International Journal of Analysis and Applications Volume 17, Number 2 (2019), 275-281 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-17-2019-275 FIXED POINT THEOREM OF ĆIRIĆ-PATA TYPE AO-LEI SIMA, FEI HE∗ AND NING LU School of Mathematical Sciences, Inner Mongolia University, Hohhot 010021, China ∗Corresponding author: hefei@imu.edu.cn Abstract. In this article, we proved a fixed point theorem of Ćirić-Pata type in metric space. This result extends several results in the existing literature. Moreover, an example is given in the support of our result. In particular, the main result provides a complete solution to an open problem raised by Kadelburg and Radenović (J. Egypt. Math. Soc. 24 (2016) 77-82). 1. Introduction Throughout this paper, (X,d) will be a complete metric space. Fix an arbitrary point x0 ∈ X and denote ‖x‖ = d(x,x0), for each x ∈ X. Also, ψ : [0, 1] → [0,∞) is an increasing function, continuous at zero, with ψ(0) = 0. Given a function f : X → X. In 2011, Pata [1] obtained the following result which is a generalization of the classical Banach contraction principle. Theorem 1.1. [1] Let Λ ≥ 0, α ≥ 1 and β ∈ [0,α] be fixed constants. If the inequality d(fx,fy) ≤ (1 −ε)d(x,y) + Λεαψ(ε)[1 + ‖x‖ + ‖y‖]β (1.1) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Received 2018-11-06; accepted 2018-12-14; published 2019-03-01. 2010 Mathematics Subject Classification. 47H10, 54H25. Key words and phrases. fixed point theorem; Pata type contraction; Ćirić type quasi-contraction; metric space. 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. 275 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-17-2019-275 Int. J. Anal. Appl. 17 (2) (2019) 276 Afterward many pata type fixed point theorems have been established by various authors; see ( [2], [3], [4], [5], [6], [7], [8], [9]). Particularly, Kadelburg and Radenović [7] proved some fixed point theorems of Pata type and raised the following open question on Pata-version of Ćirić contraction principle (see [10]). Problem 1.1. [7] Prove or disprove the following. Let f : X → X and let Λ ≥ 0, α ≥ 1 and β ∈ [0,α] be fixed constants. If the inequality d(fx,fy) ≤ (1 −ε) max{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)} + Λεαψ(ε)[1 + ‖x‖ + ‖y‖]β (1.2) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Furthermore, the sequence {fnx0} converges to z. Very recently, Jacobe et al. give the following result. Theorem 1.2. [5] Let f : X → X and let Λ ≥ 0, α ≥ 1 and β ∈ [0,α] be fixed constants. If the inequality d(fx,fy) ≤ (1 −ε) max { d(x,y), d(x,fx) + d(y,fy) 2 , d(x,fy) + d(y,fx) 2 } + Λεαψ(ε)[1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖]β (1.3) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point in X. In this paper, we give a fixed point theorem of Ćirić-Pata type in metric space. This theorem extends the main results in ( [1], [5], [7]) and provides a complete solution to the above Problem 1.1. Finally, an example is given to illustrate the superiority of the main results. 