Bull. Malays. Math. Sci. Soc. DOI 10.1007/s40840-016-0347-x

A Generalized Contraction Principle Under w-Distance for a Partially Ordered Metric Space Rakesh Batra1 · Sachin Vashistha2

Received: 23 February 2014 / Revised: 22 June 2015 © Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2016

Abstract Fixed point results for generalized weak contractions under w-distance are proved using discontinuous control functions that are more relaxed than functions used in related work for metric spaces. Later, we apply our theory to coupled coincidence point problems and existence of solution of Fredholm type integral equation. We present examples to justify our claims. Keywords w-Distance · Complete metric space · Mixed g-monotone property · Fixed point · Coincidence point Mathematics Subject Classification

Primary 47H10; Secondary 54H25 · 55M20

1 Introduction and Preliminaries Applications of fixed point theorems are very important in diverse disciplines of mathematics, statistics, chemistry, biology, computer science, engineering, and economics in dealing with problems arising in approximation theory, potential theory, game theory, mathematical economics, theory of differential equations, theory of integral equations, theory of matrix equations, etc. For example, fixed point theorems are extremely useful when it comes to prove the existence of various types of Nash equilibria in economics. One of the very popular tools of fixed point theory is the Banach contraction principle

Communicated by Tomonari Suzuki.

B

Rakesh Batra [email protected]

1

Hans Raj College, University of Delhi, Delhi 110007, India

2

Hindu College, University of Delhi, Delhi 110007, India

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which first appeared in 1922. It states that if (X, d) is a complete metric space and T : X → X is a contraction mapping, that is, d(T x, T y) ≤ kd(x, y) for all x, y ∈ X , where k is a non-negative number such that k < 1, then T has a unique fixed point. Several mathematicians have been dedicated to improvement and generalization of this principle. Fixed point theory in partially ordered metric spaces is of relatively recent origin. An early result in this direction is due to Turinici [18], in which fixed point problems were studied in partially ordered uniform spaces. Later, in 2004, Ran and Reurings [17] showed the existence of fixed points of nonlinear contraction mappings in metric spaces endowed with a partial ordering and presented applications of their results to matrix equations. Subsequently, Nieto and Rodríguez-López [16] extended the result of Ran and Reurings [17] for nondecreasing mappings and applied their results to get a unique solution for a first-order differential equation. Bhaskar and Lakshmikantham [5] introduced the notion of a coupled fixed point of a mapping F from X × X into X . They established some coupled fixed point results and applied their results to the study of existence and uniqueness of solution for a periodic bound´ c [15] introduced the concept of coupled ary value problem. Lakshmikantham and Ciri´ coincidence points and proved coupled coincidence and coupled common fixed point results for mappings F from X × X into X and g from X into X satisfying nonlinear contraction in ordered metric space. The studies of asymmetric structures and their applications in mathematics are important. One of the types of asymmetric structures on a metric space was introduced by Kada et al. [13] in 1996 known as a w-distance, and he proved some fixed point theorems using it. Since then, many fixed point results have been developed by different authors using w-distance on metric spaces or a generalized w-distance such as c-distance on cone metric spaces. For more study in this area, one may refer to [1–4,6]. Weak contraction was studied in partially ordered metric spaces by Harjani et al. [10]. In a recent result by Choudhury et al. [8], a generalization of the above result to a coincidence point theorem has been done using three control functions. Later, in 2013, Choudhury et al. [9] proved coincidence point results by assuming a weak contraction inequality with three control functions, two of which are not continuous. The results were obtained under two sets of additional conditions. By R, Q, N, and χ A , we shall mean the set of real numbers, the set of rational numbers , the set of natural numbers, and the characteristic function of a set A defined by χ (x) = 1 if x ∈ A and χ (x) = 0 otherwise, respectively. Following basic definitions can be referred in [5,7,9,11,12,14,15]. Let T and S be any two self maps on a metric space (X, d). T and S are said to be compatible if limn→∞ d(T Sxn , ST xn ) = 0 whenever {xn } is a sequence in X such that limn→∞ T xn = limn→∞ Sxn = t for some t ∈ X and weakly compatible if T and S commute at their coincidence points. Let  be a partial order relation on X and g be a self mapping on X . Then T is said to be g-nondecreasing if gx  gy ⇒ T x  T y for all x, y ∈ X and g-nonincreasing if gx  gy ⇒ T x  T y for all x, y ∈ X . Let F : X × X → X be any mapping. F is said to have mixed monotone property if for any x, y ∈ X we have x1 , x2 ∈ X, x1  x2 ⇒ F(x1 , y)  F(x2 , y) and y1 , y2 ∈ X, y1  y2 ⇒ F(x, y1 )  F(x, y2 ) and mixed g-monotone property if for any x, y ∈ X we have x1 , x2 ∈ X, gx1  gx2 ⇒ F(x1 , y)  F(x2 , y) and y1 , y2 ∈ X, gy1  gy2 ⇒ F(x, y1 )  F(x, y2 ). An element (x, y) ∈ X × X is

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called a coupled fixed point of F if F(x, y) = x and F(y, x) = y and a coupled coincidence point of F and g if F(x, y) = gx and F(y, x) = gy. The mappings F and g are said to commute if g F(x, y) = F(gx, gy) for all x, y ∈ X , compatible if limn→∞ d(F(gxn , gyn ), g F(xn , yn )) = 0, limn→∞ d(F(gyn , gxn ), g F(yn , xn )) = 0 whenever {xn } and {yn } are sequences in X such that limn→∞ F(xn , yn ) = limn→∞ gxn = x and limn→∞ F(yn , xn ) = limn→∞ gyn = y for some x, y ∈ X and weakly compatible if they commute at their coupled coincidence points. It is simple to observe that compatibility implies weakly compatibility. X is called regular if for any nonincreasing sequence {xn } in X such that xn → x ∈ X we have x  xn for all n and for any nondecreasing sequence {xn } in X such that xn → x ∈ X we have xn  x for all n. Next we give the notation of w-distance of Kada et al. [13] with some properties. Definition 1.1 ([13]) Let (X, d) be a metric space. A function p : X × X → [0, ∞) is called a w-distance on X if the following conditions hold: (w1) p(x, z) ≤ p(x, y) + p(y, z) for all x, y, z ∈ X , (w2) p(x, .) is lower semi-continuous for all x ∈ X . That is, if x ∈ X and yn → y in X then p(x, y) ≤ lim inf p(x, yn ). (w3) For all  > 0, there exists δ > 0 such that p(z, x) ≤ δ and p(z, y) ≤ δ imply d(x, y) ≤ . Example 1.2 (i) Every metric d is a w-distance on (X, d) as d(x, y) ≥ 0 for all x, y ∈ X , (w1) holds because it is nothing but the triangular inequality of the metric d on X , (w2) follows by taking z = yn in the inequality |d(x, y) − d(x, z)| ≤ d(y, z) that holds for all x, y, z ∈ X , and for (w3), one can use triangular inequality and take δ = /2. (ii) For any positive real number λ, p(x, y) = λd(x, y) is a w-distance on a metric space (X, d) and argument is similar to what is applied to (i). (iii) Consider X = [0, ∞) together with the Euclidean metric d on X . Let f : X → X be any given mapping. Define a mapping p : X × X → R by p(x, y) = f (x)+ y for all x, y ∈ X . Then p is a w-distance on X as p : X × X → [0, ∞), (w1) holds because for any x, y, z ∈ X we have p(x, y) + p(y, z) − p(x, z) = f (x) + y + f (y) + z − f (x) − z = y + f (y) ≥ 0, (w2) holds since for any sequence yn → y in X we have for any x ∈ X , f (x) + yn → f (x) + y and hence p(x, y) = f (x) + y = limn→∞ ( f (x) + yn ) = lim inf( f (x) + yn ) = lim inf p(x, yn ). Further, for any  > 0, there exists δ = /2 > 0 such that p(z, x) = f (z) + x ≤ δ and p(z, y) = f (z) + y ≤ δ together imply d(x, y) = |x − y| = |( f (z) + x) − ( f (z) + y)| ≤ | f (z) + x| + | f (z) + y| = p(z, x) + p(z, y) ≤ 2δ = . So (w3) holds. (iv) Consider X = [0, ∞) together with the Euclidean metric d on X . Define a mapping p : X × X → R by p(x, y) = y for all x, y ∈ X . Then p is a w-distance on X as it is a particular case of w-distance defined in (iii) for f = 0. (v) p(x, y) = d(u, y) is a w-distance on a metric space (X, d) where u is a fixed point in X . Clearly p(x, y) = d(u, y) ≥ 0 for all x, y ∈ X . Further for any x, y, z ∈ X , we have p(x, y) = d(u, y) ≤ d(u, z)+d(u, y) = p(x, z)+ p(z, y). So (w1) holds. (w2) holds because of continuity of the function d(u, .). Further,

