Some fixed point theorems and applications

Voniinear r\nnlwr. Theory. Pnnred m Grear Bnraln

.Merhodr

& A(ppimnom.

Vol.

10. So

II. pp. 1193-1302.

036?-5466x86 53.00 + 00 Pergamon Journals Ltd.

1986

SOiME FIXED POINT THEOREMS

AND APPLICATIONS

Guo DAWN Department

of IMathematics, Shandong University, Jinan, Shandong, The People’s Republic of China (Received

15 September 1985; received for publication 20 January 1986)

Key rtsords and phrases: Completely continuous operator, and ordered Banach space.

Leray-Schauder

degree, fixed point, cone

1. INTRODUCTION

paper we use the Leray-Schauder degree theory [l, 21 to establish some fixed point theorems and give some applications to nonlinear integral equations. In Section 2 we establish a fixed point theorem, which can be called as the fixed point theorem of domain expansion and compression (theorem 1) and which is essentially different from the Krasnosel’skii’s fixed point theorem of cone expansion and compression [3,4]. As a consequence we obtain an improvement of a theorem obtained by Cronin [5]. In Section 3 we establish a fixed point theorem for decreasing operators (theorem 2) and from it we obtain an improvement of a theorem obtained by Stuart [6].

IN THIS

2. A FIXED

LEEMA

POINT

THEOREM

1. Let Q be a bounded

OF DOMAIN

EXPANSION

open set in infinite-dimensional

A: R-,

AND COMPRESSION

real Banach space E and inf

E be completely continuous (i.e. continuous and compact). Suppose that (i) degree lpXl/ > 0 and (ii) Ax = px, x E ~3523 ,u 4 (0,11.Then the Leray-Schauder deg(l - A, Q, 0) = 0,

where I denotes the identical operator

and 8 denotes the null-element

XEJR

(1)

of E.

Proof. We first prove that

Ly= _ir&l~ux

- AXI1’ 0.

(2)

In fact, suppose that (2) is not true, then there exist x, E dQ and ,u,,E [0, l] (n = 1,2,3, . . .) such that ]l,u~,, - Ax,//+ 0 as n + m. By the complete continuity of A and the boundness of ,uu,, there exists a subsequence {nk} of {n} such that pn,, += y. E [0, l] and Axnk - y. E E. Obviously, ,uo > 0, since p. = 0 implies ]wxnk]l --, 0, which contradicts hypothesis (i). Therefore Xno + ,u;*yo = X0 E aS2 and by the continuity of A we obtain ,u~x~ = Axe, 0 < ,po s 1, in contradiction with hypothesis (ii). This shows that (2) is true. By method of Leray-Schauder we construct a completely continuous operator A, from 5 into some finite-dimensional subspace E. of E such that IjAX- &Xl/ < a,

Vx E n.

Let the dimension of E. be s and Cy,, y,, . . . , ys} be a base of E,,. Since E is infinite-dimensional, there exists y,, 1 E E\Eo such that J1ys+lII = 1. Let El be the subspace spanned by y,, y2, . . . , 1293

ys, ys,i. Evidently, we can regard A, as a completely continuous therefore. by the definition of Leray-Schauder degree. we have

operator

from a’-into E,,

deg(l-A.Q.f3)=degE,(I-i\,.R,.e).

(3)

where R, = Q fl E, and the right-hand of (3) denotes the Brouvver degree in E,. Since, by virtue of (2).

it follows from the homotopy

invariance of Brouwer degree that

deg~,(l-A,.R,,8)=deg,,(-A,,4,,8). Consider the constant operator

(5)

A 0 :AOx = Y~+~, Vx E 3,.

Obviously,

degE,(A,,, R,. 0) = 0.

(6)

We now prove that -rA,s

+ (1 - t)Aox + 0,

OSfS

VxEdQ,,

1.

(7)

Indeed. if there exist x0 E aR, and to E [0, l] such that -toA,sO

f (1 - to)Aox,,

= 0.

then observing to $ 1 (since AIxo # 8 hy (4)) and Alxo E EU. we obtain

in contradiction with ys+, E AEo. Brouwer degree, we have

Hence, by virtue of (7) and the homotopy

deg,,(-A,.

