SOME FIXED POINT THEOREMS FOR WEAKLY COMPATIBLE MAPPINGS SATISFYING AN IMPLICIT RELATION. Ishak Altun and Duran Turkoglu

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1 TAIWANESE JOURNAL O MATHEMATICS Vol. 13, No. 4, pp , August 29 This paper is available online at SOME IXED POINT THEOREMS OR WEAKLY COMPATIBLE MAPPINGS SATISYING AN IMPLICIT RELATION Ishak Altun and Duran Turkoglu Abstract. In this paper, we prove a common fixed point theorem for weakly compatible mappings satisfying an implicit relation. Our theorem generalizes many fixed point theorems. 1. INTRODUCTION AND PRELIMINARIES It is well known that the Banach contraction principle is a fundamental result in fixed point theory. After this classical result, many fixed point results have been developed (see [15, 17, 22, 23]. In [5] Branciari proved the following interesting result for fixed point theory. Theorem 1. Let (X, d be a complete metric space, λ (, 1 and T : X X be mapping such that for each x, y X one has d(tx,ty f(tdt λ d(x,y f(tdt where f :[, [, ] is a Lebesque integrable mapping which is finite integral on each compact subset of [,, non-negative and such that for each t>, t f(sds >, then T has a unique fixed point z X such that for each x X, lim n T n x = z. Theorem 1 has been generalized in [4, 21] and [31]. Again in [2], Aliouche proved a fixed point theorem using a general contractive condition of integral type on symmetric spaces. Received June 7, 26, accepted December 8, 27. Communicated by Sen-Yen Shaw. 2 Mathematics Subject Classification: Primary 54H25; Secondary 47H1. Key words and phrases: ixed point, Weakly compatible mapping, Contractive condition of integral type, Implicit relation. 1291

2 1292 Ishak Altun and Duran Turkoglu Sessa [25] generalized the concept of commuting mappings by calling selfmappings A and S of metric space (X, d a weakly commuting pair if and only if d(asx, SAx d(ax, Sx for all x X, and he and others gave some common fixed point theorems of weakly commuting mappings [24-27]. Then, Jungck [11] introduced the concept of compatibility and he and others proved some common fixed point theorems using this concept [9, 11, 12, 14, 3]. Clearly, commuting mappings are weakly commuting and weakly commuting mappings are compatible; examples in [25] and [11] show that neither converse is true. Recently, Jungck [1] gave the concept of weak compatibility the following way. Definition 2. ([1, 13]. Two maps A, S : X X are said to be weakly compatible if they commute at their coincidence points. Again, it is obvious that compatible mappings are weakly compatible; giving examples in [13] and [28] shows that neither converse is true. Many fixed point results have been obtained using weakly compatible mappings (see [1, 4, 6, 7, 13, 18] and [28]. 2. IMPLICIT RELATION Implicit relation on metric spaces have been used in many articles. (see [3, 8, 19, 2, 29]. Let R + denote the non-negative real numbers and let be the set of all continuous functions : R 6 + R satisfying the following conditions: 1 (t 1,..., t 6 is non-increasing in variables t 5 and t 6. 2 there exists an upper semi-continuous function f : R + R +, f( =,f(t <tfor t>, such that for u, v, (u, v, v, u,,u+ v or (u, v, u, v, u + v, implies u f(v. 3 (u, u,,,u,u >, u >. Example 1. (t 1,..., t 6 =t 1 α max{t 2,t 3,t 4 } (1 α[at 5 + bt 6 ], where α<1, a< 1 2, b< Obviously.