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for any  > 0, there exists δ = /2 > 0 such that p(z, x) = d(u, x) ≤ δ and p(z, y) = d(u, y) ≤ δ together imply d(x, y) = d(x, u) + d(u, y) ≤ 2δ = . So (w3) holds. Lemma 1.3 ([13]) Let (X, d) be a metric space and p be a w-distance on X . Let {xn } and {yn } be sequences in X and x, y, z ∈ X . Suppose that u n and vn are sequences in [0, ∞) converging to 0. Then the following hold: (i) If p(xn , y) ≤ u n and p(xn , z) ≤ vn , then y = z. In particular if p(x, y) = 0 and p(x, z) = 0 then y = z. (ii) If p(xn , yn ) ≤ u n and p(xn , z) ≤ vn , then yn converges to z. (iii) If p(xn , xm ) ≤ u n for m > n, then {xn } is a Cauchy sequence in X . (iv) If p(y, xn ) ≤ u n , then {xn } is a Cauchy sequence in X . Lemma 1.4 ([13]) Let p be a w-distance on metric space (X, d) and {xn } be a sequence in X such that for each  > 0 there exist N ∈ N such that m > n > N implies p(xn , xm ) <  (or limm,n p(xn , xm ) = 0 ) then {xn } is a Cauchy sequence. In this paper, we relax some conditions on control functions used by Choudhury et al. in [9] and use them to prove coincidence point results using w-distance in weak contraction inequalities. Since every metric is a w-distance, therefore, our results may be applied even when we use metric in weak contraction inequalities. Thus all results of Choudhury et al. in [9] hold even after ignoring some conditions on the control function used by them and are implied by our results for such control functions.

2 Main Results We denote by , the set of all functions ψ : [0, ∞) → [0, ∞) such that ψ is continuous and nondecreasing and ψ(t) = 0 ⇒ t = 0. Further, by  we denote the set of all functions α : [0, ∞) → [0, ∞) such that α is bounded on any bounded interval in [0, ∞). Theorem 2.1 Let (X, , d) be a partially ordered metric space equipped with a wdistance p and T, g be two self maps on X satisfying the following conditions: (i) T (X ) ⊆ g(X ) ,T is g-nondecreasing, there exists an x0 ∈ X with gx0  T x0 or T x0  gx0 and either T and g are continuous, compatible with (X, d) complete or g(X ) is complete and (X, d) is regular. (ii) There exist ψ ∈  and ϕ, θ ∈  such that for all s, t ∈ [0, ∞), we have ψ(s) ≤ ϕ(t) ⇒ s ≤ t. Further, for any sequence {xn } in [0, ∞) with xn → t > 0, we have ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0 and for all x, y ∈ X with gx  gy or gx  gy, ψ( p(T x, T y)) ≤ ϕ( p(gx, gy)) − θ ( p(gx, gy)). Then T and g have a coincidence point in X . Proof Let us first assume that (X, d) is complete, T and g are continuous and the pair (T, g) is compatible. Also assume that T x0  gx0 . Since T (X ) ⊆ g(X ), one can find x1 ∈ X such that T x0 = gx1 . Again, we can choose x2 ∈ X such that T x1 = gx2 . Continuing like this we can find xn ∈ X such that T xn = gxn+1 for all n ≥ 0. Further,

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gx1  gx0 and T is g-nondecreasing implies T x1  T x0 or gx2  gx1 and hence T x2  T x1 . Continuing this process, we obtain gxn  gxn+1 and T xn  T xn+1 for all n ≥ 0. Let Q n = p(gxn , gxn+1 ) and Rn = p(gxn+1 , gxn ) for all n ≥ 0. Then ψ( p(gxn+1 , gxn+2 )) = ψ( p(T xn , T xn+1 )) ≤ ϕ( p(gxn , gxn+1 )) − θ( p(gxn , gxn+1 )), ψ( p(gxn+2 , gxn+1 )) = ψ( p(T xn+1 , T xn )) ≤ ϕ( p(gxn+1 , gxn )) − θ( p(gxn+1 , gxn )).

That is, for all n ≥ 0, ψ(Q n+1 ) ≤ ϕ(Q n ) − θ (Q n ) and ψ(Rn+1 ) ≤ ϕ(Rn ) − θ (Rn ). This implies that ψ(Q n+1 ) ≤ ϕ(Q n ) and ψ(Rn+1 ) ≤ ϕ(Rn ) and hence Q n+1 ≤ Q n , Rn+1 ≤ Rn for all n ≥ 0. So, there exist q ≥ 0 and r ≥ 0 such that Q n = p(gxn , gxn+1 ) → q and Rn = p(gxn+1 , gxn ) → r as n → ∞. Then for all n ≥ 0, we have ψ(q) ≤ lim sup ϕ(Q n ) + lim sup(−θ (Q n )) and ψ(r ) ≤ lim sup ϕ(Rn )+lim sup(−θ (Rn )). That is, ψ(q) ≤ lim sup ϕ(Q n )−lim inf θ (Q n ) and ψ(r ) ≤ lim sup ϕ(Rn ) − lim inf θ (Rn ). or ψ(q) − lim sup ϕ(Q n ) + lim inf θ (Q n ) ≤ 0 and ψ(r ) − lim sup ϕ(Rn ) + lim inf θ (Rn ) ≤ 0. This gives a contradiction unless q = 0 and r = 0. Thus Q n = p(gxn , gxn+1 ) → 0 and Rn = p(gxn+1 , gxn ) → 0 as n → ∞. Next, we show that lim p(gxm , gxn ) = 0. m,n

