Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 DOI 10.1186/s13663-017-0603-2 RESEARCH Open Access Some applications via fixed point results in partially ordered Sb-metric spaces GNV Kishore1* , KPR Rao2 , D Panthi3 , B Srinuvasa Rao4 and S Satyanaraya5 * Correspondence: [email protected] 1 Department of Mathematics, K L University, Vaddeswaram, Guntur-522 502, Andhra Pradesh, India Full list of author information is available at the end of the article Abstract In this paper we give some applications to integral equations as well as homotopy theory via fixed point theorems in partially ordered complete Sb -metric spaces by using generalized contractive conditions. We also furnish an example which supports our main result. Keywords: Sb -metric space; w-compatible pairs; Sb -completeness 1 Introduction Banach contraction principle in metric spaces is one of the most important results in fixed theory and nonlinear analysis in general. Since , when Stefan Banach [] formulated the concept of contraction and posted a famous theorem, scientists around the world have published new results related to the generalization of a metric space or with contractive mappings (see [–]). Banach contraction principle is considered to be the initial result of the study of fixed point theory in metric spaces. In the year , Bakhtin introduced the concept of b-metric spaces as a generalization of metric spaces []. Later several authors proved so many results on b-metric spaces (see [–]). Mustafa and Sims defined the concept of a generalized metric space which is called a G-metric space []. Sedghi, Shobe and Aliouche gave the notion of an S-metric space and proved some fixed point theorems for a self-mapping on a complete S-metric space []. Aghajani, Abbas and Roshan presented a new type of metric which is called Gb -metric and studied some properties of this metric []. Recently Sedghi et al. [] defined Sb -metric spaces using the concept of S-metric spaces []. The aim of this paper is to prove some unique fixed point theorems for generalized contractive conditions in complete Sb -metric spaces. Also, we give applications to integral equations as well as homotopy theory. Throughout this paper R, R+ and N denote the sets of all real numbers, non-negative real numbers and positive integers, respectively. First we recall some definitions, lemmas and examples. 2 Preliminaries Definition . ([]) Let X be a non-empty set. An S-metric on X is a function S : X → [, +∞) that satisfies the following conditions for each x, y, z, a ∈ X: (S): < S(x, y, z) for all x, y, z ∈ X with x = y = z = x, © The Author(s) 2017. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 2 of 14 (S): S(x, y, z) = if and only if x = y = z, (S): S(x, y, z) ≤ S(x, x, a) + S(y, y, a) + S(z, z, a) for all x, y, z, a ∈ X. Then the pair (X, S) is called an S-metric space. Definition . ([]) Let X be a non-empty set and b ≥ be a given real number. Suppose that a mapping Sb : X → [, ∞) is a function satisfying the following properties: (Sb ) < Sb (x, y, z) for all x, y, z ∈ X with x = y = z = x, (Sb ) Sb (x, y, z) = if and only if x = y = z, (Sb ) Sb (x, y, z) ≤ b(Sb (x, x, a) + Sb (y, y, a) + Sb (z, z, a)) for all x, y, z, a ∈ X. Then the function Sb is called an Sb -metric on X and the pair (X, Sb ) is called an Sb -metric space. Remark . ([]) It should be noted that the class of Sb -metric spaces is effectively larger than that of S-metric spaces. Indeed each S-metric space is an Sb -metric space with b = . The following example shows that an Sb -metric on X need not be an S-metric on X. Example . ([]) Let (X, S) be an S-metric space and S∗ (x, y, z) = S(x, y, z)p , where p > is a real number. Note that S∗ is an Sb -metric with b = (p–) . Also, (X, S∗ ) is not necessarily an S-metric space. Definition . ([]) Let (X, Sb ) be an Sb -metric space. Then, for x ∈ X, r > , we define the open ball BSb (x, r) and the closed ball BSb [x, r] with center x and radius r as follows, respectively: BSb (x, r) = y ∈ X : Sb (y, y, x) < r , BSb [x, r] = y ∈ X : Sb (y, y, x) ≤ r . Lemma . ([]) In an Sb -metric space, we have Sb (x, x, y) ≤ bSb (y, y, x) and Sb (y, y, x) ≤ bSb (x, x, y). Lemma . ([]) In an Sb -metric space, we have Sb (x, x, z) ≤ bSb (x, x, y) + b Sb (y, y, z). Definition . ([]) If (X, Sb ) is an Sb -metric space, a sequence {xn } in X is said to be: () Sb -Cauchy sequence if, for each > , there exists n ∈ N such that Sb (xn , xn , xm ) < for each m, n ≥ n . () Sb -convergent to a point x ∈ X if, for each > , there exists a positive integer n such that Sb (xn , xn , x) < or Sb (x, x, xn ) < for all n ≥ n , and we denote limn→∞ xn = x. Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 3 of 14 Definition . ([]) An Sb -metric space (X, Sb ) is called complete if every Sb -Cauchy sequence is Sb -convergent in X. Lemma . ([]) If (X, Sb ) is an Sb -metric space with b ≥ , and suppose that {xn } is Sb -convergent to x, then we have (i) Sb (y, x, x) ≤ lim inf Sb (y, y, xn ) ≤ lim sup Sb (y, y, xn ) ≤ bSb (y, y, x) n→∞ n→∞ b (ii) Sb (x, x, y) ≤ lim inf Sb (xn , xn , y) ≤ lim sup Sb (xn , xn , y) ≤ b Sb (x, x, y) n→∞ n→∞ b and for all y ∈ X. In particular, if x = y, then we have limn→∞ Sb (xn , xn , y) = . Now we prove our main results. 3 Results and discussions Definition . Let (X, Sb , ) be a partially ordered complete Sb -metric space which is said to be regular if every two elements of X are comparable, i.e., if x, y ∈ X ⇒ either x y or y x. Definition . Let (X, Sb , ) be a partially ordered complete Sb -metric space which is also regular; let f : X → X be a mapping. We say that f satisfies (ψ, φ)-contraction if there exist ψ, φ : [, ∞) → [, ∞) such that (..) (..) (..) (..) f is non-decreasing, ψ is continuous, monotonically non-decreasing and φ is lower semi-continuous, ψ(t) = = φ(t) if and only if t = , ψ(b Sb (fx, fx, fy)) ≤ ψ(Mfi (x, y)) – φ(Mfi (x, y)), ∀x, y ∈ X, x y, i = , , and Mf (x, y) = max Sb (x, x, y), Sb (x, x, fx), Sb (y, y, fy), Sb (x, x, fy), Sb (y, y, fx) , Mf (x, y) = max Sb (x, x, y), Sb (x, x, fx), Sb (y, y, fy), Sb (x, x, fy) + Sb (y, y, fx) , b Mf (x, y) = max Sb (x, x, y), Sb (x, x, fx) + Sb (y, y, fy) , b Sb (x, x, fy) + Sb (y, y, fx) . b Definition . Suppose that (X, ) is a partially ordered set and f is a mapping of X into itself. We say that f is non-decreasing if for every x, y ∈ X, x y implies that fx fy. () Theorem . Let (X, Sb , ) be an ordered complete Sb metric space, which is also regular, and let f : X → X satisfy (ψ, φ)-contraction with i = . If there exists x ∈ X with x fx , then f has a unique fixed point in X. Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 4 of 14 Proof Since f is a mapping from X into X, there exists a sequence {xn } in X such that xn+ = fxn , n = , , , , . . . . Case (i): If xn = xn+ , then xn is a fixed point of f . Case (ii): Suppose xn = xn+ ∀n. Since x fx = x and f is non-decreasing, it follows that x fx f x f x · · · f n x f n+ x · · · . Now ψ b Sb fx , fx , f x = ψ b Sb (fx , fx , fx ) ≤ ψ Mf (x , x ) – φ Mf (x , x ) , where (x , x , x ), S (x , x , fx ), S (x , x , fx ) S b b b Mf (x , x ) = max Sb (x , x , f x ), Sb (fx , fx , fx ) = max Sb (x , x , fx ), Sb fx , fx , f x , Sb x , x , f x . Therefore ψ b Sb fx , fx , f x ≤ ψ max Sb (x , x , fx ), Sb fx , fx , f x , Sb x , x , f x – φ max Sb (x , x , fx ), Sb fx , fx , f x , Sb x , x , f x ≤ ψ max Sb (x , x , fx ), Sb fx , fx , f x , Sb x , x , f x . By the definition of ψ, we have that ⎧ ⎪ ⎨ ⎫ S (x , x , fx ) ⎪ ⎬ b b Sb fx , fx , f x ≤ max b Sb (fx , fx , f x ) . ⎪ ⎪ ⎩ ⎭ S (x , x , f x ) b b But Sb x , x , f x ≤ bSb (x , x , fx ) + b Sb fx , fx , f x b b ≤ max Sb (x , x , fx ), Sb fx , fx , f x . b b From () we have that Sb fx , fx , f x ≤ max Sb (x , x , fx ), Sb fx , fx , f x . b b () Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 If S (fx , fx , f x ) b b Page 5 of 14 is maximum, we get a contradiction. Hence Sb fx , fx , f x ≤ Sb (x , x , fx ). b () Also ψ b Sb f x , f x , f x = ψ b Sb (fx , fx , fx ) ≤ ψ Mf (x , x ) – φ Mf (x , x ) , where Sb (fx , fx , f x ), Sb (fx , fx , f x ), Sb (f x , f x , f x ) Sb (fx , fx , f x ), Sb (f x , f x , f x ) = max Sb fx , fx , f x , Sb f x , f x , f x , Sb fx , fx , f x . Mf (x , x ) = max Therefore ψ b Sb f x , f x , f x Sb (fx , fx , f x ), Sb (f x , f x , f x ) ≤ ψ max Sb (fx , fx , f x ) Sb (fx , fx , f x ), Sb (f x , f x , f x ) – φ max Sb (fx , fx , f x ) Sb (fx , fx , f x ), Sb (f x , f x , f x ) . ≤ ψ max Sb (fx , fx , f x ) By the definition of ψ, we have that ⎫ ⎧ ⎪ ⎬ ⎨ b Sb (fx , fx , f x ) ⎪ Sb f x , f x , f x ≤ max b Sb (f x , f x , f x ) . ⎪ ⎪ ⎭ ⎩ S (fx , fx , f x ) b b But Sb fx , fx , f x b ≤ bSb fx , fx , f x + b Sb f x , f x , f x b ≤ max Sb fx , fx , f x , Sb f x , f x , f x . b b From () we have that Sb f x , f x , f x ≤ max Sb fx , fx , f x , Sb f x , f x , f x . b b () Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 If S (f x , f x , f x ) b b Page 6 of 14 is maximum, we get a contradiction. Hence Sb f x , f x , f x ≤ Sb fx , fx , f x b ≤ Sb (x , x , fx ). (b ) Continuing this process, we can conclude that Sb f n x , f n x , f n+ x ≤ Sb (x , x , fx ) (b )n → as n → ∞. That is, lim Sb f n x , f n x , f n+ x = . () n→∞ Now we prove that {f n x } is an Sb -Cauchy sequence in (X, Sb ). On the contrary, we suppose that {f n x } is not Sb -Cauchy. Then there exist > and monotonically increasing sequences of natural numbers {mk } and {nk } such that nk > mk . Sb f mk x , f mk x , f nk x ≥ () Sb f mk x , f mk x , f nk – x < . () and From () and (), we have ≤ Sb f mk x , f mk x , f nk x ≤ bSb f mk x , f mk x , f mk + x + b Sb f mk + x , f mk + x , f nk x . So that b ≤ b Sb f mk x , f mk x , f mk + x + b Sb f mk + x , f mk + x , f nk x . Letting k → ∞ and applying ψ on both sides, we have that ψ b ≤ lim ψ b Sb f mk + x , f mk + x , f nk x k→∞ = lim ψ b Sb (fxmk , fxmk , fxnk – ) k→∞ ≤ lim ψ Mf (xmk , xnk – ) – lim φ Mf (xmk , xnk – ) , k→∞ k→∞ () Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 7 of 14 where lim Mf (xmk , xnk – ) k→∞ ⎧ ⎫ m m n – m m m + ⎪ ⎨Sb (f k x , f k x , f k x ), Sb (f k x , f k x , f k x )⎪ ⎬ = lim max Sb (f nk – x , f nk – x , f nk x ), Sb (f mk x , f mk x , f nk x ) ⎪ ⎪ k→∞ ⎩ ⎭ Sb (f nk – x , f nk – x , f mk + x ) < lim max , , ,Sb f mk x , f mk x , f nk x , Sb f nk – x , f nk – x , f mk + x . k→∞ But lim Sb f k→∞ mk x , f mk x , f x ≤ lim nk k→∞ bSb (f mk x , f mk x , f nk – x ) < b. + b Sb (f nk – x , f nk – x , f nk x ) Also lim Sb f nk – x , f nk – x , f mk + x ≤ lim k→∞ k→∞ bSb (f nk – x , f nk – x , f mk x ) < b . + b Sb (f mk x , f mk x , f mk + x ) Therefore lim Mf (xmk , xnk – ) ≤ max , b, b k→∞ = b . From (), by the definition of ψ, we have that b ≤ b , which is a contradiction. Hence {f n x } is an Sb -Cauchy sequence in complete regular Sb metric spaces (X, Sb , ). By the completeness of (X, Sb ), it follows that the sequence {f n x } converges to α in (X, Sb ). Thus lim f n x = α = lim f n+ x . k→∞ k→∞ Since xn , α ∈ X and X is regular, it follows that either xn α or α xn . Now we have to prove that α is a fixed point of f . Suppose f α = α, by Lemma (.), we have that Sb (f α, f α, α) ≤ lim inf Sb (f α, f α, f n+ x ). n→∞ b Now from (..) and applying ψ on both sides, we have that ψ b Sb (f α, f α, α) ≤ lim inf ψ b Sb f α, f α, f n+ x n→∞ ≤ lim inf ψ Mf (α, xn ) – lim inf φ Mf (α, xn ) . n→∞ n→∞ () Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 8 of 14 Here Sb (α, α, xn ), Sb (α, α, f α), Sb (xn , xn , fxn ) Sb (α, α, fxn ), Sb (xn , xn , f α) ≤ lim sup max , Sb (α, α, f α), , , Sb (xn , xn , f α) lim inf Mf (α, xn ) = lim inf max n→∞ n→∞ n→∞ ≤ max Sb (α, α, f α), b Sb (α, α, f α) ≤ b Sb (f α, f α, α). Hence from () we have that ψ b Sb (f α, f α, α) ≤ ψ b Sb (α, α, f α) – lim inf φ Mf (α, xn ) ≤ ψ b Sb (f α, f α, α) , n→∞ which is a contradiction. So that α is a fixed point of f . Suppose that α ∗ is another fixed point of f such that α = α ∗ . Consider ψ b Sb α, α, α ∗ ≤ ψ Mf α, α ∗ – φ Mf α, α ∗ = ψ max Sb α, α, α ∗ , Sb α ∗ , α ∗ , α – φ max Sb α, α, α ∗ , Sb α ∗ , α ∗ , α ≤ ψ bSb