Q,, 8) = degE,(Ao.

R,, e).

invariance of (8)

Finally, (3), (j), (8) and (6) imply (1) and our proof is complete. COROLLARY1. Let R be a bounded 0 @ 3 $2 and A : Q + E be completely

open set in infinite-dimensional continuous. Suppose that

I/A,~lt2 ttxij,

Ax + 0x3

VxEdQ.

real Banach space E, (9)

Then (1) holds. Proof It is sufficient to prove that the conditions (i) and (ii) of lemma 1 are satisfied. From (9) and 8 $. JQ, we obtain

hence, (i) is satisfied. If (ii) is not satisfied, then there exist x0 E aR and 0 < ,uo G 1 such that $z~;~;~x,,. From (9) we know ,uo f 1, and therefore lbxoll = ~o\~olj < I~,\\, in contradiction

Some fixed point

LEMMA 2. Let R be a bounded completely continuous. Suppose

open that

theorems

set in real Banach

Ax = px,

1295

dnd apphcations

E. 8 E Q and A : n-

space

xEdR+,u
E be

(10)

Then deg(Z-A,R,

0) = 1.

(11)

Proof. Let h,(x) = x - fAx. From (10) it is easy to know 8 &Zq(aQ) homotopy invariance of Leray-Schauder degree, we obtain

(0 < t s 1) and by the

deg(Z - A, ‘2, 0) = deg(Z. R. 0) = 1. COROLLARY 2. Let Q be a bounded open set in real Banach completely continuous. Suppose that llA+r]lc ti,~]I, Then

Ax f x,

space E, 8 E Q and ,-I :n --f E be

t/s E as2.

(12)

(11) holds.

Proof. It is easy to see that (12) implies

(10).

THEOREM 1. Let R, and Rz be two bounded open sets in infinite-dimensional real Banach space E such that 0 E R, and 3, C Rz. Let A : ?&\Q, - E be completely continuous. Suppose that one of the two conditions

IL4 c I/XI/? v’xE an,;

pIxl12 ljxli,

vx E dRz

WJ

and ]iAx]] 2 ]]x]]. is satisfied.

Then

v’x E aR,;

A has at least one fixed point

IM s Italic vx E a522

Wd

in Rl\Q,.

Proof. We need only to prove the theorem under condition (HI), since the proof is similar when (H,) is satisfied. By the extension theorem [7], A has a completely continuous extension (also denoted by A) from s1 into E. We may assume that A has no fixed points on an, and aQ,. By corollary 1 and corollary 2 we have deg(Z - A, R?, 0) = 0,

deg(Z - A, Q,, 0) = 1,

and therefore deg(Z-A,~~\~,,~)=deg(Z-A,~,,~)-deg(Z-A,R,,8)=O-l=-lfO, hence,

A has a fixed point

in R?\E,

and our theorem

is proved.

Remark 1. Theorem 1 was first proved in Chinese (see [lo]) and it is not true for finitedimensional E. For example, let E = R’, 52, = {(x, y)/x’ + y’ C l}, Rz = {(x, y)ix’ + y’ < 4}, AZ = zl, where z = x + iy = reie, z, = re’(ec(~T~4)),i.e. A is a rotation around the origin with angle x/4. Evidently, ]/A.z~]= ]jz I/ f or z E dR, and z E dS??. but A has no fixed points in GZ\RI.

COROLLARY3. Let E be an infinite-dimensional real Banach space and A : E continuous A8 = 8. Suppose that one of the two conditions

E be completely

W;) and

is satisfied. Then the following two conclusions hold. (a) Every ,n f 0 is an eigenvalue of A, i.e., there exists x4, E E. xu f 8, such that Ax, = px,; (b) !lXllxllxJ,lI = += under (H3) and tl$UIl = 0 under (H,). Proof. We need only to prove this corollary under condition (H,), since the proof is similar when (H,) is satisfied. For given ~14 0. by virtue of (H,), there exist R > r > 0 such that V,r E E,

I/x/l = r

and

hence, condition (H,) is satisfied for operator (l/p)A. It follows therefore from theorem 1 that operator (l/u)A has a fixed point x, in R2\R, and conclusion (a) is proved. It remains to prove JJxJJ+ +m as ,u- = (i.e. conclusion (b)). Supposing that this is not true, there exist a number c > 0 and a sequence J~,,--,x such that //_XU”]/ < c

(n = 1,2,3,.