(2.1)

Suppose (2.1) does not hold. Then there exists an  > 0 such that for all positive integers p, we get two positive integers n p , m p with n p > m p > p and p(gxm p , gxn p ) ≥ . Also, we can find a positive integer k such that p(gxn , gxn+1 ) <  for all n ≥ k. For the choice of p = k and n = m k , we get two positive integers n k , m k with n k > m k > k, p(gxm k , gxn k ) ≥ 

(2.2)

and p(gxm k , gxm k +1 ) < . Clearly n k = m k +1. We can assume that n k is the smallest positive integer such that (2.2) holds. Then p(gxm k , gxr ) <  for r ∈ {m k + 1, m k + 2, . . . n k − 1}. In particular, p(gxm k , gxn k −1 ) < . Now  ≤ p(gxm k , gxn k ) ≤ p(gxm k , gxn k −1 ) + p(gxn k −1 , gxn k ) <  + p(gxn k −1 , gxn k ). Letting k → ∞, we get limk→∞ p(gxm k , gxn k ) = . Again p(gxm k , gxn k ) ≤ p(gxm k , gxm k +1 ) + p(gxm k +1 , gxn k +1 ) + p(gxn k +1 , gxn k ) and p(gxm k +1 , gxn k +1 ) ≤ p(gxm k +1 , gxm k ) + p(gxm k , gxn k ) + p(gxn k , gxn k +1 ). Letting k → ∞ in above inequalities, we get limk→∞ p(gxm k +1 , gxn k +1 ) = . Since n k > m k , we have gxn k  gxm k . So we get ψ( p(gxm k +1 , gxn k +1 )) = ψ( p(T xm k , T xn k )) ≤ ϕ( p(gxm k , gxn k )) − θ ( p(gxm k , gxn k ))

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This gives ψ() ≤ lim sup ϕ( p(gxm k , gxn k )) + lim sup(−θ ( p(gxm k , gxn k ))) = lim sup ϕ( p(gxm k , gxn k )) − lim inf θ ( p(gxm k , gxn k )). Hence ψ()−lim sup ϕ( p(gxm k , gxn k ))+lim inf θ ( p(gxm k , gxn k )) ≤ 0. Thus we get a contradiction. So (2.1) must be true. By Lemma 1.4, the sequence {gxn } is a Cauchy sequence in X . Since X is a complete metric space, therefore, there exists an x ∈ X such that limn→∞ T xn = limn→∞ gxn = x. Since the pair (T, g) is compatible, we have limn→∞ d(T gxn , gT xn ) = 0. By triangular inequality, d(gx, T gxn ) ≤ d(gx, gT xn ) + d(gT xn , T gxn ) Taking limit n → ∞ in above inequality, and using continuities of T and g, we get d(gx, T x) = 0. That is T x = gx. Thus T and g have a coincidence point at x. Now, we consider the case that (X, d) is regular and g(X ) is a complete subspace of X . As in the first case, choose x0 ∈ X satisfying T x0  gx0 and by using it, construct a sequence {xn } in X satisfying T xn = gxn+1 such that {gxn } is a nonincreasing Cauchy sequence satisfying (2.1). Since g(X ) is complete, therefore, there exists x ∈ X such that gxn → gx as n → ∞. By regularity of X we have gxn ≥ gx for all n ≥ 0. This implies that, for all n ≥ 1 ψ( p(gxn , T x)) = ψ( p(T xn−1 , T x)) ≤ ϕ( p(gxn−1 , gx)) − θ ( p(gxn−1 , gx)) ≤ ϕ( p(gxn−1 , gx)) and hence p(gxn , T x) ≤ p(gxn−1 , gx). Since {gxn } satisfies (2.1), for any k ∈ N, there exists a positive integer m k such that p(gxm k , gxn ) < 1/k for all n > m k . This implies that for all integers k ≥ 1, p(gxm k +1 , T x) ≤ p(gxm k , gx) ≤ lim inf p(gxm k , gxn ) ≤ 1/k = αk and also p(gxm k , T x) ≤ p(gxm k , gxm k +1 ) + p(gxm k +1 , T x) ≤ p(gxm k , gxm k +1 ) + αk = βk Now, using Lemma 1.3(i) and the fact that αk → 0 and βk → 0 as k → ∞, we get T x = gx. That is, T and g have coincidence point at x.

Corollary 2.2 In the hypothesis of Theorem 2.1, if we replace (ii) by any one of the following: (ii a ) There exists ϕ ∈  such that ϕ(0) = 0 and for all x, y ∈ X with gx  gy or gx  gy, we have p(T x, T y) ≤ ϕ( p(gx, gy)). Further, assume that for any sequence {xn } in [0, ∞) with xn → t > 0, we have lim sup ϕ(xn ) < t. (ii b ) There exist ψ ∈  and k ∈ [0, 1) such that for all x, y ∈ X with gx  gy or gx  gy, we have ψ( p(T x, T y)) ≤ k ψ( p(gx, gy)).

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(ii c ) There exists an increasing function ψ ∈  and a function θ ∈  such that for all x, y ∈ X with gx  gy or gx  gy, we have ψ( p(T x, T y)) ≤ ψ( p(gx, gy))− θ ( p(gx, gy)) and for any sequence {xn } in [0, ∞) with xn → t > 0, we have lim inf θ (xn ) > 0. (ii d ) There exists θ ∈  such that for all x, y ∈ X with gx  gy or gx  gy, we have p(T x, T y) ≤ p(gx, gy)) − θ ( p(gx, gy)) and for any sequence {xn } in [0, ∞) with xn → t > 0, we have lim inf θ (xn ) > 0. (ii e ) There exists k ∈ [0, 1) such that for all x, y ∈ X with gx  gy or gx  gy, we have p(T x, T y) ≤ k p(gx, gy). Then T and g have a coincidence point in X . Proof (ii a ) Take ψ as an identity mapping on [0, ∞) and θ = 0 in Theorem 2.1. Let ψ(t) ≤ ϕ(s) for t, s ≥ 0. This implies t ≤ ϕ(s) for t, s ≥ 0. If s > 0 then take xn = s. So we have ϕ(s) < s whenever s > 0. Since ϕ(0) = 0 we get t ≤ ϕ(s) ≤ s for all s ≥ 0. Conclusion follows now by Theorem 2.1. (ii b ) Take θ = 0 and ϕ = k ψ in Theorem 2.1. (ii c ) Take ϕ = ψ in Theorem 2.1. (ii d ) Take ϕ = ψ as an identity mapping on [0, ∞) in Theorem 2.1. (ii e ) Take θ = 0, ϕ(t) = kt for all t ≥ 0 and ψ as an identity mapping on [0, ∞), in Theorem 2.1.

