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Nonlinear Analysis 43 (2001) 693 – 706 www.elsevier.nl/locate/na Multivalued nonexpansive mappings in Banach spaces Hong-Kun Xu Department of Mathematics, University of Durban-Westville, Private Bag X54001, Durban 4000, South Africa Received 13 November 1998; accepted 9 February 1999 Keywords: Multivalued nonexpansive mapping; Fixed point; Weak inwardness condition; Uniformly convex Banach space 1. Introduction Let X be a Banach space and E a nonempty subset of X . We shall denote by F(E) the family of nonempty closed subsets of E, by CB(E) the family of nonempty closed bounded subsets of E, by K(E) the family of nonempty compact subsets of E, and by KC(E) the family of nonempty compact convex subsets of E. Let H (·; ·) be the Hausdor distance on CB(X ), i.e.,   H (A; B) = max sup dist(a; B); sup dist(b; A) ; A; B ∈ CB(X ); a∈A b∈B where dist(a; B) = inf {||a − b||: b ∈ B} is the distance from the point a to the subset B. A multivalued mapping T : E → F(X ) is said to be a contraction if there exists a constant k ∈ [0; 1) such that H (Tx; Ty) ≤ k||x − y||; x; y ∈ E: (1.1) If (1.1) is valid when k = 1, then T is called nonexpansive. A point x is a xed point for a multivalued mapping T if x ∈ Tx. Banach’s Contraction Principle was extended to a multivalued contraction in 1969. (Below is stated in a Banach space setting.) E-mail address: hkxu@pixie.udw.ac.za (H-K. Xu). 0362-546X/01/$ - see front matter ? 2001 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 9 ) 0 0 2 2 7 - 8 694 H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 Theorem 1.1. (Nadler [14]). Let E be a nonempty closed subset of a Banach space X and T : E → CB(E) a contraction. Then T has a ÿxed point. The xed point theory of multivalued nonexpansive mappings is however much more complicated and dicult than the corresponding theory of single-valued nonexpansive mappings. One breakthrough was achieved by T.C. Lim in 1974 by using Edelstein’s method of asymptotic centers [4]. Theorem 1.2. (Lim [12]). Let E be a nonempty closed bounded convex subset of a uniformly convex Banach space X and T : E → K(E) a nonexpansive mapping. Then T has a ÿxed point. Lim’s original proof was later simpli ed independently by Lim himself [13] and Goebel [5]. Another important result for multivalued nonexpansive mappings was obtained by W.A. Kirk and S. Massa in 1990. Theorem 1.3. (Kirk and Massa [9]). Let E be a nonempty closed bounded convex subset of a Banach space X and T : E → KC(E) a nonexpansive mapping. Suppose that the asymptotic center in E of each bounded sequence of X is nonempty and compact. Then T has a ÿxed point. Theorem 1.3 applies to all k-uniformly rotund (k-UR) Banach spaces [17]. However, it does not apply to a nearly uniformly convex (NUC) Banach space [6] as in such a space the asymptotic center of a bounded sequence is not necessarily compact (cf. [10]). Also note that Theorem 1.3 requires that T take convex values. In order to study the xed point theory for nonself-mappings, we must introduce some terminology for boundary conditions. The inward set of E at x ∈ E is de ned by IE (x) := {x + (y − x): y ∈ E;  ≥ 0}; x ∈ E: Let IE (x) = x + TE (x); with x∈E  d(x + y; E) =0 ; TE (x) = y ∈ X : lim inf →0+   x ∈ E: Note that for a convex E, we have IE (x) = IE (x), the closure of IE (x). A multivalued mapping T : E → F(X ) is said to be inward on E if Tx ⊂ IE (x) ∀x ∈ E (1.2) and weakly inward on E if Tx ⊂ IE (x) ∀x ∈ E: The two following results are now basic in the mappings. (1.3) xed point theory of multivalued H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 695 Theorem 1.4. (Deimling [2]). Let E be a nonempty closed subset of a Banach space X and T : E → F(X ) a contraction. Assume that T is weakly inward on E and that each x ∈ E