A Fixed Point Theorem in a Generalized Metric Space for Mappings Satisfying a Contractive Condition of Integral Type

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1 Int. Journal of Math. Analysis, Vol. 3, 29, no. 26, A Fixed Point Theorem in a Generalized Metric Space for Mappings Satisfying a Contractive Condition of Integral Type Bessem Samet Ecole Supérieure des Sciences et Techniques de Tunis Département de Mathématiques 5, Avenue Taha Hussein-Tunis, B.P.:56 Bab Menara-18, Tunisie bessem.samet@gmail.com Abstract. We establish a fixed point theorem in the generalized metric space introduced by Branciari for mappings satisfying a general contractive inequality of integral type. The obtained result can be considered as an extension of the theorem of Branciari (22). Mathematics Subject Classification: 54H25, 47H1 Keywords: Generalized metric space; Fixed point 1. Introduction In [2], Branciari established the following theorem. Theorem 1. Let (X, d) be a complete metric space, c ], 1[, and let f : X X be a mapping such that for each x, y X, d(fx,fy) d(x,y) (1), where ϕ :[, + [ [, + [ is a Lebesgue-integrable mapping which is summable on each compact subset of [, + [, nonnegative, and such that ε>, >. Then, f admits a unique fixed point a X such that for each x X, lim n f n x = a.

2 1266 B. Samet Theorem 1 is a generalization of the Banach-Caccioppoli principle. In fact, putting ϕ(t) = 1 for each t, we obtain d(fx,fy) 1 dt = d(fx,fy) cd(x, y) =c d(x,y) 1 dt. Then, a Banach-accioppoli contraction also satisfies (1). The converse is not true in general (see [2]). In [3], Rhoades proved that Theorem 1 holds if we replace d(x, y) in (1) by m(x, y) = max { d(x, y),d(x, fx),d(y, fy), d(x, fy)+d(y, fx) 2 The purpose of this paper is to establish that Theorem 1 is also valid if we replace the metric space (X, d) by the generalized metric space (or shortly g.m.s) introduced by Branciari in [1]. Definition 1. Let X be a nonempty set. Suppose that the mapping d : X X [, + ), satisfies: 1. d(x, y) = x = y 2. d(x, y) =d(y, x) for all x, y X 3. d(x, y) d(x, w)+d(w, z) +d(z, y) for all x, y X and for all distinct points w, z X\{x, y} (rectangular property). Then d is called a generalized metric and (X, d) is a generalized metric space (or shortly g.m.s). Definition 2. Let (X, d) be a g.m.s, (x n ) be a sequence in X and x X. We say that (x n ) converges to x with respect to d if and only if d(x n,x) asn +. We say that (x n ) is Cauchy if and only if d(x n,x m ) asn, m +. We say that (X, d) is complete if and only if every Cauchy sequence in X is convergent in X. 2. Main result We have obtained the following result. Theorem 2. Let (X, d) be a complete g.m.s, c ], 1[, and let f : X X be a mapping such that for each x, y X, (2) d(fx,fy) d(x,y), where ϕ :[, + [ [, + [ is a Lebesgue-integrable mapping which is summable on each compact subset of [, + [, nonnegative, and such that (3) ε>, >. }.

3 Fixed point theorem in a generalized metric space 1267 Then, f admits a unique fixed point a X such that for each x X, lim n f n x = a. Proof. Let x X and consider the sequence (x n ) defined by x n = f n x for all n N. By (2), we have d(fx,f 2 x). Again Similarly d(f 2 x,f 3 x) d(f 3 x,f 4 x) d(fx,f 2 x) c Thus in general, if n is a positive integer, then d(f n x,f n+1 x) (4) n. We divide the proof into two cases. Case I. First, assume that f m x = f n x for some m, n N, m n. Let m>n, then f m n (f n x)=f n x, i.e., f k y = y where k = m n, y = f n x.nowifk>1 = d(f k y,f k+1 y) c d(f k 1 y,f k y) (by (4) ) c k. Then, (5) (1 c k ). Assume that y fy, then d(y, fy) >, and by (3), we obtain >. Since <c<1, we obtain a contradiction with (5). Hence fy = y, i.e., y is a fixed point of f.