Remark 2.3 In Theorem 2.1, compatibility of T and g can be substituted with commutativity because commutativity of T and g implies compatibility of the pair (T, g). Theorem 2.4 In addition to the hypotheses of Theorem 2.1 and Corollary 2.2 as stated above, suppose that for any x, y ∈ X there exists u ∈ X such that T u is comparable to both T x and T y, then, there is a unique point of coincidence. Further, if the pair (T, g) is weakly compatible, then, T and g have a unique common fixed point. Proof Let x and y be coincidence points of T and g. That is gx = T x and gy = T y. By assumption, there exists u ∈ X such that T u is comparable with T x and T y. Let T u  T x (proof is similar for the other case and can be outlined with a little adjustment). Let u 0 = u and since T (X ) ⊆ g(X ), there exists u 1 ∈ X such that T u 0 = gu 1 . Clearly gu 1  gx. Since T is g-increasing, we have T u 1  T x = gx. Let T u 1 = gu 2 for some u 2 ∈ X . Then gu 2  gx. Inductively, one can find a sequence {u n } such that gu n+1 = T u n for all n ≥ 0 and gu n  gx for all n ≥ 1. Let Rn = p(gx, gu n ), n ≥ 1. Since gx and gu n are comparable for all n ≥ 1, we have for all n ≥ 1, ψ(Rn+1 ) = ψ( p(gx, gu n+1 )) = ψ( p(T x, T u n )) ≤ ϕ( p(gx, gu n )) − θ ( p(gx, gu n )) = ϕ(Rn ) − θ (Rn ) ≤ ϕ(Rn ). This implies that Rn+1 ≤ Rn for all n ≥ 1. So {Rn } is a nonincreasing sequence of non-negative real numbers. Then as in the proof of Theorem 2.1, we can show that limn→∞ Rn = 0. That is, limn→∞ p(gx, gu n ) = 0. Similarly limn→∞ p(gu n , gx) = 0, limn→∞ p(gy, gu n ) = 0 and limn→∞ p(gu n , gy) = 0. By triangular inequality, p(gx, gx) ≤ p(gx, gu n ) + p(gu n , gx) for all n ≥ 1. Letting n → ∞, we get p(gx, gx) = 0. Again, p(gx, gy) ≤ p(gx, gu n ) + p(gu n , gy) for all n ≥ 1. Letting n → ∞ we get p(gx, gy) = 0. Lemma 1.3(i) now implies that gx = gy. Thus, there is a unique point of coincidence. Further, assume that the pair (T, g) is weakly

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compatible. Since gx = T x, therefore, ggx = gT x = T gx. Let gx = z. Then we have gz = T z. Due to uniqueness of point of coincidence we get gx = gz. It follows that z = gz = T z. Thus, z is a common fixed point of T and g. To prove uniqueness, let w be a common fixed point of T and g. Then T w = gw = w. By uniqueness of point of coincidence, we have z = gz = gw = w. Hence, T and g have a unique common fixed point in X .

Example 2.5 Consider X = [0, ∞), together with the Euclidean metric d and let p(x, y) = y for all x, y ∈ X . Then (X, d, ≤) is a partially ordered complete metric space with w-distance p as proved in Example 1.2(iv). Define self maps T and g on X by T x = x 2 /3 and gx = x 2 for all x ∈ X. Then T (X ) = g(X ) = [0, ∞). Further, for any x, y ∈ X we have gx ≤ gy ⇒ x 2 ≤ y 2 ⇒ (1/3)x 2 ≤ (1/3)y 2 ⇒ T x ≤ T y. So T is g-nondecreasing. Also there exists 0 ∈ X with T 0 = g0 = 0. T and g both are continuous being polynomial functions. T and g are compatible as for any sequence {xn } in X with limn→∞ T xn = limn→∞ gxn = t ∈ [0, ∞), we have limn→∞ (xn2 /3) = limn→∞ xn2 = t ⇒ limn→∞ xn2 = 0 ⇒ limn→∞ d(T gxn , gT xn ) = limn→∞ |(1/3)xn4 − (1/9)xn4 | = limn→∞ (2/9)xn4 = 0. Let ψ(t) = t 2 , ϕ(t) = 4.5χ (3,4) (t) + (4/9)t 2 χ [0,3]∪[4,∞) (t) for all t ∈ [0, ∞) and θ = χ (3,4) . Then ψ ∈  as it is continuous being a polynomial function, nondecreasing for having non-negative first derivative and ψ(t) = t 2 = 0 ⇒ t = 0. Also ϕ, θ ∈  as let I = [a, b] be any bounded interval in [0, ∞). For any t ∈ [a, b], ϕ(t) is equal to either (4/9)t 2 or 4.5. So it lies between min{(4/9)a 2 , 4.5} and max{(4/9)b2 , 4.5}. Thus ϕ is bounded on [a, b]. θ taking just two values 0 and 1 is bounded on any interval. Now let ψ(s) ≤ ϕ(t) for some s, t ∈ [0, ∞). Then s 2 ≤ (4/9)t 2 if t ∈ [0, 3] ∪ [4, ∞) and s 2 ≤ 4.5 if t ∈ (3, 4). In the former case, s 2 ≤ t 2 and in the later case, s 2 ≤ 4.5 < 9 < t 2 . Thus in both cases, we have / {3, 4}, then s ≤ t. Consider now a sequence {xn } in [0, ∞) with xn → t > 0. If t ∈ both ϕ and θ are continuous at t and hence ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) = ψ(t) − limn→∞ ϕ(xn ) + limn→∞ θ (xn ) = ψ(t) − ϕ(t) + θ (t) which is equal to t 2 − (4/9)t 2 = (5/9)t 2 > 0 if t ∈ [0, 3) ∪ (4, ∞) and to t 2 − 4.5 + 1 = t 2 − 3.5 > 0 if t ∈ (3, 4). Now consider the case when t = 3. Since xn → 3, we can choose a positive integer m such that 2.5 < xn < 3.5 for all n ≥ m. If xn ∈ (3, 3.5) for all n ≥ m then Am = sup{ϕ(xm ), ϕ(xm+1 ), . . .} = sup{4.5} = 4.5. If xn ∈ (2.5, 3] for all n ≥ m then Am = sup{ϕ(xm ), ϕ(xm+1 ), . . .} = sup{(4/9)xn2 : n ≥ m} = 4. In case, if both the sets {n ≥ m : xn ∈ (3, 3.5)} and {n ≥ m : xn ∈ (2.5, 3]} are not empty, then we have Am = sup{ϕ(xm ), ϕ(xm+1 ), . . .} = max{sup{ϕ(xn ) : n ≥ m, xn ∈ (3, 3.5)}, sup{ϕ(xn ) : n ≥ m, xn ∈ (2.5, 3]}} = max{sup{(4/9)xn2 : n ≥ m, xn ∈ (2.5, 3]}, 4.5} = 4.5. Thus Am ≤ 4.5. This implies that lim sup ϕ(xn ) = inf n≥1 supk≥n ϕ(xk ) ≤ supk≥m ϕ(xk ) = Am ≤ 4.5. Thus ψ(3) − lim sup ϕ(xn ) + lim inf θ (xn ) ≥ 9 − lim sup ϕ(xn ) ≥ 9 − 4.5 = 4.5 > 0. Consider the case when t = 4. Since xn → 4, we can choose a positive integer m such that 3.5 < xn < 4.5 for all n ≥ m. If xn ∈ (3.5, 4) for all n ≥ m then Am = sup{ϕ(xm ), ϕ(xm+1 ), . . .} = sup{4.5} = 4.5. If xn ∈ [4, 4.5) for all n ≥ m, then Am = sup{ϕ(xm ), ϕ(xm+1 ), . . .} = sup{(4/9)xn2 : n ≥ m} ≤ (4/9)(4.5)2 = 9. In case, if both the sets {n ≥ m : xn ∈ (3.5, 4)} and {n ≥ m : xn ∈ [4, 4.5)} are not empty, then we have Am = sup{ϕ(xm ), ϕ(xm+1 ), . . .} = max{sup{ϕ(xn ) : n ≥ m, xn ∈

