiem of d ( g , n , c ) = d ( ~ , %.c ) + d(.y,Rz, c) dgebra. .

48 CH2. THE BROUWrER DEGREE 2.2 Properties of th,e Degree

Proposition 2.2.2 (Excision) Let K C f i be a closed set arid let corollary 2.2.2 ~ e t f , g E C( f i ;Rn) . Assume th,at th,ere ezists b @ f ( a R ) U f ( K ) . Then H E c(an x [o, i]; R.) su,ch th,at H neuer assu,mes th,e ualue b and

such that H(.,O) = f l a n and H( . , 1) = g l a n Th,en d ( f , R, b) = d ( f , R\K, 6 ) . Proof: Choose g , a C 2 function such that d ( g ; R, 6 ) = d ( f , R, 6 ) d ( f , a , 6 ) = 4% a, b ) . and such that b @ g ( K ) . Now choose c, a regular value of g , close enough to b such that c 6 g (K) and belonging to the same Proof: By Tietze's theorem, we can extend H to H E C ( f i x connected component as b in Rn\g(aR) and Rn\g(a(R\K)). The [o, 11; R%). Set H(. , O ) = 7 and H ( . , 1) = 9. Then, by homotopy result now follows from the definition of the degree in the regular invariance of the degree, case.

d(?, R, 6 ) = d(T , R, 6 ) . The following two exercises can also be solved by reducing the The result now follows from the previous proposition since f = f

problern to the regular case. and g = on the boundary. .

Exercise 2.2.4 Let { R J } j t ~ be a family of pairwise disjoint open Remark 2.2.1 The above proposition and its corollary imply sets in Rn whose uni011 is contained in a bounded open set R. that, as long as the value b is not attained on the boundary Let f E C( f i ;R7' ) and b such that f - ' ( 6 ) c UjEJRj . Show that a homotopy, the degree is essentially determined by l l o r ~ l o t o ~ ~ d ( f , Ri, 6 ) = O for a11 biit a finite number of j E J and that classes of continiioiis functions defined on the boundary. If Sn

is the unit sphere in R>'+', and if O is not attained on it for a d ( f , R, b) = C d ( f ; R,, b).. ?,.+I

continuous function f : B -t R"+'; where Bn+' is the open j t J unit ball in R"+', we can consider 7 ( x ) = f ( x ) / l f ( x ) ( which then

Exercise 2.2.5 (Product Formula) For i = 1,2, let pj E C ( Q ; Rn' ) maps Sn into itself. We can define where R.i c RfL' is a bounded open set. Let bj $ pi(aRi). Show that d ( j ) = d(f,Bn+',O).

d ( (p i ,pz ) , Ri x R2, ( b i , b ) ) = d(pi , Ri, bl).d(rpz, R2, bz).. Then d ( . ) will be constant on homotopy classes of continuous maps Proposition 2.2.3 Let f , g E C ( f i ;Rn ) such that f = g on aR. ~f sn into itself. This gives rise to a theory of a topological degree Let b @ f (aR). Then for such maps. We can also define a degree of continuous maps 7: Sn + f L in another way. We know that the singular homology

d ( f , R, b) = d ( g , % h ) , groups of S fL are given by Proof: Define H E C ( f i x [O, 11; Rn) by Z if m = O , or n

H(x , t ) = t f ( 5 ) + (1 - t )g(x) . Then H(. , t ) = f = g on the boundary and so d(H(. , t ) , R,b) is

Thus Ygenerates a homomorphism defined and independent of t and the result follows by siiccessively setting t = O and t = 1.. f# : Hn(Sn) -t Hn(Sn)

50 CH2. THE BRO UWER DEGREE 2.9 BrouwerS Theorem

and, as H,,(Sn) Z Z1 f# is completely deteririined by f # ( l ) c Z. Proof: Assume that f has no fixed point. Then f (x) # x for every It t,iirns out that d(f) = f#( l ) . In the case n = 2, this is the z c 3. Tlie line segment starting at f (x) and goiiig to x is tlien familiar winding number for functions on S1.W well-defined and can be produced iii the same direction to nieet

at a point that we denote by p(x). Then : B" -i sn-l is Exercise 2.2.6 If n is odd: show that there does not exist. a clearly a relraclio~i and we thus get a contradiction to thr previ- ho~notopy H : Sn-' x [O, lj i Sn-' such that. H(x,O) = z and H(x , 1) = -2 for a11 x E Sn-I..