. .).

hence, the sequence {l/x,,//} contains a subsequence which converges to a number ~(0 c t c c). For simplicity of notation, assume that {I~x,~,]]} itself converges to t. If t > 0, then Ib,U,jI> t/2 for sufficiently large n (say, n > N), hence

where M = suplwxjl, which contradicts ~1,’ 3~. hl/
in contradiction

with ,u,+

Remark 2. Corollary

x too. Hence, /.vJ -+ +x as !l--, r and our proof is complete.

3 is an improvement of a theorem obtained by Cronin in 1972 (see [5, theorem 41). Cronin’s theorem asserts only that one of the two numbers ,Uand -11 must be an

Some fixed point theorems

eigenvalue of A for every ,U E (0. 11 when contain the conclusion (b). Example

1. Consider

the nonlinear

(H3) is satisfied

integral

pp(x)

1397

and Cronin’s

theorem

does

not

equation

I 4x,

=

and applications

dy,

~)lrp(~P

(13)

G

where p > 1, G is a bounded closed domain in Euclidean space Rv and kernel k(x. y) is nonnegative and continuous on G x G and JG k(x, y) do > 0 for all y E G. We prove that for every ,U + 0 the integral equation (13) has a continuous solution p!,(x), which does not equal to zero identically; moreover, we have

~imllcp,uIIc= +x U-X and

(14)

limlbPyllc = 0. u-o

Proof. Define

operator

(13

A by Aq =

iG

(16)

k(x, Y>MY>I~ dy,

then A = KF, where Kg7 =

IG

W,

YMY)

dy

and

It is evident that Fis a bounded and continuous operator from space L!‘(G) into space L(G), hence, A = KF is a completely continuous operator from LP(G) into C(G) (space of continuous on G functions). Obviously, we can regard A as a completely continuous operator from L!‘(G) into L!‘(G). It is easy to see that

II-%4LP where

M =

max

s

Ames

(3

ljp

II&, )

(17)

k(x, y), hence

(x.y)EGxG

lim -= IkwIILP le~lL-0 Il~llLP On the other

hand,

by a known

inequality P

dy

= (mes

G)(l/P)-l

\G

UP

1) du

o

(18)

.

(see [9]), we have 2 (mes G)(i/P)-1 IG

k

IG

0,

Y)]~(Y)~ dy

Iv(Y>I~ dY _/ k(x7y) h 2 P(mes G)(l~p)-lII~IIPL~. G

1298

where

GLO D.-\JLs

,f? = min Jo k(x, y) dr > 0. i.e. IfG

h4d~7

maps P(G) into trivial inequality

C(G).

2

we prove

G)”

qU( x ) is a continuous

IlRA.P Finally,

p(mes

(15) as follows.

function.

< jlq,llc.

From

P)-l

.

Moreover,

(14) is deduced

from

A a

(mes G)“‘.

(19), we obtain

/PI . ll~p,~ll~P = llAyl.li~~ Z-(mesG)(l’p)-’ll~D1&~

(P+ 01,

hence

< /f-l/(~-l) IlrFullLP

(mes

(21)

G)‘.‘PI;~l”(P-‘).

But (22)

it follows

therefore

which implies

from (21) and (22) that

(15). Our proof

is complete.

Remark 3. The conclusions of example 1 improve a theorem obtained by Cronin in 1972 (see [5, theorem 51). Cronin’s theorem asserts only that for any ,II E [- 1,0) U (0, 11, the integral equation (13) has a continuous solution, which does not equal to zero identically. Cronin’s theorem does not contain the conclusions (14) and (15) and hypotheses about kernel k(x.y) is more strong: k(x, y) is continuous and k(x, y) > 0 for all (x, y) E G x G.

3. A FIXED

POINT

THEOREM

FOR

DECREASING

OPERATORS

Let (15, P) be an ordered Banach space, where P is a cone in real Banach space E. P is called normal, if there exists a constant N > 0 such that 8
2.

