"Science, Faculty of"@en . "Mathematics, Department of"@en . "DSpace"@en . "UBCV"@en . "Ko, Hwei-Mei"@en . "2011-04-26T23:55:35Z"@en . "1970"@en . "Doctor of Philosophy - PhD"@en . "University of British Columbia"@en . "Let f be a point-to-set mapping from a topological X\r\nspace X into the family 2(X) of nonempty closed subsets of X . K. Fan [13] proved that if X is a Hausdorff locally convex linear topological space and K is a nonempty compact convex subset of X , then an upper semicontinuous mapping (abbreviated by u.s.c.) f from K into k(K), the family of nonempty closed convex subsets of K, has a fixed point in K . Our main object in this work is to weaken \"compactness\"\r\nof K to \"weak compactness\" and prove a fixed point theorem for a mapping f on K into certain subfamily of 2(K).\r\nThe definition of convex function has been extended to point-to-set mappings in Chapter I. Let I denote the identity mapping on a Banach space X. Assume that I-f is a convex mapping on a weakly compact closed convex subset K of X. Then any of the following conditions implies the existence of the fixed point of f on K: \r\n(1) f : K \u00E2\u0086\u0092 2(K) is u.s.c. and [formula omitted] d(x,f(x)) = 0. \r\n(2) f : K \u00E2\u0086\u0092 2(K) is u.s.c. and is asymptotically regular (see definition 1.3) at some point in K .\r\n(3) f : K \u00E2\u0086\u0092 cc(K) is nonexpansive and the Banach space X has a strictly convex norm.\r\nMoreover, it has been shown that if f : K \u00E2\u0086\u0092 cpt(K) (see definition 0.3) is nonexpansive and I-f is strictly convex (see definition 1.5) on K, then K has a fixed point on K . Finally, an effort has been made to investigate the properties of the set of fixed points of a point-to-set mappings.\r\nIn Chapter II, we have confined ourselves to a reflexive Banach space X which has a weakly continuous duality map J (see definition 2.3) and X has a strictly convex norm. On such a special space we are able to prove that a nonexpansive mapping f : X \u00E2\u0086\u0092 cc(X) such that f(x)\u00CA\u0097 K, for any x in a closed convex bounded subset K of X , has a fixed point. As an application of this result we prove a fixed point theorem for semicontractive mappings (see definition 2.7). F : X \u00E2\u0086\u0092 cc(X) such that F(x)\u00CA\u0097K for any x \u00CE\u00B5 K , where K and X are the same as above. .\r\nIn the last Chapter, we have proved that if f is strictly nonexpansive on a.Banach space X into cpt(X) and if there is x(o) \u00CE\u00B5 X such that [formula omitted] has a subsequence convergent to a set A \u00CE\u00B5 cpt(X) under the Hausdorff metric D on cpt(X), then f has a fixed point in A . Furthermore we prove that a nonexpansive mapping f : K \u00E2\u0086\u0092 cpt(K), where K is a weakly compact convex subset of a metrizable locally convex linear topological space X, has a fixed point in K, provided that a constant k > 0 exists such that the set E(x) = {y \u00CE\u00B5 K ; d(x,y) \u00E2\u0089\u00A5 kd(y,f(y))} is nonempty and convex and the mapping E : K \u00E2\u0086\u0092 k(K), with E(x) defined above, is weakly locally closed (see definition 3.1). Finally the comparisons of the continuities of a point-to-set mapping have been made."@en . "https://circle.library.ubc.ca/rest/handle/2429/34010?expand=metadata"@en . "FIXED POINT THEOREMS FOR ?OINT-TO-SET MAPPINGS J by HWEI-MEI KO B. Sc., Taiwan Normal University, Taiwan, 1962 M. A., University of British Columbia, 1966 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR T H E DEGREE OF DOCTOR OF PHILOSOPHY in the Department of MATHEMATICS Ccr-rKs$T\$-r^ O V \"TbvC_^ ( j > V \ V - O ^ T We accept this thesis as' conforming to t h r , required standar:. THE UNIVER.. i A OF BRITISH COLUMBIA November, 1970 In presenting this thesis in partial fulfilment of the requirements for an advanced degree at the University of Br i t ish Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the Head of my Department or by his representatives. It is understood that copying or publication of this thesis for financial gain shall not be allowed without my written permission. Department of The University of Bri t ish Columbia Vancouver 8, Canada Date Supervisor: Professor L. P. Belluce i i ABSTRACT Let f be a point-to-set mapping from a topological X space X into the family 2 of nonempty closed subsets of X . K. Fan [13] proved that i f X is a Hausdorff locally convex linear topological space and K is a nonempty compact convex subset of X , then an upper semicontinuous mapping (abbreviated by u.s.c.) f from K into k(K) , the family of nonempty closed convex subsets of K , has a fixed point in K . Our main object in this work is to weaken \"compact-ness\" of K to \"weak compactness\" and prove a fixed point theorem for a mapping f on K into certain subfamily of 2 The definition of convex function has been extended to point-to-set mappings in Chapter I. Let I denote the identity mapping on a Banach space X . Assume that I-f is a convex mapping on a weakly compact closed convex subset K of X . Then any of the following conditions-implies the existence of the fixed point of f on K : (1) f : K -y 2 K is u.s.c. and inf d(x,f(x)) = 0 . x e K . K (2) f : K ->\u00E2\u0080\u00A2 2 is u.s.c. and is asymptotically regular (see definition 1.3) at some point in K . (3) f ; K cc(K) is nonexpansive and the Banach space X has a strictly convex norm. Moreover, i t has been shown that i f f : K cpt(K) (see definition 0.3) is nonexpansive and I-f is strictly convex (see definition 1.5) on K 5 then K has a fixed point on K . Finally, an effort has been made to i i i investigate the properties of the set of fixed points of a point-to-set mappings. In Chapter II, we have confined ourselves to a reflexive Banach space X which has a weakly continuous duality map J (see definition 2.3) and X has a strictly convex norm. On such a special space we are able to prove that a nonexpansive mapping f ; X cc(X) such that f ( x ) C K , for any x in a closed convex bounded, subset K of X , has a fixed point. As an application of this result we prove a fixed point theorem for semicontractive mappings (see definition 2.7) F : X -v cc(X) such that F(x) C K for any x e K , where K and X are the same as above. . In the last Chapter, we have proved that i f f is strictly nonexpansive on a.Banach space X into cpt(X) and i f there is X q e X such that {f n(x o)} has a subsequence convergent to a set A e cpt(X) under the Hausdorff metric D on cpt(X) , then f has a fixed point in A . Furthermore we prove that a nonexpansive mapping f : K \u00E2\u0080\u00A2> cpt(K) , where K is a weakly compact convex subset of a metrizable locally convex linear topological space X , has a fixed point in, K , provided that a constant k > 0 exists such that the set E(x) = {y e K ; d(x,y) _> kd(y,f(y))} is nonempty and convex and the mapping \u00C2\u00A3 : K -*\u00E2\u0080\u00A2'k(K) , with E(x) defined above, is weakly locally closed (see definition 3.1). Finally the comparisons of the continuities of a point-to-set mapping have been made. iv TABLE OF CONTENTS Page INTRODUCTION 1 CHAPTER 0 PRELIMINARIES 4 J CHAPTER I POINT-TO-SET MAPPINGS WITH THE CONDITION OF CONVEXITY \u00C2\u00A7 1 Convex Mappings 8 \u00C2\u00A72 Strictly Convex Mappings . ' 25 \u00C2\u00A73 The Set of Fixed Points of Point-to-Set Mappings 33 CHAPTER II POINT-TO-SET MAPPINGS ON BANACH SPACES WITH A ' DUALITY MAP i l Nonexpansive Mappings 40 \u00C2\u00A72 Semicontractive Mappings 52 CHAPTER III FIXED POINT THEOREMS FOR POINT-TO-SET MAPPINGS WITHOUT THE CONDITIONS OF CONVEXITY \u00C2\u00A71 Strictly nonexpansive Mappings 60 \u00C2\u00A72 Further Restrictions on Continuity A. Netwise Continuity 66 B. Locally Closed Mappings . . . . . . . . . . . . 69 C. Complete.^ Continuity . . . . . . . 78 BIBLIOGRAPHY . \ . . . . . . . . . . 83 V ACKNOWLEDGEMENTS The author would like to express her sincere thanks to Professor Lawrence P. Belluce for his guidance and valuable suggestions during the preparation of this thesis. She would also like to thank Professor J. T. Schnute for his constructive criticism of the draft form of this work,.and Professors R. A. Adams, A. T. Bui and E. E. Granirer for their reading the final form. The generous financial support of the National Research Council of Canada and the University of British Columbia is gratefully acknowledged. \" / 1 INTRODUCTION In this work we consider point-to-set mappings in a topological space X . A point X q in X is said to be a fixed point of a mapping f from X into the family 2 of nonempty closed subsets of. X i f X q e f ( x Q ) \u00E2\u0080\u00A2 Fixed point theorems for point-to-set mappings have been studied during the past 30 years. In 1941 Kakutani [16] generalized Brouwer's [4] fixed point theorem in a Euclidean space Rn to an upper semicontinuous (u.s.c.) point-to-set mapping f from a compact convex subset K of Rn into the family k(K) of closed convex nonempty subsets of K . In 1952 K. Fan [13] showed that Kakutani's fixed point theorem is valid in a Hausdorff locally convex linear topological space. Recently Maikin [19] has proved a fixed point theorem in a. Hilbert space H . It says that a nonexpansive mapping f from H into the family cc(H) of compact convex subsets of H , such that f(x) cpt(X) , where X is a metric space, for which there is a convergent subsequence of the iterated sequence of f at some point X q e X , has a fixed point. This is an extension of Edelstein's fixed point theorem { 1 2 J . In \u00C2\u00A72 various definitions of continuity of point-to-set mapping have been given and the comparisons of continuities have been made so that one get a good understanding of how the continuities are related to each other. 4 CHAPTER 0 PRELIMINARIES Throughout this work the spaces we consider are mainly Banach spaces. In this chapter we introduce the basic notations and some preliminary results primarily on metric spaces. The space X hereby stalls for a metric space with metric d , unless otherwise stated. Notations and Definitions: 0.1 2 = the family of a l l nonempty closed subsets of X , where X is an arbitrary topological space. 0.2 k(X) = {A e 2 ; A is convex} , where X is a topological linear space. 0.3 cpt(X) = {A e 2 ; A is compact} , X an arbitrary topological space. 0.4 cc(X) = k(X)0 cpt(X) , X a topological linear space, \ 0.5 Let~ x e X and r > 0 , define u S(x,r) = {y e X ; d(y,x) < r} , and S~(x,r) = {y e X ; d(y,x) <. r} . 0.6 For x e X , A e 2 X , define d(x,A) = inf{d(x,y) ; y e A} . 0.7 Given A e 2 X and r > 0 , define V_,(A) = {x e X ; d(x,A) < r} 0.8 Let A, B in 2 be such that both are bounded. Define D(A,B) = inf{r ; A C V . ( B ) and B C V . ( A ) > 5 We see that D is a metric on the family of a l l bounded closed subsets of X', usually called the Hausdorff metric for subsets ([15], Ch. VI, \u00C2\u00A728). Lemma 0.1 Let x,y c X and A be a nonempty subset of X . Then d(x,A) <_ d(x,y) + d(y,A) . Proof. Let-----z\u00E2\u0080\u0094e\u00E2\u0080\u0094A-\u00E2\u0080\u0094such that d(y,A) = lim d(y,z ) . Then d(x,A) <_ d(x,zn) <_ d(x,y) + d(y,zn) ->\u00E2\u0080\u00A2 d(x,y) + d(y,A) . Lemma 0.2 Let A e 2 and B e cpt(X) . Assume that A is bounded. Then for each x e A , there exists y e B , suchLthat d(x,y) _< D(A,B) . Proof. Set r Q = D(A,B) . By the definition of the Hausdorff metric, we see that for each r > r , we have A C V (B) . i.e. \u00E2\u0080\u00A2 o r d(x,B) < r for any x e A and r > r Q . This implies that d(x,B) <^ r Q , for any x e A . Now i f we fix an arbitrary x in A , there is y n e B such that d(x,B) = lim d(x,y ) . n-*\u00C2\u00BB Since B is compact, there exists a subsequence {y } n i of {y } such that y \u00E2\u0080\u00A2> y e B . Then n i n . d(x,B) = lim d(x,y ) \u00E2\u0080\u00A2 lim d(x,y ) = d(x,y) . n-*\u00C2\u00BB i - x o i 6 Therefore d(x,y) = d(x,B) _< r = D(A,B) , with y e B Without the assumption that B is compact, the point y in Lemma 0.2 may f a i l to exist. For example ([20], pp. 480), let X be the Hilbert space . a = (-1, - L...) , e is the unit z z n n vector with n-th coordinate equal to 1 and 0 elsewhere. Take the set A = {a, e t; i =. 