α-(ψ, φ) Contractive mappings on quasi-partial metric spaces

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1 Karapınar et al. Fixed Point Theory and Applications (2015) 2015:105 DOI /s z RESEARCH Open Access α-(ψ, φ) Contractive mappings on quasi-partial metric spaces Erdal Karapınar1,2*, Leila Gholizadeh3, Hamed H Alsulami2 and Maha Noorwali2 * Correspondence: erdalkarapinar@yahoo.com 1 Department of Mathematics, Atilim University, I ncek, Ankara, 06836, Turkey 2 Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia Full list of author information is available at the end of the article Abstract In this paper, we consider α -(ψ, φ ) contractive mappings in the setting of quasi-partial metric spaces and verify the existence of a fixed point on such spaces. Moreover, we present some examples and applications in integral equations of our obtained results. MSC: 46T99; 46N40; 47H10; 54H25 Keywords: quasi-partial metric space; fixed point; α -admissible; contractive mapping 1 Introduction and preliminaries One of the most interesting extensions of distance function was reported by Matthews [ ] by introducing the notion of a partial metric in which self-distance need not be zero. In this celebrated report, Matthews [ ] successfully characterized the distinguished result, Banach contraction mapping, in the setting of partial metric spaces. Later, many authors have generalized some fixed point theorems on such a space, see e.g. [ ] and the related references therein. Very recently, Karapınar et al. [ ] presented quasi-partial metric spaces and investigated the existence and uniqueness of certain operators in the context of quasi-partial metric spaces. Throughout this paper, we suppose that R+ = [, + ), N = N { }, where N denotes the set of all positive integers. First, we recall some basic concepts and notations. For more information, see [, ]. Definition. A quasi-metric on a nonempty set X is a function d : X X [, + ) such that for all x, y, z X: (QM ) d(x, y) = x = y, (QM ) d(x, y) d(x, z) + d(z, y). A quasi-metric space is a pair (X, d) such that X is a nonempty set and d is a quasi-metric on X. Definition. A partial metric on a nonempty set X is a function p : X X [, + ) such that for all x, y, z X: (PM ) x = y p(x, x) = p(x, y) = p(y, y), (PM ) p(x, x) p(x, y), (PM ) p(x, y) = p(y, x), (PM ) p(x, y) p(x, z) + p(z, y) p(z, z) Karapınar et al. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( 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.

2 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 2 of 20 A partial metric space is a pair (X, p)suchthatx is a nonempty set and p is a partial metric on X. Definition 1.3 (See [13]) A quasi-partial metric space on a nonempty set X is a function q : X X [0, + ) such that for all x, y, z X: (QPM1) if q(x, x) = q(x, y) = q(y, y),thenx = y (equality), (QPM2) q(x, x) q(x, y) (small self-distances), (QPM3) q(x, x) q(y, x) (small self-distances), (QPM4) q(x, z) +q(y, y) q(x, y) +p(y, z) (triangle inequality). A quasi-partial metric space is a pair (X, q)suchthatx is a nonempty set and q is a partial metric on X. If q(x, y)=q(y, x) for all x, y X,then(X, q) becomes a partial metric space. Lemma 1.1 (See [13]) Let (X, q) be a quasi-partial metric space. Then the following holds: If q(x, y)=0then x = y. For a partial metric p on X, the function d p : X X [0, + )definedby d p (x, y)=2p(x, y) p(x, x) p(y, y) is a metric on X. For a quasi-partial metric q on X, the function d q : X X [0, + ) defined by d q (x, y)=q(x, y)+q(y, x) q(x, x) q(y, y) is a metric on X and p q (x, y)= 1 [ ] q(x, y)+q(y, x) 2 is a partial metric on X. Notice also that for a quasi-partial metric q on X, the function d q m : X X [0, + )definedby dm q q(x, y)+q(y, x) (x, y)= min { q(x, x), q(y, y) } 2 is a metric on X. Definition 1.4 (See [13]) Let (X, q) be a quasi-partial metric space. Then: (i) a sequence {x n } X converges to x X if and only if q(x, x)= lim n + q(x, x n)= lim n + q(x n, x); (ii) a sequence {x n } X is called a Cauchy sequence if and only if lim n,m + q(x n, x m ) and lim n,m + q(x m, x n ) exist (and are finite); (iii) the quasi-partial metric space is said to be complete if every Cauchy sequence {x n } X converges, with respect to τ q,toapointx X such that q(x, x)= lim q(x n, x m )= lim q(x m, x n ); n,m + n,m +