Remark 2.3.1 We c a i describe the inapping p above aiidytically Proposition 2.2.4 (Hairy Bo.11 Theorem) If n is odd, there is PLO as follows. We look for X 2 1 such that non-vanishiny vector field on Sn-', i.e., there is no continuous m0.p p : S1'-' i Rnsuch that p(x) # O and (v(x) ,x) = O (where IXx + ( i - A) f (.)I2 = 1 (., .) denotes the asual inner-product in Rn) for a11 z E Sn-l. Proof: If such a iriap were to exist, set $(x) = p(x)/lzJ. Then

I Z - f ( ~ ) 1 " ~ +2(x - f (x), f (z))X+ ( ~ f (z)i2 - 1) = 0. H(x, t ) = c o s ( ~ t ) x + s in(~ t )$(x)

Since f (x) - z # O for a11 x, this qiiadratic equation in X has defines a homotopy as in tlie preceding exercise, which is impos- exactly two roots. The product of the roots is iion-positive since sible. W 1 f (x)l < 1. Hence there are two real root,s, one positive and the

other negative. Since, at X = 1, the value of the quadratic expres- Remark 2.2.2 The map (x, y) h' (y , - x) is a non-vanishing vec- sion is (x(' - 1 5 O, t,he posit,ive root, is greater t h a ~ i or equal to tor field on S1..

.

2.3 Brouwer's Theorem and Applications Remark 2.3.2 Obviously, Brouwer's theore~n holds for any closed

Proposition 2.3.1 There is no retraction from the closed unit -n ball B zn IWn onto Sn-l , i.e., there does not exisf a continuous

?L mop ip : B i Sn.-l, such that p(x) = z for a11 z E Sn-'. Exercise 2.3.1 Show t,hat the followiiig statemeiits are equiva-

Proof If such a map existed, then O $ p(Bn) and since p = I, (i) There is i10 retraction f ro~n a closed ball iii Rn onto its bound- the ident.ity map, on Sn-I) we have

(ii) Every contiiiuous map of a closed ball in IRr' iiito itself has a O = d(p,Bn,O) = d(I,Bn,O) = 1

(iii) Let f : Rn i in be continuous. Let R > O such that for a11 which is inipossible. . 1x1 = R, we have ( f (x) ,x ) 2 O. Then tliere exists x, such t,hat

Theorem 2.3.1 (Brouwer's Fixed Point Theorem) Let f : i lx,l 5 R and f (x,) = 0.. II B ' be continuous. Then f has a fixed point.

52 GTI2. THE BROUWER DEGREE 2.3 Brovwer's Theorem

Exercise 2.3.2 Prove Brouwer's theorein directly from the prop- ~ h ~ ~ , f h% a fixed point x, E K and we have Azo = Azo wllere erties of the degree. .

Corollar~ 2.3.1 Let K C Rn be a compact and convez e~jbset. Rema& 2.3.3 The Perron - Frobenius theoreni states that if, in Let f : K + K be continuoun. Then f ha,s a fized point. addition A satisfies a condition called irreducibility, then , in f x t ,

the spectral radius is itself a (simple) eigenvalue and we liave ali Proof: If K is compact, there exists a ball B(o; R) containing K. eigenvector whose components are ali strictly positive. . Since K is closed and convex, let PK : R* -t K be the projection

map, i .e . given X E Rn, P K ( ~ ) E K is the unique point such that Example 2.3.2 (Periodic solutions, cf. Deimling [71) Let f :

12 - PK(X)] = minlx - yl. R x Rn + Rn be w - periodic in t ; i.e. f (t + w , X) = f (i, z ) for y E K

every ( t , ~ ) E ~x ~ n . Consider the system of ordinasy differentid Define f : P(O; R) K C B(o; R) by f(x) = f (PK(z)). Tlien f has a fixed point and as the iniage of this map is contained in K, ul(t) = f (t ,u). it follows that this fixed point r, is in K. But then pK(Zo) = xo

~~t us asume that f is continuous and t.hat there exists a ball and so

~ ( 0 ; ~ ) C ssuh that for every x E B ( o ; ~ ) , the initial value xo = f ( P ~ ( x o ) ) = f ( G )

which proves the result. W

We 110~ illustrate the usof Brouwer's theorem via some ex. lias a unique soiution u(t; x) OU [O, m) whicli co1ltinuousl~ depends amples.