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(3.5, 4)}, sup{ϕ(xn ) : n ≥ m, xn ∈ [4, 4.5)}} = max{4.5, sup{(4/9)xn2 : n ≥ m, xn ∈ [4, 4.5)}} = sup{(4/9)xn2 : n ≥ m, xn ∈ [4, 4.5)} ≤ (4/9)(4.5)2 = 9. Thus Am ≤ 9. This implies that lim sup ϕ(xn ) = inf n≥1 supk≥n ϕ(xk ) ≤ supk≥m ϕ(xk ) = Am ≤ 9. Thus ψ(4) − lim sup ϕ(xn ) + lim inf θ (xn ) ≥ 16 − lim sup ϕ(xn ) ≥ 16 − 9 = 7 > 0. Now we show that ψ( p(T x, T y)) ≤ ϕ( p(gx, gy)) − θ ( p(gx, gy)) for all x, y ∈ X . Let us first assume that 3 < y 2 < 4. Then ϕ( p(gx, gy)) − θ ( p(gx, gy)) = ϕ(gy) − θ (gy) = ϕ(y 2 ) − θ (y 2 ) = 4.5 − 1 = 3.5 > (16/9) > (1/9)y 4 = ψ(y 2 /3) = ψ(T y) = ψ( p(T x, T y)). Now assume that y 2 ∈ [0, 3] ∪ [4, ∞). Then ϕ( p(gx, gy)) − θ ( p(gx, gy)) = ϕ(gy) − θ (gy) = ϕ(y 2 ) − θ (y 2 ) = (4/9)y 4 ≥ (1/9)y 4 = ψ(y 2 /3) = ψ(T y) = ψ( p(T x, T y)). Thus T, g, ψ, ϕ, and θ satisfy all the conditions in Theorem 2.1. We observe that T and g have a coincidence point at 0. Further, because of comparability of any two real numbers, and compatibility (and hence weakly compatibility) of T and g all additional conditions as mentioned in Theorem 2.4 is also satisfied, and we see that 0 is a unique common fixed point of T and g. Example 2.6 Consider X = [0, ∞) together with the Euclidean metric d and let p(x, y) = λd(x, y) for all x, y ∈ X where λ is any positive real number. Then, by Example 1.2(ii), p is a w-distance on (X, d). Further (X, d, ≤) is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define self maps T and g on X by T = χ X and g = χ Q +(1/2)χ X−Q . Then T (X ) = {1} ⊆ {1/2, 1} = g(X ). T is g-nondecreasing being a constant function. Further for 0 ∈ X, T 0 = g0 = 1. g(X ) = {1/2, 1}, being finite is a closed subset of a complete space (X, d) and hence complete. Consider the functions ψ, ϕ, and θ as defined in Example 2.5 where we have already established that these functions satisfy the conditions ψ ∈ , ϕ, θ ∈ , for all s, t ∈ [0, ∞), ψ(s) ≤ ϕ(t) ⇒ s ≤ t and for any sequence {xn } in [0, ∞) with xn → t > 0, ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0. Now we show that ψ( p(T x, T y)) ≤ ϕ( p(gx, gy))−θ ( p(gx, gy)) for all x, y ∈ X . If x, y both are rational or both are irrational, then gx = gy. So, p(gx, gy) = λd(gx, gy) = 0. This implies that ψ( p(T x, T y)) = ψ( p(1, 1)) = ψ(0) = 0 = ϕ(0) − θ (0) = ϕ( p(gx, gy)) − θ ( p(gx, gy)). Now assume that exactly one of the two numbers x and y is rational. Then d(gx, gy) = |1 − (1/2)| = 1/2. So ψ( p(T x, T y)) = ψ( p(1, 1)) = ψ(0) = 0 ϕ( p(gx, gy)) − θ ( p(gx, gy)) = / (6, 8) and to 4.5 − 1 = ϕ(λ/2) − θ (λ/2) which is equal to (4/9)(λ/2)2 = λ2 /9 if λ ∈ 3.5 otherwise. In both, cases ϕ( p(gx, gy)) − θ ( p(gx, gy)) ≥ 0 = ψ( p(T x, T y)). So T, g, ψ, ϕ and θ satisfy all the conditions in Theorem 2.1. We observe that every rational number is a coincidence point of T and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of T and g, all additional conditions as mentioned in Theorem 2.4 are also satisfied and we see that 1 is a unique common fixed point of T and g. Example 2.7 Consider X = [0, ∞) together with the Euclidean metric d and let p(x, y) = f (x) + y for all x, y ∈ X where f : X → X is any arbitrary function such that f (0) = 0. Then, by Example 1.2(iii), p is a w-distance on (X, d). Further (X, d, ≤) is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define self map T on X by T = χ φ (φ denotes an empty set) and take g as in Example 2.5. Then T (X ) = {0} ⊆ [0, ∞) = g(X ).

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T is g-nondecreasing being a constant function. Further for 0 ∈ X, T 0 = g0 = 0. g(X ) = X is complete. Consider the functions ψ, ϕ, and θ as defined in Example 2.5 where we have already established that these functions satisfy the conditions ψ ∈ , ϕ, θ ∈ , for all s, t ∈ [0, ∞), ψ(s) ≤ ϕ(t) ⇒ s ≤ t and for any sequence {xn } in [0, ∞) with xn → t > 0, ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0. Now we show that ψ( p(T x, T y)) ≤ ϕ( p(gx, gy)) − θ ( p(gx, gy)) for all x, y ∈ X . We note that ϕ(t) − θ (t) is equal to (4/9)t 2 if t ∈ [0, 3] ∪ [4, ∞) and to 4.5 − 1 = 3.5 if t ∈ (3, 4). In both cases, ϕ(t) − θ (t) ≥ 0. So, ϕ( p(gx, gy)) − θ ( p(gx, gy)) ≥ 0 = ψ(0) = ψ( p(0, 0)) = ψ( p(T x, T y)) for all x, y ∈ X . Thus, T, g, ψ, ϕ, and θ satisfy all the conditions in Theorem 2.1. We observe that 0 is a coincidence point of T and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of T and g, all additional conditions as mentioned in Theorem 2.4 are also satisfied and we see that 0 is a unique common fixed point of T and g.

3 Applications 3.1 Coupled Coincidence Point Theory In this section, we establish some coupled coincidence point results in partially ordered metric spaces by using results in the previous section. These results extend some of the existing results. Let (X, d, ) be a partially ordered metric space with a partial order  and a metric d. Define a relation “” on X × X by (x, y)  (u, v) if and only if x  u and y  v. Further, define a function d1 by d1 ((x, y), (u, v)) = d(x, u) + d(y, v) for all (x, y), (u, v) ∈ X × X. Then X × X becomes a partially ordered metric space with a partial order  and a metric d1 . Now, consider two function F : X × X → X and g : X → X . Define T : X × X → X × X and G : X × X → X × X by T (x, y) = (F(x, y), F(y, x)) for all (x, y) ∈ X × X and G(x, y) = (gx, gy) for all (x, y) ∈ X × X. Now, following results have been settled in [9, Lemma 4.1, Lemma 4.2, Lemma 4.3 , and Lemma 4.4, respectively]. (i) (ii) (iii) (iv)

If If If If

F F F F

satisfies mixed g-monotone property, then T is G-nondecreasing. and g are commutative, then so are T and G. and g are compatible, then so are T and G. and g are weakly compatible, then so are T and G.

Theorem 3.1 Let (X, , d) be a partially ordered metric space equipped with a wdistance p and suppose F : X × X → X and g : X → X be two functions with F(X × X ) ⊆ g(X ) and F satisfying mixed g-monotone property. Suppose there exist x0 , y0 ∈ X with gx0  F(x0 , y0 ) and gy0  F(y0 , x0 ) or gx0  F(x0 , y0 ) and gy0  F(y0 , x0 ). Further assume that either F and g are continuous, (F, g) compatible with (X, d) as a complete metric space or (g(X ), d) is a complete subspace of a regular metric space (X, d). Let ψ, ϕ, and θ be functions having the same properties as in Theorem 2.1 except the contractive condition which is replaced by the following condition. For all x, y, u, v ∈ X with gx  gu and gy  gv or gx  gu and gy  gv we have ψ( p(F(x, y), F(u, v)) + p(F(y, x), F(v, u))) ≤ ϕ( p(gx, gu) +

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p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)). Then F and g have a coupled coincidence point in X . In addition to above, suppose for any pair of elements (x, y), (x ∗ , y ∗ ) ∈ X × X , there exists (u, v) ∈ X × X such that (F(u, v), F(v, u)) is comparable with (F(x, y), F(y, x)) and (F(x ∗ , y ∗ ), F(y ∗ , x ∗ )) and also the pair (F, g) is weakly compatible. Then there exists a unique coupled common fixed point of F and g. Proof Let S = X × X and d1 , a relation  on S and the self maps T and G on S be as defined just before Theorem 3.1. Consider p1 ((x, y), (u, v)) = p(x, u) + p(y, v) for all (x, y), (u, v) ∈ S. Then (S, d1 , ) is a partially ordered metric space equipped with a w-distance p1 , T is G-nondecreasing and T (S) ⊆ G(S). Let p0 = (x0 , y0 ). Then gx0  F(x0 , y0 ) and gy0  F(y0 , x0 ) or gx0  F(x0 , y0 ) and gy0  F(y0 , x0 ) imply that Gp0  T p0 or Gp0  T p0 for some point p0 ∈ S. Also the pair (T, G) is compatible and T and G are continuous on S if so are F and g. Also regularity of (X, d) implies regularity of (S, d1 ) and completeness of (G(S), d1 ) is implied by completeness of (g(X ), d). Let p = (x, y) and q = (u, v). Then gx  gu and gy  gv or gx  gu and gy  gv means Gp  Gq or Gp  Gq. So ψ( p1 (T p, T q)) ≤ ϕ( p1 (Gp, Gq)) − θ ( p1 (Gp, Gq)) for all p, q ∈ S with Gp  Gq or Gp  Gq. Thus all the conditions of Theorem 2.1 are satisfied. As a result, there exists a point s = (x, y) ∈ S such that T s = Gs. That is (F(x, y), F(y, x)) = (gx, gy) or F(x, y) = gx and F(y, x) = gy. Hence (x, y) is a coupled coincidence point of F and g. Since weak compatibility of T and G follows by weak compatibility of F and g, the last part of the theorem follows by Theorem 2.4.

Example 3.2 Consider X = [0, ∞) together with the Euclidean metric d and p(x, y) = y for all x, y ∈ X . Then (X, d, ≤) is a partially ordered complete metric space and as proved in Example 1.2(iv), p is a w-distance on (X, d). Consider now the function defined by F(x, y) = (1/3)(x 2 − y 2 ) if x ≥ y, F(x, y) = 0 if x < y and take g as in Example 2.5. Let x1 , x2 ∈ X with gx1 ≤ gx2 and y ∈ X . Then x12 ≤ x22 ⇒ x1 ≤ x2 . If y ≤ x1 ≤ x2 then F(x1 , y) = (1/3)(x12 − y 2 ) ≤ (1/3)(x22 − y 2 ) = F(x2 , y). If x1 ≤ y ≤ x2 then F(x1 , y) = 0 ≤ (1/3)(x22 −y 2 ) = F(x2 , y). Further, if x1 ≤ x2 ≤ y then F(x1 , y) = 0 = F(x2 , y). Thus we always have F(x1 , y) ≤ F(x2 , y). Now, let y1 , y2 ∈ X with gy1 ≤ gy2 and x ∈ X . Then y12 ≤ y22 ⇒ y1 ≤ y2 . If y1 ≤ y2 ≤ x then F(x, y1 ) = (1/3)(x 2 − y12 ) ≥ (1/3)(x 2 − y22 ) = F(x, y2 ). If y1 ≤ x ≤ y2 then F(x, y1 ) = (1/3)(x 2 − y12 ) ≥ 0 = F(x, y2 ). Further, if x ≤ y1 ≤ y2 then F(x, y1 ) = 0 = F(x, y2 ). Thus we always have F(x, y1 ) ≥ F(x, y2 ) . So F satisfies mixed g-monotone property. Further, F(X × X ) = g(X ) = [0, ∞) and g is continuous on X being a polynomial function. Next we show that F is continuous on X × X . Let (xn , yn ) → (x, y) in X × X .√This implies that xn → x and yn → y in X . If x > y then choose r ≤ (x − y)/ 2 which is nothing but the perpendicular distance of the point (x, y) from the line y = x. Choose now a positive integer m such that for all n ≥ m, (xn , yn ) belongs to the circular open neighborhood D centered at (x, y) having radius r . Then we have xn > yn for all n ≥ m. This implies that, for all n ≥ m, |F(xn , yn ) − F(x, y)| = (1/3)|(xn2 − yn2 ) − (x 2 − y 2 )| < |(xn2 − x 2 ) − (yn2 − y 2 )| ≤ |xn2 − x 2 | + |yn2 − y 2 | → 0 as n → ∞. If x < y, by the same argument as above we can find a positive integer N such that xn < yn for all n ≥ N . So, |F(xn , yn ) − F(x, y)| = 0 for all n ≥ N . Now assume that x = y. Then |F(xn , yn ) − F(x, y)| = |F(xn , yn )| ≤ |(1/3)(xn2 − yn2 )| → (1/3)|x 2 − y 2 | = 0