3 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 3 of 20 (iv) a mapping f : X X is said to be continuous at x 0 X if for every ε >0,there exists δ >0such that f (B(x 0, δ)) B(f (x 0 ), ε). Lemma 1.2 (See [13]) Let (X, q) be a quasi-partial metric space. Let (X, p q ) be the corresponding partial metric space, and let (X, d pq ) be the corresponding metric space. The following statements are equivalent: (A) The sequence {x n } is Cauchy in (X, q). (B) The sequence {x n } is Cauchy in (X, p q ). (C) The sequence {x n } is Cauchy in (X, d pq ). Lemma 1.3 (See [13]) Let (X, q) be a quasi-partial metric space. Let (X, p q ) be the corresponding partial metric space, and let (X, d pq ) be the corresponding metric space. The following statements are equivalent: (A) (X, q) is complete. (B) (X, p q ) is complete. (C) (X, d pq ) is complete. Moreover, lim d p n q (x, x n )=0 p q (x, x)= lim p q(x, x n )= n lim p q(x n, x m ) (1) n,m q(x, x) = lim q(x, x n)= lim q(x n, x m ) (2) n,m n,m = lim q(x n, x)= lim q(x m, x n ). (3) n n,m In this paper, we shall handle Definition 1.5 in the following way. Definition 1.5 (See [13]) Let (X, q) be a quasi-partial metric space. Then: (ii) a asequence{x n } in X is called a left-cauchy sequence if and only if for every ε >0 there exists a positive integer N = N(ε) such that q(x n, x m )<ε for all n > m > N; (ii) b asequence{x n } in X is called a right-cauchy sequence if and only if for every ε >0 there exists a positive integer N = N(ε) such that q(x n, x m )<ε for all m > n > N; (iii) a the quasi-partial metric space is said to be left-complete if every left-cauchy sequence {x n } in X is convergent; (iii) b the quasi-partial metric space is said to be right-complete if every left-cauchy sequence {x n } in X is convergent. Remark 1 It is clear that a sequence {x n } in a quasi-partial metric space is Cauchy if and only if it is left-cauchy and right-cauchy. Analogously, a quasi-partial metric space (X, q) is complete if and only if it is left-complete and right-complete. Very recently, Samet et al. [14] introduced the concept α-admissible mappings and established various fixed point theorems for such mappings in complete metric spaces. Later, in 2013, Karapınar et al. [15] proved the existence and uniqueness of a fixed point for triangular α-admissible mappings. For more on α-admissible and triangular α-admissible mappings, see [14, 15].

4 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 4 of 20 Definition 1.6 [14] LetT be a self-mapping on X and α : X X [0, + ) be a function. We say that T is an α-admissible mapping if x, y X, α(x, y) 1 α(tx, Ty) 1. Definition 1.7 [15] LetT be a self-mapping on X and α : X X [0, + ) be a function. We say that T is a triangular α-admissible mapping if T is α-admissible and x, y, z X, α(x, z) 1 and α(z, y) 1 α(x, y) 1. Very recently, Popescu [16] improved the notion of α-admissible asfollows. Definition 1.8 [16] LetT : X X be a self-mapping and α : X X [0, ) be a function. Then T issaid tobe α-orbital admissible if (T3) α(x, Tx) 1 α ( Tx, T 2 x ) 1. Inspired by the notion of α-admissible defined by Popescu [16], we state the following definitions. Definition 1.9 [16] LetT : X X be a self-mapping and α : X X [0, ) be a function. Then T is said to be right-α-orbital admissible if (T3) α(x, Tx) 1 α ( Tx, T 2 x ) 1, and be left-α-orbital admissible if (T3) α(tx, x) 1 α ( T 2 x, Tx ) 1. Note that a mapping T is α-orbital admissible if it is both right-α-orbital admissible and left-α-orbital admissible. Popescu [16] refinedthe notion oftriangular α-admissible asfollows. Definition 1.10 [16] LetT : X X be a self-mapping and α : X X [0, )beafunction. Then T is said to be triangular α-orbital admissible if T is α-orbital admissible and (T4) α(x, y) 1 and α(y, Ty) 1 α(x, Ty) 1. Triangular α-admissible defined by Popescu [16] imposes the following definitions. Definition 1.11 [16] LetT : X X be a self-mapping and α : X X [0, ) beafunction. Then T is said to be triangular α-orbital admissible if T is right-α-orbital admissible and (T4) α(x, y) 1 and α(y, Ty) 1 α(x, Ty) 1, and be triangular left-α-orbital admissible if T is α-orbital admissible and (T4) α(tx, x) 1 and α(x, y) 1 α(tx, y) 1.