1, 2, ...} , B = {e^ i = 1,2, ....'} . Evidently A and B are both bounded. To show that A and B are closed, we see - e || = 1/2\" for i \u00C2\u00A3 j , and \a - %t = I '-T+ ( 1 + n ) 2 = I V 1 * ! n i=i ir n i - i i z n i#n 2 2 I a || + 1 + \u00E2\u0080\u0094 , for each n _> 1 . This shows that both A and B are formed by isolated points, hence both are closed. Now / d(a,B) = inf{(||a||2 + 1 + | ) 1 / 2 ; n = 1, 2, ...} = (||a||2..+ 1 ) 1 / 2 , and d(e^,B) = 0 for each i = 1, 2, ... . Thus for any 2' 1/2 r > (||a|| + 1) , we have ACV.(B) . Similarly, we see that BCV.(A) for any r > (||a||2 + 1 ) 1 / 2 . This implies that D(A,B) \u00C2\u00BB (||a||2 + 1 ) 1 / 2 . But for any e^ e B , d(a,en) = ||a - en|| - (||a||2 + 1 + f ) 1 / 2 > (||a||2 + 1 ) 1 / 2 = D(A,B) . 7 Therefore a point y in B such that d(a,y) _< D(A,B) fails to exist. The trouble arises since both A and B are not compact, for both of them are formed by infinitely many isolated points. The fact stated above explains why in the sequel we are concerned quite often with mappings f : X -*\u00E2\u0080\u00A2 cc(X) instead of mappings from X into . k(X) . Lemma 0.3 Let X be a metric space and b(X) be the family of bounded closed subsets of X . Given f : X b(X) , then for any x,y e X , we have (1) d(x, f(x)) <_d(x,f(y)) +D(f(x), f(y)) (2) d(x, f(y)) < d(x,f(x)) +D(f(x), f(y)) (3) d(x, f(x)) < d(x, y) + d(y, f(y)) +D(f(x), f(y)) . Proof. The proofs of (1) and (2) are similar, and (3) can be obtained from (2) and Lemma 0.1. Here we prove (1). Let y n e f(y) such that d(x, f(y)) = lim d(x, y ) . Given e > 0 , there exists x e f(x-) h-x\u00C2\u00BB n n such that d( x n> v n) Jl D(f( x)> f(y)) + e \u00E2\u0080\u00A2 Thus for any n , we have d(x, f(x)) <_ d(x, x n) _< d(x, y n) + d(y n > xn> < d(x, y n) + D(f(x), f(y)) + e . Hence d(x, f(x)) < d(x, \u00C2\u00A3(y)) + D(f(x), f(y)) + e \ Since e is an arbitrary positive number, we have d(x, f(x)) < d(x, f(y)) +D(f(x), f(y)) . 8 CHAPTER I POINT-TO-SET MAPPINGS WITH THE CONDITION OF CONVEXITY \u00C2\u00A71. Convex Mappings The space X in this chapter is assumed to be a Banach space. Definition 1.1 A mapping f from X into 2 X is said to be convex i f for any x,y e X and m = Xx + (1 - X)y with 0 < A < 1 , and any x^ e f(x) , y^ e f(y) , there exists m^ e f(m) such that Kll < AKII+ ( 1 - \u00E2\u0080\u00A2 x Proposition 1.1 Let f : X 2 , I : X ->- X be the identity mapping. If I-f((I-f)(x) = {x - y ; y e f(x)}) is convex, then for any x,y e X , and m = Ax + (1 - X)y , 0 <_ X __ 1 , we have d(m,f(m)) < Xd(x,f(x)) + (1 - X)d(y,f(y)) . Proof. Let X r e f(x) such that ||xn - X||T*- d(x,f(x)) and y e f(y) such that ||y - y|| -*- d(y,f(y)) . By the convexity of I-f , there exists m___ e f(m) corresponding to x n and y such that \u00E2\u0080\u00A2'n II* \" \u00C2\u00BBni;<.X|x - xn||+ (1 - X)||y - y\u00E2\u0080\u009E|| \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \"N. Now d(m,f (m)) < inf ||m - m || < X||x - x || + (1 - X)j|y - y || for any n >_ 1. n>l n n n i.e. 9 d(m,f(m)) < x||x - x j | + (1 - X)||y - yn|| + Ad(x,f(x)) + (1 - X)d(y,f(y)), Remark 1.1 If f : X \u00E2\u0080\u00A2+ X is a linear mapping, then for any x,y e X, and m = Xx + (1 - X)y , 0 _< X _< 1 , we have ||m - f(m)||= ||xx+(l-X)y - Xf(x) - (l-X)f (y) || < x||x-f(x)||+ (l-X)||y-f (y) || . This shows that I-f is convex. Thus the convexity of I-f is more or less generalizing the notion of linearity of a point to point mapping to a kind of sublinearity of point-to-set mapping. Definition 1.2 Let X be a topological space. A mapping f : X 2 is said to be upper semicontinuous (abbreviated by u.s.c.) at X q e X , i f for any open set U , such that f(X q) C U , there exists a neighborhood V of X q such that f (y) C. U for any y e V . A mapping f i s u.s.c. in X i f i t is u.s.c. at any x in X . Theorem 1.1 Let K be a nonempty weakly compact closed convex subset of X . If f : K 2 is upper semicontinuous and inf d(x,f(x)) = 0 , and I-f is convex on K , then f has a fixed point xek in K . Proof. Let r > 0 , define = {x e K; d(x,f(x)) _< r} . We see that H ^ 0 , since inf d(x,f(x)) = 0 . \ xek (a) is convex, for let x,y e H__ , and m = Xx + (1 - X)y , then by Proposition 1.1 10 d(m,f(m)) < Ad(x,f(x)) + (1 - A)d(y,f(y)) <_ rA + r ( l - A) = r . Thus m e H . Hence H is convex, r r (b) H is closed. Let z e H , such that z -> z . We want to prove r n r ' n o p that d(z Q,f(z o)) <^ r . Suppose, on'the contrary, that d(z ,f(z )) > r . Let 6 = d(z ,f(z )) - r , then 6 > 0 . Consider o o o o the set ^5/3^^ z 0^ which is an open set containing f ( z Q ) - By the upper semicontinuity of f , there exists a neighborhood V of z such that f(z) C Vr/\u00E2\u0080\u009E(f(z )) for any z e V . We may assume o o/j o that V is contained in the open sphere S(Z q,6/3) . But since z -\u00C2\u00BB\u00E2\u0080\u00A2 z , there is an integer N such that z e V for a l l n > N. n o ' 6 n \u00E2\u0080\u0094 Therefore f(z ) CW_,Af(z )) for a l l n > N . Set U = V,,,(f(z )), n o/J o \u00E2\u0080\u0094 o/j o then d(z ,f(z )) = inf || z - z|| _> inf||z - z|| = d(z ,U) (1) n n \u00C2\u00A3 / \ U __n n zef(z ) zeU n But by Lemma 0.1 d(zo,U) < || Z Q - zj|+ d(zn,U) (2) z e U ==> || Z q - z|| >^ d(z o,f(z Q)) - d(z,f(z Q)) (again by Lemma 0.1) > d(z.f(z )) - 4 = r + 6 - 4 = r + | 6 \u00E2\u0080\u0094 o o J 3 - 3 => d(zQ,U) = inf || z - z|| > r + | 6 (3) zeU By (1), (2) and (3), we have, for n>N X 11 d(z ,f(z )) > d(z ,U) > d(z ,U) - z - z n n \u00E2\u0080\u0094 n \u00E2\u0080\u0094 o 11 o n1 2 1 1 _>r+-j6--j$ = r + - j 6 > r This contradicts the fact that d(z ,f(z )) < r for, n > 1 n n \u00E2\u0080\u0094 \u00E2\u0080\u0094 Therefore d(z ,f(z )) < r . Hence H is closed, o o \u00E2\u0080\u0094 r Thus H__ is closed and convex and hence is weakly closed for each r > 0. The family {H__; r > 0} has the finite intersection property. Therefore by the weak compactness of K , we have f\ H ^ \u00C2\u00A7 r>0 r XEH H =^ d(x,f(x)) < r for any r > 0 r>0 r => d(x,f(x)) =0 , i.e. x e f(x) , since f(x) is closed. Therefore any point in (~) H is a fixed point o f f . r>0 r The following example will show that the condition inf d(x,f(x)) =0 in Theorem 1.1 is indispensable. The construction of xek mapping f is similar to that in [1]. Example 1.1 Let X be a reflexive Banach space such chat the closed unit sphere S = {x e X ; ||x|| <_ 1} is not compact. J. Dugundji ([10], pp. 362) showed that there exists a continuous mapping to the set ^ . ^ , e* there exists a continuous mapping : S2 -*\u00E2\u0080\u00A2 S such that i|>(x) = (|)(x) for x e S . Then ty does not have fixed point either, for the fixed point of ty must be in S , but ty = ij) on S . Thus ty does not have fixed point. Now define f t -*\u00E2\u0080\u00A2 as follows: f ( x ) = X + [I i K x ) 1 - x|| ( * ( x ) - X ) f \u00C2\u00B0 r X \u00C2\u00A3 S 2 > f(x) is well defined since ||^(x) - x|| ^ 0 . f(x) e S2 > for i f ||i|>(x) - x|| _> 1 , then \u00C2\u00A3 ( X ) = ( 1 \" ||\u00C2\u00ABKx) - x | | ) x + ||*<*> - x|| * ( x ) e S2 \u00C2\u00BB since S2 is convex. On the other hand i f ^(x) - x|| < 1 , set t = ||^ (x) - x|| and y = ||^(x) _ x|[l>(x> x> \u00C2\u00BB then IMI = 1 and' x = ty(x) ~ ty . Thus ||f(x)|| - ||x + yII = ||^ (x) - ty + y|| - ||*(x) + (1 - t)y|| < \\ty(x) || + (1 - t)||y|| < 1 + 1 - t < 2 . Hence f(x) e S2 (a) f is continuous on ^ , because ijj(x) is continuous on S2 \u00E2\u0080\u00A2 (b) I-f is convex on S^, , for let x,y e S2 and m - Ax + (1 - A)y 0 <_ A <_ lx, then Hm \" f(m)H = IL(m)1- m ^ ( m ) \" m>i = 1 * 13 H x\" f ( x ) | 1 = ||\u00C2\u00BB(x) - x|| H* ( x ) - x'l = 1 = Hy - fl! \u00E2\u0080\u00A2 Hence ||m - f(m)|| = X||x - f(x)jj + (1 - X)||y - f(y)|| i.e. I-f is convex. We see that f has no fixed point, since inf ||x - f(x)|| = 1 . xeS2 Example 1.2 Let C[0,1] be the space of real valued continuous functions on the closed interval [0,1] with the uniform norm. Set K = {f e C[0,1] ; f(0) = 0 and f(1) = 1 , 0 < f(x) <_ 1} Define : K -* K as follows: <|>(f)(x) = xf(x) for f e K . It has been shown ([1], Example 4.1) that is nonexpansive (i.e. ||(f) _ (m)|| = sup |Xf(x) + (1 - X)g(x) - x[Xf (x)( + (1-- X)g(x)j: 0(f)||+ (l - x)j|g - 4>(g)|| Hence I-(j) is convex on K . But \u00C2\u00A7 has no fixed point in K , i f 14 fails to have a fixed point since K is not weakly compact. Example 1.3 Let K = {x e %^ ; ||x || <_ 1} be the closed unit sphere of the Hilbert space %^ . Then K is closed, convex and weakly compact. Define f : K -*\u00E2\u0080\u00A2 K as follows: Let x = (x^, x^, . .. ,\u00E2\u0080\u00A2 x , ...)eK, define f (x) = (1 || X |j, x.^ , X2, \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 \u00E2\u0080\u00A2, x^, \u00E2\u0080\u00A2 \u00E2\u0080\u00A2 . ) \u00E2\u0080\u00A2 Then 1 - ||f(x)||2 - 1 - [(1 - ||x||)2 + ||x||2] = 2||x||(l - ||x||) > 0, i.e., |f(x)|| _< 1 . The continuity of f can be seen from the following inequality. ||f(x) - f(y)||2..- (||x||- ||y||)2 + ||x - y||2< ||x. - y ||2 + ||x. - y ||2 ^ i.e. , ||f(x) - f(y)|| < - y|| \u00E2\u0080\u00A2 We claim that inf ||x - f(x) || = 0 . Let x ^ = (x^, X2, ...) e i i 9 be xek 1 2 such that x, = x \u00E2\u0080\u009E = ... = x \u00E2\u0080\u009E = \u00E2\u0080\u0094 and x. = 0 for i > n . Then 1 2 2 n x n ||x^ n |^| = 1 and f ( x ^ ) = (0, x^, X2, x 0, 0, ...) . We see that n II ( n ) r/ ( n ) \ II & n ||xv ' - f (x ) II = \u00E2\u0080\u0094 -* 0 as n.->\u00E2\u0080\u00A2 \u00C2\u00B0\u00C2\u00B0 Hence inf |x - f(x)|| = 0 . jjut I-f is not convex. Let xek x x = (y, 0, 0, . . . ' ) , y \u00E2\u0080\u00A2 (- j , ), 0, ...) . Then f.(x) ' ;\u00C2\u00AB*^|, j , 0, 0, ...), 15 f(y> - (f, - j , o, o, ...) ||x - f(x)|| = - ^ , ||y - f ( y ) | | - ^ \u00E2\u0080\u00A2 Take m = -|(x + y) = 0 , then f(m) = (1, 0, 0, ...) and ||m - f (*) I = 1 > \ ||x - f (x) || + \ ||y - f (y) || = f has no fixed point, for i f f(x) = x , where x = (x^, x^, ...) s K , 00 2 then x, = x\u00E2\u0080\u009E = X. = ... , and 7 x. < \u00C2\u00BB . Thus x. = 0 for i > 1. 1 2 3 x i. \u00E2\u0080\u0094 But then f(x) = (1, 0, 0, ...) 