on the initial value z . Thus the map Pt : B(0;r) -t defined by ptiz) = ~ ( t ; Z) iS continuous. Now assume further that the

Example 2.3.1 Let A be an n x n matrix such that a11 its coeffi- followiiig condition holds: cients are non-negative. Then A has a non-negative eigenvalue (H) FOI every t E [O, w], and for every x such that 1x1 = we have with a11 associated eigenvector whose components are a11 non- negative as well. To see this, set (f(t,.),x) < 0.

n

K = { z E ~ " / z ~ ~ O , I < i : < n , and xxi=l.) Then Pt : B(0: r ) -t B(0; r ) . For, if ( ~ ( t ; z)l = r, theu ,=I

This is a conipact convex set in R%. If there exists x, E K such that Azo = 0, then O is an eigenvalue and we are through. ~f not, AX # O for a11 X E Kaud so Z I = , ( A X ) ~ attains a strict positive Hence, hy the Brouwer fixed point theorem, P, (in particular) has mininium iii K. Define f : K + K by a fixed point, i.e. there exists x, E B(O; r ) such tliat ~ ( d ; 550) = zo.

fi(.) = (Az)i L;=I(Ax)j ' v ( t ) = u(t - lw;x,) for t E [ h , (h: + l)wl.

54 CH2. THE BROGWER DEGREE 2.3 Broz~wer's Theomm

BY the w-periodicity of f ; it follows that v is a w-periodic soliition of (2.3.1). L is called the Poincw operator associated (2.3.1).

We conclude with an example of the Galerkin method, ~ h i ~ method is very useful in constructing solutions of nonlinear equa- hions. The theorem which loilows is an ahstrart resiilt with appli- (TV)~ = (v, tili) + (Av, UJ.~) - (f, lu i ) . cations to nonlinear partia1 differential equation (cf. ~i~~~ [16]),

~ l ~ ~ ~ , T is continuous and (denobing the usual iri~ier product in Theorem 2.3.2 Lei H be a .~cpa.~a~ble Hilhert space toith, scalnr hy (,., .)k) we have prod~ct (., .) and let A : H -t H he o rnap such that (i) A i s 7nonoto~e, i . e . foi e ~ e v ~ E H , (TV, v)k = llwll ' + (Av,") - ( f .0)

= /(t,llQ (Av - 1 1 0 , ~ ) - (f - AO,w) (Av-Av+-v ) 2 O; 2 IIv1I2 - IIAO - f ll.IIvII.

A H 4(.u + Av) i s confinuuus; setting R = ((AO - f 11, we have that ( T v , v ) ~ 2 O for (v1 = (iiij A mops bounded sets into bounded sct,,-. j j v j j = R. H ~ ~ ~ ~ , by Broiiwer's theorem (cf. Exercise 2.3.11, there Then, given of?! f E H, th.ere ezists o unique soiution H of esists a 5 E ~k ~ U C ~ I that 15 5 R aiid TU = O. Thus, 2 E * the equotion,

u + A v = f (U, "oi) + (AU, wi) = (f, ~ i ) , 1 5 i 5 and, further,

JluJJ L IIAO - f 11. ( ) + ( 4 , I = (.fr v) for a11 v E Proof: Step 1. (Uniqiienrss) If v1 and v i were two oliitions of (2.3.4), we have and Ililll I - f 11.

v i - v2 + Av1 - Auz = O. step 4. ~~t {w,) be a complete orthonormal basis for H (wh'lch is separable) aitl se(, Wk = ~pa7~{'Uli, ui2, ..., 71~k). Lct Un E W L

Thus, i Ilui - 7jzll + (AUI - AUZ, u1 - %) = O iIunll IIAO - f 1 1

and heiice, hy virtue of (2.3.3), we have vi - = 0.

S t e ~ 2. ( A peori Estimate) If t i C H is a soliition of (2.3.4), then (un,v) + (AIL,,~) = ( f , ~ ) ~OI . v E + ( A - A , ) = (f - >40,u)

as guaranteed by the result of Step 3. Thus, upto the extraction and the estiniate (2.3.5) follows, again thaliks to (2.3.3). of a subsequence, un L 71 weAly in H .