5 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 5 of 20 Notice that a mapping T is triangular α-orbital admissible if it is both triangular rightα-orbital admissible and triangular left-α-orbital admissible. It was noted in [16]thateachα-admissible mapping is an α-orbital admissible mapping and each triangular α-admissible mapping is a triangular α-orbital admissible mapping. The converse is false, see e.g. [16], Example 7. Definition 1.12 [16] Let(X, d) be a b-metric space, X is said α-regular if for every sequence {x n } in X such that α(x n, x n+1 ) 1 for all n and x n x X as n,thereexists asubsequence{x n(k) } of {x n } such that α(x n(k), x) 1 for all k. Lemma 1.4 [16] Let T : X Xbeatriangularα-orbital admissible mapping. Assume that there exists x 0 Xsuchthatα(x 0, Tx 0 ) 1. Define a sequence {x n } by x n+1 = Tx n for each n N 0. Then we have α(x n, x m ) 1 for all m, n N with n < m. The following result can be easily derived from Lemma 1.4. Lemma 1.5 Let T : X Xbeatriangularα-orbital admissible mapping. Assume that there exists x 0 Xsuchthatα(Tx 0, x 0 ) 1. Define a sequence {x n } by x n+1 = Tx n for each n N 0. Then we have α(x m, x n ) 1 for all m, n N with n < m. In this paper, we investigate and extend the existence of a fixed point of (ψ, φ)- contractive mappings on quasi-partial metric spaces via α-admissibility. 2 Main results In this section, we shall present the main theorem of the paper. For our aim, we need to define the following class of auxiliary mappings: Let be set of functions ϕ :[0,+ ) [0, + )suchthatϕ 1 ({0})={0}: = {ψ ψ is continuous, non-decreasing} and = {φ φ is lower semi-continuous}. Let (X, q) be a quasi-partial metric space. We consider the following expressions: M(x, y)=max { q(x, y), q(x, Tx), q(y, Ty) }, (4) N(x, y)=min { d q m (x, Tx), dq m (y, Ty), dq m (x, Ty), dq m (y, Tx)} (5) for all x, y X. Definition 2.1 Let (X, q) be a quasi-partial metric space where X is a nonempty set. We say that X is said to be α-left-regular if for every sequence {x n } in X such that α(x n+1, x n ) 1 for all n and x n x X as n,thereexistsasubsequence{x n(k) } of {x n } such that α(x, x n(k) ) 1 for all k. Analogously, a quasi-partial metric space X is said to be an α-rightregular if for every sequence {x n } in X such that α(x n, x n+1 ) 1 for all n and x n x X as n,thereexistsasubsequence{x n(k) } of {x n } such that α(x n(k), x) 1 for all k.wesay that X is regular if it is both α-left-regular and α-right-regular. Ourfirstresultisthefollowing.

6 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 6 of 20 Theorem 2.1 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbeaselfmapping. Assume that there exist ψ, φ, L 0 and a function α : X X [0, + ) such that for all x, y X, α(x, y)ψ ( q(tx, Ty) ) ψ ( M(x, y) ) φ ( M(x, y) ) + LN(x, y). (6) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Proof We construct a sequence {x n } in X in the following way: x n = Tx n 1 for all n N. If q(x n0, x n0 +1)=0forsomen 0 0, then we have x n0 = x n0 +1 = Tx n0,thatis,x n0 is the fixed point of T.Consequently,wesupposethatq(x n, x n+1 ) > 0 for all n N 0. By (ii), we have α(x 0, Tx 0 ) 1andα(Tx 0, x 0 ) 1. On account of (i), we derive that α(x 0, x 1 )=α(x 0, Tx 0 ) 1 α(x 1, x 2 )=α(tx 0, Tx 1 ) 1, α(x 1, x 0 )=α(tx 0, x 0 ) 1 α(x 2, x 1 )=α(tx 1, Tx 0 ) 1. Recursively, we obtain that α(x n, x n+1 ) 1 and α(x n+1, x n ) 1, n N 0. (7) Regarding (6)and(7), wefindthat ψ ( q(x n, x n+1 ) ) = ψ ( q(tx n 1, Tx n ) ) α(x n 1, x n )ψ ( q(tx n 1, Tx n ) ) ψ ( M(x n 1, x n ) ) φ ( M(x n 1, x n ) ) + LN(x n 1, x n ), (8) where N(x n 1, x n )=min { d q m (x n 1, x n ), d q m (x n, x n+1 ), d q m (x n 1, x n+1 ), d q m (x n, x n ) } =0 (9) and M(x n 1, x n )=max { q(x n 1, x n ), q(x n 1, Tx n 1 ), q(x n, Tx n ) } = max { q(x n 1, x n ), q(x n 1, x n ), q(x n, x n+1 ) } = max { q(x n 1, x n ), q(x n, x n+1 ) }. (10)