4 (0, 0, ...) . This example shows that in Theorem 1.1 the condition that I-f is convex is indispensable. Remark 1.2 Example 1.3 is a special case of the example given by W. A. Kirk [18]\u00E2\u0080\u00A2 He uses his example to emphasize that the nonexpansiveness of the mapping concerned in his theorem can no.t^ be removed. Here we only require f to be continuous but emphasize that the convexity of I-f can not be removed. If f is a point to point mapping, then theorem 1.1 can be stated in the following which is Theorem 4.1 in [1]. Corollary 1.1.1 Let K be a nonempty closed convex weakly compact subset of X . If f : K -*- K is continuous and inf|x - f(x)|| = 0 and I-f is convex, then f has a fixed point in K xek \ 16 2\u00C2\u00A3 Definition 1.3 f : X 2 is said to be asymptotically regular at x e X i f there exists a sequence of points x such that o n x e f (x .) and II x - x n II -* 0 as n ->\u00E2\u0080\u00A2 \u00C2\u00B0\u00C2\u00B0 n n-l 11 n n-l\" One immediate result of Theorem 1.1 is the following: Corollary 1.1.2 If f : K -*\u00E2\u0080\u00A2 2 is asymptotically regular at some point in K , where K is a nonempty closed convex weakly compact subset of X , and if f is upper semicontinuous in K such that I-f is convex, then f has a fixed point in K , Proof. Assume f to be asymptotically regular at e K , then there exists x e K such that x ef(x ,) , n > 1 , and n n n-l \u00E2\u0080\u0094 II x - x , I -\u00C2\u00BB- 0 . Since d(x ,f(x )) < II x - x II -> 0 , we have 11 n n-l\" n n \u00E2\u0080\u0094 11 n+1 n\" inf d(x,f(x)) = 0 . Theorem 1.1 implies that f has a fixed point in K. xek Definition 1.4 Let (X, d) be a metric space. A mapping f : X -*\u00E2\u0080\u00A2 b(X) is said to be nonexpansive on X i f D(f(x),f(y)) < d(x,y) for any x,y in X . Proposition 1.2 If f : X -\u00C2\u00BB\u00E2\u0080\u00A2 cpt(X) is nonexpansive, then i t is upper semicontinuous on X where X is a metric space with metric d . Proof. Let X Q e X and U be an open set containing f(,x ) . Set A = X - U , then A is closed and f ( X Q ) f\ A = (j) . Let 17 .6 = inf d(x,A) , then 6 > 0 , since f(x ) is compact and A is xef(x ) o closed. Furthermore we have inf d(y,f(x )) = inf d(x,A) = 6 > 0 . yeA xef(x Q) Take V = S(Xq,<5) , we want to prove that f (x) C U , for x z V . Notice that x e V => D(f (x) ,f (xo>) <_d(x ,X Q ) < 6. Thus x e V and y e f(x) => there is z e f(x ) such that o d(y,z)\u00C2\u00A3 D(f(x),f (X Q ) ) < 6 (the existence of z follows Lemma 0.2) =-> d(y,f(x Q)) \u00C2\u00A3d(y,z) < o . = i > y e X - A = U. This shows that x e V f(x)C U . Hence f is upper semicontinuous at x . Therefore f is u.s.c. on X . o The condition that the values of f be compact sets is not removable in above proposition. The following example shows that a x \u00E2\u0080\u00A2 nonexpansive mapping f : X 2 may f a i l to be upper semicontinuous i f the values of f are not compact sets. Example 1.4 Let X = {(x,y) ; x _> .0 and y j> 0} , d be the usual metric on X . Define a bounded metric e on X as follows: Let P,Q e X , define e ( P 0 ) = d(P,Q.) e U % Q ; 1 + d(P,Q) ' Then the topology of (X,e) i s identical with that of (X,d) ( [ l 7 j v p i 3\" and the . Hausdorff \"metric is well defined on the class of closed subsets of X . Define f : X -\u00C2\u00BB\u00E2\u0080\u00A2 2 as follows: ,x f(x,y) = {(y,z); a l l z _> 0 } , (x,y) e X Then (1) f is e-nonexpansive, i.e.,nonexpansive with respect to the metric e . To show this, let P = (x 7.y^ , Q = (x 2,y 2) a l l in X , and let Dg be the Hausdorff metric with respect to e. Then \u00E2\u0080\u009E | x x~] De(f(P),f(Q)) 1~- < 1 + I|x 1 - x 2| d(g>Q) = e(P Q) 1 + d(P,Q) e^>^ since the function ^ is a strictly increasing function. The; : tfore f is e-nonexpansive. (2) f is not upper semicontinuous with respect to both e and d Let (x ,y ) e X .. Set f ( V V 7 > X x \ \, (r~-,0) (-^ -,0) 19 x U = {(x,y) ; (x,y) e X and 0 <_ x <_ -y-} U2 = {(x,y) ; (x,y) e X and y ~ - < 0} . o x - T (See figure on page 18) Ther. the set U = {J is open (here we do not mention which topology we are referring to, because both topologies on X are equivalent) and f(x o,y Q) C U . But for any (x,y) e X such that x > X q , we see that (f, + i) e f <*.y> n [x - u] t 4> . o i.e. f(x,y)^U i f x > X q . This shows that there does not exist a neighborhood V of ( x 0> v 0) such that f(x,y) C U for any (x,y) e V . Hence f is not upper semicontinuous. Definition 1.5 The norm \" | | j | \" of a normed linear space X is strictly convex i f |x + y|| < ||x|| + ||y|| unless x and y are linearly dependent. Theorem 1.2 Let K be a nonempty weakly compact closed convex subset of a Banach space X which has strictly convex norm. If f : K -r cc(K) is nonexpansive and i f I-f is convex on K , then f has a fixed point in K . We need the following results for the proof o f the theorem. Lemma 1.1 Let X be a normed linear space,. If \u00E2\u0080\u00A2> ra, rru c X are such that x, = ^(m+riu) and 20 r ijm - jj = , then d (x^CCSCa,^))'; = \u00E2\u0080\u0094 , where C(S(m,ro)) = X - S(m,ro) . r o Proof. First we have d(x.. ,C(S(m,r ))) < -r- since nu e C(S(m,r l o \u00E2\u0080\u0094 J. l o r r and j|x1 - m2|| = ||m - x .J | = -y . Suppose dCx^CCSOn.r ))) < -y- , then r there exists V q e C(S(m,ro)) such that ||x^ ~ y fl < ~Y \u00E2\u0080\u00A2 Now j|m y j l = |l2x1-\u00C2\u00AB2 \" y j l < ||x1-m2|| + ||Xl - yjj < f + f = r Q . This implies that y e S(m.r ) , a contradiction. Hence c Jo ' o ' r d(x1,C(S(m,ro))) = -f . Lemma 1.2 Let X be a strictly convex norm space and points x^ , m, m2 satisfy the conditions stated in Lemma 1.1. Then m2 is the r unique point in C(S(m,ro) such that [|x^ - m2 |j = d(x^, C(S(m,ro))) = -~r. Proof. Suppose there exists y^ e C(S(m,r )) such that y^ / m2 r and ||xx - y j = -f . Then !i m - yj = l l 2 x i - m 2 ~ yxII < l l x i - m 2 H + Hxi - yjl i f x^ j^ - m2 ? - y 1) for any real number A (a) otherwise we have ||m - y^l' = ||x - || + Hx.^ - y || (b) r r In case (a), we get ||m - y.J| < \u00E2\u0080\u0094 H\u00E2\u0080\u0094\u00E2\u0080\u0094 = . This implies y^ e Sfavrr) a contradiction. In case (b), there is A 0 such that 21 X l ~ m2 = ^ x l ~ y l ^ * r But ||x1 - m21| = \u00E2\u0080\u0094 = || - y 1|| , thus |x|= 1 . This implies m2 = y 1 or m = y . . Which contradicts the assumption that m_ \u00C2\u00B1 y., and y., e C(S(m,r )) i z 1 l o Hen J . ce m2 is the unique point such that \\x1 - m2 || = d(x ,C(S(m,r ))) = -y Lemma 1.3 Let X be a normed linear space with strictly convex norm, K be a closed convex subset of X . Given a nonexpansive mapping f : K -> cc(K) and given X q e K , i f x^e f ( x Q ) is a point such that ||Xq - x^ || = d(x o,f(x o)) , then the point m^ \u00C2\u00A3 f (m) such that j|m - m^|| = d(m,f(m)) is collinear with X Q , X^ and m , and i(m+m-^ ) , where m= i ( x + .x, ) and d(x ,f(x ))=d(m,f(m))=r ^ 0 . 2. o \u00E2\u0080\u00A2 i o o o Proof. Let = 2x^ - m . Then r (a) d(x1,C(S(m,ro))) = -y (by Lemma 1.1) (b) m2 is the unique point in C(S(m,ro)) such that ||x1 - m21| = d(x ,C(S'(m,r ))) (by Lemma 1.2) . We want to prove that m^ = m^ . Suppose on the contrary that m^ ^ m l + m2 (i) If e f(m) , then e f (m) , as f(m) is convex . 1 m - 1 2 2H < ||p - + ||m..- = | r Q + j*CQ = r Q provided m - m^ and m - are linearly independent. This implies that [(m^ + n^)^] e S(m,rQ) which is absurd., since 22 m + m\u00E2\u0080\u009E || m ^ \u00E2\u0080\u0094 - | | >. d(m,f(m)) = r i 0 . On the other hand i f m - nu = \(m - n^) for some A f 0 , then m^ + m2 ||m - m1|| = r Q = ||m - m2|| ^ | \|= 1 ==> m1 = m2 or m = This contradicts the assumption that ^ m2 and in. tj: f(m) . Hence case (i) is impossible. (ii) Assume m2 \ f(m) . Since d(m,f(m)) = r Q , we have. f (m) C C(S(m,rQ)) . Then d( X ; L,f (m)) > dCx^CCSGu.r )) = r Q/2 (by (a)). r r We claim that d(Xj,f(m)) > \u00E2\u0080\u0094 , for i f d(x^,f(m)) = \u00E2\u0080\u0094 , then there is z e f(m) , such that ||x., - z || = d(xn,f(m)) . But o \" i o\" 1 Z Q $ m2 (since m2 {\u00E2\u0080\u00A2 f(m)), so the equality K \" Zoi = \" X l \" M = \"T = d(Xl,C(S(m,ro)) r contradicts (b). Heic.-' d(x^^J^a)) > . On the other hand 9(f (X q) ,f (m)) >. d(x. f(m)),because X ; L e f(x Q), there is z e f(m) , 2 3 such that || x - z || < D(f (x ) ,f (m) ) , h e n c e D(f(xo),f(m)) > ||xx - z|| > d(x1,f(m)) . r Thus D(f(x ),f(m)) >. d(x ,f(m)) > -y = || x - m|| . This contradicts the nonexpansiveness of f . So case (ii) is impossible. Hence = . Therefore = = 2x^ - m . This shows that Xq, m, x^ and m^ are collinear and x^ = -|-(m + m^ ) Proof of Theorem 1.2: f is nonexpansive, so i t is upper semicontinuous (by Proposition 1.2). Let H^ be the set defined as in Theorem i . i . Then H^ is closed and convex and hence weakly closed. Let r = inf {r; H <(>} . o r The family H ^ has finite intersection property. By the weak compactness of K , we have H = f l H 4 ty o. njty r We want to prove that H i s the set of fixed points of f . i.e. to o prove that r = 0 . Assume r \t 0 . Let x e H . Since f(x ) is r o o o r o o compact, there is x^ \u00C2\u00A3 f(x Q) such that ||xQ - x j l = d(x o,f(x o)) . Set m = -^ (x + x..) , we claim that m e H . Let mn e f(m) be such 2 o 1 r 1 o that J!K \u00E2\u0080\u0094 mn || = d(m,f (m)) . Now since x. e f (x ) , there is m' e f (in) 1 1 O 2. (by Lemma 0.2) such that 24 I I x ! \" m 2 i l 1 D(f(x o),f(m)) <_ jjx Q - m|j . Thus Jm - m-J = d(m,f(m)) <_ ||m - m\\\ <_ ||m - x1|| + ||Xl - ji < || m - x1|| + || X q - m|| = I X q - x1|| = r Q . By definition of r , we conclude that d(m,f(m)) = r . Hence m e H ' o o r o Lemma 1.3 shows that m ^ , X q , m and x^ are collinear and x^ = ^ - ( m + m Replace the points X q , x^, m by m , m^ , x^ respectively! and applying such the same argument to m, m^ , x^, we will get a point x^ e f(x^) that m^ = + x2^ * Therefore we have HXo \" X2H = 2 l X o \" XlH * By induction, we get a sequence of points x^ in K such that || x - x || = n||x - x- || = nr II 0 N II II 0 i II Q r ^ 0 implies [Ix - x || -> <\u00C2\u00BB as n \u00E2\u0080\u00A2> \u00C2\u00AB . This contradicts the o Ir o n\" boundedness of K , as K vis weakly compact. Hence the assumption that r ^ 0 is false. Therefore the set H , with r = 0 , is a set of o r o o fixed point of f . Theorem 1.2 extends Theorem 4.2 of Belluce and Kirk ll] to point-to-set mapping. Remark 1.3 We are \u00E2\u0080\u00A2 unable to find a counterexample, to show the strict convexity of the nor2. :>i Bana'ea space is essential in Theorem 1.2. But we are able to remove It b y strenghening the convexity of mapping 25 I-f . This is the topic we want to study in next section. \u00C2\u00A72. Strictly Convex Mappings Definition 1.6 A map f : X -*\u00E2\u0080\u00A2 2 is called strictly convex on X i f the following conditions satisfied: For any x,y e X \u00E2\u0080\u009E m = Xx + (1 - X)y, 0 X < 1 , and any x^ e f (x) , y^ e f(y) , there exists m.( e f(m) , such that (1) i!xn_3 [j < max {||x ||, ||y^||} if not both f(x) and f(y) contain 0 . (2) . ||m., jj \u00C2\u00A3 max'.{||Xl||,-||y ||} i f both f(x) and f(y) connain 0 . Remark 1.4 A s t r i c t l y convex mapping is not necessarily convex, take the mapping f(x) = /ic, x e [0,1] , for instance. The. map f strictly convex because i t is strictly increasing. But f .is not convex, 1 1 3 for example take m = ^ - = -^--i+^-'0 , then f(l) = 1 , f(0; = 0 , so Proposition 1.3 Let f : X -> cpt(X) be a mapping with I-f strictly convex. For any x,y e X and m = Xx -,- (I - X)y , 0 < X \u00E2\u0080\u00A2< 1 , (1) i f not both d(x,f(x)) , d(y,f(y)) are 0 , then d(m,f(m)) < max {d(x,f(x)),d(y,f(y))} . (2) i f d(x,f(x)) = d(y,i- = max {d(x, f(>:)), d(y, f(y))} .\u00E2\u0080\u00A2 If x = yy for some y f 1 , then d(m, f (m)) = \ J Xx + (1 - X)y|| < \ max {||x||, ||y|j> = max {d(x, f(x)), d(y, f(y))} . If d(x, f(x)) = d(y, f(y)) =0 , then x = y = 0 , therefore m = 0 and d(ms f (in)) = 0 . Hence I-f is strictly convex. Theorem 1.3 Let K be a weakly compact convex closed subset of a Banach space X . If f : K -> cpt (K) is nonexpansive and I-f is strictly convex, then f has a fixed point in K . Proof: Let r > 0 , define H = {x e K ; d(x, f(x)) <_ r} . Since f is nonexpansive, i t is upper semicontinuous; a by Proposition 1.2. Hence H is closed as i t was shown in Theorem 1.1. To show that H r r is convex, we may assume that ^ .. Let x, y s H , ia = Xx + (1 - X)y , for 0 <_ X <_ 1 . By strict convexicy of I-f . -ve have 29 d(m, f(m)) < max { d ; \ . f(x)), d(y, f(y))} < r , since d(x, f (x)) <_ r and d(y, f (y)) \u00C2\u00A3 r . Thus m e . Therefore H is convex. Now H is closed and convex, hence i t is weakly closed, r r ' \u00E2\u0080\u00A2J Th e family {H ; f has finite intersection property. Therefore, by weak compactness of K , we have H ^ . Let r r \u00E2\u0080\u00A2- inf {r ; H ^ d>} . Then o ' r T r = inf \"(d(x, f(x))} and H H i \u00C2\u00A7 \u00C2\u00B0 xeK ro H H r r Y We claim that r^ = 0 . Suppose on the contrary that r Q > 0 . Let x F. H , then d ( x , f ( x ) ) = r Since f(x ) is compact, there o r o o o o f \u00C2\u00BB o exists x, e f(x ) such that II x -\u00E2\u0080\u00A2 x, [| = d (x , f (x )) , Let x\u00E2\u0080\u009E e f(x, ) l o \" o 1\" o o Z 1 be such that' || - x^\\ = D C f ^ ) , f(x Q)) . Then d(x1, f(x1>) <. ||x1 - x9|| <_ D(f (x 1), f (X q)) \u00C2\u00A3 ||x1 - x^jj , since f is nonexpansive. Thus d(x,,f(x,)) < [Ix, - x ij = ct (x , f(x )). 