Fixed Point Theorems with Implicit Function in Metric Spaces

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1 Int. J. Contemp. Math. Sciences, Vol. 8, 013, no. 17, HIKARI Ltd, Fixed Point Theorems with Implicit Function in Metric Spaces A. Muraliraj P.G. and Research Department of Mathematics Urumu Dhanalashmi College Kattur, Tiruchirappalli Tamilnadu, India R. Jahir Hussain P.G. and Research Department of Mathematics Jamal Mohamed College(Autonomous Tiruchirappalli Tamilnadu, India Copyright c 013 A. Muraliraj and R. Jahir Hussain. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original wor is properly cited. Abstract In this paper, wealy compatible and occasionally wealy compatible mappings are discussed in detailed and generalized common fixed point theorems of two and four mappings are proved in metric space with implicit function. Our results extends and generalists many well nown results. Mathematics Subject Classification: Primary 54H5; Secondary 47H10, 47H09 Keywords: Fixed point, coincidence point, dislocated metric linear space 1 Introduction In recent years, Popa [7] used implicit functions rather than contraction conditions to prove fixed point theorems in metric spaces whose strength lies in its

2 834 A. Muraliraj and R. Jahir Hussain unifying power as an implicit function can cover several contraction conditions at the same time which includes nown as well as unnown contraction conditions. This fact is evident from examples furnished in Popa [7]. They also proved a generalized version of Banach contraction mapping theorem which was applied to obtain fixed point semantics for logic programs. In this paper we have discussed the dislocated topologies associated with a metric space and also proved a common fixed point theorem of Presic type which extends and generalises the well nown Banach contraction principle and nown results. Preliminary Definition.1. Let T and S be self maps of a set X. Maps T and S are said to be commuting if STx = TSx for all x X. Definition.. Let T and S be self maps of a set X. If w = Tx = Sx for some x X, then x is called a coincidence point of T and S, and w is called a point of coincidence of T and S. Example.3. Tae X =[0, 1], Sx= x,tx= x. It is clear that {0, 1 } is the set of coincidence points of S and T and 0 is the unique common fixed point. Definition.4. The mappings S and T are said to be wealy compatible if and only if they commute at their coincidence points. Definition.5. The mappings S and T are said to be occasionally wealy compatible if and only if they commute at some coincidence point of S and T, i.e. STu = TSu for some coincidence point u. Lemma.6. If a wealy compatible pair (S, T of self maps has a unique point of coincidence, then the point of coincidence is a unique common fixed point of S and T. Example.7. Tae X =[0, 1], Sx= x,tx= x. It is clear that {0, 1} is the set of coincidence points of S and T, ST0=TS0 but ST 1 TS1 and so S and T are occasionally wealy compatible but not wealy compatible. Definition.8. Let X be a non-empty set, and suppose the mapping d : X X X is said to be a metric space if it satisfies (i 0 d(x, y for all x, y X and d(x, y =0if and only if x = y. (ii d(x, y =d(y, x for all x, y X.

3 Fixed point theorems with implicit function 835 (iii d(x, y d(x, z+d(z, y for all x, y, z X. Example.9. Let E = R,P = {(x, y E : x, y 0},X = R and d : X X X defined by d(x, y =( x y,α x y where α 0 is a constant. Then (X, d is a metric space. Definition.10. Let (X, d is said to be a complete metric space, if every cauchy sequence is convergent in X. Definition.11. Implicit function: Let F denote the family of all continuous functions ϕ :[0, 1] 4 R satisfying the following conditions: (F1 For every u>0, v 0 with ϕ(u, v, u, v 0 or ϕ(u, v, v, u 0, we have u<v. (F ϕ(u, u, 0, 0 < 0, for all u>0. Example.1. Define ϕ :[0, 1] 4 R as ϕ(t 1,t,t 3,t 4 =t 1 δ(max{t,t 3,t 4 }, where δ : [0, 1] [0, 1] is a continuous function such that δ(s > s for 0 <s<1. Then (F1 ϕ(u, v, u, v =u δ(max{v, u, v} If u v, then u δ(u 0 imply u δ(u >u,a contradiction. Hence u<v. (F ϕ(u, u, 0, 0 = u δ(max{u, 0, 0} =u δ(u < 0, u >0. Example.13. Define ϕ :[0, 1] 4 R as ϕ(t 1,t,t 3,t 4 =t 1 α max{t,t 3,t 4 }, where α>1. Then (F1 ϕ(u, v, u, v =u α max{v, u, v} If u v, then u αu u, a contradiction. Hence u<v. (F ϕ(u, u, 0, 0 = u α max{u, 0, 0} =u(1 α < 0, u >0. 3 Main results Theorem 3.1. Let T 1,T,T 3 and T 4 be four self-mappings of a metric space (X, d satisfying the condition: ϕ(d(t 1 x, T y,d(t 3 x, T 4 y,d(t 3 x, T 1 x,d(t y, T 4 y 0 for all distinct x, y X where ϕ F. If T 1 (X T 4 (X, T (X T 3 (X and one of T 1 (X,T (X,T 3 (X or T 4 (X is a complete subspace of X. Then pair (T 1,T 3 and (T,T 4 has a point of coincidence. Moreover, if the pairs (T 1,T 3 and (T,T 4 are wealy compatible, then T 1,T,T 3 and T 4 have a unique common fixed point.