3 ixed Point Theorem Let u>and (u, v, v, u,,u+v=u αmax{u, v} (1 αb(u+v. If u v, then u [α+2b(1 α] u<uwhich is a contradiction. Thus u<v and so u [α+2b(1 α]v. Similarly, let u>and (u, v, u, v, u+v,. Ifu v, then u [α +2a(1 α]u <uwhich is a contradiction. Thus u<vand so u [α +2a(1 α]v. If u =then u max{[α +2a(1 α], [α +2b(1 α]}v. Thus 2 is satisfied with f(t =max{[α +2a(1 α], [α +2b(1 α]}t. 3 (u, u,,,u,u=u(1 α(1 a b >, u >. Thus. 1 2 Example 2. (t 1,..., t 6 =t 1 k max{t 2,t 3,t 4, 1 2 [t 5 + t 6 ]}, where k (, 1. Obviously. Let u> and (u, v, v, u,,u+ v =u k max{u, v}. If u v, then u ku, which is a contradiction. Thus u<vand so u kv. Similarly, let u> and (u, v, u, v, u + v, then we have u kv. If u =, then u kv. Thus 2 is satisfied with f(t =kt. 3 (u, u,,,u,u=u ku >, u >. Thus. Example 3. (t 1,..., t 6 = t 1 ψ(max{t 2,t 3,t 4, 1 2 [t 5 + t 6 ]}, where ψ : R + R + right continuous and ψ( =,ψ(t <tfor t>. 1 Obviously. 2 Let u> and (u, v, v, u,,u+ v =u ψ(max{u, v}. If u v, then u ψ(u, which is a contradiction. Thus u<vand so u ψ(v. Similarly, let u> and (u, v, u, v, u + v, then we have u ψ(v. If u =then u ψ(v. Thus 2 is satisfied with f = ψ. 3 (u, u,,,u,u=u ψ(u >, u >. Thus. Example 4. (t 1,..., t 6 =t 2 1 t 1(at 2 + bt 3 + ct 4 dt 5 t 6, where a>, b, c, d, a+ b + c<1 and a + b + d< Obviously. Let u> and (u, v, v, u,,u+ v =u 2 u(av + bv + cu. Then u ( a+b 1 c v. Similarly, let u>and (u, v, u, v, u + v, then we have u ( a+c a+c 1 b v. If u =, then u ( 1 b v. Thus 2 is satisfied with f(t =max{( a+b a+c 1 c, ( 1 b }t. 3 (u, u,,,u,u=u 2 (1 a d >, u >.

4 1294 Ishak Altun and Duran Turkoglu Thus. Example 5. (t 1,..., t 6 =t 3 1 α t2 3 t2 4 + t2 5 t2 6, where α (, 1. t 2 + t 3 + t Obviously. 2 Let u> and (u, v, v, u,,u+ v =u 3 αv2 u 2, which implies u +2v +1 αv 2 u u +2v +1. But αv 2 αv, thus u αv. Similarly, let u> u +2v +1 and (u, v, u, v, u +v,, then we have u αv. If u =, then u αv. Thus 2 is satisfied with f(t =αt. 3 (u, u,,,u,u= u4 (1 α+u 3 >, u >. u +1 Thus. 3. COMMON IXED POINT THEOREMS We need the following lemma for the proof of our main theorem. Lemma 1. ([16]. Let f : R + R + be an upper semi-continuous function such that f(t <tfor every t>, then lim f n (t =, where f n denotes the n composition of f, n-times with itself. Now we give our main theorem. Theorem 2. Let A, B, S and T be self-maps defined on a metric space (X, d satisfying the following conditions: (i S(X B(X, T(X A(X, (ii for all x, y X, ( d(sx,ty d(sx,by d(ax,by d(ty,ax ϕ(tdt d(sx,ax d(ty,by where and ϕ : R + R + is a Lebesque integrable mapping which is summable, (3.1 a+b ϕ(tdt a ϕ(tdt + b ϕ(tdt