1 1 \u00E2\u0080\u0094 \" 1 o\" o o Let. m = -j (X q + x^) , then m e K and d(m, f(m)) < max {d(x , f(x )), d(x , f(x ))} = r . o o \u00C2\u00B1 i o This contradicts the fact that r = inf d(x, f(x)) . 0 x\u00C2\u00A3K Hence r = 0 . Therefore H ^ is the set of fixed points o f f . o So far we have seen fckit 'die convexity\" of the mapp..:ig I-f 3 0 provides a good cool to pry up fixed point of certain mappings f . Later we will s e e that i t also gives us some idea of the shape of the set of fixed points of f . \u00C2\u00A73. The Set of Fixed Points of Point-to-Set Mappings Let K be a closed convex subset of a Banach apace X . Denote by W the set of fixed points of a mapping f : K -> cc(K) Throughout this section we assume that W is nonempty. Definition 1.7 \u00E2\u0080\u00A2 Let (X, d) be a metric space. A mapping f : X ~r b(X) is said to be strictly nonexpansive on X if D(f(x),f(y)) < d(x, y) for any x,y e X and x ^ y . If f is a point to point mapping, then the following properties are true. Proposition 1.5 If f is strictly nonexpansive, then W is a singleton. Proposition 1.6 If f is nonexpansive and the norm of t h e Banach space is strictly convex, then W is convex. The two propositions above are no longer true i f f is a point-to-set mapping. 31 Exc.-:;t,la 1.6 Let K = [0,1] x [0,1] , a subset of R\" under the usual topology. Define f : K cc(K) as follows: X l X l f((x^,x 2)) = the triangle with vertices ( - y-jO), (-\u00E2\u0080\u0094-,1) and (1,0), Let x = (x^ x 2) , y = (y^ y^) e K , w i t h x ^ y, then D(f(x),f(y)) = f |x1 - y j < d(x,y) . Hence f is strictly nonexpansive. But W contains a l l the points (0,x9) , for 0 _< x^ 1 . That is , W is not a singleton. The. afore Proposition 1.5 does not hold for point-to-set mappings. Remark i.5 Proposition 1.5 can be rewritten 3.3 follows: (*) If f is strictly nonexpansive, then there exists x^ e W such chat f(x ) = W . o Although statement (*) implicitly says that 7J is a \u00E2\u0080\u00A2singleton for point-to-point mapping, but i t does not imply chat W is a singleton for point-to-set mappings. So i t seems that statement (*) holds for point-to-set mappings. Unfortunately this is not true either. For instance, the mapping f stated in Example 1.6 is strictly nonexpansive. By definition of f , f((x^,x 2)) is a triangle for any (x 1 x^) \u00C2\u00A3 K . But the set W of fixed points of f is bounded by positive x, y axes 2x + 2y - xy - 2 = 0 and a branch of hyperbola 2x + 2y - xy - 2 = 0 . W = {(x,y) e R ; 32 Hence f((x,y)) ^ W for any (x,y) e K . Furthermore for any (X ,y) \u00C2\u00A3 .. such that x =f 0 , we have f((x,y))^W since (y, 1) e f((x,y)) , but \. W . Therefore i t i s not true that if f is strictly nonexpansive, then f(x)C W for any x z. W . Example 1.7 Let K be the set defined in Example 1.6. Define f as follows: Let (x^,x2) e K . (i) If x j x 2 ^ ^ * t^ i e n f ((x^>x2)) = t n e triangle with v e r t i c e s (0,0), (x^O) and (0,x?) . (ii) If = 0 , then f((0,0)) (0,0) . f((0,x o)) = the set '{(0,x): x z L'0,xoj> , where ^ 0 . f((x 1,0)) = the set {(x,0); x e [O.x^} , where x^ ^ 0 \u00E2\u0080\u00A2 Let x = (x x,x 2) , y = ( y ^ y ^ e K j then D(f(x),f(y)) = inax{|x1 - y j , ^ - y \u00C2\u00A3 | } < [ (x\u00C2\u00B1 - y^2 + (x 2 - y 2 ) 2 ] 1 / 2 - l l * - y | | . 2 Thus f is nonexpansive* Evidently the norm in R is strictly convex. But the set W of the fixed points of f is W = {(x1,x2) : (x i s>. ?) E K and x ^ = 0} which is not convex. Hence Proposition .^,6 is false for point-to-set, 33 mappings. Therefore the strict convexity of norm plays no role in the ca.ivexity of W i f f is a point-to-set mapping. Rather the roiie has been taken by the convexity of I-f . Proposition 1.7 Let f : K -* cpt(K) be such that I-f is convex or strictly convex on K , then W is convex. Proof. If I-f is convex on K , then from Proposition 1.1. we have for any x,y e K and m = Xx + (1 - X)y , 0 _< X <_ 1 , d(m,f(m)) < Xd(x,f(x)) + (1 - X)d(y,f(y)) . Now i f both x and y are in W , then the above inequality implies that d(m,f(m)) = 0 for any m = Xx + (1 - X)y , 0 _< X j< 1 . i . e . m e W . Hence W is convex. If I-f is strictly convex\u00E2\u0080\u009E then Proposition 1.3 implies that W is convex. From this proposition we see that the trouble with Example 1.7 is that I-f is not convex. Let x = (xQ>yo) e K in Example 1.7, then x y d(x,f(x)) = 2 \u00C2\u00B0 g 1/2 ' f o r x * ( 0> 0 ) \u00E2\u0080\u00A2 ( xo + yo } Take a = (|\",1) , b = , m = -| a + -j b , then d(a,f(a)) - d(b,f(b)) = ^ , d(m,f(m)) = . d(m,f(m)) = ^ ~ > ^ r \ d(a,f(a)) + \ d(b,f(b.)) . 34 Hence I-f is not convex. The question arises: Is W a singleton i f I-f is strictly convex? The answer is no. Let us consider the following example: Example 1.8 Let K = { (r ,9) ; 0 <_ r \u00C2\u00A3 1 , 0 <_ 9 <_ -j} be a 2 sublet of R with the usual topology. The point has been expressed in the polar coordinates. Define a point to point mapping f :!<-\u00C2\u00BB\u00E2\u0080\u00A2 K as follows: f((r,9)) = (r,|) for (r,9) e K . We see that f is nonexpansive. We want to prove that I-f is strictly convex. Let d denote the metric on K . Given P = (r,9) e K > then d(P,f(P)) = Jl r [ l - cos(| - 9 ) ] 1 / 2 . Set g(r,9) ='d(P,f(P)) . Then we have (a) g(r,9) is a strictly increasing function of r for each fixed 9 in [0, f] . (b) g(r,9) is a strictly decreasing function of 9 for each fixed r in [0,1] . Let P 1 = ( 1^,9]L) , P 2 = (r 2,6 2) both in K such that dCP^f (P )) , and d(P 2,f(P 2)) not both 0 and P 1 f P 2 . Take 0 <\u00E2\u0080\u00A2 X < 1 , M = XP1 + (1 - X)?2 = (r 3,9 3) . Consider the following four cases: (1) r 1 = r 2 . Then minO^e^ f ( P 2 ) ) } (2) 8 . , = 8 2 . Then r 3 < max {r^,^} , hence (by (a)) d(M,f(M)) < max ( d ( P 1 , f ( P 1 ) ) , d ( P 2 > f ( P 2 ) } . (3) r.^ < r 2 and 6 1 > 0 2 . Define M' = ( r 2 > 9 3 ) , then r., < r 2 and 3 3 > 92 ' Therefore (a), (b) imply d(M,f(M)) < d(M',f(M')) < d ( P 2 , f ( P 2 ) ) . (4) r.^ < r 2 and 0 ^ < 0 2 . Then the required i n e q u a l i t y can be proved by means of the following diagram. Let r 3 = Xr^ + (1-A)r 2 . I t IT can be shown that r 3 < r1^ . Let M\" = ( r 3 , \u00E2\u0080\u0094) . Let P^? 2 denote the segment j o i n i n g P., and P 0 . Define Q x such that P , f ( P 1 ) , M\" form a parallelogram; define Q 2 such that P^, f ( P 2 ) , M\" Q 2 form a parallelogram. Then we have MP ! M\"f (P^) P iQ ] MP M\"f (P ) P Q 2^ 2 angle MP = angle MP Q . 3 6 This shows that the triangles MP^ Q-^ and MP2Q? are similar. Thus angle P-^ MQ^ = angle P2MQ2 . Hence Q , M, are collinear. Consider the triangle M','Q2 , we see that M: M\"' < max {Q M\", Q2M\" } = max {P f (P ) , P f (P2>} , i.e., d(M,f(M)) b(X) is called Lipschitz mapping with Lipschitz constant X >_ 0 i f D(f(x),f(y)) _< Xd(x,y) , for any x,y e X . f is called contractive i f 0 <_ X _< 1 , strictly cor-.trac.tive i f 0 <_ A < 1 . Ve assume that empty set i s not i n b(X). Proposition 2.1 Let X be a metric space and f ; X -\u00C2\u00BB- b(X) be a Lipschitz mapping with Lipschitz constant A . If. x^ -\u00E2\u0080\u00A2- x^ , then d(x .fCx^)) -y d(x o >f(x Q)) . i.e. d(x,f(x)) is a continuous function of x . Proof. By Lemma 0.-3 .' we have jd(x n,f(x n)) - d(x o,f(x Q))| < d(xn,xo) + D(f(x n),f - x = ! > d ( x , f ( x ) ) ^ d ( x , f ( x ) ) . n o n n o' o Similar to Banach\"s fixed point theorem for strictly contractive point-to-point mapping, we have the following fixed point theorem for strictly contractive point-to-set mappings. Theorem 2.1 Let (X,d) be a complete metric space and 1 f : X -> cpt(X) be strictly contractive; then f has a fixed point i; X . Proof. Let x e X and x^ e f(x) , then there exists x,, <*: such that d(x2,x1) _< D(f & ) ,f (x)) < Ad(x ,x) . Similarly there is x^ e f(x,>) such that 2 d(x3,x2) <_ D(f (x2) ,f (x 1)) <_ Xd(x2,x1) <_ X d(x ,x; . By induction, we. get a sequence of points x n , such that Xn+1 \u00C2\u00A3 f ( x n } a n d ^ ^ I ' V \u00C2\u00B1 dC f<* n>\u00C2\u00BB f ( V l \u00C2\u00BB i X d ( V V i ; Thus d(x , ,,x ) < Xd(x ,x .)<...< Xnd(x.,x) n+1' n \u00E2\u0080\u0094 n n-l \u00E2\u0080\u0094 \u00E2\u0080\u0094 1' d ( x ,x) < d(x,x .) + ... + d(x ,x ) m n \u00E2\u0080\u0094 m m-I n+1 n \u00C2\u00A3 (X m _ 1 + ... + X n)d( X l,x) = ^\u00E2\u0080\u0094 d(x ,x) 0 as m,n -* \u00C2\u00B0\u00C2\u00B0 X \u00E2\u0080\u0094 X i Hence - x n^ ^ s a Cauchy sequence, therefore x^ -\u00C2\u00BB\u00E2\u0080\u00A2 x^ say, X q e X Now d(x ,f(x )) < d(x ,x n) . By Proposition 2.1, we have n' n \u00E2\u0080\u0094 n n-l J r 39 d(x ,r(x j) = lim d(x ,f(x )) < lim d(x ,x .) = 0 o o n n \u00E2\u0080\u0094 nn-1 n-H\u00C2\u00BB n-*\u00C2\u00BB Hence x s f(x ) . i.e. x is a fixed point of f . o o o Remark 2.1 For a more general form of the fixed point theorem for strictly contractive point-to-set mapping, the reader is referred to Nadler ([20], pp. 479), and Covitz and Nadler [8]. Definition 2.2 A mapping J on X into its dual space X* is called a duality map i f the following conditions are satisfied . (1) (x,Jx) = jjjx||-||x|| , for any x s X , where (x,Jx) means the value at x of the linear functional Jx . (2) ||JxJ = u(||x||) , for any x e X , where u(r) is a continuous monotonically increasing function on R such that u { 0 ) = 0 and u(r) co as r -> oo . u is called a gauge function of J . Definition 2.3 A duality map J : X ->\u00E2\u0080\u00A2 X* is weakly continuous if i t is continuous from the weak topology of X to the weak* topology \u00E2\u0080\u00A2 of X* . . < X Definition 2.4 Let f : X -> 2 and J be a duality map on X. f is said to be J-monotone i f for any x,y e X and x e f(x) , there exists y e f(y) such that (x - y, J(x - y)) >^ 0 . A mapping \"T on X into X* is said to b s Definition 2.5 monotone if 4 0 (x - y, Tx - Ty) _> C , for any x,y \u00C2\u00A3 X . Definition 2.6 A mapping T : X X* is said to be hemi-continuous i f i t is continuous from every segment in X to the weak* topology of X* . Theorem 2.2 Let X be a reflexive Banach space with a weakly continuous duality map J and strictly convex norm. Let K be a bounded closed convex subset of X . If f : X -> cc(X) is nonexpan-sive and f ( x ) d K for a l l x e K \u00E2\u0080\u009E then f has a fixed point in K . As a special case of this theorem, we h a v e t h e following: Corollary 2.1 (MarJcin [19]) Let K be a bounded closed convex subset of a Hilbert space H . If f : H ->\u00E2\u0080\u00A2 cc(H) is nonexpansive, such that f ( x ) C K for x e K , then f has a fixed point in K . The following is a proof that Hilbert space has a strictly convex norm and a weakly continuous duality map. Therefore Corollary 2.1 follows Theorem 2.2. Let ( , ) denote both the inner product and the bilinear mapping from H x H* into 7\ (real numbers) such that (x,w*) = v*(x) , 1/2 where x e H , w* e H* . Sir;:.e = (x,x) , then by the property of inner product, we h e . - i l l * + y|| < N I + |y|| \u00E2\u0080\u00A2\u00E2\u0080\u00A2 41 if x and y are not linearly dependent. This shows that H has a strictly convex norm. Define J : H -\u00C2\u00BB\u00E2\u0080\u00A2 H* by J(w) = w* , where (x,w*) - (x,w) for a l l x e H . Then ||j(w) j| = j|w*| = j|wj| for any w e H . Define u : R r R + to be the identity map on RT (non-negative real numbers);, then (1) (w,Jw) = (w,w*) = (w,w) = ||w||-||w|| = ||w||.