4 836 A. Muraliraj and R. Jahir Hussain Proof. Let x 0 be an arbitrary point in X. Construct sequences {x n } and {y n } in X such that y n = T 4 x n+1 = T 1 x n and y n+1 = T 3 x n+ = T x n+1. The sequences {x n } and {y n } in X are such that x n x, y n y, implies d(x n,y n d(x, y. Now, ϕ(d(t 1 x n,t x n+1,d(t 3 x n,t 4 x n+1,d(t 3 x n,t 1 x n,d(t x n+1,t 4 x n+1 0 or ϕ(d(y n,y n+1,d(y n 1,y n,d(y n 1,y n,d(y n,y n+1 Hence in view of (F 1, we have d(y n,y n+1 <d(y n 1,y n. Thus {d(y n,y n+1 }, n>0 is a decreasing sequence of positive real numbers in [0, 1] and therefore tends to 0. Therefore for every n N, using analogous arguments one can also show that {d(y n+1,y n+ }, n 0 is a sequence of positive real numbers in [0, 1] which converges to 0. Therefore for every n N d(y n,y n+1 <d(y n 1,y n and d(y n,y n+1 0. Now for any positive integer p, d(y n,y n+p d(y n,y n d(y n+p 1,y n+p. it follows that d(y n,y n+p 0 which shows that {y n } is a Cauchy sequence in X. Now suppose that S(X is a complete subspace of X, then the subsequence {y n+1 } must converge in S(X. Call this limit to be u. Then Sv = u. As {y n } is a Cauchy sequence containing a convergent subsequence {y n+1 }, therefore the sequence {y n } also converges implying thereby the convergence of {y n } being a subsequence of the convergent sequence {y n }. If T 1 v T 3 v, then on setting x = v and y = x n+1 we gets ϕ(d(t 1 v, T x n+1,d(t 3 v, T 4 x n+1,d(t 3 v, T 1 v,d(t x n+1,t 4 x n+1 0 which on letting n reduces to ϕ(d(t 1 v, u,d(t 3 v, u,d(t 3 v, T 1 v,d(u, u 0 ϕ(d(t 1 v, T 3 v, 0,d(T 3 v, T 1 v, 0 0 There for d(t 1 v, T 3 v < 0, a contradiction. Hence, T 1 v = T 3 v which shows that the pair (T 1,T 3 has a point of coincidence. As T 1 (X T 4 (X and T 1 v = u implies that u T 4 (X. Let w T4 1 u, then T 4 w = u. Suppose that T 4 w T w. By using (1, we have ϕ(d(t 1 x n,t w,d(t 3 x n,t 4 w,d(t 3 x n,t 1 x n,d(t w, T 4 w 0