5 ixed Point Theorem 1295 for all a, b R + and such that (3.2 ε ϕ(tdt > for each ε>. If one of A(X,B(X,S(X or T (X is a complete subspace of X, then (1 A and S have a coincidence point,or (2 B and T have a coincidence point. urther, if S and A as well as T and B are weakly compatible, then (3 A, B, S and T have a unique common fixed point. Proof. Let x X be an arbitrary point of X. rom (i we can construct a sequence {y n } in X as follows: y 2n+1 = Sx 2n = Bx 2n+1 and y 2n+2 = Tx 2n+1 = Ax 2n+2 for all n =, 1,... Define d n = d(y n,y n+1. Suppose that d 2n =for some n. Then y 2n = y 2n+1 ; that is, Tx 2n 1 = Ax 2n = Sx 2n = Bx 2n+1, and A and S have a coincidence point. Similarly, if d 2n+1 =, then B and T have a coincidence point. Assume that d n for each n. Then by (ii, we have ( d(sx2n,t x 2n+1 d(ax2n,bx 2n+1 d(sx2n,ax 2n d(tx2n+1,bx 2n+1 Thus we have (3.3 ( d2n+1 d2n+1 d(sx2n,bx 2n+1 On the other hand, from (3.1 we have (3.4 d2n +d 2n+1 d(tx2n+1,ax 2n d2n d2n d2n +d 2n+1, ϕ(tdt ϕ(tdt Now from (3.3, (3.4 and 1, we have ( d2n+1 d2n ϕ(tdt + d2n d2n+1 d2n ϕ(tdt + d2n ϕ(tdt. d2n+1. ϕ(tdt. d2n+1, ϕ(tdt.

6 1296 Ishak Altun and Duran Turkoglu rom 2, there exists an upper semi-continuous function f : R + R +,f( =, f(t <tfor t>, such that d2n+1 ( d2n ϕ(tdt f ϕ(tdt Similarly we can have d2n ϕ(tdt f In general, we have for all n =1, 2,..., (3.5 rom (3.5, we have dn dn ϕ(tdt f ϕ(tdt f ( d2n 1 ( dn 1 ( dn 1 f 2 ( dn 2. ( d f n ϕ(tdt. ϕ(tdt. ϕ(tdt ϕ(tdt ϕ(tdt and taking the limit as n we have, from Lemma 1, for d >, dn ( d lim ϕ(tdt lim f n ϕ(tdt =, n n which from (3.2 implies that lim d n = lim d(y n,y n+1 =. n n We now show that {y n } is Cauchy sequence. or this it is sufficient to show that {y 2n } is a Cauchy sequence. suppose that {y 2n } is not Cauchy sequence. Then there exists an ε>such that for an even integer 2k there exist even integers 2m(k > 2n(k > 2k such that (3.6 d(y 2n(k,y 2m(k ε. or every even integer 2k, let 2m(k be the least positive integer exceeding 2n(k satisfying (3.6 such that (3.7 d(y 2n(k,y 2m(k 2 <ε.

7 ixed Point Theorem 1297 Now < δ := ε ϕ(tdt d(y2n(k,y 2m(k Then by (3.6 and (3.7 it follows that (3.8 lim k ϕ(tdt d(y2n(k,y 2m(k 2 +d 2m(k 2 +d 2m(k 1 d(y2n(k,y 2m(k ϕ(tdt = δ. ϕ(tdt. Also, by the triangular inequality, we have d(y 2n(k,y 2m(k 1 d(y 2n(k,y 2m(k d 2m(k 1 and d(y2n(k+1,y 2m(k 1 d(y 2n(k,y 2m(k d2m(k 1 + d 2n(k. Thus we have and d(y2n(k,y 2m(k 1 d(y 2n(k,y 2m(k ϕ(tdt d(y2n(k+1,y 2m(k 1 d(y 2n(k,y 2m(k ϕ(tdt By using (3.8 we get d2m(k 1 ϕ(tdt d2m(k 1 +d 2n(k ϕ(tdt. (3.9 and (3.1 as k. Now we get d(y2n(k,y 2m(k 1 d(y2n(k+1,y 2m(k 1 ϕ(tdt δ ϕ(tdt δ d(y 2n(k,y 2m(k d 2n(k + d(y 2n(k+1,y 2m(k d 2n(k + d(sx 2n(k,Tx 2m(k 1