7 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 7 of 20 Thus,weconcludefrom(8)that ψ ( q(x n, x n+1 ) ) ψ ( max { q(x n 1, x n ), q(x n, x n+1 ) }) φ ( max { q(x n 1, x n ), q(x n, x n+1 ) }) (11) by taking (10) and(9) into account. If for some n we have max{q(x n 1, x n ), q(x n, x n+1 )} = q(x n, x n+1 ), then (11)yieldsthat ψ ( q(x n, x n+1 ) ) ψ ( q(x n, x n+1 ) ) φ ( q(x n, x n+1 ) ). Consequently, we conclude that φ(q(x n, x n+1 )) = 0. Since φ 1 ({0})={0},wegetq(x n, x n+1 )= 0, which contradicts the assumption that q(x n, x n+1 ) > 0 for all n N 0.Thus,wehave M(x n 1, x n )=q(x n 1, x n ). (12) Hence, (8)turnsinto ψ ( q(x n, x n+1 ) ) ψ ( q(x n 1, x n ) ) φ ( q(x n 1, x n ) ) ψ ( q(x n 1, x n ) ) for all n N. Due to the property of the auxiliary function ψ,we have q(x n, x n+1 ) q(x n 1, x n ) for all n N. (13) Eventually, we observe that the sequence {q(x n, x n+1 )} is non-increasing. So, there exists δ 0suchthat lim n + q(x n, x n+1 )=δ. If δ >0,takinglim sup n + in inequality (11), by keeping (10) and(12) inthemind,we obtain that lim sup n + ψ ( q(x n, x n+1 ) ) lim sup ψ ( q(x n 1, x n ) ) lim inf φ( q(x n 1, x n ) ). n + n + By continuity of ψ and lower semi-continuity of φ,weobtainψ(δ) ψ(δ) φ(δ), which is a contradiction. So, lim n + q(x n, x n+1 )=0. (14) Analogously, we derive that lim q(x n+1, x n )=0. (15) n + Now, we shall show that {x n } is a Cauchy sequence in the quasi-partial metric space (X, q), that is, the sequence {x n } is left-cauchy and right-cauchy.

8 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 8 of 20 Suppose that {x n } is not a left-cauchy sequence in (X, q). Then there is ε >0suchthat for each integer k there exist integers n(k)>m(k)>k such that q(x n(k), x m(k) ) ε. (16) Further, corresponding to m(k), we can choose n(k) so that it is the smallest integer with n(k)>m(k)satisfying(16). Consequently, we have q(x n(k) 1, x m(k) )<ε. (17) Due to the triangle inequality, we have ε q(x n(k), x m(k) ) q(x n(k), x n(k) 1 )+q(x n(k) 1, x m(k) ) q(x n(k) 1, x n(k) 1 ) < q(x n(k), x n(k) 1 )+ε. (18) Letting k and taking (14) into account, we get that lim k q(x n(k), x m(k) )=ε. (19) On the other hand, again by the triangle inequality, we find that q(x n(k), x m(k) ) q(x n(k), x n(k) 1 )+q(x n(k) 1, x m(k) 1 )+q(x m(k) 1, x m(k) ) q(x n(k) 1, x n(k) 1 ) q(x m(k) 1, x m(k) 1 ) q(x n(k), x n(k) 1 )+q(x n(k) 1, x m(k) 1 )+q(x m(k) 1, x m(k) ) (20) and q(x n(k) 1, x m(k) 1 ) q(x n(k) 1, x n(k) )+q(x n(k), x m(k) )+q(x m(k), x m(k) 1 ) q(x n(k), x n(k) ) q(x m(k), x m(k) ) q(x n(k) 1, x n(k) )+q(x n(k), x m(k) )+q(x m(k), x m(k) 1 ). (21) Letting k andtaking (14), (15), (19), (20), (21) into account, we derive that lim q(x n(k) 1, x m(k) 1 )=ε, (22) k q(x n(k) 1, x m(k) ) q(x n(k) 1, x n(k) )+q(x n(k), x m(k) ) q(x n(k), x n(k) ) q(x n(k) 1, x n(k) )+q(x n(k), x m(k) ). (23) Letting k and taking (14), (18), (20), (23) into account, we derive that lim k q(x n(k) 1, x m(k) )=ε. (24)