5 Fixed point theorems with implicit function 837 which on letting n reduces to ϕ(d(t 4 w, T w, 0, 0,d(T 4 w, T w 0 which implying, d(t 4 w, T w < 0, a contradiction. Hence T 4 w = T w. Thus we have u = T 1 v = T 3 v = T w = T 4 w which amounts to say that both the pairs have point of coincidence. If we assumes T (X to be complete, then analogous arguments establish this claim. The remaining two cases pertain essentially to the previous cases. Indeed, if T 1 (X is complete then u T 1 (X T 4 (X and if T (X is complete then u T (X T 3 (X. Moreover, if the pairs (T 1,T 3 and (T,T 4 are wealy compatible at v and w respectively, then T 1 u = T 1 (T 3 v= T 3 (T 1 v=t 3 u and T u = T (T 4 w=t 4 (T w=t 4 u. If T 1 u u, then ϕ(d(t 1 u, T w,d(t 3 u, T 4 w,d(t 3 u, T 1 u,d(t w, T 4 w 0 ϕ(d(t 1 u, u,d(t 1 u, u, 0, 0 0 which contradicts (F. Hence T 1 u = u. Similarly we can show that T u = u. Thus u is a common fixed point of T 1,T,T 3 and T 4. The uniqueness of common fixed point follows easily. Also u remains the unique common fixed point of both the pairs separately. This completes the proof. Corollary 3.. Let S and T be two self-mappings of a metric space (X, d satisfying the condition: ϕ(d(sx,sy,d(tx,ty,d(tx,sx,d(sy,ty 0 for all x, y X, where ϕ F. If S(X T (X and one of S(X and T (X is complete subspace of X. Then the pair (S, T has a point of coincidence. Moreover, if the pair (S, T is wealy compatible, then S and T have a unique common fixed point. Proof. The proof of this corollary follows by setting T = T 1 = S and T 4 = T 3 = T in the above theorem. A Class of Implicit Relation : Definition 3.3. Let F denote the family of all continuous functions : [0, 1] 4 R satisfying the following conditions: (G1 For every u>0, v 0 with (u, v, v, u 0 or (u, v, v, v 0 or (u, v, v, 0 0 we have u v. (G is decreasing in nd, 3rd and 4th variable, Example 3.4. Define :[0, 1] 4 R as (t 1,t,t 3,t 4 =t 1 max{t,t 3,t 4 }, where δ : [0, 1] [0, 1] is a continuous function such that δ(s > s for 0 <s<1. Theorem 3.5. Let (X, d be a complete metric space and let T 1,T,T 3 and T 4 be the self mappings of X, satisfying the following condition

6 838 A. Muraliraj and R. Jahir Hussain (1 T 1 (X T 4 (X and T (X T 3 (X ( The pairs (T 1,T 4 and (T,T 3 are wealy compatible. (3 T 3 (X or T 4 (X is complete. (4 There exists (0, 1 such that ( d(t1 x, T y, d(t 3x, T 1 x+d(t 1 x, T 4 y,d(t 4 y, T 3 x, d(t 4y, T y 0...(1 For some F, every x, y X, then T 1,T,T 3 and T 4 have a unique common fixed point in X. Proof. Let x 0 be any arbitrary point. Since T 1 (X T 4 (X, T (X T 3 (X. So there must exists points x 1,x X such that T 1 x 0 = T 4 x 1 and T x 1 = T 3 x. In general, we get sequence {y n } and {x n } in X such that y n = T 4 x n+1 = T 1 x n and( y n+1 = T 3 x n+ = T x n+1. 0 d(t1 x n,t x n+1, d(t 3x n,t 1 x n +d(t 1 x n,t 4 x n+1,d(t 4 x n+1,t 3 x n, d(t 4x n+1,t x n+1 ( d(yn,y n+1, d(y n 1,y n +d(y n,y n,d(y n,y n 1, d(y n,y n+1 0 ( d(yn,y n+1, d(y n 1,y n,d(y n,y n 1, d(y n,y n+1 0 is non- increasing function, ( d(yn,y n+1,d(y n 1,y n,d(y n,y n 1, d(y n,y n+1 0 d(y n,y n+1 d(y n 1,y n. Similarly d(y n+1,y n+ d(y n+1,y n. There fore for all n we get d(y n,y n+1 d(y n 1,y n. Hence d(y n,y n+1 d(y n 1,y n d(y n,y n 1 d(y 0,y 1 There fore Lt n d(y n,y n+1 =0. n. Now for any positive integer p, d(y n,y n+p d(y n,y n d(y n+p 1,y n+p. It follows that d(y n,y n+p 0 which shows that {y n } is a Cauchy sequence in X which is complete. Therefore {y n } converges to z, for some z X. So it follows that {T 1 x n }, {T 3 x n }, {T x n+1 } and {T 4 x n+1 } also converges to z. That is, there exists a finite sequence Now if T 4 (X is complete then, if we tae z T 4 (X, so there exist u X, such that z = T 4 u ( d(t1 x n,t u, d(t 3x n,t 1 x n +d(t 1 x n,t 4 u,d(t 4 u, T 3 x n, d(t 4u, T u