8 1298 Ishak Altun and Duran Turkoglu and so d(y2n(k,y 2m(k ϕ(tdt d2n(k +d(sx 2n(k,T x 2m(k 1 Letting k both of the last inequality, we have ϕ(tdt. (3.11 δ lim d(sx2n(k,t x 2m(k 1 k d(y2n(k+1,y 2m(k = lim ϕ(tdt ϕ(tdt k d(y2n(k+1,y 2m(k 1 +d 2m(k 1 lim k = δ. ϕ(tdt On the other hand, from (ii, we have ( d(sx2n(k,t x 2m(k 1 d(sx2n(k,ax 2n(k d(sx2n(k,bx 2m(k 1 and so ( d(y2n(k+1,y 2m(k d(ax2n(k,bx 2m(k 1 d(tx2m(k 1,Bx 2m(k 1 d(tx2m(k 1,Ax 2n(k d(y2n(k,y 2m(k 1 ϕ(tdt (3.12 d2n(k d(y2n(k,y 2m(k 2 d2m(k 1 ϕ(tdt d(y 2n(k+1,y 2m(k 1 rom (3.12, considering 1, (3.6, (3.7, (3.8, (3.9, (3.1 and (3.11, letting k we have the following, (δ, δ,,,δ,δ which is a contradiction with 3. Thus {y 2n } is a Cauchy sequence and so {y n } is a Cauchy sequence.

9 ixed Point Theorem 1299 Now, suppose that A(X is complete. Note that the sequence {y 2n } is contained in A(X and has a limit in A(X. Call it u. Let v A 1 u. Then Av = u. We shall use the fact that the sequence {y 2n 1 } also converges to u. To prove that Sv = u, let r = d(sv,u >. Then taking x = v and y = x 2n 1 in (ii, ( d(sv,tx2n 1 d(av,bx2n 1 d(sv,av d(tx2n 1,Bx 2n 1 d(sv,bx2n 1 d(tx2n 1,Av ϕ(tdt and so (3.13 ( d(sv,y2n d(y2n,y 2n 1 d(u,y2n 1 d(sv,y2n 1 d(sv,u d(y2n,u ϕ(tdt Since lim n d(sv,y 2n = lim n d(sv,y 2n 1 =r and lim n d(u, y 2n 1 = lim n d(y 2n,y 2n 1 = lim n d(y 2n,u=, we have from (3.13 ( r r,, r which is a contradiction with 2. Hence from (3.2 we have Sv = u. This proves (1 Since S(X B(X, Sv = u implies that u B(X. Let w B 1 u. Then Bw = u. Hence by using the argument of the previous section, it can be easily verified that Tw = u. This proves (2. The same result holds if we assume that B(X is complete instead of A(X. Now if T (X is complete, then by (i, u T (X A(X. Similarly if S(X is complete, then u S(X B(X. Thus (1 and (2 are completely established. To prove (3, note that S, A and T, B are weakly compatible and (3.14 u = Sv = Av = Tw = Bw then (3.15 Au = ASv = SAv = Su (3.16 Bu = BTw = TBw = Tu.

10 13 Ishak Altun and Duran Turkoglu If Tu u, then from (ii, (3.14, (3.15 and (3.16 we have ( d(sv,tu d(tu,bu and so ( d(u,t u d(u,t u d(av,bu d(sv,bu,, d(sv,av d(tu,av d(u,t u ϕ(tdt d(tu,u ϕ(tdt which is a contradiction with 3.SoTu = u. Similarly Su = u. Then, evidently from (3.15 and (3.16, u is a common fixed point of A, B, S and T. Now let u and v be two common fixed points of A, B, S and T. Then from (ii, we have ( d(su,tv d(tv,bv and so ( d(u,v d(u,v d(au,bv d(su,bv,, d(su,au d(tv,au d(u,v ϕ(tdt d(v,u which is a contradiction with 3. Thus u = v. This completes the proof. ϕ(tdt If ϕ(t =1in Theorem 2, we obtain Theorem 2.1 of [8] and a generalization of Theorem 1 of [2]. If we combine Example 1 with Theorem 2 we obtain the following result. Corollary 1. Let A, B, S and T be self-maps defined on a metric space (X, d satisfying the following conditions: (i S(X B(X, T(X A(X, (ii for all x, y X,