56 CH2. THE BROUWER DEGREE 2.4 Borsuk's Theorem

Step 5 . Given v E H, the sequence {v,} defined by 2.4 Borsuk's Theorem n

Let 0 C R" be a bounded open set which is symmetric with v, = C(V,W,)W~

i=l respect to tlie origin, i.e. if x E 0, then -a: E 0. Let f E ~ ' ( n ; Rn) be an odd function, i.e. f ( -2) = - f ( x ) for a11 x E 0 .

is such that v, E W, for each n and v , + v strongly in H . From Assume that O $ f ( S ) U f(aC2). Assume furtlier tliat O $ n. If (2.3.7); we get f - ' (O) is empty, then d ( f , 0 , O ) = O . If not, the solution set has

to be of the form ( ~ L ? L , un) + (Au,, v,) = ( f , v,). u ~ l { x i , -x,}.

As {Av,) is bounded, we can also assume (after taking a further Since f ' is now an even function, we have Jf(-:L') = J f ( x ) . Thus, subsequence if necessary) that Aun - x we&ly in H. passiiig to the liniit as n + cu in (2.3.8), we get d ( f , R, O ) = C ~ $ l ( s g n ( J f ( x i ) ) + s g ? a ( ~ ~ ( - z i ) ) )

= 2 CEl s g n ( J j ( x z ) ) . (u , v ) + (x , v ) = ( f , v ) for a11 11 E H.

Thus, if O $ n, ilie degree is an even integer. Step 6. By (2.3.3) we have for any v E H , If O E Q, we do have f ( O ) = O. Thus the solution set is now of

0 5 X n = ( A u n - A t ~ , u n - v ) {O) 01 {O) ugl { ~ i , -xi>. = un) - (AfLn, 7 1 ) - (a4t1, un, - v ) In the fornier case? the degree is f 1 and in the latter it is f l + = ( ~ , u ~ ) - I I u ~ ~ I I ~ - ( ~ ~ L , , , ~ I ) - ( A u , u , - u ) ,

usiiig (2.3.8). Thus, -

O 5 limn+coXn = ( f , u ) - hn+,~~un,I12 - ( x , V ) - (Ao , u - o) 5 ( f , U ) - 1 1 ~ 1 1 ~ - ( f , v ) + (u, V ) - ( A v , u - v ) nical lemmas which essentially deal with the extension of functions

to larger sets retaining special properties. using (2.3.9). Thus,

Lemma 2.4.1 Let K C Rn be compact. Let p E C ( K ; R m ) where ( f - - 411, u - v ) + ( A u - Av, - v ) 2 0. m > n. Let O $ ip(K). Then, i f Q is any cube containing K , there

ezists pg E C(Q;Rn,) extending ip and such that O $ ipQ(Q). Lei X > 0 and w E H. Set v = u - Xw in (2.3.10) to get

Proof: Step 1. Since K is compact, aiid O $ p ( K ) , we have ( f - u - Au, w ) + ( A u - A(U - X W ) , W ) 2 o. cu = inf 1p(x)1 > 0.

As + 0 , by the hemi-contiiiuity of A, tlie second term on the TEK left-hand side tends to zero. Thiis ( f - u - AU, w ) 2 O for Let O < E < n /2 . Let + E C1(Q; IRm) such that w E H and, by considering -w in place of w , we conclude that u satisfies (2.3.4). .

58 CH2. T H E BROUWER DEGREE 2.4 Borsuk's Theorem

If 0 ?b(Q), set $1 = ?b. If, on the other hand, O E $(Q), define proofi We will proceed by induction on n. Let n = 1. We set Q E C1(Q x Rm-n;IWm) by -

R = [ - 6 , - E ] U [ E , ~ ] , O < E < 6. Q(xi Y ) = $ (x ) for x E Q, y E By the preceding lemma, we can extend 9 to 91 on [E, 61 siich that

it is non-vanishing. Now define Then for ali z = ( x , Y ) E Q x Rm-n, Jq(z) = O . Hence, by Sard>s theorem, the range of 1C. is of measure zero. Thus, there exists a ? I ( & ) with Ia1 < ~ / 2 . Set $1 = ?I - a.

- v l ( -x ) , x E [-6, - E ] . Thus, in either case, there exists ?Ii E C1(Q;Rm) such that

11?bi - ( P ~ I ~ , K E and O ?bl(Q). Thus iP has the required properties. We now assume the result to hold for a11 dimensions 5 ( n - 1).