4 1268 B. Samet Case II. Assume that f m x f n x for all m, n N, m n. Let n N, we have (6) d(f n x,f n+2 x) c d(f n 1 x,f n+1 x) d(f n 2 x,f n x) c 2 Taking the limit of (4), as n +, gives lim n + d(f n x,f n+1 x) d(x,f 2 x) c n. = which, from (3), implies that (7) lim n,x n+1 )=. n + We now show that (f n x) is Cauchy. Suppose that it is not. Then, there exists an ε>such that for each p N there are m(p),n(p) N, with m(p) >n(p) >p, such that (8) d(f m(p) x, f n(p) x) ε, d(f m(p) 1 x, f n(p) x) <ε. Hence ε d(f m(p) x, f n(p) x) d(f m(p) x, f m(p) 2 x)+d(f m(p) 2 x, f m(p) 1 x) +d(f m(p) 1 x, f n(p) x) < d(f m(p) x, f m(p) 2 x)+d(f m(p) 2 x, f m(p) 1 x)+ε. Then, using (4), (6) and tacking p +, we get (9) d(f m(p) x, f n(p) x) ε + as p +. This implies that (1) k N p>k d(f m(p)+1 x, f n(p)+1 x) <ε. In fact, if there exists a subsequence (p k ) N, p k >k, d(f m(pk)+1 x, f n(pk)+1 x) ε, we obtain ε d(f m(pk)+1 x, f n(pk)+1 x) d(f m(pk)+1 x, f m(pk) x)+d(f m(pk) x, f n(pk) x)+d(f n(pk) x, f n(pk)+1 x) ε as k +, and from (2) d(f m(p k )+1 x,f n(p k )+1 x) d(f m(p k ) x,f n(p k ) x),

5 letting now k +, we get Fixed point theorem in a generalized metric space 1269, which is a contradiction since c ], 1[ and >. Then, (1) holds. Let us prove now that (11) σ ε ],ε[,p ε N p>p ε d(f m(p)+1 x, f n(p)+1 x) <ε σ ε. If (11) is not true, by (1), there exists a subsequence (p k ) N such that Then, from d((f m(p k)+1 x, f n(p k)+1 x) ε as k +. d(f m(p k )+1 x,f n(p k )+1 x) d((f m(p k ) x,f n(p k ) x) tending k +, we obtain also a contradiction that. Hence, (11) holds. Now, the Cauchy character is easy to obtain. In fact, for each naturel number p>p ε, we have ε d(f m(p) x, f n(p) x) d(f m(p) x, f m(p)+1 x)+d(f m(p)+1 x, f n(p)+1 x) +d(f n(p)+1 x, f n(p) x) < d(f m(p) x, f m(p)+1 x)+(ε σ ε )+d(f n(p)+1 x, f n(p) x) ε σ ε as p +. Thus ε ε σ ε which is a contradiction. We conclude that (f n x) is Cauchy. By the completeness of X, there is a X such that f n x a as n +. We shall now show that fa = a. We divide this proof into two parts. First, assume that f r x a, fa for any r N. Then, we have d(a, fa) d(a, f n x)+d(f n x, f n+1 x)+d(f n+1 x, fa) asn +. In fact, by the definition of the limit, we have d(a, f n x) asn +. By (7), d(f n x, f n+1 x) asn +. On the other hand, we have d(f n+1 x,fa) d(f n x,a) asn + which gives by (3) that d(f n+1 x, fa) asn +. Hence, we conclude that d(a, fa) =, i.e., a = fa. Next, assume that f s x = a or f s x = fa for some s N. Obviously x a. Now, one may easily verify that (f n a) is a sequence with the following properties (a) f n a a as n +. (b) f p a f r a for any p, r N, p r.,

6 127 B. Samet So, (12) d(f n+1 a,fa) d(f n a,a) asn +. On the other hand, we have d(f n+1 a, fa) d(a, fa) d(f n+1 a, f n+2 a)+d(f n+2 a, a) asn +. Hence, (13) d(f n+1 a,fa) d(a,fa). Now, from (12) and (13) it follows that d(a,fa) = which implies that d(a, fa) =, i.e., a = fa. Suppose now that there are two distinct points a, b X such that fa = a and fb = b.by (2), we obtain the following contradiction < d(a,b) = d(fa,fb) d(a,b) < d(a,b). Then, the fixed point of f is unique, that is the limit of f n x as n +. Now, we give a simple example that illustrate Theorem 2. Let X = {1, 2, 3, 4}. Define d : X X R as follows: d(1, 2) = d(2, 1) = 3 d(2, 3) = d(3, 2) = d(1, 3) = d(3, 1) = 1 d(1, 4) = d(4, 1) = d(2, 4) = d(4, 2) = d(3, 4) = d(4, 3) = 4. Then (X, d) is a complete generalized metric space but (X, d) is not a metric space because it lacks the triangular property: 3=d(1, 2) >d(1, 3) + d(3, 2)=1+1=2. Now define a mapping f : X X as follows: { fx =3 if x 4 fx =1 if x =4. Then, f satisfies (2) with ϕ(t) =e t and c = e 3. Hence, f admits a unique fixed point that is a =3.

7 Fixed point theorem in a generalized metric space 1271 References 1. A. Branciari. A fixed point theorem of Banach-Caccippoli type on a class of generalized metric spaces. Publ. Math. Debrecen, (2), A. Branciari. A fixed point theorem for mappings satisfying a general contractive condition of integral type. Int. J. Math. Math. Sci. 29 (22), no. 9, B. E. Rhoades. Two fixed point theorems for mappings satisfying a general contractive condition of integral type. IJMMS. 63 (23), Received: April, 29