Fixed Point Theorems in Partial b Metric Spaces

Applied Mathematical Sciences, Vol. 12, 2018, no. 13, 617-624 HIKARI Ltd www.m-hikari.com https://doi.org/10.12988/ams.2018.8460 Fixed Point Theorems in Partial b Metric Spaces Jingren Zhou College of Mathematics and Information Science Guangxi University, Nanning, 530004, China Dingwei Zheng College of Mathematics and Information Science Guangxi University, Nanning, 530004, China Gengrong Zhang College of Mathematics and Information Science Hunan First Normal University, Changsha, China Copyright c 2018 Jingren Zhou, Dingwei Zheng and Gengrong Zhang. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract In this paper, we investigate some contraction mapping in partial b metric space and prove the existence of fixed point of this mapping in partial b metric space under some conditions. We improve and generalize some results in partial b metric space. Mathematics Subject Classification: 05C25 Keywords: Partial b metric space, Cauchy sequence, Fixed point 1 Introdution The fixed point theory in metric space is an important part of nonlinear functional analysis. In the 1920s, Banach proposed and proved the famous

618 Jingren Zhou, Dingwei Zheng and Gengrong Zhang Banach Contraction Principle. Since then, many scholars have proposed a series of new concepts of contraction mapping and new fixed point theorems. In 1994, S.G.Matthews introduced the concept of partial metric space and proved the Banach Contraction Principle in the partial metric space[10]. The biggest difference between partial metric space and metric space is that it is may not zero from its own distance. More fixed point theory of partial metric space have been proposed, for example, see [1,4,12]. In 1993, Bakhtin [3] introduced the concept of b metric space which is a generalization of metric space, he proved the famous Banach Contraction Principle in the b metric space, also see [11]. In recent years, scholars discuss various types of non-expansive mappings, single and multi-value mappings in b metric space [2,7]. In 2014, S.Satish introduced the concept of partial b metric space [13], and the fixed point theorem of Banach Contraction Principle and Kannan type mapping was proved in partial b metric space. In this paper, we prove some fixed point theorem for C-contractive mappings and Meir-Keeler mappings in partial b metric space which generalize and extend the result of S.K.Chatterjea [11] and S.Satish [13], respectively. Definition 1.1. Let X be a nonempty set, s 1 be a given real number and let b : X X [0, ) be a mapping such that for all x, y, z X, the following conditions hold: (Pb1) x = y if and only if b(x, x) = b(x, y) = b(y, y) (Pb2) b(x, x) b(x, y) (Pb3) b(x, y) = b(y, x) (Pb4) b(x, y) s[b(x, z) + b(z, y)] b(z, z) Then the pair (X, b) is called a partial b metric space. The number s is call the coefficient of (X, b, s). Remark 1.2. In a partial b-metric space (X, b, s), if x, y X and b(x, y) = 0, then x = y but the converse may not be true. Remark 1.3. It is clear that every partial metric space is a partial b metric space with coefficient s = 1 and every b metric space is a partial b metric space with the same coefficient and zero self-distance. However, the converse of this fact need not hold. Remark 1.4. (see [14]) Every partial b metric b defines a b metric d, where d(x, y) = 2b(x, y) b(x, x) b(y, y), x, y X. Definition 1.5. Let (X, b, s) be a partial b metric space. Let x n be any sequence in X and x X. Then: (i) The x n sequence is said to be convergent and converges to x if lim b(x n, x) exists and is finite. (ii) The x n sequence is said to be Cauchy sequence in (X, b, s) if lim n,m b(x n, x m ) exists and is finite.

Fixed point theorems in partial b metric spaces 619 (iii) (X, b, s) is said to be a complete partial b metric space if for every Cauchy sequence x n in X there exists x X such that lim b(x n, x m ) = lim b(x n, x) = b(x, x). n,m Note that in a partial b metric space the limit of convergent sequence may not be unique. Example 1.6. Let X = [0, + ), A > 0 be any constant and define a function b : X X [0, + ) by b(x, y) = max{x, y} + A for all x, y X Then (X, b, s) is a partial b metric space with arbitrary coefficient s 1. Now, define a sequence {x n } in X by x n = 1 for all n N. If y 1, we have b(x n, y) = y + A = b(y, y); therefore, lim b(x n, y) = b(y, y) for all y 1. Thus, the limit of convergent sequence in partial b metric space need not be unique. 2 Main results Theorem 2.1. Let (X, b, s) be a complete partial b metric space with coefficient s 1 and f : X X be a mapping satisfying the following condition: b(fx, fy) λ[b(x, fy) + b(y, fx)] x, y X, (2.1) where λ [0, 1 ). Then f has unique fixed point z X and b(z, z) = 0. 2s P roof. First prove that z X is the fixed point of f, that is fz = z, from (2.1), that b(z, z) = b(fz, fz) λ[b(z, fz) + b(z, fz)] = λ[b(z, z) + b(z, z)] = 2λb(z, z) < 1 b(z, z) s < b(z, z), a contradiction. Thus, we have b(z, z) = 0. If fixed point of f exists, then it is unique. Let z, v X be two distinct fixed points of f, that is, z = fz fv = v. then we have b(z, z) = b(v, v), from (2.1), we have b(z, v) = b(fz, fv) λ[b(z, fv) + b(v, fz)] = λ[b(z, v) + b(v, z)] = 2λb(z, v) < 1 b(z, v), s