||Jw|| . (2) ||Jw|| - ||w|| = y (||w||) . Furthermore y satisfies a l l the conditions to be a gauge function. Kence J defined above is a duality map on H . That J is weakly continuous follows definition of J . We conclude that H is a rer_exive Banach space with strictly convex norm and a weakly continuous duality map J on H onto H* . Let 1 < p < oo . Consider the Banach space . Let xvy e Jl*5 ; i f x and y are not linearly dependent, then Minkowski's inequality implies ll* + y|| < ! i * ! l + lly| \u00E2\u0080\u00A2 Hence Jl*5 has a strictly convex norm. Browder ([7], pp. 1:67) showed that there is a weakly continuous duality map from ir into its dual space , where -jjj- + = 1 . Therefore we get the following corollary which is a direct result of Theorem 2.2. Corollary 2.2 !.<:\u00E2\u0080\u00A2\". K be a bounded closed -convex subset: of 42 SL j 1 < p < 00 . Given f : & ->- cc(A ) If f is nonexpansive and f(x) CT K for any x e K , then f has a fixed point in K . Now we come back to Theorem 2.2. To prove i t , we need following preliminaries. First we introduce a theorem by Browder. Browder's Theorem ([5], pp. 368) Let J : X -> X* be hemicontinuous monotone mapping on a reflexive Banach space X . If there exist a function c(r) on R with cfr) -> +\u00C2\u00B0\u00C2\u00B0 as r +\u00C2\u00B0\u00C2\u00B0 , and (x,J(x)) >_ c(||x.||).\u00C2\u00AB||x|| , for any x e X , then J maps X onto X* . Lemma 2.1 If X is a reflexive Banach space and J is a weakly continuous duality map of X into X* , then J maps Z onto X* . ' v Proof. Let x -> x along a segment, then x -> x weakly, and n o o > n o J then Jx -v Jx in w*-topology of X* , since J is weakly continuous, n o Therefore J is hemicontinuous. I n order to apply Browder's theorem, we need to prove that J is monotone. Let x,y e X , then (x - y, Jx - Jy) = (x,Jx) + (y,Jy) - (x,Jy) - (y,Jx) \u00E2\u0080\u00A2 . ^ l -BJx I + l ly l l l lJyl lHWIII ' jy l l - IHII lJxI i -= (UJ,,! - ||Jy|)(||x.| H M D \u00E2\u0080\u00A2 43 Let u be the gauge function of J ; then i t is monotone increasing. So i f ||x|| >_||y|| , then ||jx|| = y(||x||) > y(||y|t> = ||jy|| . On the other hand i f || y|[ >. ||x|| , then || Jy || _> || Jx|| . Hence (x - y, Jx - Jy) >(||jx|| - ||Jy||)(||x|| - ||yj|) > 0 , i.e., J is monotone. Therefore, by Browder's theorem, J maps X onto X* . Lemma 2.2 Let X be a reflexive, strictly convex Banach space, J a duality map of X into X* . Let f : X cc(X). If the mapping \" f is continuous from the norm topology of X to the Hausdorff metric topology of cc(X) and i f for any x , x e X and y e X , there exists y e f(y) such that (x - y, J(x - y)) _> 0 , then x e f(x) , provided that J is weakly continuous. To prove this we need the following lemma by Browder. Browder's Lemma ([.5], pp. 371): (a) For any Banach space X , there exists a duality map J of X into X* . (b) If X is strictly convex, then there is exactly one duality map J for every function y(r) of Definition 2.2 (i.e., J(Ax) = y(A)J(x)). (c) If X is reflexive and strictly convex, then the duality map J , corresponding to each y(r) , is a continuous mapping of X into 4 4 the weak topology of I* . Proof of Lemma 2.2: By part (.?.) of Browder's lemma, we see that the duality map J is weakly continuous. Lemma 2.1 implies that J maps X onto X* . Suppose x \u00C2\u00A3 f(x) , then there is w* e X* such that (x,w*) < (z,w*) , for a l l z z f(x) (see [11], p. 417, Theorem 10), since f(x) is compact. As J maps X onto X* , there is w E X such that J(w) = w>'c . Hence we have (1) (x.,Jw) < (z,Jw) . for a l l z e f(x) Se*: x = x.- -- w : by assumption there exists x e f (x ) such chat n n n n ( x - x , J ( x - x ) ) > 0 , i.e. . (x - x ,J(\u00E2\u0080\u0094 w))>0 n n \u00E2\u0080\u0094 * n n \u00E2\u0080\u0094 But X is strictly convex, therefore part (b) of Browder's lemma, implies thst J ( n W ) = V ( _ _ ) J M ' whars p(r) is the gauge function of J . Hence we have u(^)(x - x n , J(w)) = ,(x - x n, J(^ w)) >_ 0 . Thus (x - x , J(w)) > 0 , since u(\u00E2\u0080\u0094) > 0 . Therefore n \u00E2\u0080\u0094 n (2) (x, J(w)) > (x , J(w)) , for a l l n > 1 . \u00E2\u0080\u0094 n \u00E2\u0080\u0094 Since x x and f is continuous, we have f(x ) ->\u00E2\u0080\u00A2 f(x) . Now n n 0 0 consider the set A = Co(f(x,[J U {x }) (the notion Co means fcKe. n=l closed convex hull). A :/ \u00E2\u0080\u00A2 closed and convax, hence i t is weakly elose.1. 45 Furthermore A i s bounded, since f(x) bounded and x e f(x ) \u00E2\u0080\u00A2+ f(x). rt n A _ being bounded and weakly closed in a reflexive Banach space, is weakly compact (see [UJ, p. 425). Thus we may assume that {x } -> x e A n o weakly. By (2), we have (x,J(w)) > (x , J(w)) -> (x , J(w)) . \u00E2\u0080\u0094 n o Hence (3) (x, J(w)) _> (x , J(w)) . We claim that X q e f(x) . Suppose X q \u00C2\u00A3 f(x) , then there is u e X such that (4) (X q, Ju) < (z, Ju) , for a l l z e f(x) Choose the sequence z \u00C2\u00A3 f(x) such that lim || z - x || = 0 . This is n n-*\u00C2\u00BB n ^ possible since x e f(x ) and f (x ) -> f (x) . We may assume that n n n J z^ -> Z Q e f(x) weakly, since f(x) is weakly compact. (4) implies (x ,Ju) < (z ,Ju) o o 0 < (z - x , Ju) = (z - z , Ju) + (z - x , Ju) + (x - x . J u ) o o o n' n n n o' C as n -> \u00C2\u00B0\u00C2\u00B0 This is absurd. Hence x e \u00C2\u00A3{x) . But then (1) implies o (x, J(w)> < (x^, J(w)) Thi1. contradicts (3) . Hence x e f(x) . Lemma 2.3 Let X be a reflexive Banach space with a strictly convex norm and a duality map .;' , Let K be a closed convex,--weekly c.-^^ct subset o f X . If f : X cc(X) is continuous (using the Hausderi*' 46 metric topology on cc(X)) and J-monotone, then f(K) is closed in X , where f(K) = U f(x) > provided that J is a weakly continuous.' xeK Proof. Let {x } be a sequence in f(K) such that x -*\u00E2\u0080\u00A2 x .. n n n o We want to prove that X q e f(K) . For each n , there is X r e K such that x e f(x ) . We may assume that x -\u00C2\u00BB- x weakly in K , since K n N n J n o J '. is weakly compact. We claim that X q e f(x Q) . From Lemma 2.2, we see that i t is sufficient to prove that for each y E X , there, exists y c f(y) such that ' \" ( X q - y, J ( X q - y)) > 0 . Suppose on the contrary that there is y Q e X such that (1) (X\"q - y Q, J ( X q - y Q)) -< 0 , for a l l y~Q e f(y Q) . Consider y , x -and x As x e f(x ) and f is J-monotone, ; o n n n n . ' ' there exists y \u00C2\u00A3 f(y ) such that /on J o (x - y , J(x - y )) > 0 . n J on n J o \u00E2\u0080\u0094 Now f(y Q) is compact, so we may assume that y -> y E f (y ) \u00E2\u0080\u00A2 'on 'oo 'o Therefore x - y -> x - y and n on o oo x - y x \u00E2\u0080\u0094 y weakly . n 'o o Jo J Then J ( X r - y^) -y J ( X q - y Q) under the w*-topology of X* (w* means weak*), since J is weakly continuous. Hence 4 7 0 < ( x - y > J(x - y )) (x - y , J(x - y )) \u00E2\u0080\u0094 n Jon' n \u00E2\u0080\u00A2'o n \u00E2\u0080\u00A2'onj o Jo \u00E2\u0080\u00A2+ (x - y , J(x - y )) . o oo o o Hence (x - y , J(x - y )) > 0 , where y e f(y ) . This contradicts o oo o o \u00E2\u0080\u0094 oo o (1). Hence for any y e X , there exists y e f(y) such that (x - y, J(x - y)) > 0 . Then Lemma 2.2 imolies that x e f(x ) C f(K). o o \u00E2\u0080\u0094 o o Hence f(K) is closed. Lemma 2.4 Let X be a Banach space which has a duality map J and let f : X -> cc(X) be nonexpansive. Then I-f is J-monotone, where I is the identity map on X . Proof. Let x,y e X and x e f(x) . Then there is y e f(y) such that l|x - y|[ < D(f(x),f(y)) < ||x - y|| . Thr-; (x - x - (y - y) , J(x - y)) = (x - y, J(x - y)) - (x - y, J(x - y)) > ||x - y||||j(x - y)|| - ||x.- 7|| BJU - y)|| > ||x - y||||j(x - y)|| - ||x - y|| ||J(x - y) || = 0 . Hence I-f is J-monotone. Proof of Theorem 2.2: First we want to prove that 0 e closure of-(I-f)(K) Given 0 < k < 1 define : X -> cc(X) as follows: Pick an arbitrary z e K . Set J o K = {x - ZQ ; x e K} . 4 8 Thc-n K i s a bounded closed c o n v i i subset tf\u00C2\u00A3 X ) hence i t i s weakly compact. K contains the zero element of X . Define J? : X \u00E2\u0080\u00A2> cc(X) o o by f Q ( x - Z q) = f (>:) - Z q - {y - Z q; y e fCx)} -Then f (x - z ) CL K , f o r any x - z e K . f i s nonfix-oansive on o o o o o o X ._, since f i s nonexpansive. on X . Nov/ define f (x) = kf (x) , for x e X . Then f, i s k s t r i c t l y contractive ( i . e . D(x (x),\u00C2\u00A3 ( y ) ) < k x \u00E2\u0080\u0094 y ) oc X . and f. (x) CT K for any x e K , because K i s convex and ic o o o contains 0 - 5Jow applying Theorem 2.1 to f. , we a;ec a f i x e d point of , where x^ e K q . That i s , there e x i s t s y^ <_. K , such that x, ~ y. - z and y, - z e kf (y, - z ) = k(f (y, ) - z. ) . Ic 'k o Jk o o ; k o Jk o Thj...-; there i s y, e f (y, ) such th&c y. - z = k(y, - z ) . Then 'k k 'k o w k o d ( y k , f ( y k ) ) = i n f ||y k- z|| < | | y k - 7 k ! l zef(y-.j IX. -!|z 0 + M 7 K - zo> -? k|| - { 1\" k> il^k-2oH \u00E2\u0080\u00A2 Therefore i n f d(y,f(y)) < i n f d(y, ,.?(y, ;) \u00C2\u00A3 i n f (1 - k) ||7, - 0, yeK 0 cpt (X) is called completely continuous i f i t is continuous from the weak topology of X to the Hausdorff metric topology on cpt(X) . Definition 2.8 f : X -> cpt(X) is said to be semicontractive i f there is a mapping S. on . XxX to cpt(X) such that f(x) = S(x,x) for each x e X , and S has the following properties: (1) for each x e X , S(*,x) is nonexpansive in the first variable. (2) for each x e X , S(x,\u00C2\u00AB) is completely continuous in the second V - . variable. 50 Let us call . the above mapping S the corresponding mapping of f . Theorem 2.3 Let K be a bounded closed convex subset of a reflexive Banach space X . Assume that X has strictly convex norm and there is a weakly continuous duality map J : X -*\u00E2\u0080\u00A2 X* . If F is a semicontractive mapping from X into cc(X) and S : X x X -*- cc(X) is the corresponding mapping of F such that S(x,y) C K for any x,y in K and the set T(x) = {y; y e S(y,x)} is convex for each x e X , then F has a fixed point in K . Remark 2.2 If F is a point-to-point semicontractive mapping and X has strictly convex norm, then the set T(x) defined above is a. convex set, because S(',x) is a nonexpansive mapping and X has strictly convex norm. Thus Theorem 2.3 implies the following result which was proved by Browder ([7], pp. 261). Corollary 2.3 Let X and K be the same as stated in Theorem 2.3. If F is a point-to-point semicontractive mapping and S : X x X ->\u00E2\u0080\u00A2 X is the corresponding mapping of F such that S(x,y) e K for any x, y in K , then F has a fixed point in K . Let us prove the following results which are useful in the proof of Theorem 2.3. 51 Lemma 2.5 Let (X,d) be a metric space, and let A e cpt(X) for each a, in a directed set E . Assume a {A } -*\u00E2\u0080\u00A2 A e cpt(X) under the Hausdorff metric D . If x e A , a aeE o a a then there exists a subnet {x } of {x }' such that x y say, a. a aeE a. ;o } x x y G . A . Proof. As A is compact and x e A , there exists (by Lemma a a a 0.2) y z A such that d(x'y') < D(A ,A ) . Now {y } \u00E2\u0080\u009E C A which a o a a \u00E2\u0080\u0094 a o Ja aeE o is compact, then there is a convergent subnet {y } of {y } , say a i 1 2 y y e A . Given e > 0 , there exists a , a e E such that a. o o x 1 e a _> a => D(AW,AQ) < \u00E2\u0080\u0094 a. > a 2 => d(y ,y ) < -f . x \u00E2\u0080\u0094 a^ o z 3 3 1 3 2 Take a e E such that a _> a and a _> a . Then d(x ,y ) < d(x ,y ) + d(y ,y ) < D(A ,A ) + d(y ,y ) a. o \u00E2\u0080\u0094 a. 'a. a. o \u00E2\u0080\u0094 a. o a. o x x x x 1 x < \u00E2\u0080\u00A2!\u00E2\u0080\u00A2 + -f- = e for a. > a? . 