2. Main results Our result of this paper are stated as follows. Theorem 2.1. Let Λ ≥ 0, α ≥ 1 be fixed constants. For x,y ∈ X, we denote M(x,y) = max{d(x,y),d(x,fx),d(y,fy),d(x,fy),d(y,fx)}. If the inequality d(fx,fy) ≤ (1 −ε)M(x,y) + Λεαψ(ε)[1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖]α (2.1) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Furthermore, the sequence {fnx0} converges to z. Int. J. Anal. Appl. 17 (2) (2019) 277 Proof. Starting from x0, construct a sequence {xn} such that xn = fxn−1 = fnx0. If xn0 = xn0+1 for some n0, then xn0 is a fixed point of f. Thus, we always assume that xn 6= xn+1 for all n ∈ N. We prove that xn 6= xm for all m,n ∈ N and n 6= m. Assume that there exist n0,m0 ∈ N such that n0 < m0 and xn0 = xm0. Denote A = max{d(xi,xj) : n0 ≤ i < j ≤ m0} and B = max{‖xi‖ : n0 ≤ i ≤ m0 + 1}. It is obvious that A = max{d(xi,xj) : n0 + 1 ≤ i < j ≤ m0} and A > 0. For each i,j ∈ N such that n0 + 1 ≤ i < j ≤ m0, we have d(xi,xj) ≤ (1 −ε)M(xi−1,yj−1) + Λεαψ(ε)[1 + ‖xi−1‖ + ‖xj−1‖ + ‖xi‖ + ‖xj‖]α ≤ (1 −ε)A + Λεαψ(ε)(1 + 4B)α. It follows that A ≤ (1 −ε)A + Λεαψ(ε)(1 + 4B)α and A ≤ Λεα−1ψ(ε)(1 + 4B)α. Letting ε → 0, we can see A ≤ 0. This is a contradiction with A > 0. Denote Dn = max{d(xi,xj) : 0 ≤ i < j ≤ n} and δn = sup{d(xi,xj) : n ≤ i < j}. In order to prove {xn} is a Cauchy sequence, we divide into the following three steps. Step 1. We show that d(fx,fy) < M(x,y) for all x,y ∈ X and x 6= y. Let ε = 0 in (2.1), we have d(fx,fy) ≤ M(x,y) for all x,y ∈ X. Assume that there exist x0,y0 ∈ X and x0 6= y0 such that d(fx0,fy0) = M(x0,y0). Using (2.1), we get M(x0,y0) = d(fx0,fy0) ≤ (1 −ε)M(x0,y0) + Λεαψ(ε)[1 + ‖x0‖ + ‖y0‖ + ‖fx0‖ + ‖fy0‖]α. It follows that M(x0,y0) ≤ Λεα−1ψ(ε)[1 + ‖x0‖ + ‖y0‖ + ‖fx0‖ + ‖fy0‖]α. Passing to the limit as ε → 0, we see M(x0,y0) ≤ 0, a contradiction. Step 2. We prove that {Dn} is bounded. By step 1, we see that d(xi,xj) = d(fxi−1,fxj−1) < M(xi−1,xj−1) ≤ Dn, for all i,j ∈ N such that 0 < i < j ≤ n. Thus there exists `n ∈ N such that 1 ≤ `n ≤ n and Dn = d(x0,x`n ). Using (2.1), we have Dn = d(x0,x`n ) ≤ d(x0,x1) + d(x1,x`n ) ≤ d(x0,x1) + (1 −ε)M(x0,x`n−1) + Λε αψ(ε)[1 + ‖x0‖ + ‖x`n−1‖ + ‖x1‖ + ‖x`n‖] α ≤ (1 −ε)Dn + Λεαψ(ε)(1 + 3Dn)α + d(x0,x1). Int. J. Anal. Appl. 17 (2) (2019) 278 This implies that εDn ≤ Λεαψ(ε)(1 + 3Dn)α + d(x0,x1). Suppose that {Dn} is unbounded. Then there exists a subsequence {Dnk} with {Dn} such that Dnk → ∞ (k →∞) and Dnk ≥ 1 + d(x0,x1). Let ε = εk = 1+d(x0,x1) Dnk . Then we get 1 + d(x0,x1) Dnk ·Dnk ≤ Λ[ 1 + d(x0,x1) Dnk ]αψ(εk)(1 + 3Dnk ) α + d(x0,x1) and 1 ≤ Λ( 1 Dnk + 3)αψ(εk)[1 + d(x0,x1)] α. Letting k →∞, we have εk → 0 and Λ( 1 Dnk + 3)αψ(εk)[1 + d(x0,x1)] α → 0. This is a contradiction. Thus {Dn} is bounded and there exists a constant M > 0 such that Dn ≤ M. Step 3. We show that δn → 0. Observe that d(xi,xj) = d(fxi−1,xj−1) < M(xi−1,xj−1) ≤ δn for every i,j ∈ N with n + 1 ≤ i < j. Thus we get δn+1 ≤ δn ≤ ··· ≤ δ0 ≤ M. It is easy to see that {δn} is decreasing and bounded sequence. It follows that lim n→∞ δn = δ for some δ ≥ 0. Assume that δ > 0. From (2.1), it holds for each i,j ∈ N with n + 1 ≤ i < j, d(xi,xj) ≤ (1 −ε)M(xi−1,xj−1) + Λεαψ(ε)(1 + 4M)α. This implies that δn+1 ≤ (1 −ε)δn + Λεαψ(ε)(1 + 4M)α. (2.2) Letting n →∞ in (2.2), we get δ ≤ (1 −ε)δ + Λεαψ(ε)(1 + 4M)α and δ ≤ Λεα−1ψ(ε)(1 + 4M)α. From Λεα−1ψ(ε)(1 + 4M)α → 0 (ε → 0), we see δ ≤ 0, a contradiction. For each p ∈ N, we get d(xn,xn+p) ≤ δn → 0 (n →∞). Hence, {xn} is Cauchy sequence. Since X is complete, there exists z ∈ X such that xn → z (n →∞). Now, we show that fz = z. Using (2.1), we get d(fz,xn+1) ≤ (1 −ε) max{d(z,xn),d(z,fz),d(xn,xn+1),d(z,xn+1),d(xn,fz)} + Λεαψ(ε)(1 + 4M)α. Int. J. Anal. Appl. 17 (2) (2019) 279 By taking limits on both sides when n →∞, we obtain d(fz,z) ≤ (1 −ε)d(fz,z) + Λεαψ(ε)(1 + 4M)α. Then d(fz,z) ≤ Λεα−1ψ(ε)(1 + 4M)α → 0 (ε → 0). This implies that d(fz,z) = 0 and fz = z. Finally, we prove the uniqueness of z. If fu = u, fv = v for any two fixed u,v ∈ X, then we can write (2.1) in the form d(u,v) = d(fu,fv) ≤ (1 −ε) max{d(u,v),d(u,fu),d(v,fv),d(u,fv),d(v,fu)} + Λ�αψ(ε)[1 + ‖u‖ + ‖v‖ + ‖fu‖ + ‖fv‖]α ≤ (1 −ε)d(u,v) + Λεαψ(ε)[1 + 2‖u‖ + 2‖v‖]α. Therefore d(u,v) ≤ Λεα−1ψ(ε)[1 + 2‖u‖ + 2‖v‖]α → 0 (ε → 0), which implies that d(u,v) = 0 and u = v. Hence, f has a unique fixed point z ∈ X. � Remark 2.1. It is easy to see that the condition (2.1) is weaker than the condition (1.2). Hence, Theorem 2.1 provides a solution to Problem 1.1. From Theorem 2.1 we get the following Corollaries. Corollary 2.1. Let f : X → X and Let Λ ≥ 0, α ≥ 1 be fixed constants. If the inequality d(fx,fy) ≤ (1 −ε)d(x,y) + Λεαψ(ε)[1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖]α (2.3) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Furthermore, the sequence {fnx0} converges to z. Remark 2.2. It is easy to see that the condition (2.3) is weaker than the condition (1.1). Thus Corollary 2.1 is an extension of Theorem 1.1. Corollary 2.2. Let f : X → X and Let Λ ≥ 0, α ≥ 1 be fixed constants. If the inequality d(fx,fy) ≤ 1 −ε 2 (d(x,fy) + d(y,fx)) + Λεαψ(ε)[1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖]α (2.4) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Furthermore, the sequence {fnx0} converges to z. Int. J. Anal. Appl. 17 (2) (2019) 280 Remark 2.3. It is easy to see that the condition 2.4 is weaker than the condition 2.1 in [7]. Thus Corollary 2.2 is an extension of Theorem 2.1 in [7]. Corollary 2.3. Let f : X → X and Let Λ ≥ 0, α ≥ 1 be fixed constants. If the inequality d(fx,fy) ≤ (1 −ε) max { d(x,y),d(x,fx),d(y,fy), d(x,fy) + d(y,fx) 2 } + Λεαψ(ε)[1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖]β (2.5) is satisfied for every ε ∈ [0, 1] and all x,y ∈ X, then f has a unique fixed point z ∈ X. Furthermore, the sequence {fnx0} converges to z. Remark 2.4. It is easy to see that the condition (2.5) is weaker than the condition (1.3). Thus Corollary 2.3 is an extension of Theorem 1.2. The following is an example which can apply to Theorem 2.1 but not Corollary 2.3 or Theorem 1.2. Example 2.1. Let X = {0, 1, 2, 4, 8, 9} ⋃ { 1 2n : n = 1, 2, · · ·} with the usual metric. It is easily to check that X is a complete metric space. Define f : X → X by fx =   8 x = 9,1 2 x others . Then mapping f satisfies the condition (2.1) with Λ = 8 9 , β = 1 and ψ(ε) = ε 1 8 (for all ε ∈ [0, 1] ). Moreover, it is worth mentioning that 8 9 − 1 + ε ≤ 8 9 [1 + 9 8 (ε− 1)] ≤ 8 9 ε 9 8 ≤ 8 9 εε 1 8 . Thus we have the following two cases. (1) If x = 9 and y 6= 9, then d(fx,fy) = d(8, 1 2 y) = 8 − 1 2 y ≤ 8 9 (9 − 1 2 y) = 8 9 d(9,fy) ≤ 8 9 M(9,y) = (1 −ε)M(9,y) + ( 8 9 − 1 + ε)M(9,y) ≤ (1 −ε)M(9,y) + ( 8 9 − 1 + ε)[1 + ‖9‖ + ‖y‖ + ‖f9‖ + ‖fy‖] ≤ (1 −ε)M(9,y) + 8 9 εε 1 8 [1 + ‖9‖ + ‖y‖ + ‖f9‖ + ‖fy‖] Int. J. Anal. Appl. 17 (2) (2019) 281 (2) If x 6= 9 and y 6= 9, then d(fx,fy) = 1 2 (x−y) ≤ 8 9 (x−y) = 8 9 d(x,y) ≤ 8 9 M(x,y) = (1 −ε)M(x,y) + ( 8 9 − 1 + ε)M(x,y) ≤ (1 −ε)M(x,y) + ( 8 9 − 1 + ε)[1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖] ≤ (1 −ε)M(x,y) + 8 9 εε 1 8 [1 + ‖x‖ + ‖y‖ + ‖fx‖ + ‖fy‖] Hence, f satisfies all conditions of Theorem 2.1. This leads to f has a unique fixed point. Indeed, 0 is the fixed point for the mapping f. 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