Suppose

that (i) P is normal

and A : P + P is decreasing

and completely

continu-

Some fixed pomr theorems

ous; (ii) A0 > f3, A’8 2 E&O, where 17= ~(x, t) > 0 such that

s [r(l + rj)]-‘AX.

-_K*~/-+o

(fZ+

on account

of the decreasing 8

=

~1~ s

11 ?n

=).

(24)

11, = Au,_ I

=

s ..

. s

LL:,

A’u~,_~,

S.

. . s

LL~“+~

s.

. s

u~,,+~ = A2uln-,

-

Ll,,II

c

AjILl*

u3

s

Lll

(12

s

&,

s

(27) (26). (27) and subsequence

(n ’ n!c),

-

II,,,//

ll*

(26)

. .).

(II

>

n,),

N denotes the normal constant of P, hence uln-+ U* (n-, x). x), and so by hypothesis ,} is convergent: uznTI - u* (n=

Ae,

it follows from a convergent

ILL?,,+

8 < &,,A0 s A%

=

(n = 1.2.3,.

8 s hnk s kn s II* IILl*

(25)

.).

of A. it is easy to see

property

L12

(n = 1.2,3,.

Since P is normal, the ordered interval [e, Af?] is bounded, continuity of A’ that {uzn} contains the complete {uZnk}:~llnk ---f u*(k+ x). By virtue of (26), we have

Taking

successively

Putting U0 = 8,

where know

exists

(23)

fixed point X* > 0. Moreover, constructing . .) for any initial _yuE P. we have ji.K,

Proof.

1799

.sO> 0 and (iii) for any x > 8 and 0 < r < 1, there

A(R) Then A has exactly one positive sequence x, =Ax,_,(n= 1,2.3..

and applicarions

s

U*

s

LIJn,,

(n

In the same way, we (ii) and (26), . .).

1,2,3,.

=

W)

limit in (27) and (25), we obtain 11*

By virtue

=

A?u*,

u*

=

A?“*,

u*

=

ALl*.

u*

=

Au*.

(2%

of (28), we see I(*

5

&OAe = E()U, 5 &()U+,

it hence, putting t, = sup{t > O]u* 2 tu+}. we find 0 < &oc to < +m and u* 2 tou*. Moreover, must be to G 1, since LL* S u*. We now prove to = 1. If otherwise 0 < to < 1, by hypothesis (iii), there exists q. > 0 such that u* = Au* < A(t,u*)

s [r,(l

hence, IL* 2 r,(l + n,Ju*, in contradiction Therefore, by (28), we obtain

+ ‘lo)]-iAu*

= [to(l + vo)]%l*,

with the definition U

*-

-

UC,

of t,,. Thus,

to = 1 and u* 2 u*.

(30)

and so, by (29), u* = Au*, i.e. u* is a positive fixed point of A. Finally, we prove that (24) holds for any positive fixed point x* of A and any initial x0 E P, which at the same time implies the uniqueness of positive fixed points of A. Since x0 2 8, we

GLO D.AJL>

1300

have t9~Axo~AB, i.e. u,~~x,~u,: continuing this process. we obtain, U?fl s x2, s IIJn-,

.

llZn

operating in general. s.r:,,i

c

this inequality

uzn+,

s

by A. we find u1 G x1 G ui;

llln-,

(n=

1.2.3,..

(31)

.).

In the same way, we obtain U?fl s x* s

llZn_,

It follows

from (30), (31) and (32) that

Similarly,

I~.Y~~+,- x*lj + 0 (n+

Example

2. Consider

x),

(n

hence,

the nonlinear

=

1.2.3..

(24) is true and our proof

integral

is complete.

equation

j-a

1 = Y’(X)+ Y(X)

(32)

. .).