has a nearest point in Tx. Then T has a ÿxed point. Theorem 1.5. (Downing and Kirk [3] and Reich [16]). Let E be a nonempty closed bounded convex subset of a uniformly convex Banach space X and T : E → K(X ) a nonexpansive mapping. Assume T is inward on E. Then T has a ÿxed point. The following consequence of Theorem 11:5 of Deimling [2] will be often used throughout the paper. Theorem 1.6. Let E be nonempty bounded closed convex subset of a Banach space and T : E → KC(X ) a contraction. Assume Tx ∩ IE (x) 6= ∅ for all x ∈ E. Then T has a ÿxed point. It is the objective of this paper to prove some xed point theorems for multivalued mappings. Among other things, we extend Theorem 1.3 to nonself-mappings. Also a simple proof of Theorem 1.3 is presented. Moreover, we give an armative answer to a question of Deimling. A negative answer to a question of Downing and Kirk [3] is included as well. Fixed point theorems in Banach spaces with Opial’s property [15] can be found in [11]. 2. The method of asymptotic centers Let K be a weakly compact convex subset of a Banach space X and {x n } a bounded sequence in X . De ne a function f on X by f(x) = lim sup ||x n − x||; n→∞ x ∈ X: Let r ≡ rK ({x n }) := inf {f(x): x ∈ K} and A ≡ AK ({x n }) := {x ∈ K: f(x) = r}: Recall that r and A are, respectively, called the asymptotic radius and center of {x n } relative to K. As K is weakly compact convex, we see that AK ({x n }) is nonempty, weakly compact and convex. Deÿnition 2.1. Let {x n } and K be as above. Then {x n } is called regular w.r.t. K if rK ({x n }) = rK ({x ni }) for all subsequences {x ni } of {x n }; while {x n } is called asymptotically uniform if AK ({x n }) = AK ({x ni }) for all subsequences {x ni } of {x n }. The method of asymptotic centers plays an important role in the xed point theory of both single- and multi-valued nonexpansive mappings, due to the fundamental lemma below. 696 H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 Lemma 2.1. (Geobel [5] and Lim [13]). Let {x n } and K be as above. Then we have (i) There always exists a subsequence of {x n } which is regular w.r.t. K; (ii) if K is separable; then {x n } contains a subsequence which is asymptotically uniform w.r.t. K Remark 2.1. If X is uniformly convex in every direction (especially uniformly convex), then AK ({x n }) consists of exactly one point so every regular sequence in such a space is always asymptotically uniform w.r.t. K. Let now E be a weakly compact convex subset of a Banach space X and T : E → K(E) a nonexpansive self-mapping. For each integer n ≥ 1, the contraction Tn : E → K(E) de ned by   1 1 Tx; x ∈ E; (2.1) Tn (x) := x0 + 1 − n n where x0 ∈ E is a xed point, has a xed point x n ∈ E. Let r and A be the asymptotic radius and center of {x n } w.r.t. E, respectively. It is easily seen that 1 diam(E) → 0: n Since T is compact-valued, we can take yn ∈ Tx n such that dist(x n ; Tx n ) ≤ ||yn − x n || = dist(x n ; Tx n ); n ≥ 1: Since T is a self-mapping, we may assume that E is separable (otherwise, we can construct a closed convex subset of E that is invariant under T , see [10]). Then by Lemma 2.1 we may assume that {x n } is asymptotically uniform. Take any z ∈ A, as Tz is compact, we can nd zn ∈ Tz satisfying ||yn − zn || = dist(yn ; Tz) ≤ H (Tx n ; Tz): It follows from the nonexpansiveness of T that ||yn − zn || ≤ ||x n − z||: Because of the compactness of Tz, we may also assume that {zn } (strongly) converges to a point z̃ ∈ Tz. It then follows that lim sup ||x n − z̃|| = lim sup ||yn − zn || ≤ lim sup ||x n − z||: This shows that z̃ ∈ A. Hence we can de ne a multivalued self-map T̃ : A → A by setting for each z ∈ A T̃ z := A ∩ Tz: This map T̃ is in general neither nonexpansive, nor lower semicontinuous. However, it is upper semicontinuous, which is observed by Kirk and