9 Karapınar etal. Fixed Point Theory and Applications (2015) 2015:105 Page 9 of 20 Since T is triangular α-orbital admissible, from Lemma 1.4 and Lemma 1.5 we derive that α(x n, x m ) 1 and α(x m, x n ) 1, n > m N 0. (25) Regarding (6)and(25), wefindthat ψ ( q(x n(k), x m(k) ) ) = ψ ( q(tx n(k) 1, Tx m(k) 1 ) ) α(x n(k) 1, x m(k) 1 )ψ ( q(tx n(k) 1, Tx m(k) 1 ) ) ψ ( M(x n(k) 1, x m(k) 1 ) ) φ ( M(x n(k) 1, x m(k) 1 ) ) + LN(x n(k) 1, x m(k) 1 ), (26) where N(x n(k) 1, x m(k) 1 ) = min { d q m (x n(k) 1, x n(k) ), d q m (x m(k) 1, x m(k) ), d q m (x n(k) 1, x m(k) ), d q m (x m(k) 1, x n(k) ) } (27) and M(x n(k) 1, x m(k) 1 )=max { q(x n(k) 1, x m(k) 1 ), q(x n(k) 1, x n(k) ), q(x m(k) 1, x m(k) ) }. (28) Regarding (14) and(15), we note that lim N(x n(k) 1, x m(k) 1 )=0. (29) k On the other hand, we get that lim k M(x n(k) 1, x m(k) 1 )=ε (30) due to the limits (14), (15), (19). From the above observation, letting k in (26), we obtain ψ(ε) ψ(ε) φ(ε). So, φ(ε) = 0, which is a contradiction with respect to the fact that ε >0.Thus{x n } is a left-cauchy sequence in the metric space (X, q). Analogously, we derive that {x n } is a right- Cauchy sequence in the metric space (X, q). Since (X, q) is complete, then from Lemma 1.3 (X, d pq ) is a complete metric space. Therefore, the sequence {x n } convergestoapointu X in (X, d pq ), that is, lim n + d p q (x n, u)=0. Again, from Lemma 1.3, p q (u, u)= lim n + p q(x n, u)= lim n + p q(x n, x n ).

10 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 10 of 20 On the other hand, by (14) and the condition (QPM2) from Definition 1.3, lim q(x n, x n )=0. (31) n + So, it follows that 1 [ q(u, u)= lim q(xn, u)+q(u, x n ) ] = lim n + 2 q(x n, x n )=0. (32) n + Now, for proving fixed point of T,firstwesupposethatT is continuous, then we have Tu = lim n + Tx n = lim n + x n+1 = u. So, u is a fixed point of T. As the last step, suppose that X is α-regular. Hence it is α-right-regular, then there exists a subsequence {x n(k) } of {x n } such that α(x n(k), u) 1 for all k. Now, we show that q(u, Tu) = 0. Assume this is not true, then from (6) weobtain ψ ( q(x n(k)+1, Tu) ) = ψ ( q(tx n(k), Tu) ) α(x n(k), u)ψ ( q(tx n(k), Tu) ) ψ ( M(x n(k), u) ) φ ( M(x n(k), u) ) + L min { d q m (x n(k), Tx n(k) ), d q m (u, Tu), dq m (x n(k), Tu), d q m (u, Tx n(k)) }, where M(x n(k), u)=max { q(x n(k), u), q(x n(k) Tx n(k) ), q(u, Tu) } (33) = max { q(x n(k), u), q(x n(k) x n(k)+1 ), q(u, Tu) }. (34) It is obvious that lim k + q(x n(k), Tu)=q(u, Tu). Therefore, using (14) and(32), we deduce that Also, lim K + M(x n(k), u) max { 0, 0, q(u, Tu) } = q(u, Tu). lim K + N(x n(k), u)=0 because (14), (15)and(31)givelim n + d q m(x n(k), Tx n(k) )=0.Now,byusingtheproperties of ψ and φ and taking the upper limit as n +,weobtain ψ ( q(u, Tu) ) ψ ( q(u, Tu) ) φ ( q(u, Tu) ). Then φ(q(u, Tu)) = 0, i.e., q(u, Tu)=0,andsoTu = u.now,weconcludethatt has a fixed point u X and q(u, u)=0. As a consequence of Theorem 2.1, we may state the following corollaries. First, taking L =0inTheorem2.1, wehavethefollowing.

11 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 11 of 20 Corollary 2.1 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbeaselfmapping. Suppose that there exist ψ, φ and a function α : X X [0, + ) such that for all x, y X, α(x, y)ψ ( q(tx, Ty) ) ψ ( M(x, y) ) φ ( M(x, y) ). (35) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Corollary 2.2 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbea self-mapping. Suppose that there exist k [0, 1), L 0 and a function α : X X [0, + ) such that for all x, y X, α(x, y)q(tx, Ty) km(x, y)+ln(x, y). (36) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Proof It followsby taking ψ(t)=t and φ(t)=(1 k)t in Theorem 2.1. Corollary 2.3 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbea self-mapping. Suppose that there exist k [0, 1), L 0 and a function α : X X [0, + ) such that for all x, y X, α(x, y)q(tx, Ty) km(x, y). (37) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Proof It is sufficient to take L =0inCorollary2.2. Corollary 2.4 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbea self-mapping. Suppose that there exist k [0, 1), L 0 and a function α : X X [0, + ) such that for all x, y X, α(x, y)q(tx, Ty) kq(x, y). (38) Also, suppose that the following assertions hold:

12 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 12 of 20 (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Proof By following the lines in the proof of Theorem 2.1, we derive the desired result. We skip the details to avoid repetition. Denote by the set of functions λ :[0,+ ) [0, + ) satisfying the following hypotheses: (1) λ is a Lebesgue-integrable mapping on each compact subset of [0, + ), (2) for every ɛ >0,wehave ɛ 0 λ(s) ds >0. We have the following result. Corollary 2.5 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbeaselfmapping. Suppose that there exist λ, β, L 0 and a function α : X X [0, + ) such that for all x, y X, α(x,y)q(tx,ty) 0 λ(s) ds M(x,y) 0 λ(s) ds M(x,y) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Proof It followsfromtheorem2.1by taking and ψ(t)= t 0 λ(s) ds 0 β(s) ds + LN(x, y). (39) φ(t)= t 0 β(s) ds. Taking L =0inCorollary2.5, we obtain the following result. Corollary 2.6 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbeaselfmapping. Suppose that there exist λ, β and a function α : X X [0, + ) such that for all x, y X, α(x,y)q(tx,ty) 0 λ(s) ds M(x,y) 0 λ(s) ds M(x,y) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; 0 β(s) ds. (40)

13 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 13 of 20 (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Now, let F be the set of functions ϕ :[0,+ ) [0, + ) satisfying the following hypotheses: (ϕ 1 ) ϕ is non-decreasing, (ϕ 2 ) + n=0 ϕn (t) converges for all t >0. Note that if ϕ F, ϕ is said to be a (C)-comparison function. It is easily proved that if ϕ is a (C)-comparison function, then ϕ(t)<t for any t >0.Oursecondmainresultisas follows. Theorem 2.2 Let (X, q) be a complete quasi-partial metric space. Let T : X Xbea mapping such that there exist ϕ F, L 0 and a function α : X X [0, + ) such that for all x, y X, α(x, y)q(tx, Ty) ϕ ( M(x, y) ) + LN(x, y). (41) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible; (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. Proof By following the lines in the proof of Theorem 2.1,wederivethat α(x n, x n+1 ) 1 and α(x n+1, x n ) 1, n 1. (42) By combining (41)and(42), wefind that q(x n, x n+1 ) α(x n 1, x n )q(x n, x n+1 ) ϕ ( M(x n 1, x n ) ) + LN(x n 1, x n ). So, q(x n, x n+1 )=q(tx n 1, Tx n ) ϕ ( M(x n 1, x n ) ) + L min { d q m (x n 1, Tx n 1 ), d q m (x n, Tx n ), d q m (x n 1, Tx n ), d q m (x n, Tx n 1 ) }. Now,similartoprovingTheorem2.1,weobtain min { d q m (x n 1, Tx n 1 ), d q m (x n, Tx n ), d q m (x n 1, Tx n ), d q m (x n, Tx n 1 ) } =0 and M(x n 1, x n )=max { q(x n 1, x n ), q(x n, x n+1 ) }.

14 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 14 of 20 Therefore q(x n, x n+1 ) ϕ ( max { q(x n 1, x n ), q(x n, x n+1 ) }). (43) If, for some n 1, we have q(x n 1, x n ) q(x n, x n+1 ). So, from (43), we obtain that q(x n, x n+1 ) ϕ ( q(x n, x n+1 ) ) < q(x n, x n+1 ), which is a contradiction. Thus, for all n 1, we have M(x n 1, x n )=max { q(x n 1, x n ), q(x n, x n+1 ) } = q(x n 1, x n ). (44) Using (43)and(44), we get that q(x n, x n+1 ) ϕ ( q(x n 1, x n ) ) for all n 1. Iteratively, we find that q(x n, x n+1 ) ϕ n( q(x 0, x 1 ) ) for all n 0. By the triangle inequality, we get, for m > n, k=m 1 k=m 1 q(x n, x m ) q(x k, x k+1 ) q(x k, x k ) (45) k=n k=m 1 k=n k=+ k=n+1 q(x k, x k+1 ) (46) q(x k, x k+1 ) (47) k=n ϕ k( q(x 0, x 1 ) ) (48) k=+ k=n = ϕ n( q(x 0, x 1 ) ) k=+ k=0 ϕ n( q(x 0, x 1 ) ). (49) Since + k=0 ϕk (q(x 0, x 1 )) <, thenlim n,m + q(x n, x m )=0.So,{x n } is a right-cauchy sequence in (X, q). Similarly, since α(tx 0, x 0 ) 1, we get lim n + q(x m, x n ) = 0. (50) So, {x n } is a left-cauchy sequence in (X, q). By part (ii) of Definition 1.5, {x n } is a Cauchy sequence in a complete quasi-partial metric space (X, q), then {x n } converges, with respect to τ q,toapointu X such that q(u, u)= lim n + q(x n, u)= lim n + q(u, x n) = lim q(x n, x m )= lim q(x m, x n )=0. (51) n,m + n,m +

