International Journal of Analysis and Applications Volume 17, Number 4 (2019), 586-595 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-586 COMMON FIXED POINT THEOREM FOR ĆIRIĆ TYPE QUASI-CONTRACTIONS IN RECTANGULAR b-METRIC SPACES SHU-FANG LI, FEI HE∗ AND NING LU School of Mathematical Sciences, Inner Mongolia University, Hohhot 010021, China ∗Corresponding author: Email address: hefei@imu.edu.cn Abstract. The purpose of this paper is to give positive answers to questions concerning Ćirić type quasi- contractions in rectangular b-metric spaces proposed in George et al. (J. Nonlinear Sci. Appl. 8 (2015), 1005-1013). 1. Introduction and preliminaries In [1], George et al. introduced the concept of rectangular b-metric spaces as a generalization of metric space, rectangular metric space and b-metric space (see also [2, 3]). Since then many fixed point theorems for various contractions were established in rectangular b-metric spaces (see [4–12]). Definition 1.1. ( [1]) Let X be a nonempty set and the mapping d : X ×X → [0,∞) satisfies: (1) d(x,y) = 0 if and only if x = y; (2) d(x,y) = d(y,x) for all x,y ∈ X; (3) there exists a real number s ≥ 1 such that d(x,y) ≤ s[d(x,u) + d(u,v) + d(v,y)] for all x,y ∈ X and all distinct points u,v ∈ X\{x,y}. Then d is called a rectangular b-metric on X and (X,d) is called a rectangular b-metric space (in short RbMS) with coefficient s. Received 2019-03-28; accepted 2019-04-30; published 2019-07-01. 2010 Mathematics Subject Classification. 47H10, 54H25. Key words and phrases. Common fixed point theorem; Ćirić type quasi-contractions; rectangular b-metric space. The research was supported by the National Natural Science Foundation of China (11561049, 11471236). c©2019 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 586 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-586 Int. J. Anal. Appl. 17 (4) (2019) 587 Definition 1.2. ( [1]) Let (X,d) be a RbMS, {xn} be a sequence in X and x ∈ X. Then (1) The sequence {xn} is said to be convergent in (X,d) and converges to x, if for every ε > 0 there exists n0 ∈ N+ such that d(xn,x) < ε for all n > n0 and this fact is represented by limn→∞xn = x or xn → x as n →∞. (2) The sequence {xn} is said to be Cauchy sequence in (X,d) if for every ε > 0 there exists n0 ∈ N+ such that d(xn,xn+p) < ε for all n > n0 and p > 0. (3) (X,d) is said to be a complete RbMS if every Cauchy sequence in X converges to some x ∈ X. In the setting of RbMS, limit of a convergent sequence is not necessarily unique and also every convergent sequence is not necessarily a Cauchy sequence. For details, we can see [1]. However, we have that the following result. Lemma 1.1. ( [3]) Let (X,d) be a RbMS with s ≥ 1, and let {xn} be a Cauchy sequence in X such that xn 6= xm whenever n 6= m. Then {xn} can