Massa [9]. With this observation they are able to prove Theorem 1.3 by using the Bohnenblust–Karlin xed point theorem (cf. [19]) that is of topological rather than metric nature. In addition, if we assume that X is uniformly convex (uniformly convex in every direction is enough), the asymptotic center A consists of exactly one point z. Then the above argument shows H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 697 that we must have z̃ = z and therefore z is a xed point of T . This is the idea of the simpli ed proof of Lim’s Theorem given independently by Goebel [5] and Lim [13]. If T is a nonself-mapping from E into K(X ), then the result of [10] is not valid and we may not assume the separability of E. So we have to use a universal net. Deÿnition 2.2. A net {x } in a set S is called a universal net if for each subset U of S, either {x } is eventually in U or {x } eventually in S \U . The following facts are pertinent [7, p. 81]: (a) Every net in a set has a universal subnet. (b) If f: S1 → S2 is a map and if {x } is a universal net in S1 , then {f(x )} is a universal net in S2 . (c) If S is compact and if {x } is a universal net in S, then lim x exists. Now for a nonself-mapping T : E → K(X ) we can de ne the contraction by the same formula (2.1). Suppose for each integer n ≥ 1; Tn has a xed point x n ∈ E. Let {x n } be a universal subnet of {x n }. Then we can work out the asymptotic center A as above just by replacing the sequence {x n } by the net {x n }. This technique will be employed to show our extension of Theorem 1.3 to nonself-mappings. 3. Fixed point theorems We begin this section with an extension of Kirk–Massa’s result to a nonself-mapping. Theorem 3.1. Let E be a nonempty closed bounded convex subset of a Banach space X and T : E → KC(X ) a nonexpansive nonself-mapping which satisÿes the inwardness condition: Tx ⊂ IE (x) for x ∈ E. Suppose that the asymptotic center in E of each bounded sequence of X is nonempty and compact. Then T has a ÿxed point. Proof. Fix x0 ∈ E and de ne for each integer n ≥ 1 the contraction Tn : E → KC(X ) by   1 1 Tx; x ∈ E: Tn (x) := x0 + 1 − n n Then Tn satis es the inwardness condition, i.e., Tn x ⊂ IE (x) for all x ∈ E. Thus by Theorem 1.4, Tn has a xed point x n ∈ E. By Lemma 2.1, we may assume that {x n } is regular. Let yn ∈ Tx n be a constructed as in Section 1, i.e., ||x n − yn || = dist(x n ; Tx n ). Let {x n } be a universal subnet of {x n } and de ne a function g by g(x) = lim sup ||x n − x||; Let C := {x ∈ E: g(x) = r}; x ∈ E: 698 H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 where r = inf x∈E g(x). Then by assumption (see [8, Proposition 6]), C is nonempty and compact. The key to proof is that the inwardness of T on E implies a weaker inwardness of T on C, i.e., Tx ∩ IC (x) 6= ∅; x ∈ C: (3.1) Indeed, if x ∈ C, by compactness, we have for each n ≥ 1, some zn ∈ Tx such that ||yn − zn || = dist(yn ; Tx) ≤ H (Tx n ; Tx) ≤ ||x n − x||: Let z = lim zn ∈ Tx. It follows that g(z) = lim ||x n − z|| = lim ||yn − zn || ≤ lim ||x n − x||: Hence g(z) ≤ g(x) = r: (3.2) It remains to show z ∈ IC (x). As Tx ⊂ IE (x), we have some  ≥ 0 and v ∈ E such that z = x + (v − x): If  ≤ 1, then by the convexity of E; z ∈ E and hence by (3.2), z ∈ C ⊂ IC (x) and we are done. So assume that  ¿ 1. Then we can write 1 v = z + (1 − )x with  = ∈ (0; 1):  By the convexity of g, we have by (3.2), g(v) ≤ g(z) + (1 − )g(x) ≤ r: Since v ∈ E, it follows that v ∈ C and thus z = x + (v − x) belongs to IC (x). Now, we have a nonexpansive mapping T : C → KC(X ) which satis es the boundary condition (3.1). The following lemma shows that T has a xed point in C. Lemma 3.1. If C is a compact convex subset of a Banach space X and T : C → KC(X ) is a nonexpansive mapping satisfying the boundary condition: Tx ∩ IC (x) 6= ∅ ∀x ∈ C: Then