15 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 15 of 20 Suppose that T is continuous, then we have Tu = lim n + Tx n = lim n + x n+1 = u. So, u is a fixed point of T. Now, suppose that X is α-regular. Hence it is both α-left-regular and α-right-regular. Thus, there exists a subsequence {x n(k) } of {x n } such that α(x n(k), u) 1 for all k. Now,we claim that q(u, Tu) =0. Supposethecontrary, then q(u, Tu) >0. By(41), we have q(u, Tu) q(u, x n(k)+1 )+q(tx n(k), Tu) q(x n(k)+1, Tx n(k) ) (52) q(u, x n(k)+1 )+q(tx n(k), Tu) (53) q(u, x n(k)+1 )+α(x n(k), u)ϕ ( M(x n(k), u) ) (54) + L min { dm q (x n(k), Tx n(k) ), dm q (u, Tu), dq m (u, Tx n(k)), dm q (x n(k), Tu) }, (55) where M(x n(k), u)=max { q(x n(k), u), q(x n(k), Tx n(k) ), q(u, Tu) } (56) = max { q(x n(k), u), q(x n(k), x n(k)+1 ), q(u, Tu) }. (57) By (51), we have and lim min{ dm q (x n(k), Tx n(k) ), dm q (u, Tu), k + dq m (u, Tx n(k)), dm q (x n(k), Tu) } =0 lim k + M(x n(k), u)=q(u, Tu). Therefore q(u, Tu) ϕ ( q(u, Tu) ) < q(u, Tu), which is a contradiction. That is, q(u, Tu)=0.Thus,weobtainedthatu is a fixed point for T and q(u, u)=0. The following corollary is a generalization of Theorem 17 in [22]. Corollary 2.7 Let (X, q) be a complete quasi-partial metric space. Let n : X Xbea mapping such that there exist ϕ F and a function α : X X [0, + ) such that for all x, y X, α(x, y)q(tx, Ty) ϕ ( M(x, y) ). (58) Also, suppose that the following assertions hold: (i) T is triangular α-orbital admissible;

16 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 16 of 20 (ii) there exists x 0 X such that α(x 0, Tx 0 ) 1 and α(tx 0, x 0 ) 1; (iii) T is continuous or X is α-regular. Then T has a fixed point u Xandq(u, u)=0. We give the following two examples making effective our obtained results. Example 2.1 Let X =[0,+ )andq(x, y)= x y + x for all x, y X.Then(X, q)isacomplete quasi-partial metric space. Consider T : X X defined by Tx = x 2. Define α : X X [0, + )by α(x, y)= { 1 ifx y, 2 3 otherwise. Take ψ(t)= 2t 3 and φ(t)= t for all t 0. Note that ψ and φ.takex y,then 3 α(x, y)ψ ( q(tx, Ty) ) = α(x, y)ψ ( Tx Ty + Tx ) ( x = ψ 2 y 2 + x ) = y 2 3 y 2 = y y 2 = ψ ( M(x, y) ) φ ( M(x, y) ) ( since M(x, y)= 3y ) 2 ψ ( M(x, y) ) φ ( M(x, y) ) + LN(x, y) for all L 0. Now, let y < x,then α(x, y)ψ ( q(tx, Ty) ) = α(x, y)ψ ( Tx Ty + Tx ) = 2 ( 3 ψ x 2 y 2 + x ) = x 2 9 y. We have two possibilities for M(x, y). Case 1: if M(x, y)=2x y then α(x, y)ψ ( q(tx, Ty) ) = 4 9 x 2 9 y 2 3 x y 3 = ψ( M(x, y) ) φ ( M(x, y) ) ψ ( M(x, y) ) φ ( M(x, y) ) + LN(x, y) for all L 0. Case 2: if M(x, y)= 3x 2 then α(x, y)ψ ( q(tx, Ty) ) = 4 9 x 2 9 y x 2 = ψ( M(x, y) ) φ ( M(x, y) ) ψ ( M(x, y) ) φ ( M(x, y) ) + LN(x, y) for all L 0.