2 2 / x \u00E2\u0080\u0094 Hence x -> y e A a. o o x Definition 2.9 Let X, Y be two topological spaces. A mapping y f : X ->\u00E2\u0080\u00A2 2 is called netwise continuous i f for any net {x } in X a which converges to x , and any y e f(x') , there exists a subnet \ o a a {y } of {y } such that y converges to a point in f(x ) . a i a a i \u00C2\u00B0 52 Proposition 2.2 Let X, Y be two topological spaces. If Y f : X -*\u00E2\u0080\u00A2 2 is netwise continuous, then i t is upper semicontinuous. Proof. Suppose f is not upper semicontinuous, then there exists x e X and an open set U , such that f(x ) CZ U and for any o o neighborhood V of x there is x e V and y e f(x ) with o v 'v v y^ \u00C2\u00A3 U . Let [/ denote the neighborhood system of Xq . Then ( x v ; v e (/} is a net which converges to Xq . Since f ( x D ) c H and U is open, and y^ \u00C2\u00A3 U for any V e \J , we see that no subnet of {y^ } will converge to a point in f ( x Q ) \u00E2\u0080\u00A2 Hence f is not netwise continuous. This completes the proof. Proof of Theorem 2.3 Let x e K be fixed. Then (1) r(x) is nonempty for each x e K , for S(*,x) is nonexpansive in the first variable and X is reflexive, strictly convex in norm and X has a weakly continuous duality map J : X ->- X* , Theorem \ 2.2 implies that S( \u00E2\u0080\u00A2 ,x) has a fixed point which, by definition of r(x) , is a point of the set T(x) . (2) r(x) is closed. Let y n e T(x) such that y^ ->\u00E2\u0080\u00A2 jQ > w e want to prove that y Q e S(yQ,x) and hence y Q e T(x) . By Lemma. 0 . 3 d(yo,S(yo,x)) < ||y o - y Q || + d(y n >S(y n,x)) + D(S(yn,x) ,S(yQ,x)) = ||yo \" y n|| + D(S(yn,x),S(yo,x)) (since e S(y n >x)) \u00E2\u0080\u00A2+ 0 as n :\u00E2\u0080\u00A2>\u00E2\u0080\u00A2<\u00C2\u00BB . Hence y e S(y ,x) . -. 5 3 r(x) i s closed and convex, so i t i s weakly closed. Consider the mapping T : x -> r(x) , fo r . x s K . We claim that T i s upper semicontinuous under the weak topology on K . As a matter of fac t we w i l l prove that F i s netwise continuous under the weak topology o f K . Therefore Proposition 2.2 implies that T i s upper semicontinuous. Let { x a^ D e a net i n K such that x -> x weakly, denoted by x S x . Given a o J J a o y e r(x^) ,< we want to f i n d a weakly convergent subnet of {y } a a a converging weakly to a point of T ( X q ) . Define S 1(y,x) = y - S(y,x) = (I - S(.,x))(y) . Then : X x X -> cc(X) i s well defined and 0 e S^(y,x) i f and only i f y e S(y,x) . Lemma 2.4 shows that S^(y,x) i s J-monotone f or each f i x e d x e K . Now f i x x^ , l e t z be an a r b i t r a r y point i n K . Since y e S(y ,x ) , then 0 e S,(y ,x ) . By J-monotone, there ex i s t s a a a 1 ^ a a x \u00C2\u00A3 ST(Z,X\u00E2\u0080\u009E) such that (x - 0 , J(z - y )) > 0 . az I u az a \u00E2\u0080\u0094 . x 5 x => S n ( z , x ) = z - S(z,x ) z - S(z.x ) = S,(z,x ) . a o ' 1 \u00E2\u0080\u00A2 a . a o l o x e S., (z,x ,) and S, (z,x ) -> S, (z,x ) imply that there ex i s t s a az . 1 a l a l o subnet {x } of {x } such that x -> x say, x E S, (z,x ) a.z J az a.z oz J ' oz 1 o l l (by Lemma 2.5). By the weak compactness of K , we may assume that w m i . w rr ' T / \ w* T , N y \u00E2\u0080\u00A2> y say. Then z - y -> z - y . Hence J ( z - y ) \u00E2\u0080\u0094> J(z - y ) a o y J a Jo J a o (here \u00E2\u0080\u0094 r means convergence i n w> topology: Actually \u00E2\u0080\u0094>- and ->-are equivalent when X i s r e f l e x i v e ) , and then 0 < ( x a z , J(z - y a ) ) -> ( x o z , J(z - y o \u00C2\u00BB . i i Hence (x , J ( z - y ) ) > 0 . oz' J o \u00E2\u0080\u0094 5 4 Therefore for each z e X , there is x e S..(z,x ) such that oz 1 o Let v be an arbitrary element in X . Denote z = y + tv : then t Jo ' there is x e S.. (z .x ) such that oz 1 t o 0 < ( x ^ , J ( z t - y Q)) = ( x ^ , J(tv)) . Then y(t)(x , J(v)) _> 0 , where u is the gauge function of J and O Z t u(t) > 0 for t > 0 . Hence (x , J(v)) > 0 with x e S-(z_,x ) . oz^ _ \u00E2\u0080\u0094 oz^ 1 t o Set t = \u00E2\u0080\u0094 , then z_ = y + \u00E2\u0080\u0094 v -> y a s n -\u00C2\u00BB-<\u00C2\u00BB\u00E2\u0080\u00A2 . and then n n t o n o n S l ^ Z t ' Xo^ S l ^ y ' x ^ ' ^PPly i ng Lemma 2.5 to the sequence n {S 1(z ,x )} and x e S..(zt ,x ) , there exists a subsequence l t o oz i t o n t n n \u00E2\u0080\u00A2 -{x } of {x } such that O Z t \u00C2\u00B0 zt . n. n l x -> x say , x E S, (y ,x ) , as i \u00C2\u00AB> . oz^ oov oov 1 Jo o n. l Thus 0 < (x ,J(v)) (x ,J(v)) . We conclude that for any v e X, \u00E2\u0080\u0094 oz^ oov J n. i there is x e S n(y ,x ) such that oov l o o ( + ) : ' \ ( X , J ( V ) ) > 0 oov Suppose 0 \u00C2\u00A3 S i ( y o ' x Q ) \u00C2\u00BB then'there is an open convex neighborhood W of 0 , such that W f] S,(y ,x ) = 4 . Now both W and Sn (y ,x ) are l o o T 1 Jo o 55 convex and W i s open, Hahn-Banach's separation theorem [ 22 ] implies that there exists a constant c and a point v* e X* such that (x,v*) > c , for a l l x e W , and (x,v*) < c , for a l l x e S (y , x ) . J. o o \u00E2\u0080\u00A2 Since 0 e W , we have 0 = (o,v-'0 > c , T-herefore (x,v\") _< c < 0 , for a l l x e S^(y o,x o) . But J maps X onto X* , so there i s v e X such that J(v) = v* , thus (x,J(v)) _< c < 0 for any x e S^(y ,x ) \u00E2\u0080\u00A2 This contradicts (+). Hence 0 e S,(y ,x ) . That i s y e S(y ,x ) . i o o o - o o Therefore, for any x ->\u00E2\u0080\u00A2 x and y' e T(x_,) , there i s subnet {y } J a o J a 01 Ja. l of {y } such that y y e T(x ) . Hence T i s upper semicontinuous Ci ct. o o l \" on K under the weak topology. Therefore K. Fan's f i x e d point theorem [13] implies that there i s x e K such that x e T(x ) , i . e . , o o o x e 3(x ,x ) = F(x ) . Hence x i s a fi x e d point of F . o o o o o Coroll a r y 2.4 Let X be a Banach space defined i n Theorem 2.3, and K be a bounded closed convex subset of X . If F : X -> cc(X) i s a semicontractive mapping with a corresponding mapping S : X x X -> cc(X) such that S(x,y) CZ K for x,y e K and for each x e K , I - S(',x) i s a convex mapping of K into cc(K) , then F has a fi x e d point i n K (where I i s the i d e n t i t y mapping and (I - S(-,x))(y) = {y - z; z e S(y,x)}) . 56 Proof. Since I - S(*,x) i s a convex mapping, the set r(x) = {y; y e S(y,x)} i s convex (by Proposition 1.7). Then the proof follows Theorem 2 . 3 . 57 CHAPTER III FIXED POINT THEOREMS FOR POINT-TO-SET MAPPINGS WITHOUT THE CONDITIONS OF CONVEXITY . \u00C2\u00A71. S t r i c t l y Nonexpansive Mappings In this section we assume X to be a metric space with metric d . Let f : X -> cpt(X) . Define f(A) = [ J f(x) , for A \u00C2\u00A3 cpt(X) . xeA Lemma 3.1 Let f : X cpt(X) be a L i p s c h i t z mapping, then f(A) e cpt(X) for any A e cpt(X) . Proof. Let y^ \u00C2\u00A3 f(A) , then there i s x n \u00C2\u00A3 A such that y s f ( x ) . There e x i s t s a subsequence {x } of {x } such that n n n. n x x -> x \u00C2\u00A3 A , since A i s compact. In order to s i m p l i f y the notation, i we may assume from the\u00E2\u0080\u009Ebeginning that y e f ( x ) and x -> x e A . e ' n n n o Now for each y e f ( x ) , there e x i s t s z e f ( x ) (by Lemma 0.2) such Jn n n o that d ( y n , z n ) < D ( f ( x n ) , f ( x Q ) ) . The sequence { z n} 1 S i n ^ ^ x 0 ^ which i s compact, hence there exists a subsequence {z } of- {z } such that z -+ z e f ( x ) . Consider n. n n. o o l i the subsequence {y } c f {y'} . Given e > 0 , there are p o s i t i v e i integers N^ and N such that 58 d(z , z ) <' -r- for nJ > N, n . o 2 i \u00E2\u0080\u0094 i l D(f (x ) ,f (x ) ) < \u00E2\u0080\u0094 for n > N\u00E2\u0080\u009E , since f i s a n o 2 \u00E2\u0080\u0094 2 L i p s c h i t z mapping. Let N = max {N^, N^} , we have n. > N => d(y ,z ) < d(y ,z ) + d(z ,z ) i \u00E2\u0080\u0094 n . o \u00E2\u0080\u0094 n . n . n . o I i i i < D(f(x ),f(x )) + | < | + | = e . \u00E2\u0080\u0094 n. o 2 2 2 l Hence y -> z e f ( x ) CZ f (A) . This proves that any sequence i n f(A) i has a convergent subsequence i n f(A) . Hence f(A) i s compact. Now we can consider f as a mapping from cpt(X) i n t o cpt(X) . Lemma 3.2 If f : X -> cpt(X) i s s t r i c t l y nonexpansive, then f : cpt(X) -> cpt(X) i s s t r i c t l y nonexpansive. Proof. Let A,B e cpt(X) and A f B . We want to prove that (1) for any a e f(A) , there ex i s t s b e f(B) (or for any b e f(B) , there i s a e f(A)) such that d(a,b) < D(A,B) . Since a e f(A) , there i s a' e A such that a e f ( a ' ) . As a' e A , there i s b' e B such that d(a'.,b') <_ D(A,B) . If b' = a' , then a e f(a') = f ( b 1 ) C f(B) . 5v In this case choose b = a , then we are through. If b' 4 a' , we have D ( f ( a ' ) , f ( b ' ) ) < d(a',b') , since f i s s t r i c t l y nonexpansive. Consider a , f(a') , f(b') . As a e f(a') , there i s . b e f(b') such that d(a,b) < D( f ( a ' ) , f ( b ' ) ) . Therefore d(a,b) < D( f ( a ' ) , f ( b ' ) ) < d(a',b') < D(A,B) . ( 1 ) i s then proved f o r any a e f(A) . sup d(x,f(B)) < D(A,B) and sup d(y,f(A)) < D(A,B) . Here we xef(A) yef(B) prove the f i r s t i n e q u a l i t y . Let c = sup d(x,f(B)) . x e f ( A ) Given an a r b i t r a r y x E f(A) , ( 1 ) implies the existence of a, point y \u00C2\u00A3 f(B) such that d(x,y) < D(A,B) . d(x,f(B)) _< d(x sy) < D(A,B) . Hence c = sup d(x,f(B)) <_ D(A,B) . Since d(x,f(B)) i s a xef(A). continuous function of x and f(A) i s compact, there e x i s t s X q e f(A) , such that d ( x Q , f ( B ) ) = c . Suppose c = D(A,B) , then, by ( 1 ) , there e x i s t s y^ e f(B) such that d ( x o , y Q ) < D(A,B) . 60 Thus d( x Q , f ( B ) ) <_ d ( x Q , y o ) < D(A,B) = c , and we have a con t r a d i c t i o n . Hence c < D(A,B) . S i m i l a r l y , one can prove that sup d(y,f(A)) < D(A,B) . yef(B) R e c a l l the d e f i n i t i o n of the notation V (S) , where r > 0 and S i s a subset of X , we can rewrite (2) as follows: i ( A ) C V D ( A , B ) ( i ( B ) ) a n d i ( B ) C V D ( A , B ) ( i ( A ) ) \u00E2\u0080\u00A2 Therefore D(f(A),f(B)) = i n f {r > 0; f(A) C V. (f(B)) and f ( B ) C (f(A))} < D(A,B) . The proof i s complete. Theorem 3.1 Let K be a compact subset of a metric space X .\u00E2\u0080\u00A2 If f : K 2 i s . s t r i c t l y nonexpansive, then f has a f i x e d point i n K . Proof. Let c = i n f d(x,f(x)) . If c > 0 , then there i s . xeK x^ e K such that d ( x n , f ( x n ) ) -> c . By the compactness of K , we may assume that x \u00E2\u0080\u00A2+ x \u00C2\u00A3 K . Proposition 2.1 implies that n o r c = lim d(x ,f(x )) = d(x ,f(x )) . Again by the compactness of f ( x ) , n n o o o n-x\u00C2\u00BB there i s x, e f ( x ) such that I o d(x ,x ) = d(x ,f(x )) = c . o I o o Since x^ e f ( x Q ) > there exists x^ e f(x^) with d ( X ; L , x 2 ) < D(f ( x Q ) , f ( X ]_)) . Then we have d ( X l , f ( x . ) ) < d(x 1 5x\u00E2\u0080\u009E) < D(f(x ),f(x.)) < d(x ,x.) = d(x ,f(x )) = c. l l \u00E2\u0080\u0094 1 z \u00E2\u0080\u0094 o 1 o l o o This contradicts the assumption that i n f d(x,f(x)) = c . Hence c = 0 , x eK then the compactness of K implies that there i s Xq e K with d(x ,f(x )) = 0 . o o Hence x i s a fixed point of f i n K . o r Remark 3.1 The s t r i c t nonexpansiveness i s a very strong condition of continuity. Therefore i t allows us to take closed sets as values of f which i s much weaker than taking closed convex subsets as values of f . In [131 K. Fan proved a f i x e d point theorem for point-to-set mapping, i t said \" I f K i s a compact convex subset of a Hausdorff l o c a l l y convex l i n e a r t o p o l o g i c a l space, then an upper semicontinuous point-to-set mapping from K into the family of closed convex subsets of K has a fi x e d point i n K\". Because the values, of a point-to-set mapping play an important r o l e i n fixed point theorems, Theorem 3.1 by no means can be considered as a c o r o l l a r y to K. Fan's f i x e d point theorem mentioned above, even i f we l e t X and K s a t i s f y the hypotheses of K. Fan's theorem. The following example i l l u s t r a t e the importance of the values of point-to-set mappings. 