T has a ÿxed point. Proof. Fix an x0 ∈ C and de ne for each integer n ≥ 1 a mapping Tn : C → KC(X ) by   1 1 Tx; x ∈ C: Tn (x) := x0 + 1 − n n Then Tn is a contraction satisfying the same boundary condition as T does, i.e., we have Tn (x) ∩ IC (x) 6= ∅ ∀x ∈ C: H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 699 Hence by Theorem 1.6, Tn has a xed point x n ∈ C. Since C is compact, we may assume x n → x ∈ C. Also it is easily seen that dist(x n ; Tx n ) ≤ 1 diam C → 0 n as n → ∞: Taking limit as n → ∞ yields dist(x; Tx) = 0 and hence x ∈ Tx. Remark 3.1. If T satis es a stronger condition: Tx ∩ IC (x) 6= ∅ for all x ∈ C, Lemma 3.1 follows from a xed point theorem of F.E. Browder (cf. [2]). However, in our nonexpansive case, the proof is constructive. Remark 3.2. It is unclear whether the conclusion of Theorem 3.1 remains valid if the inwardness of T is weakened to the weak inwardness. Remark 3.3. Theorem 1.3 applies to Banach spaces which are uniformly convex or more general k-uniformly rotund Banach spaces [17] since the asymptotic center of a bounded sequence w.r.t. a bounded closed convex subset of such spaces is compact [8]. However, Theorem 1.3 does not apply to a nearly uniformly convex Banach space since in such a space, the asymptotic center of a bounded sequence w.r.t. a closed bounded convex subset is not necessarily compact (cf. [10]). Recall that a Banach space X is said to be nearly uniformly convex (NUC) [6] if X is re exive and if for any  ¿ 0, there a  = () ∈ (0; 1) such that {x n } ⊂ BX ; x n * x; sep(x n ) ≥  ⇒ ||x|| − ; where BX is the closed unit ball of X; * denotes the weak convergence, and sep(x n ):= inf {||x n −xm ||: n 6= m}. It is thus a natural question whether the conclusion of Theorem 1.3 holds in a nearly uniformly convex Banach space. This question still remains open. Next, we present a simple proof of Theorem 1.3, in which we do not use the Bohnenblust–Karlin xed point theorem. The proof is only involved with the multivalued contraction principle. (For convenience and completeness, Theorem 1.3 is restated below.) Theorem 3.2. (Kirk and Massa [9]). Let E be a nonempty closed bounded convex subset of a Banach space X and T : E → KC(E) a nonexpansive mapping. Suppose that the asymptotic center in E of each bounded sequence of X is nonempty and compact. Then T has a ÿxed point. Proof. Since T is self-mapping, we may assume that E is separable. As in the proof of Theorem 3.1, we have a regular sequence {x n } ⊂ E such that lim dist(x n ; Tx n ) = 0. We also have a sequence {yn } with yn ∈ Tx n for each n such that ||x n −yn ||=dist(x n ; Tx n ). By Lemma 2.1 we may assume that {x n } is asymptotically uniform with respect to E. Set A = AE ({x n }) and r = rE ({x n }). We have already shown that Tx ∩ A 6= ∅ ∀x ∈ A: (3.3) 700 H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 For each integer m ≥ 1 de ne a contraction Sm : A → KC(E) by   1 1 Tx; x ∈ A; Sm (x) = x0 + 1 − m m where x0 ∈ A is a (3.3), i.e., Sm (x) ∩ A 6= ∅ xed point. Then each Sm satis es the same boundary condition ∀x ∈ A: (3.4) As Sm is a compact and convex-valued contraction, by Theorem 1.6, Sm has a point vm ∈ A. It follows that the sequence {vm } satis es the property: lim dist(vm ; Tvm ) = 0: xed (3.5) m→∞ Now since A is compact and {vm } ⊂ A; {vm } has a subsequence {vmj } converging strongly, by (3.5), to a xed point of T . Remark 3.4. Kirk and Massa [8] asked whether the conclusion of Theorem 1.3 remains valid if T only takes compact values. The answer is armative if the space X is uniformly convex [12] or X satis es Opial’s property [11]. But in general, the question is still unanswered. Our proof is involved with a nonself-contraction Sm which satis es (boundary) condition (3.4). So in order to answer the above question, one would rst ask this question