Step 2. Let q : R+ pt R+ be defined by ~ e t 2 c R ~ . Set

a+ = { X E R I X N > o}, a- = { X E a / 2 N < 0). N ~ ~ , a(a n ~ n - 1 ) = aR n Rnn-' and, by induction, we can extend p to @ on n n IRn-', with 6 odd and non-vanishing. Now, let Q Define lpi(x) = ?Ii(x)/q(1$~1(~)1) for x E Q. By definition, I(o1(x)l >

and so O $4 pi(Q). If x E K , be a cube in R$ containing a+ and n n Rn-' . Consider "

l ? I l ( ~ ) l 2 l9(x)1 - Ilp(x) - ?I1(x)I > a - E 2 012.

Hence ~(1dJi ( x ) 1) = 1 and so rpl = d J l on K . Then Ipl can be extended t.o a uon-vanishing function PQ on Q. Step 3. Let 6' = 91 - 9. By the Tietze extension theorem, we can extend 6' to gon Q such that 14 < E on Q (since 16') = l d J i < E on K ) . Set I ~ Q = 91 - 8. Then I ~ Q extends p and It is now immediate to see that has the required properties. W

I P Q I > J p i J - 16') > a/2 - E > O. Lemma 2.4.3 Let a C Rn be a bo'unded open set which is sym- metric with respect to the origin and such that O @ 2. Let 9 E

Hence O pg(Q), which completes the proof. W c ( a a ; R n ) be an odd and non-vanishing mapping. Then, there

Lemma 2.4.2 Let a C Rn be a bounded open set symmetric with respect to the origin and such that O a. Let 9 E C (a0 ;Rm) where m > n . Assume that p ia odd and that O 9(aCl). Then Proofi Consider ip restricted to aC2nRn-'. It is odd and noiivan- there exista @ E C(n;IWm) extending p which is odd and non- ishing and belongs to C ( a ( a n ~ ~ - l ) ; Rn)). Thus, by the preceding uanishing. lemma, we can extend it to a continuous, odd and nonvanishing

60 C'H2. THE BROUWER DEGREE 2.4 Borsok's Theorein

maPping on fi n E-' taking values in R". Let 6 be equal to this proof: Since O E R , let B ( 0 ; r ) be a ball of (sufficielitl~ small) functio~l on m R n - ' and to 9 on 8O+ n 861. B~ ~ i e t ~ ~ ' ~ theorem, radius contaiiied in R. By Tietze's tlieorem, let ? E C ( n ; p ) we can exterid it to a function o11 a+ aarid then, as risual, extend it as an odd map to a11 of 1.. Lemma 2.4.4 Let R be as in the precediny lemnaa and let E C ( a ; R n ) be odd and non-vanishiny on 8 0 . ~ h ~ ~ , d(9 , R, O ) is ~ 1 , ~ ~ ? does not vanish for 1x1 = r and so hy PropositioIl 2.2.3, etle??,. excisioii and additivity wc: have

Proof: BY the preceding lenima, we can fiiid a c )hich d(ip,R,O) = d(2ii7R,0) is eqilal to 9 on 8Q, is odd and which does not vanish on R n p - l , = d($; R\C)B(O; r ) , O ) Then, (cf. Proposition 2.2.3) = d($, R\B(o; r ) , 0) + d($, B ( 0 ; r ) , 0).

B~ Lemma 2.4.4, the first term on the right is an even iiiteger and d ( 9 , a, 0 ) = d ( a , n,o).

siiice S, = I on the houndary of B ( 0 ; r ) , the second terin is unity. There exists E > O such tliat if ( (a -@\I, < E , tlieii the degrees of This proves the theorem. . @ and 4 in R with respect to O are the sanie. choose 7jl CZ Corollary 2.4.1 Let R be as i n the preceding theorem and lei

E c.'(fi;Rn) be odd on the boundnry. Then there ezist X , Y E D such that 9 ( x ) = O and P ( Y ) = Y. proof: If 9 vanishes on the boundary, we liave x E 8O such that