62 Example 3.1 L e t K be the c l o s e d u n i t d i s k of t h e p l a n e w i t h the u s u a l m e t r i c . D e f i n e f K -> 2 K as f o l l o w s : f ( ( 1 , 6 ) ) (0,0) f o r any 0 _< 6 <_ 2ir . f ( ( r , 8 ) ) the c l o s e d a r c , l y i n g on the c i r c l e w i t h r a d i u s 1 - r and c e n t e r a t (0,0) , subtended by an a n g l e 2TT(1 - r ) such t h a t the p o i n t (1 - r , i r + 6) l i e s a t the m i d d l e of the a r c . f ( ( 0 , 0 ) ) u n i t c i r c l e . The map f i s upper s e m i c o n t i n u o u s but i t has no f i x e d p o i n t . We observe t h a t the s e t K and the map f s a t i s f y the c o n d i t i o n s of K. Fan's f i x e d p o i n t theorem e x c e p t t h a t f ( r , 6 ) i s not a convex s u b s e t of K f o r r 4 1 , and t h a t makes f. a f i x e d p o i n t f r e e mapping. Theorem 3.1 and one o f E d e l s t e i n ' s theorem [ 1 2 ] . E d e l s t e i n ' s theorem s a y s : I f f : X ->\u00E2\u0080\u00A2 X i s a s t r i c t l y n onexpansive mapping' on a m e t r i c space X and i f t h e r e i s x e X such t h a t { f n ( x )} has a subsequence o o c onvergent to z e X , then z i s a f i x e d p o i n t of f and i s u n i q u e . From t h i s we can see t h a t Theorem 3.2 below was done i n d e p e n d e n t l y by F r a s e r and N a d l e r [14]. Theorem 3.2 L e t f : X -> c p t ( X ) be s t r i c t l y n o n e x p a n s i v e . I f t h e r e i s a p o i n t x e X such that { f n ( x )} has a subsequence o o convergent to a s e t A e c p t ( X ) , then f has a f i x e d p o i n t i n A (we The f o l l o w i n g Theorem 3.2, i s an immediate a p p l i c a t i o n of 63 d e f i n e f 2 ( x ) = U f ( y ) and f n ( x ) = U f ( y ) f o r n _> 2) y e f ( x ) ^ ( n - l ) . v o y e f (x ) o P r o o f . By Lemma 3.2, f i n d u c e s a mapping f : c p t ( X ) ->- c p t ( X ) which i s s t r i c t l y nonexpansive. L e t B = f ( x Q ) \u00E2\u0080\u00A2 Then f r \" ( x Q ) = f n \"'\"(B) = f n \"'\"(B) f o r . n _> 2 . By assumption the sequence { f n ( x Q ) } has a convergent subsequence convergent to A (under the H a u s d o r f f m e t r i c D), hence so does the sequence (f l\"(B)} . E d e l s t e i n ' s theorem i m p l i e s t h a t A i s the unique f i x e d p o i n t of f . Hence f ( A ) = A . T h i s g i v e s us a compact s u b s e t which i s i n v a r i a n t under f . i . e . f : A c p t ( A ) . T h e r e f o r e the e x i s t e n c e of a f i x e d p o i n t o f f i n A f o l l o w s from Theorem 3.1. \u00C2\u00A72. F u r t h e r R e s t r i c t i o n s on C o n t i n u i t y A. Netwise c o n t i n u i t y L e t X, Y be two t o p o l o g i c a l s p a c e s . R e c a l l t h a t Y f : X -> 2 i s n e t w i s e c o n t i n u o u s i f f o r any n e t x -> x and any a o . y e f ( x ) , t h e r e e x i s t s a subnet {y } of {y } such t h a t a a a. a I y -> y say, and y e f ( x ) . We showed t h a t ( i . e . P r o p o s i t i o n 2.2) a . o o o -l Y i f f : X -r 2 i s n e t w i s e c o n t i n u o u s then i t i s upper s e m i c o n t i n u o u s . The c o n v e r s e i s not t r u e , . f o r i n s t a n c e , 2 Example 3.2 L e t X = R be the p l a n e w i t h u s u a l t o p o l o g y . 6 4 D e f i n e g : X -\u00C2\u00BB\u00E2\u0080\u00A2 k(X) by g(x,y) = ' { ( 0 , z ) ; z e R,} f o r (x,y) e R 2 . g i s a c o n s t a n t mapping, hence i t i s upper s e m i c o n t i n u o u s . But g i s no t n e t w i s e c o n t i n u o u s , f o r l e t (x ,y ) (x ,y ) . Choose P = (0,n), n Jn oo n 2 then P e g(x ,y ) , but {P } has no convergent subsequence i n R. n n n n \u00C2\u00B0 n T h e r e f o r e upper s e m i c o n t i n u i t y does not imply n e t w i s e c o n t i n u i t y . We can make the c o n v e r s e o f P r o p o s i t i o n 2.2 to be t r u e by o n l y r e s t r i c t i n g the v a l u e s o f mapping t o a s m a l l e r c l a s s o f s u b s e t s , namely c p t ( Y ) . P r o p o s i t i o n 3.1 ..Let X, Y be two t o p o l o g i c a l s p a c e s . G i v e n f : X -> c p t ( Y ) , i f f i s upper s e m i c o n t i n u o u s , then i t i s n e t w i s e c o n t i n u o u s . P r o o f . L e t {x }' be a n e t i n X such t h a t x ->\u00E2\u0080\u00A2 x e X . a aeD a o L e t y e f(x\u00E2\u0080\u009E) . Denote the s e t {y : a e D} = A . a CL a (1) I f A f l f ( x o ) = cj) , then X - A 3 f (x ) and X - A i s open. T h e r e f o r e by upper s e m i c o n t i n u i t y o f f , t h e r e e x i s t s aQ e D such t h a t a > a => f (x ) C l X - A . \u00E2\u0080\u0094 o a Then a > a = > y e f ( x ' ) C X - A \u00E2\u0080\u0094 o a a V7hich i s absurd. Hence we have (2) A /] f ( x ) 4- d) . I f t h e r e i s a subnet {y } of {y } such that o ' a . a x 65 {y } C f ( x ) , then there i s a subnet of {y } converging to a a. o a. 1 i point of f ( x Q ) > since f ( x Q ) 1 S compact. Thus we are through. If no subnet of {y } i s i n f (x ) , then there i s a' e D such a o that a > a ' = > y i f ( x ) . Set B = {y ; a > a'} , then \u00E2\u0080\u0094 a T o a \u00E2\u0080\u0094 B f\ f (x ) i , then s i m i l a r to case (1) we o T o w i l l get a co n t r a d i c t i o n ) . As B f| f ( x Q ) = f , we are able to f i n d a subnet of {y } , converging to a point i n f ( x ) . That i s , \" a a> a o there i s a subnet {y } of {y } such that {y } converges to a noint of f ( x ) . o From the above two propositions we see that netwise continuity and upper semicontinuity of a mapping f : X -> cpt(Y) are equivalent. Therefore we have Y Proposition 3.2 If f : X ->- 2 i s a c t u a l l y a point-to-point mapping, then f i s netwise continuous i f f i t i s upper semicontinuous or continuous. Remark 3.2 If a t o p o l o g i c a l space s a t i s f i e s the f i r s t axiom of c o u n t a b i l i t y , then the topology can be described i n terms of sequence. Therefore i f both t o p o l o g i c a l spaces X and Y s a t i s f y the f i r s t axiom of c o u n t a b i l i t y , then netwise continuity can be replaced by sequence-wise continuity i n Propositions 2.2 and 3.1. That i s , a l l the terms \"net\" can be replaced by \"sequence\". 66 B. L o c a l l y Closed Mappings D e f i n i t i o n 3.1 Let X, Y be two t o p o l o g i c a l spaces. A map . Y ' f : X -> 2 i s said to be closed at X q e X i f for any neighborhood V of x , there exists a neighborhood U of x such that o . o f(U) C f(V) with f as defined i n \u00C2\u00A71 . f i s l o c a l l y closed on X i f i t i s closed at every point of X . Here we quote without proof one of the known r e s u l t s by Belluce and Kirk ([3], theorem 1.1) about the r e l a t i o n between upper semicontinuous map and l o c a l l y closed map. Theorem (1) I f X i s compact and regular, and i f Y i s Y Hausdorff, then every upper semicontinuous mapping f : X -> 2 i s l o c a l l y closed. Combining above theorem and Proposition 2.2, we get Proposition 3.3 If X i s compact and regular, and i f Y i s Y Hausdorff, then a netwise continuous mapping f : X -> 2 i s l o c a l l y closed. So f a r we have four kinds of continuity f or point-to-set 67 mappings f : X 2 , ria.mely upper semicontinuity, nonexpansiveness (it is well defined i f both X and Y are metric spaces and values of f are bounded set), netwise continuity, and locally closed map. We have seen Y in Proposition 2.2 that a netwise continuous map f : X -> 2 is upper semicontinuous. The following examples will show that there are no implications among the others unless we specify either the topological spaces (e.g. Theorem (1) above) or the values of f (e.g. Proposition 3.1). First let us look back Example 3.2 where we defined a constant 2 R2 mapping g on R into 2 such that i t is not netwise continuous, g , being constant, is locally closed. Hence a locally closed mapping is not necessarily netwise continuous. One might think that netwise continuity is stronger than being locally closed. The following example shows that this is not true. Example 3.3 Let X be the set of positive integers. Define J to be the family of subsets U of X , including X itself, such that X - U is a finite subset of X . Then T is a topology on X . We see that any finite subset of X is closed and any infinite subset of X is dense in X because i t intersects every open set. Furthermore we see that ( X , T ) is compact but is neither Hausdorff nor regular. Define h : X ->- X as follows:. h(n) = n + 1 for n e X . Then h is continuous, for let n e X and U = X - {n, , ..., n } be o 1 r such that n Q + 1 4 n_^ , i = 1, r ; then U is an open neighborhood of h(nQ) = n Q + 1 . Set V = X - {n^l, n rl}. . (if .n-l = 0 for 68 some 1 <_ i _< r , then take V = X - {n^-1, . .., n_^ -|_-x> ni+l~\"'\"' nr~-'-^\u00C2\u00BB then n e V and V is a neighborhood of n . Now we have h(n) e U o o for a l l n e V . Hence h is continuous. We note that i f a mapping Y f : X -> 2 is actually a point-to-point mapping, then f is netwise continuous i f and only i f i t is continuous (by Proposition 3.2). Therefore h is netwise continuous. But h is not locally closed, for if W is a neighborhood of n^ such that W ^ X , then f(W) is an infinite proper subset of X . Now take any neighborhood V of n Q > we see that h(V) is an infinite subset of X , hence h(V) = X . Therefore no neighborhood V of n^ exists such that h(V) Ch(W) . Hence h is not locally closed. Since h is upper semicontinuous but not locally closed, this shows that the conditions on topological spaces X and Y of Theorem (1) of Belluce and Kirk {3J are not removable. The following theorems indicate further relations between upper semicontinuity and locally closed maps for certain point-to-set mappings Theorem 3.3 Let X be a locally convex linear topological space, and K- be a weakly compact convex closed subset of X . Let (j>:KxK->-2 , f : K ->- 2 . , where D is a closed bounded subset of the positive real numbers R+ . Assume and f satisfy the following conditions: 69 (1) for each f i x e d y e K , (*>y) i s upper semicontinuous from the weak topology of X to R + . ( 2 ) q>(x,x) = 0 for a l l x.e K . ( 3 ) the set E(x) = {y e K ; there exists r z (x,y) and s z f(y) such that r _> s} i s nonempty, closed and convex. If E : K -y k(K) i s weakly l o c a l l y closed i n K ( i . e . , E i s l o c a l l y closed.under the weak topology of K ) , then E i s upper semicontinuous i n the weak topology of K , and there i s some x e K such that o 0 e f ( x ) . o The mapping i s motivated by the corresponding mapping S of a semicontractive mapping. The conditions on are weaker than those on S . Only upper semicontinuity on the f i r s t v a r i a b l e i s required for instead of nonexpansiveness. The condition ( 2 ) on cj) only to imply that 0 e f(x^) for some x^ e K . The use of cj) has been introduced by Belluce and Kirk [ 3 ] ; here we extend i t to a point-to-set mapping. Theorem 3 . 3 can be obtained by the following more general r e s u l t s . Theorem 3.4 Let X be a Hausdorff space and Y a compact Y space. Suppose f : X -> 2 and : X x X ->\u00E2\u0080\u00A2 cpt(Y) s a t i s f y the following conditions: 70 (1) <)>(\u00E2\u0080\u00A2,x) i s upper semicontinuous for each f i x e d x e X ; (2) for each x e X , the set E(x) = {y e X ; c|>(x,y)fi f(y) 4 4>} i s nonempty and closed. Y If E : X ->\u00E2\u0080\u00A2 2 i s l o c a l l y closed, where E(x) i s the set defined above, then E i s upper semicontinuous. Proof. Let x e X , and l e t 1/ be the neighborhood system of x . We claim that E(x) = D E(V) . VelA I t i s clear that E(x) C fl E(V) . To prove f) E(V) C E(x) , l e t Vel/ Vel/ y e H E(V) . Then there i s a e V such that y e E(a ) , f o r V e l / . Vel/ V V i . e . , (av,y) f\"i f(y) 4 \u00E2\u0080\u00A2 Now { a v) x a n a (av>y) D f (y) ; Proposition 3.1 implies that there i s a subnet {b } of {b } such that b -> b v v v o v a a and b Q e (x,y) . But b e f(y) and f(y) i s closed, hence a b^ -> b Q e f(y) . i . e . , b e tj)(x,y)n f(y) \u00E2\u0080\u00A2 Thus, by d e f i n i t i o n of a E(x) , we see y e E(x) . Hence E(x) = f l E(V) . Vel/ The remainder of the proof i s the same as the proof of Theorem 1.3 i n [3]. That i s , l e t A be open i n Y such that E(x) C A . Since E i s l o c a l l y closed, there i s W el/.', f o r each V e l / , such that v E(W ) C E(V) , then 71 E(x) = H E(V) C f l E(W) C D E(V) Ve!