α, α, α ∗ , which is a contradiction. Hence α is a unique fixed point of f in (X, Sb ). Example . Let X = [, ] and S : X ×X ×X → R+ by Sb (x, y, z) = (|y+z –x|+|y–z|) and by a b ⇐⇒ a ≤ b, then (X, Sb , ) is a complete ordered Sb -metric space with b = . Define f : X → X by f (x) = x√ . Also define ψ, φ : R+ → R+ by ψ(t) = t and φ(t) = t . ψ b Sb (fx, fx, fy) = b |fx + fy – fx| + |fx – fy| x y = b √ – √ b Sb (x, x, y) b ≤ Mf (x, y) ≤ ψ Mf (x, y) – φ Mf (x, y) , = where Mf (x, y) = max Sb (x, x, y), Sb (x, x, fx), Sb (y, y, fy), Sb (x, x, fy), Sb (y, y, fx) . Hence, all the conditions of Theorem . are satisfied and is a unique fixed point of f . Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 9 of 14 Theorem . Let (X, Sb , ) be an ordered complete Sb metric space, and let f : X → X satisfy (ψ, φ)-contraction with i = or . If there exists x ∈ X with x fx , then f has a unique fixed point in X. Proof Follows along similar lines of Theorem . if we take Mf (x, y) or Mf (x, y) in place of Mf (x, y) in Theorem .. Theorem . Let (X, Sb , ) be an ordered complete Sb metric space, and let f : X → X satisfy b Sb (fx, fx, fy) ≤ Mfi (x, y) – ϕ Mfi (x, y) , where ϕ : [, ∞) → [, ∞) and i = or or . If there exists x ∈ X with x fx , then f has a unique fixed point in X. Proof The proof follows from Theorems . and . by taking ψ(t) = t and φ(t) = ϕ(t). Theorem . Let (X, Sb , ) be an ordered complete Sb metric space, and let f : X → X satisfy Sb (fx, fx, fy) ≤ λMfi (x, y), where λ ∈ [, b ) and i = , , . If there exists x ∈ X with x fx , then f has a unique fixed point in X. 3.1 Application to integral equations In this section, we study the existence of a unique solution to an initial value problem as an application to Theorem .. Theorem . Consider the initial value problem x (t) = T t, x(t) , t ∈ I = [, ], x() = x , () where T : I × [ x , ∞) → [ x , ∞) and x ∈ R. Then there exists a unique solution in C(I, [ x , ∞)) for initial value problem (). Proof The integral equation corresponding to initial value problem () is x(t) = x + b t T s, x(s) ds. Let X = C(I, [ x , ∞)) and Sb (x, y, z) = (|y+z–x|+|y–z|) forx, y ∈ X. Define ψ, φ : [, ∞) → [, ∞) by ψ(t) = t, φ(t) = t . Define f : X → X by f (x)(t) = x + b t T s, x(s) ds. () Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 10 of 14 Now ψ b Sb fx(t), fx(t), fy(t) = b fx(t) + fy(t) – fx(t) + fx(t) – fy(t) = b fx(t) – fy(t) t t b = T s, x(s) ds – y – b T s, y(s) ds x + b b = x(t) – y(t) = S(x, x, y) ≤ Mf (x, y) = ψ Mf (x, y) – φ Mf (x, y) , where Mf (x, y) = max Sb (x, x, y), Sb (x, x, fx), Sb (y, y, fy), Sb (x, x, fy), Sb (y, y, fx) . It follows from Theorem . that f has a unique fixed point in X. 3.2 Application to homotopy In this section, we study the existence of a unique solution to homotopy theory. Theorem . Let (X, Sb ) be a complete Sb -metric space, U be an open subset of X and U be a closed subset of X such that U ⊆ U. Suppose that H : U × [, ] → X is an operator such that the following conditions are satisfied: (i) x = H(x, λ) for each x ∈ ∂U and λ ∈ [, ] (here ∂U denotes the boundary of U in X), (ii) ψ(b Sb (H(x, λ), H(x, λ), H(y, λ))) ≤ ψ(Sb (x, x, y)) – φ(Sb (x, x, y)) ∀x, y ∈ U and λ ∈ [, ], where ψ : [, ∞) → [, ∞) is continuous, non-decreasing and φ : [, ∞) → [, ∞) is lower semi-continuous with φ(t) > for t > , (iii) there exists M ≥ such that Sb H(x, λ), H(x, λ), H(x, μ) ≤ M|λ – μ| for every x ∈ U and λ, μ ∈ [, ]. Then H(·, ) has a fixed point if and only if H(·, ) has a fixed point. Proof Consider the set A = λ ∈ [, ] : x = H(x, λ) for some x ∈ U . Since H(·, ) has a fixed point in U, we have that ∈ A. So that A is a non-empty set. We will show that A is both open and closed in [, ], and so, by the connectedness, we have that A = [, ]. As a result, H(·, ) has a fixed point in U. First we show that A is closed in [, ]. To see this, let {λn }∞ n= ⊆ A with λn → λ ∈ [, ] as n → ∞. Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 11 of 14 We must show that λ ∈ A. Since λn ∈ A for n = , , , . . . , there exists xn ∈ U with xn = H(xn , λn ). Consider Sb (xn , xn , xn+ ) = Sb H(xn , λn ), H(xn , λn ), H(xn+ , λn+ ) ≤ bSb H(xn , λn ), H(xn , λn ), H(xn+ , λn ) + b Sb H(xn+ , λn ), H(xn+ , λn ), H(xn+ , λn+ ) ≤ Sb H(xn , λn ), H(xn , λn ), H(xn+ , λn ) + M|λn – λn+ |. Letting n → ∞, we get lim Sb (xn , xn , xn+ ) ≤ lim Sb H(xn , λn ), H(xn , λn ), H(xn+ , λn ) + . n→∞ n→∞ Since ψ is continuous and non-decreasing, we obtain lim ψ b Sb (xn , xn , xn+ ) ≤ lim ψ b Sb H(xn , λn ), H(xn , λn ), H(xn+ , λn ) n→∞ n→∞ ≤ lim ψ Sb (xn , xn , xn+ ) – φ Sb (xn , xn , xn+ ) . n→∞ By the definition of ψ, it follows that lim b – Sb (xn , xn , xn+ ) ≤ . n→∞ So that lim Sb (xn , xn , xn+ ) = . n→∞ () Now we prove that {xn } is an Sb -Cauchy sequence in (X, dp ). On the contrary, suppose that {xn } is not Sb -Cauchy. There exists > and monotone increasing sequences of natural numbers {mk } and {nk } such that nk > mk , Sb (xmk , xmk , xnk ) ≥ () Sb (xmk , xmk , xnk – ) < . () and From () and (), we obtain ≤ Sb (xmk , xmk , xnk ) ≤ bSb (xmk , xmk , xmk + ) + b Sb (xmk + , xmk + , xnk ). Letting k → ∞ and applying ψ on both sides, we have that ψ b ≤ lim ψ b Sb (xmk + , xmk + , xnk ) . n→∞ () Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 12 of 14 But lim ψ b Sb (xmk + , xmk + , xnk ) n→∞ = lim ψ Sb b H(xmk + , λmk + ), H(xmk + , λmk + ), H(xnk , λnk ) n→∞ ≤ lim ψ Sb (xmk + , xmk + , xnk ) – φ Sb (xmk + , xmk + , xnk ) . n→∞ It follows that lim b – Sb (xmk + , xmk + , xnk ) ≤ . n→∞ Thus lim Sb (xmk + , xmk + , xnk ) = . n→∞ Hence from () and the definition of ψ, we have that ≤ , which is a contradiction. Hence {xn } is an Sb -Cauchy sequence in (X, Sb ) and, by the completeness of (X, Sb ), there exists α ∈ U with lim xn = α = lim xn+ , n→∞ () n→∞ ψ b Sb H(α, λ), H(α, λ), α ≤ lim inf ψ b Sb H(α, λ), H(α, λ), H(xn , λ) n→∞ ≤ lim inf ψ Sb (α, α, xn ) – φ Sb (α, α, xn ) n→∞ = . It follows that α = H(α, λ). Thus λ ∈ A. Hence A is closed in [, ]. Let λ ∈ A. Then there exists x ∈ U with x = H(x , λ ). Since U is open, there exists r > such that BSb (x , r) ⊆ U. Choose λ ∈ (λ – , λ + ) such that |λ – λ | ≤ Mn < . Then, for x ∈ Bp (x , r) = {x ∈ X/Sb (x, x, x ) ≤ r + b Sb (x , x , x )}, Sb H(x, λ), H(x, λ), x = Sb H(x, λ), H(x, λ), H(x , λ ) ≤ bSb H(x, λ), H(x, λ), H(x, λ ) + b Sb H(x, λ ), H(x, λ ), H(x , λ ) ≤ bM|λ – λ | + b Sb H(x, λ ), H(x, λ ), H(x , λ ) ≤ b + b Sb H(x, λ ), H(x, λ ), H(x , λ ) . n– M Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 13 of 14 Letting n → ∞, we obtain Sb H(x, λ), H(x, λ), x ≤ b Sb H(x, λ ), H(x, λ ), H(x , λ ) . Since ψ is continuous and non-decreasing, we have ψ Sb H(x, λ), H(x, λ), x ≤ ψ b Sb H(x, λ), H(x, λ), x ≤ ψ b Sb H(x, λ ), H(x, λ ), H(x , λ ) ≤ ψ Sb (x, x, x ) – φ Sb (x, x, x ) ≤ ψ Sb (x, x, x ) . Since ψ is non-decreasing, we have Sb H(x, λ), H(x, λ), x ≤ Sb (x, x, x ) ≤ r + b Sb (x , x , x ). Thus, for each fixed λ ∈ (λ – , λ + ), H(·, λ) : Bp (x , r) → Bp (x , r). Since also (ii) holds and ψ is continuous and non-decreasing and φ is continuous with φ(t) > for t > , then all the conditions of Theorem (.) are satisfied. Thus we deduce that H(·, λ) has a fixed point in U. But this fixed point must be in U since (i) holds. Thus λ ∈ A for any λ ∈ (λ – , λ + ). Hence (λ – , λ + ) ⊆ A and therefore A is open in [, ]. For the reverse implication, we use the same strategy. Corollary . Let (X, p) be a complete partial metric space, U be an open subset of X and H : U × [, ] → X with the following properties: () x = H(x, t) for each x ∈ ∂U and each λ ∈ [, ] (here ∂U denotes the boundary of U in X), () there exist x, y ∈ U and λ ∈ [, ], L ∈ [, b ) such that Sb H(x, λ), H(x, λ), H(y, μ) ≤ LSb (x, x, y), () there exists M ≥ such that Sb H(x, λ), H(x, λ), H(x, μ) ≤ M|λ – μ| for all x ∈ U and λ, μ ∈ [, ]. If H(·, ) has a fixed point in U, then H(·, ) has a fixed point in U. Proof Proof follows by taking ψ(x) = x, φ(x) = x – Lx with L ∈ [, b ) in Theorem (.). 4 Conclusions In this paper we conclude some applications to homotopy theory and integral equations by using fixed point theorems in partially ordered Sb -metric spaces. Kishore et al. Fixed Point Theory and Applications (2017) 2017:10 Page 14 of 14 Acknowledgements Authors are thankful to referees for their valuable suggestions. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, K L University, Vaddeswaram, Guntur-522 502, Andhra Pradesh, India. 2 Department of Mathematics, Acharya Nagarjuna University, Guntur-522 502, Andhra Pradesh, India. 3 Department of Mathematics, Nepal Sanskrit University, Valmeeki Campus, Exhibition Road, Kathmandu, Nepal. 4 Department of Mathematics, Dr. B. R. Ambedkar University, Srikakulam-532 410, Andhra Pradesh, India. 5 Department of Computing, Adama Science and Technology University, Adama, Ethiopia. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 24 April 2017 Accepted: 22 May 2017 References 1. Banach, S: Theorie des Operations Lineaires. Manograic Mathematic Zne, Warsaw (1932) 2. 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