- O. ~f not, thc degree d(

62 CH2. THE BROUWER DEGREE 2.5 The Genus

Proofi Defilie H ( z , t ) = ~ ( x ) - tp(-x) for (x, t) E 8n x [O, 11. B~ Example 2.4.2 (Sandwich theorem) Given three regions in R3, hypothesis, H does not vanish on the bouiidary and the degree is there exists a single plane which divides each region into two thus well-defiiied and independent of t. We have H(., O) = p while parts of equal volume. (A single knife stroke cai1 halve a piece H(. , 1) is odd. The sesult now follows fsom Borsuk's theorem. . of bread, a piece of cheese and a piece of ham, placed arbitrarily

in space!!) The result is true for any n regions in IESn . Consider Coroiiary 2.4.4 Let n be as in the preceding corollary and let z. = ( ~ ' , z , + ~ ) E Sn, where x1 E Rn, and the hyperplane defined p E C(aR;IWn) be odd ond non-uanishin,g. Then, theve does not ezist (I hom.otopy H E C(8n x [O, 11; Rn) which is non-oaraishing H, = {y E PSn / y.xl = xn+l}. and such that H(.,O) = p and H(. , 1) _= z, E Rn\{O}.

Proofi If such a H existed, we can extend it, by Tietze's theorern, H$ = {y E Rn I y.zl > zn+i}. to H E x [O, 11; Rn) and while d(H(. , O), n, 0) is odd, we will If p is the n-dimensional Lebesgue ineasure, define have d(H(., l), n, O) = 0, which is impossible. .

pi(z) = p(Ai n H$), 1 I i I n Exercise 2.4.1 Show that no sphese in R" can be deformed within where {Ai};=, are the given regions. By the Borsuk - Ulam the- itself to a singlc point. . orem, there exists x, E Sn such that pi(x,) = pi(-x,) for a11

1 I i I n which gives the resiilt. . Corollary 2.4.5 Let n be os in the pvecediny corollury and let p E C(8n; Rn) be odd an,d such that its im.age is contained in Exercise 2.4.2 Show that there does not exist an odd continuous a proper. subs~uce of Rn. Then there evists x, E 8n such that

map f : Sn + Sm for m < n.. ip(x0) = o. Proofi If not, p would be odd and ~ion-vanishing on the boundary R e m a r k 2.4.1 We can te11 two finite sets apart by counting their

elements. Two finite dimensional spxes cai1 be coni~ared by look- and hence its image would be a neighbourhood of the origin wliich is not possible. . ing at their dime~isions. We have dim E > dim F if, aiid only if,

there is no injective linear map from E into F . The ahove exercise Corollary 2.4.6 (BorsuA - Ulam,) Let n be as aboue and let p E is, in sonie sense. a result in this spirit, to compare two spheres. C(30; Rn) be such that its emage is contained in a proper. subspace More generally, we can compare two sets that are symmetric with of Rn. Then there exists t E 8n suxh that p ( t ) = p(-t) . respect to the origin and not containing it by examining the exis-

tente of continuous odd maps from orie to the other. This leads Proofi Apply the preceding corollary to $(z) = p(z) - v(-x)..

in the next section. . Example 2.4.1 Assuiue that the surface of the earth is spherical and that the temperature and atmospheric pressure vary contin- 2.5 The Genus uously on it. Then there exist a pair of antipodal points with the sanie temperatiire and the same pressure. . Let E be a (real) Banach space and let C(E) denote the collection

of all closed subsets in E that are symmetric with respect to the

64 2 . T H E BROUWER DEGREE 2.5 The Genus

origiii aiid not contaiiiiiig it. (vi) If A is compact, then ?(A) < m. (uii) Let A be cornpact. Define

Definition 2.5.1 Let A E C(E) . W e denote by ?(A) the genus of A th.: smallest positivr integer n such that there ezists NA(A) = {x E E 1 P(X, A) L 6). a continuous odd map of A into Rn,\{O). We set y(@) = j and qf no such n ezists for A, we set ?(A) = m.. Then, for suflciently small 6, Example 2.5.1 Let E = Wnfi and let Cl be a bounded open set, y(N(A)) = d A ) . synimetric with respect to tlie origin and containing jt.