11 ixed Point Theorem 131 d(sx,ty ϕ(tdt α [ +(1 α a max{d(ax,by,d(sx,ax,d(ty,by} d(sx,by ϕ(tdt + b d(ty,ax ϕ(tdt ] ϕ(tdt where α<1, a< 1 2, b< 1 2 and ϕ is as in Theorem 2. If one of A(X,B(X,S(X or T (X is a complete subspace of X, then (1 A and S have a coincidence point, or (2 B and T have a coincidence point. urther, if S and A as well as T and B are weakly compatible, then (3 A, B, S and T have a unique common fixed point. If ϕ(t =1in Corollary 1, we get Theorem 2.1 of [1] for single-valued mappings. If we combine Example 2 with Theorem 2 we obtain the following result. Corollary 2. Let A, B, S and T be self-maps defined on a metric space (X, d satisfying the following conditions: (i S(X B(X, T(X A(X, (ii for all x, y X, d(sx,ty ϕ(tdt k max{ max{d(ax,by,d(sx,ax,d(ty,by} 1 2 [ d(sx,by ϕ(tdt + d(ty,ax ϕ(tdt]} where <k<1 and ϕ is as in Theorem 2. If one of A(X,B(X,S(X or T (X is a complete subspace of X, then (1 A and S have a coincidence point, or (2 B and T have a coincidence point. urther, if S and A as well as T and B are weakly compatible, then (3 A, B, S and T have a unique common fixed point. By Corollary 2, we have a generalized version of Theorem 1 in this paper. If we combine Example 3 with Theorem 2 we obtain Theorem 2.1 of [4]. If ϕ(t =1in Theorem 2 and combine with Example 3, we have Theorem 1 of [9] and Theorem 2.1 of [28]. Also by Theorem 2, we have a different version of Theorem 3.1 of [6].

12 132 Ishak Altun and Duran Turkoglu Remark 1. We can have some new fixed point results if we combine Theorem 2 with some examples of. ACKNOWLEDGMENT The authors thank the referees for their valuable comments and suggestions. REERENCES 1. M. A. Ahmed, Common fixed point theorems for weakly compatible mappings, Rocky Mountain J. Math. 33(4 (23, A. Aliouche, A common fixed point theorem for weakly compatible mappings in symmetric spaces satisfying a contractive condition of integral type, J. Math. Anal. Appl., 322(2 (26, I. Altun, H. A. Hancer and D. Turkoglu, A fixed point theorem for multi-maps satisfying an implicit relation on metrically convex metric spaces, Mathematical Communications, 11 (26, I. Altun, D. Turkoglu and B. E. Rhoades, ixed points of weakly compatible maps satisfying a general contractive condition of integral type, ixed Point Theory and Appl., (27, Article ID 1731, doi:1.1155/27/ A. Branciari, A fixed point theorem for mappings satisfying a general contractive condition of integral type, Int. J. Math. Math. Sci., 29(9 (22, R. Chugh and S. Kumar, Common fixed points for weakly compatible maps, Proc. Indian Acad. Sci. Math. Sci. 111(2 (21, Lj. B. Ćirić and J. S. Ume, Some common fixed point theorems for weakly compatible mappings, J. Math. Anal. Appl. 314(2 (26, M. Imdad, S. Kumar and M. S. Khan, Remarks on some fixed point theorems satisfying implicit relations, Rad. Math., 11(1, (22, G. S. Jeong and B. E. Rhoades, Some remarks for improving fixed point theorems for more than two maps, Indian J. Pure Appl. Math., 28(9 (1997, G. Jungck, Common fixed points for non-continuous non-self maps on non-metric spaces, ar East J. Math. Sci., 4(2 (1996, G. Jungck, Compatible mappings and common fixed points, Int. J. Math. Math. Sci., 9 (1986, G. Jungck, Compatible mappings and common fixed points-ii, Int. J. Math. Math. Sci., 11 (1988, G. Jungck and B. E. Rhoades, ixed points for set valued functions without continuity, Indian J. Pure Appl. Math., 29(3 (1998,

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14 134 Ishak Altun and Duran Turkoglu Ishak Altun Department of Mathematics, aculty of Science and Arts, Kirikkale University, 7145 Yahsihan, Kirikkale, Turkey Duran Turkoglu Department of Mathematics, aculty of Science and Arts, Gazi University, 65 Teknikokullar, Ankara, Turkey