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as n → ∞. Thus F(xn , yn ) → F(x, y) whenever (xn , yn ) → (x, y) in X × X . So F is a continuous function. Consider the functions ψ, ϕ, and θ as defined in Example 2.5 where we have already established that these functions satisfy the conditions ψ ∈ , ϕ, θ ∈ , for all s, t ∈ [0, ∞), ψ(s) ≤ ϕ(t) ⇒ s ≤ t and for any sequence {xn } in [0, ∞) with xn → t > 0, ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0. In [9] (Example 4.1), it is proved that (F, g) is a compatible pair of mappings. Further, if x0 = 0 and y0 = c > 0, then F(x0 , y0 ) = F(0, c) = 0 = gx0 and F(y0 , x0 ) = F(c, 0) = c2 /3 ≤ c2 = g(y0 ). For all x, y, u, v ∈ X we have ψ( p(F(x, y), F(u, v)) + p(F(y, x), F(v, u))) = ψ(F(u, v) + F(v, u)) = ψ(|u 2 − v 2 |/3) = (u 2 − v 2 )2 /9 and ϕ( p(gx, gu) + p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)) = ϕ(gu + gv) − θ (gu + gv) = ϕ(u 2 + v 2 ) − θ (u 2 + v 2 ) = 3.5 if 3 < u 2 + v 2 < 4 and (4/9)(u 2 + v 2 )2 otherwise. It is easy to observe that ψ( p(F(x, y), F(u, v)) + p(F(y, x), F(v, u))) ≤ ϕ( p(gx, gu) + p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)) for all x, y, u, v ∈ X . So all the conditions of Theorem 3.1 are satisfied. We observe that F and g have a coupled coincidence point at (0, 0). Further, all additional conditions as mentioned in Theorem 3.1 are also satisfied and we see that (0, 0) is a unique coupled common fixed point of F and g. Example 3.3 Consider X = [0, ∞) together with the Euclidean metric d and let p(x, y) = f (x) + y for all x, y ∈ X where f : X → X is any arbitrary function such that f(0)=0. Then, by Example 1.2(iii), p is a w-distance on (X, d). Further (X, d, ≤) is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define F : X × X → X by F = χ φ (φ denotes an empty set) and take g as in Example 2.5. Then F(X × X ) = {0} ⊆ [0, ∞) = g(X ). F is mixed g-monotonic being a constant function. Further for (0, 0) ∈ X × X , F(0, 0) = g0 = 0. g(X ) = X is complete. Consider the functions ψ, ϕ and θ as defined in Example 2.5 where we have already established that these functions satisfy the conditions ψ ∈ , ϕ, θ ∈ , for all s, t ∈ [0, ∞), ψ(s) ≤ ϕ(t) ⇒ s ≤ t and for any sequence {xn } in [0, ∞) with xn → t > 0, ψ(t)−lim sup ϕ(xn )+lim inf θ (xn ) > 0. Now we show that ψ( p(F(x, y), F(u, v))+ p(F(y, x), F(v, u))) ≤ ϕ( p(gx, gu) + p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)) for all x, y, u, v ∈ X . As proved in Example 2.7, ϕ(t) − θ (t) ≥ 0 for all t ≥ 0. So, ϕ( p(gx, gu) + p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)) ≥ 0 = ψ(0) = ψ( p(0, 0) + p(0, 0)) = ψ( p(F(x, y), F(u, v)) + p(F(y, x), F(v, u))) for all x, y, u, v ∈ X . Thus, F, g, ψ, ϕ, and θ satisfy all the conditions in Theorem 3.1. We observe that 0 is a coupled coincidence point of F and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of F and g, all additional conditions as mentioned in Theorem 3.1 are also satisfied and we see that 0 is a unique coupled common fixed point of F and g. Example 3.4 Consider X = [0, ∞) together with the Euclidean metric d and let p(x, y) = λd(x, y) for all x, y ∈ X where λ is any positive real number. Then, by Example 1.2(ii), p is a w-distance on (X, d). Further (X, d, ≤) is a partially ordered regular metric space. Regularity is because of monotone convergence theorem for real sequences. Define F : X × X → X by F = χ X ×X and consider g as defined in Example 2.6. Then F(X × X ) = {1} ⊆ {1/2, 1} = g(X ). F is mixed g-monotonic being a constant function. Further for (0, 0) ∈ X × X, F(0, 0) = g0 = 1. Now,

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g(X ) = {1/2, 1} being finite is a closed subset of a complete space (X, d) and hence complete. Consider the functions ψ, ϕ and θ as defined in Example 2.5 where we have already established that these functions satisfy the conditions ψ ∈ , ϕ, θ ∈ , for all s, t ∈ [0, ∞), ψ(s) ≤ ϕ(t) ⇒ s ≤ t and for any sequence {xn } in [0, ∞) with xn → t > 0, ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0. Now we show that ψ( p(F(x, y), F(u, v)) + p(F(y, x), F(v, u))) ≤ ϕ( p(gx, gu) + p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)) for all x, y, u, v ∈ X . As proved in Example 2.7, ϕ(t) − θ (t) ≥ 0 for all t ≥ 0. So, ϕ( p(gx, gu) + p(gy, gv)) − θ ( p(gx, gu) + p(gy, gv)) ≥ 0 = ψ(0) = ψ( p(1, 1) + p(1, 1)) = ψ( p(F(x, y), F(u, v)) + p(F(y, x), F(v, u))) for all x, y, u, v ∈ X . So F, g, ψ, ϕ, and θ satisfy all the conditions in Theorem 3.1. We observe that every point (x, y) of the set Q×Q is a coupled coincidence point of F and g. Further, because of comparability of any two real numbers, and commutativity (and hence compatibility and weakly compatibility) of F and g, all additional conditions as mentioned in Theorem 3.1 are also satisfied and we see that (1, 1) is a unique coupled common fixed point of F and g. 3.2 Integral Equations Let I = [a, b] be any interval of the real line. Consider the componentwise partial order on Rn . Let X = C(I, Rn ) be the space of all continuous functions defined on I with values in Rn and pointwise partial order. Let d : X × X → R defined by d(x, y) = ||x − y||∞ for any x, y ∈ X , be a complete metric on X . By Example 1.2(v), p : X × X → R defined by p(x, y) = d(0, y) = ||y||∞ for any x, y ∈ X , is a wdistance on X . Let h : I × I × Rn → Rn be continuous. Consider the Fredholm type integral equation  x(t) =

h(t, s, x(s))ds,

(3.1)

I

for t ∈ I . Definition 3.5 A function α ∈ C(I, Rn ) is said to be a lower solution of the integral equation (3.1) if α(t) ≤ I h(t, s, α(s))ds for all t ∈ I and an upper solution if α(t) ≥ I h(t, s, α(s))ds for all t ∈ I . Now we consider the following assumptions: (a) h(t, s, .) : Rn → Rn is nondecreasing for each t, s ∈ I , (b) There exist ψ ∈  and ϕ, θ ∈  such that for all s, t ∈ [0, ∞) we have ψ(s) ≤ ϕ(t) ⇒ s ≤ t and ψ(t) ≤ t for all t ≥ 0. Further, for any sequence {xn } in [0, ∞) with xn → t > 0, suppose that ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0 and for all x ∈ X, t, s ∈ I , |h(t, s, x(s))| ≤ (1/(b − a))(ϕ(||x||∞ ) − θ (||x||∞ )). Next we give an existence theorem for the solution of the integral equation (3.1). Theorem 3.6 Suppose that assumptions (a) and (b) hold. Then the existence of a lower or an upper solution for the integral equation (3.1) provides the existence of a solution of the equation (3.1).