620 Jingren Zhou, Dingwei Zheng and Gengrong Zhang a contradiction. Thus, we have z = v. Next, we prove the existence of fixed point, let x n = f n x 0 and b n = b(x n, x n+1 ), where x 0 is arbitrary point of X. If x n+1 = x n for some n N, then x = x n is a fixed point of f. Therefore, we can suppose x n+1 x n,b n > 0 for each n N, from (2.1), b n = b(x n, x n+1 ) = b(fx n 1, fx n ) λ[b(x n 1, fx n ) + b(x n, fx n 1 )] = λ[b(x n 1, x n+1 ) + b(x n, x n )]. By b(x, y) s[b(x, z) + b(z, y)] b(z, z), we can get λ[b(x n 1, x n+1 ) + b(x n, x n )] λ[s[b(x n 1, x n ) + b(x n, x n+1 )] b(x n, x n ) + b(x n, x n )] = λ[s[b(x n 1, x n ) + b(x n, x n+1 )] = λs[b n 1 + b n ] Let µ = λs, λ [0, 1 ), then b 2s n µ[b n 1 + b n ], where µ [0, 1 ). Therefore, 2 b n αb n 1, where α = µ < 1, On repeating this process we obtain 1 µ b n α n b 0, therefore lim b n = 0. Next, we shall that x n is a Cauchy sequence in X. Let b n = b(x n, x m ), from (2.1) that for n, m N with n < m b(x n, x m ) = b(f n x 0, f m x 0 ) λ[b(x n 1, fx m 1 ) + b(x m 1, fx n 1 )] = λ[b(x n 1, x m ) + b(x m 1, x n )]. By b(x, y) s[b(x, z) + b(z, y)] b(z, z), we can get b(x n, x m ) λ[b(x n 1, x m ) + b(x m 1, x n )] = [s[b(x n 1, x n ) + b(x n, x m )] b(x n, x n ) + s[b(x m 1, x m ) + b(x n, x m )] b(x m, x m )] = λsb(x n 1, x n ) + 2λsb(x n, x m ) + λsb(x m 1, x m ), that is b(x n, x m ) β[b n 1 + b m 1 ], where β = λs. Therefore, {x 1 2λs n} is a Cauchy sequence in X and lim n,m b(x n, x m ) = 0. By the completeness of X there exists z X such that lim b(x n, z) = lim b(x n, x m ) = b(z, z) = 0. n,m Now we shall prove that z is a fixed point of f. Let d n = b(fx n, fu), b n = (z, fx n )), for each n N, from (2.1), we have d n = b(fx n, fz) λ[b(x n, fz) + b(fx n, z)] = λ[b(fx n 1, fz) + b(fx n, z)] = λ(d n 1 + b n ).

Fixed point theorems in partial b metric spaces 621 We take upper limit on both sides to the above inequality, lim sup d n λ lim sup d n 1 + lim sup b n, since then, thus, That is lim b n = 0, lim sup d n = 0, lim d n = 0. d n = b(fx n, fz) λ[b(x n, fz) + b(fx n, z)] = λ[b(fx n 1, fz) + b(fx n, z)] = λ(d n 1 + b n ). We take limit on both sides to the inequality, then b(z, fz) 0, that is, fz = z. Thus, z is a fixed point of f and it is a unique fixed point of f. Remark 2.1.1. [11] Let (X,d) be a metric space, a mapping f : X X is said to be a C-contraction if there exists α (0, 1 ) such that 2 d(fx, fy) α(d(x, fy) + d(y, fx)), holds for all x, y X. Taking s = 1 in Theorem 2.1., we can get the C-contraction fixed point theorem in partial metric spaces[1,4,12], taking b(z, z) = 0 in Theorem 2.1., we can get the C-contraction fixed point theorem in b metric spaces[6,9]. Theorem 2.2. Let (X, b, s) be a complete partial b metric space with coefficient s > 1 and f : X X be a mapping satisfying the following condition: for each ε > 0 there exists δ > 0 such that ε b(x, z) < ε + δ sb(fx, fz) < ε. (2.2) Then f has a unique fixed point z X and b(z, z) = 0. P roof. First of all, by (2.2), we point out that : for x, y X, and x y, sb(fx, fy) < b(x, y). (2.3) Suppose x 0 X be an arbitrary point, we can choose sequence {x n } in X such that x n+1 = fx n = f 2 x n 1 = = f n+1 x 0, n = 0, 1, 2, 3