converge to at most one point. George et al. [1] raised the following problems. Problem 1.1. ( [1]) In [1, Theorem 2.1], can we extent the range of λ to the case 1 s < λ < 1? Problem 1.2. ( [1]) Prove analogue of Chatterjea contraction, Reich contraction, Ćirić contraction and Hardy-Rogers contraction in RbMS. In [6], Mitrović has given a positive answer to Problem 1.1. In [7], Mitrović et al. obtained an analogue of Reich’s contraction principle in RbMS and thus give a partial solution to Problem 1.2. For further results, the reader can refer to [13, 14]. In this paper, we proved a common fixed point theorem for Ćirić type quasi-contractions in RbMS. It is well known that Ćirić contraction is more general than other contractions in Problem 1.2. Thus, we give a complete solution to the above Problem 1.2. 2. Main Results The following lemma is crucial in this paper. Lemma 2.1. Let (X,d) be a RbMS with coefficient s ≥ 1 and f,g : X → X be two self maps such that f(X) ⊆ g(X). Assume that there exists λ ∈ [0, 1 s ) such that d(fx,fy) ≤ λ max{d(gx,gy),d(gx,fx),d(gy,fy),d(gy,fx),d(gx,fy)}. (2.1) Taking x0 ∈ X, we construct a sequence {yn} by yn = fxn = gxn+1. If yn 6= yn+1 for all n ∈ N+, then Int. J. Anal. Appl. 17 (4) (2019) 588 (1) For m ∈ 0 ∪N+ and p ∈ N+, there exists 1 ≤ k(p) ≤ p such that δ(O(ym,m + p)) = d(ym,ym+k(p)), where O(ym,m + p) = {ym,ym+1, · · · ,ym+p},δ(A) = supx,y∈A d(x,y). (2) yn 6= ym whenever n 6= m. (3) δ(O(y0,n)) ≤ s1−sλ[d(y0,y1) + d(y1,y2)]. (4) δ(O(y0,∞)) ≤ s1−sλ[d(y0,y1) + d(y1,y2)], where O(y0,∞) = {y0,y1, · · · ,yn, · · ·}. (5) {yn} is a Cauchy sequence. Proof. (1) Let m ∈ {0, 1, 2, · · · ,} and p ∈ N+. Using (2.1), for any i,j ∈ N+ with m < i < j ≤ m + p, we have that d(yi,yj) = d(fxi,fxj) ≤ λ max{d(gxi,gxj),d(gxi,fxi),d(gxj,fxj),d(gxi,fxj),d(gxj,fxi)} = λ max{d(yi−1,yj−1),d(yi−1,yi),d(yj−1,yj),d(yi−1,yj),d(yj−1,yi)} ≤ λδ(O(ym,m + p)) < δ(O(ym,m + p)). This implies that max{d(yi,yj) : i,j ∈ N+ and m < i < j ≤ m + p} < δ(O(ym,m + p)). Since δ(O(ym,m + p)) = max{d(yi,yj) : i,j ∈ N+ and m ≤ i < j ≤ m + p}, there exists k(p) with 1 ≤ k(p) ≤ p such that δ(O(ym,m + p)) = d(ym,ym+k(p)). (2.2) (2) Suppose that yn = yn+p for some n,p ∈ N+. Then, by (2.1) we obtain that δ(O(yn,n + p)) = d(yn,yn+k(p)) = d(yn+p,yn+k(p)) = d(fxn+p,fxn+k(p)) ≤ λ max{d(gxn+p,gxn+k(p)),d(gxn+p,fxn+p),d(gxn+k(p),fxn+k(p)), d(gxn+k(p),fxn+p),d(gxn+p,fxn+k(p))} Int. J. Anal. Appl. 17 (4) (2019) 589 = λ max{d(yn+p−1,yn+k(p)−1),d(yn+p−1,yn+p),d(yn+k(p)−1,yn+k(p)), d(yn+k(p)−1,yn+p),d(yn+p−1,yn+k(p))} ≤ λδ(O(yn,n + p)), which implies δ(O(yn,n + p)) = 0. However, this is impossible because δ(O(yn,n + p)) ≥ d(yn,yn+1) > 0. Therefore, yn 6= ym whenever n 6= m. (3) Let n ∈ N+. Then, using (2.1) and (2.2), we get that δ(O(y0,n)) = d(y0,yk(n)) ≤ s[d(y0,y1) + d(y1,y2) + d(y2,yk(n))] = s[d(y0,y1) + d(y1,y2)] + sd(fx2,fxk(n)) ≤ s[d(y0,y1) + d(y1,y2)] + sλ max{d(gx2,gxk(n)),d(gx2,fx2),d(gxk(n),fxk(n)), d(gx2,fxk(n)),d(gxk(n),fx2)} = s[d(y0,y1) + d(y1,y2)] + sλ max{d(y1,yk(n)−1),d(y1,y2),d(yk(n)−1,yk(n)), d(y1,yk(n)),d(yk(n)−1,y2))} ≤ s[d(y0,y1) + d(y1,y2)] + sλδ(O(y0,n)). This implies that δ(O(y0,n)) ≤ s 1 −sλ [d(y0,y1) + d(y1,y2)]. (2.3) (4) Note that limn→∞δ(O(y0,n)) = δ(O(y0,∞)). Thus, from (2.3) we see that δ(O(y0,∞)) ≤ s 1 −sλ [d(y0,y1) + d(y1,y2)]. (5) For any n,p ∈ N+, d(yn,yn+p) ≤ λδ(O(yn−1,n + p)) ≤ λ2δ(O(yn−2,n + p)) ≤ ··· ≤ λnδ(O(y0,n + p)) ≤ λnδ(O(y0,∞)) ≤ λn · s 1 −sλ [d(y0,y1) + d(y1,y2] → 0(n →∞). Therefore, {yn} is a Cauchy sequence in X. � Int. J. Anal. Appl. 17 (4) (2019) 590 Theorem 2.1. Let (X,d) be a RbMS s ≥ 1 and f,g : X → X be two self maps such that f(X) ⊆ g(X), one of these two subsets of X being complete. If there exists λ ∈ [0, 1 s ) such that d(fx,fy) ≤ λ max{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)}, (2.4) for all x,y ∈ X, then f and g have a point of coincidence in X. Moreover, if f and g are weakly compatible (i.e., they commute at their coincidence points), then they have a unique common fixed point. Proof. Let x0 be an arbitrary point of X. Choose x1 ∈ X such that fx0 = gx1. Now, we can construct a sequence {yn} defined by yn = fxn = gxn+1, for n = 0, 1, 2, . . . (2.5) If yk = yk+1 for some k ∈ N+, then fxk+1 = yk+1 = yk = gxk+1 and f and g have a point of coincidence. Suppose, further, that yn 6= yn+1 for all n ∈ N+. By Lemma 2.1, we can obtain {yn} is a Cauchy sequence in X. Suppose, e.g., that the subspace g(X) is complete (the proof when f(X) is complete is similar). Then {yn} tends to some ω ∈ g(X), where ω = gu for some u ∈ X. Suppose that fu 6= gu. Then d(fu,yn) = d(fu,fxn) ≤ λ max{d(gu,gxn),d(gu,fu),d(gxn,fxn),d(gu,fxn),d(gxn,fu)} = λ max{d(gu,yn−1),d(gu,fu),d(yn−1,yn),d(gu,yn),d(yn−1,fu)}. Note that d(gu,yn−1) → 0, d(yn−1,yn) → 0 and d(gu,yn) → 0 as n → ∞. Then, for sufficiently large n ∈ N+, max{d(gu,yn−1),d(gu,fu),d(yn−1,yn),d(gu,yn),d(yn−1,fu)} = max{d(gu,fu),d(yn−1,fu)} and d(fu,yn) ≤ λ max{d(gu,fu),d(yn−1,fu)}. (2.6) Denote M(xn,u) = max{d(gu,fu),d(yn−1,fu)} for n ∈ N+. Then we can consider the following cases. Case 1. If there exists a subsequence {M(xnk,u)} of {M(xn,u)} such that M(xnk,u) = d(gu,fu), then d(fu,ynk ) ≤ λd(gu,fu). Note that d(yn,yn−1) → 0, d(yn,gu) → 0 and 1 s d(fu,gu) ≤ d(fu,ynk ) + d(ynk,ynk−1) + d(ynk−1,gu). (2.7) Thus, taking upper limit as k →∞ in (2.7), we obtain that 1 s d(fu,gu) ≤ lim sup k→∞ d(fu,ynk ) ≤ λd(gu,fu). This implies that d(gu,fu) ≤ sλd(fu,gu), which is a contradiction with sλ < 1 and fu 6= gu. Int. J. Anal. Appl. 17 (4) (2019) 591 Case 2. If there exists N ∈ N+ such that M(xn,u) = d(yn−1,fu) for all n > N, then (2.6) implies that d(fu,yn) ≤ λd(yn−1,fu) ≤ λ2d(yn−2,fu) ≤ ···≤ λn−Nd(yN,fu) = λn( 1 λN d(yN,fu)) → 0(n →∞), that is d(fu,yn) → 0 as n →∞. Since d(gu,yn) → 0 as n →∞, by Lemma 1.1 we have that fu = gu. This is a contradiction. Thus, we prove that fu = gu = ω, that is u is a point of coincidence of f and g. If f,g are weakly compatible, then, by fu = gu = ω, we obtain that fω = fgu = gfu = gω, and hence that ω is a point of coincidence of f and g. Let us prove that ω = fω = gω. Using (2.1), we get that d(ω,fω) = d(fu,fω) ≤ λ max{d(gu,gω),d(gu,fu),d(gω,fω),d(gu,fω),d(gω,fu)} = λ max{d(ω,fω), 0, 0,d(ω,fω),d(fω,ω)} = λd(ω,fω). Since λ < 1, we have that d(ω,fω) = 0, which implies that ω = fω = gω. Therefore, ω is a common fixed point of f and g. Let us prove that the common fixed point of f and g is unique. Suppose that ω1 and ω2 are two common points of f and g, that is ω1 = fω1 = gω1 and ω2 = fω2 = gω2. Using (2.1), we get that d(ω1,ω2) = d(fω1,fω2) ≤ λ max{d(gω1,gω2),d(gω1,fω1),d(gω2,fω2),d(gω1,fω2),d(gω2,fω1)} = λd(ω1,ω2). Since λ < 1, we have that d(ω1,ω2) = 0, that is ω1 = ω2. Thus, the common fixed point of f and g is unique. � Taking g = IX (identity mapping of X) in Theorem 2.1 we obtain the following. Corollary 2.1. (Ćirić type contraction) Let (X,d) be a RbMS with coefficient s ≥ 1 and f : X → X be a mapping. Assume that there exists λ ∈ [0, 1 s ) d(fx,fy) ≤ λ max{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)} for all x,y ∈ X. Then f has a unique fixed point. From Corollary 2.1, the following corollaries immediately follow. Int. J. Anal. Appl. 17 (4) (2019) 592 Corollary 2.2. (Chatterjea type contraction) Let (X,d) be a RbMS with coefficient s ≥ 1 and f : X → X be a mapping. Assume that there exists k ∈ [0, 1 s ) such that d(fx,fy) ≤ k 2 (d(x,fy) + d(y,fx)), for all x,y ∈ X. Then f has a unique fixed point. Corollary 2.3. (Reich type contraction) Let (X,d) be a RbMS with coefficient s ≥ 1 and f : X → X be a mapping. Assume that there exist λ,µ,δ ∈ [0, 1) with λ + µ + δ < 1 s such that d(fx,fy) ≤ λd(x,y) + µd(x,fx) + δd(y,fy), for all x,y ∈ X. Then f has a unique fixed point. Corollary 2.4. (Hardy-Rogers type contraction) Let (X,d) be a RbMS with coefficient s ≥ 1 and f : X → X be a mapping. Assume that there exist αi ∈ [0, 1)(i = 1, 2, 3, 4, 5) with α1 + α2 + α3 + α4 + α5 < 1s such that d(fx,fy) ≤ α1d(x,y) + α2d(x,fx) + α3d(y,fy) + α4d(x,fy) + α5d(y,fx), for all x,y ∈ X. Then f has a unique fixed point. Remark 2.1. From Corollary 2.1-Corollary 2.4, we see that Problem 1.2 has been fully answered. Finally, we give an example to illustrate our main result. Example 2.1. Let X = A ⋃ B, where A = { 1, 1 2 , 1 4 , 1 8 } and B = {0, 2}. Define d: X ×X → [0, +∞) such that d(x,y) = d(y,x) for all x,y ∈ X and d(x,y) =   0, x = y; |x−y|, x,y ∈ A; 