whether a nonself-contraction S: A → K(X ) satisfying (3.4), where Sm is replaced by S, has a xed point. The following example shows a negative answer. Example. Let A := [0; 1] × [0; 1] be the unit square in the plane. Let T : A → K(R2 ) be de ned by T (a; b) := {(a − h; b); (a + h; b)}; (a; b) ∈ A; 1 2 ). Then it is easily seen that T is nonexpansive and where h is a xed constant in (0; T (a; b) ∩ A 6= ∅ for all (a; b) ∈ A. It is obvious that T is xed point free. Now set T := T (0 ¡  ¡ 1). Then T : A → K(R2 ) is a contraction and T (a; b) ∩ A 6= ∅ for all (a; b) ∈ A. However, if 1=(1 + h) ¡  ¡ 1; T does not have any xed point. Deimling [2, p. 161] asked the following Question. Does Theorem 1.4 remain valid if the condition Tx ⊆ IE (x) for all x ∈ E is replaced by Tx ⊆ x + {(y − x):  ≥ 1; y ∈ E}? To answer this question we need Caristi’s xed point theorem [1]. Lemma 3.2. (J. Caristi’s Theorem [1]). Assume (M; d) is a complete metric space and g: M → M is a mapping. If there exists a lower semicontinuous function ’: M → R+ := [0; ∞) such that d(x; g(x)) ≤ ’(x) − ’(g(x)); x ∈ M; then g has a ÿxed point. We are now in a position to supply with an armative answer to the above question. H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 701 Theorem 3.3. Assume X is a Banach space, E is a closed subset of X, and T : E → 2X \{∅} is a contraction with closed values and satisfying the condition Tx ⊂ x + {(y − x):  ≥ 1; y ∈ E} ∀x ∈ E: (3.6) Assume that each x ∈ E has a nearest point in Tx. Then T has a ÿxed point. Proof. First observe that condition (3.6) is equivalent to the condition: Tx ⊆ x + {(y − x):  ¿ 1; y ∈ E} ∀x ∈ E: (3.7) Next choose q ∈ (0; 1) and  ∈ (0; 1) such that 1− : k ¡q¡ 1+ Let 1 dist(x; Tx); x ∈ E: ’(x) = q−k For x ∈ E, by assumption, we have some f(x) ∈ Tx satisfying ||x − f(x)|| = dist(x; Tx): It follows from (3.7) that there exist n ¿ 1 and yn ∈ E such that ||f(x) − (x + n (yn − x))|| → 0 as n → ∞: Suppose T is xed point free in E. Then we have an integer N ≥ 1 big enough such that ||f(x) − (x + N (yN − x))|| ¡  dist(x; Tx): Let   1 1 f(x) x+ z= 1− N N and g(x) = yN : This de nes a mapping g: E → E. We now claim that g satis es ||x − g(x)|| ¡ ’(x) − ’(g(x)); ∀x ∈ E: (3.8) Caristi’s theorem then implies that g has a xed point which contradicts the strict inequality in (3:8) and the proof is nished. So it remains to prove (3:8). As f(x) ∈ T (x) we have dist(g(x); T (g(x))) ≤ ||g(x) − z|| + dist(z; Tx) + H (Tx; T (g(x))) ≤ ||g(x) − z|| + ||z − f(x)|| + k||x − g(x)|| = ||yN − z|| + ||z − f(x)|| + k||x − ||g(x)||: Since ||f(x) − (x + N (yN − x))|| ¡  dist(x; Tx); 702 H-K. Xu / Nonlinear Analysis 43 (2001) 693 – 706 we have  ||yN − z|| = yN − Also     1 1  f(x) + 1 − dist(x; Tx): x ¡ N N N 1 ||z − f(x)|| = 1 − N   1 ||x − f(x)|| = 1 − N  dist(x; Tx): It follows that 1 dist(g(x); T (g(x))) q−k k −1 dist(x; Tx) + ||x − g(x)|| ≤ ’(x) + N (q − k) q−k k 1− ||z − x|| + ||x − g(x)||: = ’(x) − q−k q−k ’(g(x)) = It remains to show −(1 − )=(q − k)||z − x|| + k=(q − k)||x − g(x)|| ¡ − ||x − g(x)|| or equivalently ||x − g(x)|| ¡ 1− 1− dist(x; Tx): ||z − x|| = q N q In fact, by the choice of the integer N , we see that  dist(x; Tx) ¿ ||(f(x) − x) − N (yN − x)|| ≥ ||N (yN − x)|| − ||f(x) − x|| = N ||yN − x|| − dist(x; Tx): Thus by the choice of q we have ||g(x) − x|| = ||yN − x|| 1+ dist(x; Tx) ¡ N 1− dist(x; Tx); ¡ qN as required. Now recall that a Banach space X is uniformly convex if   ||x + y|| : x; y ∈ BX ; ||x − y|| ≥  ¿ 0 X () := inf 1 − 2 ∀ ∈ (0; 2]: A useful characteristic of uniform convexity is the following. Lemma 3.3. (cf. Xu [18]). Let X be a Banach space. Then X is uniformly convex if and only if for any given number  ¿ 0; the norm || · || of X is uniformly convex on B ; the closed ball centered at the origin with radius ; namely; there exists a