622 Jingren Zhou, Dingwei Zheng and Gengrong Zhang If x n+1 = x n for n N, then f have a point of coincidence. If x n+1 x n for n N, making use of the inequality (2.3) with x = x n 1 and y = x n, we can get sb(x n, x n+1 ) < b(x n 1, x n ). As s > 1, {b(x n, x n+1 )} is a decrease sequence, it is easy to prove that lim b(x n, x n+1 ) = 0. Now, we shall show that {x n } is a Cauchy sequence in X. We can choose an N (large enough) such that when n > N, b(x n, x n+1 ) < ε ε s s + s 2. Put K(x N, ε) = {y X : b(y, x N ) ε}. If x m K(x N, ε) with m > N then x m x N. Making use of the inequality b(x, y) s[b(x, z) + b(z, y)] b(z, z), we have b(f 2 x m, x N ) s[b(f 2 x m, f 2 x N ) + b(f 2 x N, x N )] b(f 2 x N, f 2 x N ) s[b(f 2 x m, f 2 x N ) + b(f 2 x N, x N )] s[ 1 s b(x m+1, x N+1 ) + b(f 2 x N, x N )] b(x m+1, x N+1 ) + s 2 [b(x N+2, x N+1 ) + b(x N+1, x N )] sb(x N, x N ) 1 s b(x m, x N ) + (s + s 2 )b(x N+1, x N ) ε s + (s + s2 ) ε ε s s + s 2 = ε. Therefore f 2 x m K(x N, ε), f 2 maps K(x N, ε) into itself.since x N+1 K(x N, ε), then x N+3, x N+5 K(x N, ε). By b(x, y) s[b(x, z)+b(z, y)] b(z, z), we have b(x N+2, x N ) s[b(x N+2, x N+1 ) + b(x N+1, x N )] b(x N+1, x N+1 ) sb(x N+1, x N ) + s[ 1 s b(x N, x N+1 )] s ε ε s s + s + ε ε s 2 s + s 2 < ε. Therefore, x N+2 K(x N, ε), then x N+2, x N+4 K(x N, ε). That is to say, {x n : n N} K(x N, ε). For n > m > N, since x n, x m K(x N, ε), we have b(x n, x m ) s[b(x n, x N ) + b(x N, x m )] b(x N, x N ) 2sε.

Fixed point theorems in partial b metric spaces 623 Therefore, {x n } is a Cauchy sequence in X and lim n,m b(x n, x m ) = 0. By completeness of X there exists z X such that lim b(x n, z) = lim b(x n, x m ) = b(z, z) = 0. (2.4) n,m We shall show that, z is a fixed point of f. In order to prove fz = z, we suppose that fz z. By b(x, y) s[b(x, z) + b(z, y)] b(z, z), and (2.4) b(fz, z) s[b(z, fx n ) + b(fx n, fz)] b(fx n, fx n ) s[b(z, x n+1 ) + 1 s b(x n, z)] = sb(z, x n+1 ) + b(x n, z), Passing to limit as n, we have b(fz, z) 0, which is a contradiction. Thus, fz = z. If z is a fixed point of f and b(z, z) > 0, from (2.4) we can get b(z, z) = b(fz, fz) 1 b(z, z) < b(z, z), a contradiction. Therefore b(z, z) = 0. s Using (2.4) in above inequality we can have b(fz, z) = 0, that is, fz = z. Therefore z is a fixed point of f and it is unique fixed point of f. Remark 2.2.1. Let (X, d, s) be a complete b metric space with coefficient s 1, f : X X be a mapping satisfying the following condition: for each ε > 0 there exists δ > 0 such that ε d(x, z) < ε + δ sd(fx, fz) < ε. Then f has a unique fixed point z X. Acknowledgements. This research is supported by National Natural Science Foundation of China (Nos.11461002), Guangxi Natural Science Foundation (2016GXNSFAA380317) and The Basic Ability Improvement Project (2018KY0042). References [1] Altumi and A. Erduran, Fixed point theorems for monotone mappings on partial metric spaces, Fixed Point Theory Appl., 2011 (2011), no. 12, 1-10. https://doi.org/10.1155/2011/508730 [2] H.S. Ding, M. Imdad, S. Radenovic and J. Vujakovic, On some fixed point results in b metric,rectangular and b rectangular metric spaces, Arab. J. Math. Sci., 22 (2016), 151-164.

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