FIXED POINT THEOREMS FOR COMPATIBLE MULTI-VALUED AND SINGLE-VALUED MAPPINGS

The notion of compatibility for point-to-point mappings recently defined by Jungck is generalized to include multi-valued mappings. This idea is used to establish a fixed point theorem for a generalized contractive multi-valued mapping and a single-valued napping.

THEOREM 1.A continuous self-mapping f of a complete metric space (X,d) has a fixed point iff there exists a mapping g X X which commutes with f and such that 9(X) C_ f(X), d(gx, gy) < hd(fx, fy) for all x, yeX, where 0 < h < 1. Futhermore, f and g have a unique common fixed point.
The present authors generalized this theorem in two different directions.Sessa [8], gen- eralizing the notion of commutativity for point-to-point mappings, established the idea of weak commutativity for two self-mappings f and g of a metric space (X,d), i.e. d(fgx, gfx) < d(fx, gx) for all x in X.Under this concept, he extended theorem 1.On the other hand, Kaneko  [5] proved that theorem can be extended to the setting of multi-valued mappings, generalizing a famous result of Nadler [7].Recently Jungck [3] made an extension of weak commustivity in the following way.DEFINITION 1. Two self-mappings f and g of metric space (X,d) are compatible iff d(fgxn,gfxn) 0 whenever xn is a sequence in X such that fxn t,gx,, for some point tinX.
It can be seen that two weakly commuting mappings are compatible but the converse is false.Examples supporting this fact can be found in [3].
The purpose of this paper is to extend the definition of compatibility to include multi- valued mappings.
2. RESULTS Let (X,d) be a metric space and CB(X) the family of all closed bounded subsets of X.Let H be the Hausdorff metric on CB(X) induced by d, i.e.
H(A,B) max {supd(x,B) xeA,supd(x,A) xeB} for all A,B in CB(X), where d(x,A) inf{d(x,y): yeA}.It is well known [10] that (CB(X),H) is a metric space.It is indeed a complete metric space in the event that (X,d) is also a complete metric space.DEFINITION 2. The mappings f X X and T X CB(X) are compatible iff fTxeCB(X) for all xeX and H(Tfx,, fTxn) 0 whenever x, is a sequence in X such that Txn MeCB(X) and fxn teM.Definition 2 extends definition 1 above.We are now in a position to prove our results.Theorem 2. Let (X,d) be a complete metric space, f" X X and T" X CB(X) be compatible continuous mappings such that T(X) C_ f(X) and { l[d(fx, Ty)+d(fy Tx)]} (1) H(Tx, Ty) <_ hmax d(fx, fy),d(fx, Tx),d(fy, Ty),for all x,y in X, where 0 _< h < 1.Then there exists a point teX such that fteTt.
Remark 1.A nonempty subset S of X is proximinal if, for each xeX, there exists a point yes such that d(x, y) d(x, S).Let PB(X) be the family of all bounded proximinal subsets of X.If T X PB(X), then the interative process {y,,} in the above proof can be simplified in the following way: if x, has been selected, let x,+leX be such that y,t fx,+leTxn and d(fx,,y,) d(fx,Tx,).Note that an interation scheme of Smithson [9], where Wx is compact (hence proximinal) for all xeX, is included here.Since a proximinal set is closed, we have PB(X) C_ CB(X).The results of [4] and [5] follow as corollaries.
COROLLARY 1.Let (X,d) be a complete metric space, f X X and T X PB(X) be continuous mappings such that fTxePB(X) and H(Tfx, fTx) < d(fx, Tx) for all xeX.If (1) is satisfied for all x,y in X, 0 <_ h < 1, and T(X) C_ f(X), then there exists reX such that fteTt.COROLLARY 2. Let (X,d) be a complete metric space, T X CB(X) and f be a continuous self-mapping of X such that H(Tx, Ty) < hd(fx, fy) and for all x,y in X, where 0 < h < 1 and Tfx fTx.If T(X) C_ f(X), then there exists teX such that fteTt.
Note that the continuity of implies the continuity of T in corollary 2. The following example shows theorem 2 is indeed a proper generalization over corollaries 1 and 2. EXAMPLE Let X [1, 03) be endowed with the Euclidean metric d.Let fx 2x 4- and Tx [1, x 2] for each x > 1. W and f are clearly continuous and T(X) f(X) X.Since fx,t 1 and Tx, {1} iff x, 1, H(fTx,Tfx,) 2(x 1) 0 iff xn 1, fTx [1,(2z4-1) 2] for all z >_ 1, fand T are compatible.Since H(Tz, Ty) =l z2-Y <-21z2-Y II but the above z2 + y2 /2 d(fz, fy)/2, all the conditions of theorem 2 hold with h , corollaries are not applicable since f and T do not weakly commute (for x 2), hence do not commute either.
In the sequel, we use the following lemma which is a slight generalization of proposition 2.2 (part 1) of [3].
LEMMA.Let T" X CB(X) and f X X be compatible.If fweTw for some weX, then f Tw T f w.
PROOF.Let z, w for eachn.Then f x, fw f w and Tzn M Tw.Hence if fweTw, then H(fTw, Tfw) H(fTx,,Tfxn) 0 by the compatibility.Hence we must have fTw T f w.

Q.E.D.
In order to obtain a fixed point result, we need additional assumptions as those given in [4] and [5].THEOREM 3. Let f and T have the same meanings as in theorem 2. Assume also that for each xeX either (i) f x -# f2x implies f x Tx or (ii) f xeTx implies that fnx z for some zeX.Then f and T have a common fixed point in X. PROOF.By Theorem 2, fteTt for some teX and fTt Tft by Lemma.Assuming (i), we deduce that ft f2tefTt T ft.Assuming (ii), it is clear that fz z by the continuity of f.We claim that fnteTfn-lt for each n.To see this, we have that fzt fftefTt Tft.By lemma (w ft), f3t ff2tefTft Tf2t.Repeating this argument, we obtain fnteTfn-t and the continuity of T implies that d(z, Tz) < d(z, fnt) + d(fnt, Tz) < d(z, fnt) + H(Tfn-,Tz) O, i.e. zeTz since Tz is closed.Hence z is a common fixed point of f and T. Q.E.D.
REMARK 2. Simple examples prove that the conditions "T(X) C_ f(X)" and the compatibility of f and T are necessary in theorem 2. Unfortunately, it is not yet known if the continuity of both mappings f and T is necessary in theorem 2. However in the case that f and T are single-valued mappings, it suffices only the continuity of at least one of them.
Moreover, the inequality (1) can be weakened as it is proved in the following results, which extends theorem 2.1 of [1].
THEOREM 4. Let (X,d) be a complete metric space and f,T X X be two compatible mappings (def.1)such that T(X) C_ f(X) and for-all x,y in X, where 0 < h < 1.If one of f or T is continuous, then there exists a unique common fixed point of f and T.
A similar proof can be made if we assume the continuity of f instead of T. The uniqueness follows easily from (2).
Q.E.D. REMARK 3. Note that our proof is different from that of [1], in which it is proved that fz Tz is the unique fixed point (only if f is continuous).In this paper, it is just the case that f z Tz z.REMARK 4. If neither T nor f is continuous, e.g.X [0,11, TO 1/4, f0 1/2,Tx fx for x # 0, theorem 3 fails.Similar remarks can be made on the results of [8] (2) d(Tx, Ty) < Hmax{d(fx, fy),d(fx, Tx),d(fy, Ty),d(fx, Ty),d(fy, Tx)}