17 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 17 of 20 Moreover, T is triangular α-orbital admissible, α(0, T0) 1andα(T0, 0) 1. Thus, by applying Theorem 2.1, n has a fixed point, which is u =0. Example 2.2 Let X =[0,+ )andq(x, y)=max{x, y} for all x, y X.Then(X, q)isacomplete quasi-partial metric space. Consider n : X X defined by Tx = x 1+x. Define α : X X [0, + )by α(x, y)= { 1 ifx, y [0, 1], 0 otherwise. Take ψ(t)=1andφ(t)= 1 for all t 0. Note that ψ and φ.takex y,then 1+t α(x, y)ψ ( q(tx, Ty) ) ( ) y = α(x, y) 1+y y 1+y =1 1 1+y = ψ ( M(x, y) ) φ ( M(x, y) ) ( since M(x, y)=y ) ψ ( M(x, y) ) φ ( M(x, y) ) + LN(x, y) for all L 0. Moreover, T is triangular α-orbital admissible, α(0, T0) 1andα(T0, 0) 1. Thus, by applying Theorem 2.1, n has a fixed point, which is u =0. Example 2.3 Let X =[0,2]andq : X X R + be defined by q(x, y) =max{x, y}. Then (X, q) is a complete quasi-partial metric space. Define T : X X by x 2 if x [0, 1), x+1 T(x)= 0 ifx [1, 2), 4 if x =2 3 and α : X X [0, + )by 1 ifx, y [0, 1), 1 α(x, y)= if x, y [1, 2], 2 0 otherwise. Take ϕ :[0,+ ) [0, + )definedby ϕ(t)= t2 t +1. By induction, we have ϕ n (t) t( t 1+t )n for all n 1, so it is clear that ϕ is a (C)-comparison function. Now, we show that (41) is satisfied for all x, y X.Itsufficestoproveitforx y. Consider the following six cases.

18 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 18 of 20 Case 1. Let x, y [0, 1), then ( y 2 ) α(x, y)q(tx, Ty) =α(x, y) y +1 (59) = y2 y +1 = ϕ( q(x, y) ) (60) ϕ ( M(x, y) ). (61) Case 2. Let x, y [1, 2), then α(x, y)q(nx, ny) = 1 q(0, 0) = 0 (62) 2 ϕ ( M(x, y) ). (63) Case 3. Let x = y =2,then α(x, y)q(tx, Ty) = 1 2 q ( 4 3, 4 3 ) = 4 6 (64) 4 = ϕ(2) (65) 3 ϕ ( M(x, y) ). (66) Case 4. Let x [0, 1) and y [1, 2), then ( x 2 ) α(x, y)q(nx, ny) =α(x, y)q x +1,0 =0 (67) ϕ ( M(x, y) ). (68) Case 5. Let x [0, 1) and y =2,then ( x 2 α(x, y)q(tx, Ty) =α(x, y)q x +1, 4 ) =0 3 (69) ϕ ( M(x, y) ). (70) Case 6. Let x [1, 2) and y =2,then α(x, y)q(nx, ny) = 1 ( 2 q 0, 4 ) = (71) 4 = ϕ(2) (72) 3 = ϕ ( q(x, y) ) (73) ϕ ( M(x, y) ). (74) Since, for all x, y X, LN(x, y) 0, so we have α(x, y)q(tx, Ty) ϕ ( M(x, y) ) + LN(x, y).

19 Karapınar etal.fixed Point Theory and Applications (2015) 2015:105 Page 19 of 20 Moreover, T is triangular α-orbital admissible, α(0, T0) 1andα(n0, 0) 1. Then all the required hypotheses of Theorem 2.2 are satisfied. So, T has a fixed point, which is u =0. 3 Consequences and final remarks Now, we will show that many existing results in the literature can be deduced easily from our Corollary Standard fixed point theorems Taking in Corollary 2.7 α(x, y) = 1 for all x, y X, we derive immediately the following fixed point theorem. Corollary 3.1 Let (X, q) be a complete quasi-partial metric space and T : X Xbea given mapping. Suppose thatthere existsa function ϕ F such that q(tx, Ty) ϕ ( M(x, y) ) for all x, y X. Then T has a unique fixed point. Corollary 3.2 Let (X, q) be a complete quasi-partial metric space and T : X Xbea given mapping. Suppose thatthere exists k [0,1) such that q(tx, Ty) km(x, y) for all x, y X. Then T has a unique fixed point. Proof It is sufficient to take ϕ(t)=kt,wherek [0, 1), in the above corollary. Corollary 3.3 Let (X, q) be a complete quasi-partial metric space and T : X Xbea given mapping. Suppose thatthere exists k [0,1) such that q(tx, Ty) kq(x, y) for all x, y X. Then T has a unique fixed point. Remark 2 As we show in Corollaries , we can list some more results as a consequence of our main theorems by choosing the auxiliary function α : X X [0, )asit was done in e.g. [14, 17]. 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, Atilim University, İncek, Ankara, 06836, Turkey. 2 Nonlinear Analysis and Applied Mathematics Research Group (NAAM), King Abdulaziz University, Jeddah, Saudi Arabia. 3 Young Researchers and Elite Club, Roudehen Branch, Islamic Azad University, Roudehen, Iran. Received: 8 April 2015 Accepted: 14 June 2015

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