@R(X,Y) -Y(Y)

dy.

xE

[a,p].

which is of interest in nuclear physics (see [6]). For simplicity, assume N = 0. p = 1. Suppose is continuous and does not equal to zero identically on 0 s x, y G 1 and that (i) R(x,y) R(x, y) 2 0 for x > y and R(x. y) < 0 for x < y; (ii) there esists Y > 0 such that I+, where

C = const.

y)l =Z CIX - Yl “.-G,

vosx._ys

Y>,

and S(x, y) is a nonnegative

bounded

1, function

X fy, on 0 c x, y < 1 satisfying

lim S(X,Y) -<+=. n.y-++O x + y Then, the integral equation (33) has exactly over, 0 < v*(x) G 1, v*(x) $1 and

one positive

II&l - ~*llc-+o for any initial

vo(x) E C[O, l] satisfying

continuous

(n+r)

solution

v’(x).

More-

(34)

0 < vo(x) s 1. where

(33

Proof.

Putting y(x)

equation

1 = - I, Y(X)

(36)

(33) becomes pi(x) =

Il

1 R(x, Y> ~. o x2 - y? 1 + q(y)

dy

(37)

Some fixed point theorems

and 0 < v(x) s 1 corresponds

and applications

1301

to q(x) 5 0. Let dY

and P = (97 E C[O, l]]q?(x) > 0}, then it is easy to know that P is normal and A: P+ decreasing and completely continuous. Moreover, A8 > 6’ and A28 2 E&O, where

P is

NOW, suppose 9, > 8 and 0 < f < 1, we have (38) Putting

t-’ + dY> ,:z,

=1+q,

1 + q(y)

(39)

we see that 17> 0. It follows from (38) and (39) that

A(wW

y&j

l R(x1Y)

~. a x2 - Y2

1

dy = [r(l + q)]-‘A&K).

1 + V(Y)

Thus, all conditions of theorem 2 are satisfied, and hence by theorem 2, equation (37) has exactly one positive solution cp* > 8 and for any q. E P the successive sequence q,, = Aq,_, (n = 1,2,3, . . .) converges to q*, i.e. (40) By transformation

(36), we see that the function v*(x)

=

l

1 + v,*(x)

(41)

is the unique positive continuous solution of equation (33) and 0 < q*(x) S 1, y”(x) % 1. For given rjo(x) E C[O, 11, 0 < qa(x) S 1, put qo(x) =

-L - 1, Vo(x)

then cp, E P, and it is easy to know by induction that

vn(x>=

l

1 + vn(x>

(n = 0, 1,2,. . .),

where ~1, =Aq,_, (n= 1,2,3, . . .) and v’n is given by (35). From (40), (41) and (42) we know that (34) is true and the proof is complete.

1302

GKO D.AJUN

Remark

by Stuart 4. The conclusions of example 2 improve a theorem obtained [6. theorem 21). Stuart’s theorem does not contain the conclusion of convergence ” 1s ’ more restrictive: (i.e. (34)) and th e h ypotheses (II)

in 1971 (see of IJI~to I+!J*

REFERENCES 1. LLOYD N. G.. Degree Theor!, Cambridge University Press. Cambridge (1978). 2. NIRESBERG L.. Topics in Sonlinrnr Funcrional Analysis, Courant Institute. New York (1974) 3. KRASSOSEL’SKII IM. A., Posiriue Solutions of Operator Equarions. Noordhoff, Groningen. The Netherlands (1964). 1. AMANN H.. Fixed point equations and nonlinear eigenvalue problems in ordered Banach spaces, SMM Rev. 18. 620-709 ( 1976). 5. CROW J.. Eigenvalues of some nonlinear operators, J. marh. Analysis @pk. 38. 659-667 (1972). 6. STUART C. A., Positive solution of a nonlinear integral equation. :Clarh. ~nnln. 192. 119-12-l (1971). 7. DUCXNDJI J.. An extension of Tietze’s theorem, Pacif. /. Mnfh. 1. 353-367 (1951). 8. KR..XSNOSELSKII M. A., Topological Methods in the Theory of !Vonlinear Inregral Equations. Pergamon Press, Oxford (1964). 9. HARDV G. H.. LITTLEWOOD J. E. & Porut\ G., Inequaliries, Second Edition, Cambridge University Press, Cambridge (1952). 10. Gvo D.+JuN, A new fixed point theorem, Acfa Marh. Sinicn 21. -W-450 (1981). (In Chinese.)