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 Proof Define a mapping T : X → X by T (x)(t) := I h(t, s, x(s))ds for all t ∈ I . Clearly T is a well-defined mapping. Take g as the identity function on X . By our assumption (a), we observe that T is nondecreasing. Now for any x, y ∈ X and t ∈ I , we have, by assumption (b) 

 h(t, s, y(s))ds|) ≤ ψ( |h(t, s, y(s))|ds)

ψ(|T (y)(t)|) = ψ(| I I   ≤ |h(t, s, y(s))|ds ≤ (1/(b − a))(ϕ(||y||∞ ) − θ (||y||∞ )) ds I

I

= ϕ(||y||∞ ) − θ (||y||∞ ). So for all x, y ∈ X , ψ( p(T (x), T (y))) = ψ(||T (y)||∞ ) = ψ(maxt∈I |T (y)(t)|) = maxt∈I ψ(|T (y)(t)|) ≤ ϕ(||y||∞ )−θ (||y||∞ ) = ϕ( p(gx, gy))−θ ( p(gx, gy)). From the existence of a lower or an upper solution of the equation (3.1), there exists an x0 ∈ X such that x0 ≤ T x0 or x0 ≥ T x0 . Further, X is regular as Rn is so. The conclusion follows by Theorem 2.1.

By Example 1.2(i), q : X × X → R defined by q(x, y) = d(x, y) for any x, y ∈ X , is a w-distance on X . Let k : I × I × Rn → Rn be continuous. For t ∈ I , consider the Fredholm type integral equations  x(t) =

k(t, x(s), y(s))ds, 

I

y(t) =

k(t, y(s), x(s))ds.

(3.2)

I

Definition 3.7 An element (α, β) ∈ C(I, Rn ) × C(I, Rn ) is said to be a coupled lower and upper solution of the integral equation (3.2) if α(t) ≤ β(t), α(t) ≤   k(t, α(s), β(s))ds for all t ∈ I and β(t) ≥ k(t, β(s), α(s))ds for all t ∈ I . I I Now we consider the following assumptions: (a) For any x, y ∈ X and t ∈ I , we have x1 , x2 ∈ X, x1 ≤ x2 ⇒ k(t, x1 , y) ≤ k(t, x2 , y) and y1 , y2 ∈ X , y1 ≤ y2 ⇒ k(t, x, y1 ) ≥ k(t, x, y2 ) (b) There exist ψ ∈  and ϕ, θ ∈  such that for all s, t ∈ [0, ∞), we have ψ(s) ≤ ϕ(t) ⇒ s ≤ t and ψ(t) ≤ t for all t ≥ 0. Further, for any sequence {xn } in [0, ∞) with xn → t > 0, suppose that ψ(t) − lim sup ϕ(xn ) + lim inf θ (xn ) > 0 and for all x, y, u, v ∈ X with x ≤ u, y ≥ v and t, s ∈ I , |k(t, x(s), y(s)) − k(t, u(s), v(s))| ≤ (1/2)(1/(b − a))(ϕ(||x − u||∞ + ||y − v||∞ ) − θ (||x − u||∞ + ||y − v||∞ )). Next we give an existence theorem for the solution of the integral equation (3.2). Theorem 3.8 Suppose that assumptions (a) and (b) hold. Then the existence of a lower and upper solution for the integral equation (3.2) provides the existence of a solution of the equation (3.2).

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A Generalized Contraction Principle Under w-Distance...

 Proof Define a mapping F : X × X → X by F(x, y)(t) := I k(t, x(s), y(s))ds for all t ∈ I . Clearly F is a well-defined mapping. Take g as the identity function on X . Let x, y ∈ X with x1 , x2 ∈ X, x1 ≤ x2 and y1 , y2 ∈ X , y1 ≤ y2 . By our assumption (a), k(t, x1 ,y) ≤ k(t, x2 , y) and k(t, x, y1 ) ≥ k(t, x, y2 ) for all  t ∈ I . This implies that I k(t, x1 (s), y(s))ds ≤ I k(t, x2 (s), y(s))ds and I k(t, x(s), y1 (s))ds ≥ I k(t, x(s), y2 (s))ds. That is F(x 1 , y) ≤ F(x 2 , y) and F(x, y1 ) ≥ F(x, y2 ). So, F is mixed g-monotonic. Now for any x, y, u, v ∈ X with x ≤ u, y ≥ v and t ∈ I , we have, by assumption (b), |F(x, y)(t) − F(u, v)(t)|       =  k(t, x(s), y(s))ds − k(t, u(s), v(s))ds  I  I ≤ |k(t, x(s), y(s)) − k(t, u(s), v(s))|ds I

≤ (1/2)(1/(b − a))(ϕ(||x − u||∞ + ||y − v||∞ )  − θ (||x − u||∞ + ||y − v||∞ )) ds I

= (1/2)(ϕ(d(x, u) + d(y, v)) − θ (d(x, u) + d(y, v))) = (1/2)(ϕ(q(gx, gu) + q(gy, gv)) − θ (q(gx, gu) + q(gy, gv))). This implies that for all x, y, u, v ∈ X with x ≤ u, y ≥ v, q(F(x, y), F(u, v)) = d(F(x, y), F(u, v)) = sup |F(x, y)(t) − F(u, v)(t)| t∈I

≤ (1/2)(ϕ(q(gx, gu) + q(gy, gv)) − θ (q(gx, gu) + q(gy, gv)))

(3.3)

Since x ≤ u, y ≥ v is equivalent to v ≤ y, u ≥ x, we also have q(F(v, u), F(y, x)) ≤ (1/2)(ϕ(q(gv, gy) + q(gu, gx)) − θ (q(gv, gy) + q(gu, gx))). That is q(F(y, x), F(v, u)) ≤ (1/2)(ϕ(q(gx, gu) + q(gv, gy)) − θ(q(gx, gu) + q(gv, gy))) (3.4)

Adding (3.3) and (3.4) we get for all x, y, u, v ∈ X with x ≤ u, y ≥ v, q(F(x, y), F(u, v)) + q(F(y, x), F(v, u)) ≤ ϕ(q(gx, gu) + q(gv, gy)) − θ (q(gx, gu) + q(gv, gy)).

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Now for all x, y, u, v ∈ X with gx ≤ gu, gy ≥ gv, ψ(q(F(x, y), F(u, v)) + q(F(y, x), F(v, u))) ≤ q(F(x, y), F(u, v)) + q(F(y, x), F(v, u)) ≤ ϕ(q(gx, gu) + q(gv, gy)) − θ (q(gx, gu) + q(gv, gy)). From the existence of a lower and upper solution of the equation (3.2), there exists a point (x0 , y0 ) ∈ X × X such that x0 ≤ F(x0 , y0 ) and y0 ≥ F(y0 , x0 ). Further, X is

regular as Rn is so. The conclusion follows by Theorem 3.1.

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