~h~~ ~ ( 3 0 ) < n siiice we have the identity map I : 812 -, R ~ \ { o ) which proof: (i) ~~t = n < m (otherwise the resiilt is trivially is odd. But by Co~.olla~y 2.4.5, there is no odd noii-vaiiishing map '

truej. ~f p : -i Rn \{O} is odd, then ip o f : A -3 Rn\{O} is odd into a proper subspace of RVL. Thus, $30) = IL. In particular, arid the result follows.

y(S"-') = 711.. (ii) Set f = I, tlie identity rnap, in (i). (iii) Follows froin (2.5.1); we liave y(A) 5 y(B) 5 '7(A). Example 2.5.2 Let E be a Banach space and let 2 E E , # 0. (iv) ~f either y ( ~ ) or y(B) is infinite, tlien the residt is trividly Set A = B(x; r ) 'J B(-x; r) wliere O < r <

06 CH2. THE BROUWER DEGREE 2.5 The Genus

P r o ~ o s i t i o n 2-5.1 Lei 0 C R'' be a bou,nded open set ,which, Proof: Apply the preceding proposition to lhe function P(X) = tains the origin and i8 symmetric with resyect to it. ~~t : +

be a continums an.d odd map, and let m < n. Let Lemma 2.5.1 There exists a coziering of Sn-' by n closed antipo-

A = {.x t dR 1 p(x) = 0). Z=I ,where B; = c~u(-C;), C;n(-Ci) = da1 sets, i.e., Sn-l = U"

Then y(A) 2 n - m. Proofi If n = 1, we have S0 = {-1,+1) and so Bl = (-1) U

Proof: Let Ns(A) be such that (2.5.5) holds. We claim that for {+I). I fn = 2, S1 = BlUB2 where Bi = {(x, y) E / 1x1 2 112) some E > O, Nre have Z, C Na(A), where and B~ = { ( x , ~ ) E S1 / (yl 2 1/21, Assume the result upto R,.

set s"-1 = ulLIB:. Let x = (xt,x,+l) E R?+', with X' E p. 2, = {X d o 1 lp(z)1 1 E). Identify the hyperplane {x,+I = 0) with Rn. Define

If not, we liave a sequence E,, decreasing to zero and x, E 2," {(Z ' ,Z~,+I) E sn I X,+l 2 1/41. ~ i t h 19(xn)l 5 E , and XG Ni(A). Thus, p(z,, A) 2 6. since 80 is conlpact, foi. a subsequence, x, + x and so p ( ~ , A) 2 6. ou For 1 5 i 5 n, define the hand, 9 ( ~ ) = O aiid 80 x E A, a coritradiction aiid so lhe claim holds.

Thust A c ZE C NdA) and so y(Z,) = y(A). NOW, for 7 > o, C: = {(Z', x . + ~ ) t sn I I X , + ~ I 5 i/2, xl/ JZ E c:} let

where B; = C,' u (-C;), C: n (-C:) = 0. Then the Ci for 1 2 i I C, = {X c 8 2 , 1 kf(x)l 2 9). n+l =e closed, Ciii(-C,) = 0 and the Bi = Ciu(-Ci) cover Sn..

If P(Y) = ylllyll is the radial projectiori in Rm, then The above lemma is used to prove a result which will allow us

P o p : C, +Rm\{O) to calculate the genus of a se1 made up of sets of genus unity. is odd and cotitinuous and so y(C,) < ~ n . Thus Theorem 2.5.2 Let A E C(E). Then y(A) = n i% and only

if , is the least integer such that there exist sets A. E C ( E ) for - 7(an \cq) 2 ~ ( 2 0 ) -$C,) L n - m. 1 5 i 5 n such that y(Ai) = 1 for a11 such i and A C u f ~ ~ A i .

Proofi ~f y(A) = n, then there exist D1, ..., D , in C ( E ) covering A and each of them having genus unity. For, if (o is lhe odd non-

C o r o l l a r ~ 2.5.1 Let 0 be as ahove and let m < R,. ~~t ?I E vanishing map into R" from A, and if P is lhe radial projectiotl c(dn;iwm). ~f in Rn, then P o (o maps A into 5'''-'. If Bi and Ci are as in the

stateirient of the preceding lemma, then { ( P o p)-'(Bi)) covers A = {x E 80 I $(x) = li/(-x))

then y(A) 2 n - m. oi = (P O (o)-'(&) = ( P o (o)-l(~,,) U ( P o (o)-'(-Ci)

68 CH2. THE BRO UWER DEGREE

and so, as the two sets on the right are disjoint, y(Di ) = 1.

Sufficiency: I the { A i ) exist as in the statement of the theorem, then clearly ? ( A ) I n. If y ( A ) = m < n, then by the preceding argument, there exist m srts Dj witli the same properties as the Ai, contradicting the minimality of n.