13 6 , x,y ∈ B; 3 4 , x ∈ A\{1}, y ∈ B; 2, x = 1, y ∈ B. Let f: X → X be a map defined by f(x) =   1, x ∈ B; x 2 , x ∈ A\{1 8 }; 1 8 , x = 1 8 . and g be an identity mapping on X. Then the following hold: (a) (X,d) is a complete rectangular b-metric space with coefficient s = 4 3 ; Int. J. Anal. Appl. 17 (4) (2019) 593 (b) (X,d) is neither a metric space nor a rectangular metric space; (c) All conditions in Theorem 2.1 are satisfied with λ = 1 2 ; (d) f and g have a unique common fixed point x = 1 8 . Proof. First, let us prove (a). Clearly, conditions (1) and (2) of Definition 1.1 hold. To see (3), for all x,y ∈ X and all distinct points u,v ∈ X \{x,y}, we consider the following three cases. Case 1. If x,y ∈ A or x,y ∈ B, we only need to consider the case of x,y ∈ B with u,v ∈ A\{1}. In this case, d(u,v) ≥ d( 1 4 , 1 8 ) = 1 8 . So we have d(x,y) = 13 6 = 4 3 ( 3 4 + 1 8 + 3 4 ) ≤ 4 3 [d(x,u) + d(u,v) + d(v,y)]. Case 2. If x ∈ A\{1} and y ∈ B, then d(x,y) = 3 4 . Let us consider the following three cases. • If v ∈ B ⋃ {1}, then d(x,y) = 3 4 < d(v,y) ≤ d(x,u) + d(u,v) + d(v,y). • If u ∈ B, then d(x,y) = 3 4 = d(x,u) ≤ d(x,u) + d(u,v) + d(v,y). • If u,v ∈ A and v 6= 1, then d(x,y) = 3 4 = d(v,y) ≤ d(x,u) + d(u,v) + d(v,y). Case 3. If x = 1 and y ∈ B, then we consider the following two cases. • If u ∈ B or v ∈ B, then d(x,u) = 2 or d(v,y) = 13 6 . So we have d(x,y) = 2 ≤ d(x,u) + d(v,y) ≤ d(x,u) + d(u,v) + d(v,y). • If u,v ∈ A, then v 6= 1. It follows that d(x,u) + d(u,v) ≥ d(1, 1 2 ) + d( 1 2 , 1 4 ) = 3 4 . So we have d(x,y) = 2 = 4 3 ( 3 4 + 3 4 ) ≤ 4 3 [d(x,u) + d(u,v) + d(v,y)]. Additionally, in this case, we can also check that (b) holds. Hence, from the above three cases, we prove that (X,d) is a rectangular b-metric space with coefficient s = 4 3 . Since X is a finite set, we know that (g(X),d) = (X,d) is complete. Now we prove (c). It is sufficient to prove that (2.4) holds with λ = 1 2 . Since d(x,y) = d(y,x), we consider the following three cases. Case 1. If x,y ∈ B. In this case, d(fx,fy) = 0. So (2.4) holds. Int. J. Anal. Appl. 17 (4) (2019) 594 Case 2. If x ∈ B and y ∈ A, then fx = 1, d(gx,fx) = 2 and fy ∈ A. In this case, we have d(fx,fy) ≤d(1, 1 8 ) = 7 8 < 1 2 d(gx,fx) ≤ 1 2 max{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(fx,gy)}. Case 3. If x,y ∈ A, it is clear that d(fx,fy) = 1 2 d(gx,gy) for all x,y ∈ A\{1 8 }, which follows that (2.4) holds. So we assume that x = 1 8 . In this case, we have d(fx,fy) = 1 2 y − 1 8 < 1 2 ( y − 1 8 ) ≤ 1 2 max{d(gx,gy),d(gx,fx),d(gy,fy),d(gx,fy),d(fx,gy)}. From the above three cases, we show that (c) holds. Obviously, f and g have a unique common fixed point fx = gx = x = 1 8 . � References [1] R. George, S. Radenović, K. P. Reshma and S. 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