/. Vel/ V Ve(/ i . e . , E(x) = O E(W ) . Vel/ V Then we have f l ' E(W ) = E(x) C A . Thus Vel/ V O E(W ) f] (Y - A) = cf> . Vel/ By the compactness of Y , there e x i s t s f i n i t e number of neighborhoods W ..... W of x such that v, v 1 n n O E(Wv )f] (Y - A) = cj, . i = l i Let W = C~\ W , then W i s a neighborhood of x such that v. i = l i \u00E2\u0080\u00A2 n y e W => E (y) C D E (W ) C A . i=i- v i : Hence E i s upper semicontinuous. Now we can replace Y i n the above theorem by a closed bounded subset D of the r e a l numbers, and enlarge the set E(x) without damaging the. theorem. Theorem 3.5 Let X be a Hausdorff space, D be a closed bounded subset of r e a l numbers. Let cj) : X x X -\u00C2\u00BB\u00E2\u0080\u00A2 2^ . , f : X -> 2^ s a t i s f y the following conditions: 72 (1) (x,y) and s e.f(y) such that r >_ s} i s nonempty and closed. I f E : X 2 i s l o c a l l y closed, then E i s upper semicontinuous. Proof. Let x e X and \J be the neighborhood system of x . Since D i s compact, thus i f we can prove that E(x) = D E(V) , then Vel/ the rest of the proof can be c a r r i e d out i n the same manner as that of Theorem 3.4. Let y e {\"} E(V) , then there i s x e V such that Vel/ V y s E(xv) > f \u00C2\u00B0 r V e I/, and there e x i s t s r ^ e <|>(xv\u00C2\u00BBy) > s e f(y) such that _> , for V e l / . Suppose for any r e (x,y) we have r < s for a l l s \u00C2\u00A3 f(y) , then take an a r b i t r a r y s e f(y) , we have sup {r ; r e cj>(x,y)} < s , since (x,y) i s compact. Thus there e x i s t s an open set G3(x,y) and r e G => r < s . But since x^ -> X q , the upper semicontinuity of 4>(*,y) implies E(x ) C G for some V e l / . For t h i s V , we have r e (x >y) => r e G => r < s . As s i s a r b i t r a r y , we see that r e (J>(x ,y) => r < s for any s e f ( y ) . This contradicts the f a c t that y e E(x^) . Hence there i s r E (x,y) and s e f(y) , such that r _> s , i . e . , y e E(x) . Hence H E(V) C E(x) . Vel/ The reverse i n c l u s i o n i s t r i v i a l , therefore E(x) = f~] E(V) . Vel/ 73 The proof of Theorem 3.3: As E(x) i s closed and convex, i t i s weakly closed. Theorem 3.5 implies that E i s upper semicontinuous i n the weak topolo-gy of K . Since K i s weakly compact and E : K ->\u00E2\u0080\u00A2 k(K) i s upper semicontinuous under the weak topology of, K , K. Fan's f i x e d point theorem [13] implies the existence of a fi x e d point of E , say X Q . Now x e E(x ) means that there i s r e (x ,x ) and s e f(x ) such o o o o o that r > s . But d> (x ,x ) = 0 , hence 0 > s > 0 , i . e . , 0 e f ( x ) . \u00E2\u0080\u0094 o o \u00E2\u0080\u0094 \u00E2\u0080\u0094 o The following f i x e d point theorem i s an a p p l i c a t i o n of Theorem 3.3. Theorem 3.6 Let X be a metrizable l o c a l l y convex l i n e a r t o p o l o g i c a l space, K be a weakly compact convex closed subset of X . Let f : K ->\u00E2\u0080\u00A2 cpt(K) be nonexpansive and l e t : K x K -> D + = [0,r] , for some r > 0 , s a t i s f y the following conditions: (1) f o r each fixed x e K , ct)(x,>) i s n o n e x p a n s i v e ; (2) for each fixed y e K , ()>(\u00E2\u0080\u00A2,y)' i s continuous from the weak topology of K' to R + ; (3) <}>(x,x) = 0 for each x e K . (4) there e x i s t s a constant k > 0 such that the set E(x) = {y e K ; ^ kd(y,f(y))} i s nonempty and convex f o r each x e K . If the mapping E :. K -\u00C2\u00BB\u00C2\u00BB k(K) i s weakly l o c a l l y closed, then f has a fix e d point i n K . . 74 Proof. Let d be the metric on X . F i r s t we prove that E(x) i s closed for each x e K . Let x e K and y s E(x ) such that y o \u00E2\u0080\u00A2 ' n o ; n converges to y . y e E(x ) => kd(y ,f(y )) . n o \u00E2\u0080\u00A2 o n \u00E2\u0080\u0094 n n Since d(y,f(y)) i s a continuous function of y (by Proposition 2.1), therefore condition (1) and the above i n e q u a l i t y imply < K x 0 > y 0 > >. k d ( y o , f ( y o ) ) . Hence y Q e ^ ^ x 0 ^ ' i - e - > ^ (x ) i s closed. Let g(x) = kd(x,f(x)) , and l e t g play the role of f i n Theorem 3 . 3 . Then a l l the conditions on Theorem 3.3 are s a t i s f i e d . Hence there i s x e K such that o g(x ) = 0 . Then kd(x ,f(x )) = 0 , and thus d(x ,f(x )) =0 , since o o o o o k > 0 . Therefore x i s a f i x e d point of f i n K . o 1 2 Example 3.4 Take K = [O, -^] , f : K -> K given by f (x) = x . Evidently f . i s nonexpansive on K . Define : K x K ->- K by cj)(x,y) = |x - y| , we see that conditions (1), (2) and (3) of Theorem 3.6 s a t i s f i e d , since the weak topology and usual topology on K are the same. Here we want to show that condition (4) i n Theorem 3.6 i s s a t i s f i e d 1 1 for k = 2 . Given X q e [O, -^] , l e t y e [O, -^] s a t i s f y i n g 2 3 \" /9 - ,8x |y -' x | _> 2(y - y ) . Then y _< . Since 3 - /9 - 8x i i 0 j< < for X q e IOJ-J] , we see that the set . 3 - 79 - 8x E ( X Q ) - {y e K; |y - X Q | > 2d(y,f(y))} = 10, \u00E2\u0080\u0094 -] 75 i s nonempty, closed and convex. Hence condition (4) of Theorem 3.6 s a t i s f i e d . To show that_ the mapping E : K k(K) i s weakly l o c a l l y closed, f i r s t we see that E i s upper semicontinuous because vf \ rn 3 -.79 - 8 x , , , _ 3 - /9 - 8x E(x) = [0, ; 1 and the function : i s continuous. 4 4 Therefore Belluce and Kirk's theorem (1) implies that E i s l o c a l l y closed. But the weak topology on R coincides with usual topology. Hence E i s weakly l o c a l l y closed. Thus the example meets a l l the requirements stated i n Theorem 3.6. C e r t a i n l y the f i x e d point of f ex i s t s \u00E2\u0080\u00A2;' i t i s z e r o . Now we study one more notion of continuity of point-to-set mappings concerning weak topology. C. Complete Continuity R e c a l l that f : X ->\u00E2\u0080\u00A2 cpt(X) i s completely continuous i f i t i s continuous from the weak topology of X to the Hausdorff metric topology on cpt(X) . We s t i l l use D as the Haudorff metric on cpt(X) . Proposition 3.4 Let X be a Banach space. If f : X -> cpt(X) i s completely continuous, then i t i s sequencewis'e continuous, hence upper semicontinuous. Proof. Let x e X such that x -> x . Given y e f (x ) , we n n o \u00E2\u0080\u00A2'n n ' 76 want to f i n d a subsequence of ^-Y^ converging to a point of f(x^) . Since x -> x , x -> x weakly. Thus by complete continuity of n o n o f , we have D(f (x n) ,f (XQ)) + 0 as n -> o o As y e f(x ) , there e x i s t s z e f(x ) such that n n n o lly - z || < D(f(x ),f(x )) . Since f(x ) i s compact, there e x i s t s a 11'n n11 \u00E2\u0080\u0094 n o o r subsequence {z } of { z } such that z z say, and z e f ( x ). n n. n n. o o o i x Then My - z | < l ly - z II + II z - z || < D(f(x ),f(x )) + ||.z. - z II H'n. o11 \u00E2\u0080\u0094 ' I'n. n. 11 11 n. o N \u00E2\u0080\u0094 n. ' o 11 n. o11 X 1 X 1 i x -> 0 as i o o Hence y n -> Z q e f ( x 0 ) \u00E2\u0080\u00A2 Therefore f i s sequencewis-e c o n t i n u o u s i By Proposition 2.2, i t i s upper semicontinuous. Theorem 3.7 Let K be a weakly compact subset of a Banach space X . If f : K -> cpt(K) i s completely continuous, and i f f i s asymptotically regular at some point x^ e K (see d e f i n i t i o n 1.3), then f has a f i x e d point i n K . Proof. Since f i s asymptotically regular at Xq , there e x i s t s x e f (x n) such that lim fx - x _ II = 0 Now K i s weakly n n - l 11 n n - l 1 1 J n-x\u00C2\u00BB compact and {x } C K , so there e x i s t s a subsequence {x } of {x } i such that x y e K' weakly. Since f i s completely continuous, we i 7 7 have f(x ) -> f (y ) . As x e f (x .) then n. o n n - l d (x . , f (x n ) ) < x. - x II -> 0 as n -> \u00C2\u00B0\u00C2\u00B0 n - l ' n - l \u00E2\u0080\u0094 \" n - l n\" By Lemma 0.3-,'. we have d ( x n , f ( y Q ) ) 1 d ( x n ,f(x n )) + D(\u00C2\u00A3(xn ) , f ( y )) ^ 0 as i i i i i Therefore we can fxnd y e f ( y ) such that n. J o l x - y < d(x ,f(y )) -> 0 1 n. \u00E2\u0080\u00A2'n. \" \u00E2\u0080\u0094 n.. ' ^Jo x x x Sxnce f ( y ) xs compact, we may assume that y -*\u00E2\u0080\u00A2 z e f (y ) . Then o Jn. o o x x - z < x - y + | y - z -> 0 as i 1 n. o11 \u00E2\u0080\u0094 \" n. n. 11 1 n. o\" x x i x Hence x -> z e f ( y ) . Therefore x -> z weakly, and then n . o o n . o i \ x d(z f ( z )) \u00C2\u00A3 ||z - x n || .+ d(x ,f(x )) + D(f(x ) , f ( z )) -\u00C2\u00BB\u00E2\u0080\u00A2 0 i i - i n i \u00C2\u00B0 Thus z e f ( z ' ) , i . e . , z i s a f i x e d point of f i n K o o o r I t i s i n t e r e s t i n g to notice that the condition of asymptotic r e g u l a r i t y i n above theorem can be deleted i f the values of f are closed convex subsets of K . Here we state i t as follows: 78 Theorem 3.8 Let K be a weakly compact subset of a Banach space X . If f : K -> k(K) i s completely continuous, then f has a fixed point i n K . To prove the theorem, we need the following: Proposition 3.5. Let K be a weakly compact convex closed subset of a Banach space X . I f f : K -> k(K) i s completely continuous, then i t i s netwise continuous i n the weak topology of X , and hence i t i s upper semicontinuous i n the weak topology of X . Proof. Let x e K and x 5 x (5 means weakly convergent). a a o Given y \u00C2\u00A3 f(x ) , we w i l l prove that there i s a subnet of {y } a ct a converging-weakly to a point of f ( x Q ) . As f i s completely continuous, we have x 5 x => D(f (x ) ,f (x )) -> 0 . a o o Now y e f ( x ) , then there i s z e f ( x ) such that a a a o lly - 2 II < D(f (x ) , f (x )) . Since f ( x ) i s weakly compact (because \" a a\" \u00E2\u0080\u0094 a o o f(x ) i s a weakly closed subset of K), there i s a subnet {z } o a. x of {z } , such that z 5 z say, and z \u00C2\u00A3 f ( x ) . Let h \u00C2\u00A3 X* , a o o o then |h(y - z ) | < |h(y - z ) I + |h(z - z ) | 1 a. . o 1 \u00E2\u0080\u0094 1 ' a. a . 1 1 a. o 1 x \u00E2\u0080\u00A2 x x x \u00E2\u0080\u00A2. < ||h|| ||y - z || + |h(z - z ) | a. a.\" ' a. o 1 X X ' x < ||h|| D(f (x ) ,f (x )) + |h(z - z ) | -> 0 \u00E2\u0080\u0094 II II a_ 0 1 a o ' i x 79 Hence y ~r z \u00C2\u00A3 f (x ) . thus f xs netwise contxnuous xn the weak Ja. o o x topology of X , and therefore i t is upper semicontinuous i n the weak topology of X . Proof of Theorem 3.8 By the above proposition, f i s upper semicontinuous i n the weak topology of X . As K i s weakly compact, K. Fan's fixed poin theorem [13] implies the existence of a fixed point of f i n K . 80 BIBLIOGRAPHY [1] Belluce, L. P. and Kirk, W. A. Some fixed point theorems, i n metric and Banach spaces. [2] \u00E2\u0080\u00A2 and' Fixed point theorem for c e r t a i n c l a s s of non-expansive mappings, Proc. Amer. Math. Soc. Vol. 20 (1968), 141-146. 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Monthly, Vol. 72 (1965), 1004-1006. [19] Markin, J . T. A f i x e d point theorem for set valued mappings, B u l l . Amer. Math. S o c , Vol. 74 (1968), 639-640. [20] Nadler, S. B. J r . Multi-valued contractive mappings, P a c i f . Jour, cf Math., Vol. 30 (1969), 475-488. [21] Robertson, A. P. and Robertson, W. J. Topological vector spaces, Cambridge Uni v e r s i t y Press, 1964. "@en . "Thesis/Dissertation"@en . "10.14288/1.0080488"@en . "eng"@en . "Mathematics"@en . "Vancouver : University of British Columbia Library"@en . "University of British Columbia"@en . "For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use."@en . "Graduate"@en . "Fixed point theorems for point-to-set mappings"@en . "Text"@en . "http://hdl.handle.net/2429/34010"@en .