Necessity: By our initial argunient, we know that A can be cov- ered by n sets of genus unity. If n were not minimal, then A The Leray - Schauder Degree would be covered by m, sets o genus unity, where m < n and then ? ( A ) < nL, a contradiction. 8

Inspired by the above theorem, we can define a notion analo- 3.1 Preliminaries gous to the genus in topological spaces.

~~t x be a (real) Banach space. Henceforth, unless otherwise Definition 2.5.2 Let X be a topological space and A C X a stated, a11 mappings of X into itself, or any other space, will be closed subset. A is said to be of category 1 i n X (catx(A) = 1) assunied to be coiitinuous and mapping bounded sets into bounded i f it can be deformed continuously to a single point, i.e.; there ex- ist H E C ( A x [O, 11; X ) soch that H ( z , O ) = x for a11 x E A and

Definition 3.1.1 Let X and Y be Banach spaces. Let fl be an H ( z , 1) = z, E A for a11 z E A.. open set in X. Let T : R + Y be continuous. Then T is said to be

Definition 2.5.3 Let X he a topological space and let A C X a compact if it maps bounded sets (in X) into relativek comPaCt closed subset. We say that c a t x ( A ) = n if. and only if, n is the least integer such that there exst closed sets Ai for 1 < i < n covering A and such that c a t x ( A i ) = 1 for each such i. If no such Example 3.1.1 By Ascoli's theorem, the injection C'([O, 11; R) + n exists, we say that ca tx ( i l ) = cw.8 C([O, 11; R) is compact. .

The category defined above, called the Lyusternik - Schnirelman Example 3.1.2 Let K E C([O, li x [O, 11; R). Let f E c([@ 11; Category, lias properties analogous to the genus. It is more flexi- ble and more general thaii the genus. But its properties are more difficult to prove. The genus and the category give information on the size of solution sets of nonlinear rquations.

Then T is a compact linear operator on C([0 ,1] ; R). To see this. notice that K is uniformly continuous. Hence, given E > 0, there exists 6 > O such that (zl - $21 < 6 imples that, for Y E [O, 11,

K Y - K Y < C / C

70 C H S. T H E L E R A Y - S C H A U D E R D E G R E E

where C > 0 is fixed. Hence for ai1 1 1 f , 5 C , we have

and so the the image under T o the ball of radius C is equicon- tinuous. Clearly it is also bounded. Thus, the result follows, once again, froin Ascoli's theorem.

Example 3.1.3 Let R C Rn be a bounded open set. Then, by tlie Rellich - Kondrasov theorem (c. for instance, Kesavan [13]) we have that the injection

is compact. .

A11 the above examples deal with conipact linear operators

Example 3.1.4 Let X and Y be Banach spaces and let T : X -t Y be such that T ( X ) is contained in a finite dimensional subspace o Y . Clearly such a map is compact. Such maps are called maps o finite rank. 1

Exercise 3.1.1 Let T,, T : X + Y be bounded linear inaps such that a11 the T, are o finite rank and IIT, - TI1 -t 0 as 71. 3 W. Show tliat T is coinpact. .

Henceforth, throughout this chapter, R will denote a bounded open subset of a Banach space X. The identity operator on X will, as usual, be denoted by I.

Definition 3.1.2 Let T : + X be a compact rnap. T h e map- ping p = I - T is called a compact perturbation o the iden- tity. .

Proposition 3.1.1 A compact perturbation of the identity i n X is closed (i.e. maps closed sets in to closed sets) and proper (i.e. in,uerse images of compact sets are compact).

Proof: Let p = I - T be a coinpact perturbation o the identity. Let A c X be closed. Let y, = ~ ( z , ) E p ( A ) and let y, -t y iii X . Tlius, y, = .c, - Tx,. Since {z,} is bounded, we have, for a subsequence, T x , -t z and so x, -t y + z and y + z E A, since A is closed. It then follows that y = p(y + z ) and thus p is closed.

Let A C X be compact. Let {x,} be a sequence in p- ' (A) . Thus, y, = x, - Tr, E A and since A is compact, we have , for a sub- sequene, y, -t y E A. Since R is bounded, again, for a further

, subsequeiice, T x , -t z . Again, it follows that, for that subse- quence i11 question, x, -t y + z and thus v- ' (A) is