7 Fixed point theorems with implicit function 839 As n ( d(z, T u, d(z, z+d(z, z,d(z, z, d(z, T u ( d(z,t u, 0, 0, d(z,t u So d(z, T u 0. Hence z = T u = T 3 u = T 4 u. Now (T,T 3 is wealy compatible, so T T 3 u = T 3 T u, there by T z = T 3 z. Now, Put x = x n and y = z in (1 we get, ( d(t1 x n,t z, d(t 3x n,t 1 x n +d(t 1 x n,t 4 z,d(t 4 z, T 3 x n, d(t 4z, T z Taing limit as n ( d(z, T z, d(z, z+d(z, T 4z,d(T 4 z, z, d(t 4z, T z ( d(z, T z, d(z, T 4z,d(T 4 z, z, d(t 4z, T z Since z = T 4 z, z T 4 (X, so we have ( d(z, T z, ( d(z, T z d(z, z,d(z, z, d(z, T z, 0, 0, d(z, T z Since is non- increasing. Hence T z = T 4 z = z. As T (X T 3 (X Let there exists v X, such that z = T z = T 3 v. Now put x = v, y = z in (1 ( d(t1 v, T z, d(t 3v, T 1 v+d(t 1 v, T 4 z,d(t 4 z, T 3 v, d(t 4z, T z ( d(t1 v, z, d(t 3v, T 1 v+d(t 1 v, z,d(z, T 3 v, ( d(t1 v, z, d(z, T 1v+d(T 1 v, z,d(z, z, ( d(t1 v, z,d(z, T 1 v, 0, 0 d(z, z d(z, z This implies that T 1 v = z. Now since T 1 T 4 so z = T 1 v T 4. Therefore z = T 1 v = T 4 v. Now as (T 1,T 4 is wealy compatible so T 1 T 4 v = T 4 T 1 v such that T 1 z = T 4 z. So combining all the results, we have T 1 z = T 4 z = T z = T 3 z = z.

8 840 A. Muraliraj and R. Jahir Hussain Similarly we can show that T 1 z = T 4 z = T z = T 3 z = z by taing T 3 (X is complete. Thus z is the common fixed point of T 1,T,T 3 and T 4. Uniqueness:- Let p and z be the two common fixed points of maps T 1,T,T 3 and T 4. Put x = z and y = p in condition (1 we get ( d(t1 z, T p, d(t 3z, T 1 z+d(t 1 z, T 4 p,d(t 4 p, T 3 z, d(t 4p, T p 0 ( d(z, p, ( d(z, p d(z, z+d(z, p d(p, p,d(p, z, 0 d(z, p,,d(p, z, 0 0 Hence p = z. So z is the unique common fixed points of T 1,T,T 3 and T 4. Corollary 3.6. Let (X, d be a complete metric space and let S and T be the self mapping of X, satisfying the following condition ( The pairs (S, T are wealy compatible. (3 S(X or T (X is complete. (4 There exists (0, 1 such that ( d(sx,sy, d(tx,sx+d(sx,ty,d(ty,tx, d(ty,sy 0...(1 For some F, every x, y X, then S and T have a unique common fixed point in X. Proof. The proof follows by taing T 3 = T 4 = T and T 1 = T = S in the above theorem. References [1] Branciari, A: A fixed point theorem for mapping satisfying a general contractive condition of integral type. Int J Math Math Sci. 9, (00. [] L. B. Ciric and S. B. Presic, Presic type generalisation of banach contraction mapping principle, Acta. Math. Univ. Comenian. (NS LXXVI( (007, [3] Dragan Doric, Zoran Kadeburg and Stojan Radenovic, A note on occassionally wealy compatible mappings and and common fixed points, preprint.

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