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Generalized α-ψ-contractive type mappings of integral type and related fixed point theorems

Abstract

The aim of this paper is to introduce two classes of generalized α-ψ-contractive type mappings of integral type and to analyze the existence of fixed points for these mappings in complete metric spaces. Our results are improved versions of a multitude of relevant fixed point theorems of the existing literature.

MSC:54H25, 47H10, 54E50.

1 Introduction and preliminaries

Recently, Samet et al. [1] introduced a very interesting notion of α-ψ-contractions via α-admissible mappings. In this paper, the authors [1] proved the existence and uniqueness of a fixed point for such a class of mappings in the context of complete metric spaces. Furthermore, the famous Banach [2] fixed point result was observed as a consequence of their main results. Following this initial paper, several authors have published new fixed point results by modifying, improving and generalizing the notion of α-ψ-contractions in various abstract spaces; see, e.g., [38]. Very recently, Shahi et al. [9] gave the integral version of α-ψ-contractive type mappings and proved some related fixed point theorems. As a consequence of the main results of this paper [9], the well-known integral contraction theorem of Branciari [10] and hence the celebrated Banach contraction principle were obtained.

In the present work, we introduce two classes of generalized α-ψ-contractive type mappings of integral type inspired by the report of Karapınar and Samet [7]. Also, we analyze the existence and uniqueness of fixed points for such mappings in complete metric spaces. Our results generalize, improve and extend not only the results derived by Shahi et al. [9], Samet et al. [1] and Branciari [10] but also various other related results in the literature. Moreover, from our fixed point theorems, we will derive several fixed point results on metric spaces endowed with a partial order.

We recall some necessary definitions and basic results from the literature. Throughout the paper, let denote the set of all nonnegative integers.

Berzig and Rus [4] introduced the following definition.

Definition 1.1 (see [4])

Let NN. We say that α is N-transitive (on X) if

x 0 , x 1 ,, x N + 1 X:α( x i , x i + 1 )1

for all i{0,1,,N}α( x 0 , x N + 1 )1.

In particular, we say that α is transitive if it is 1-transitive, i.e.,

x,y,zX:α(x,y)1andα(y,z)1α(x,z)1.

As consequences of Definition 1.1, we obtain the following remarks.

Remark 1.1 (see [4])

  1. (1)

    Any function α:X×X[0,+) is 0-transitive.

  2. (2)

    If α is N transitive, then it is kN-transitive for all kN.

  3. (3)

    If α is transitive, then it is N-transitive for all NN.

  4. (4)

    If α is N-transitive, then it is not necessarily transitive for all NN.

Let Ψ be a family of functions ψ:[0,)[0,) satisfying the following conditions:

  1. (1)

    ψ is nondecreasing.

  2. (2)

    n = 1 + ψ n (t)< for all t>0, where ψ n is the n th iterate of ψ.

In the literature, such mappings are called in two different ways: (c)-comparison functions in some sources (see, e.g., [11]), and Bianchini-Grandolfi gauge functions in some others (see, e.g., [1214]).

It can be easily verified that if ψ is a (c)-comparison function, then ψ(t)<t for any t>0.

Define Φ={φ:φ: R + R} such that φ is nonnegative, Lebesgue integrable and satisfies

0 ϵ φ(t)dt>0for each ϵ>0.
(1)

Shahi et al. in [9] introduced the following new concept of α-ψ-contractive type mappings of integral type.

Definition 1.2 Let (X,d) be a metric space and T:XX be a given mapping. We say that T is an α-ψ-contractive mapping of integral type if there exist two functions α:X×X[0,+) and ψΨ such that for each x,yX,

α(x,y) 0 d ( T x , T y ) φ(t)dtψ ( 0 d ( x , y ) φ ( t ) d t ) ,
(2)

where φΦ.

In what follows, we recollect the main results of Shahi et al. [9].

Theorem 1.1 [9]

Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is an α-ψ-contractive mapping of integral type and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    T is continuous.

Then T has a fixed point, that is, there exists zX such that Tz=z.

Theorem 1.2 [9]

Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is an α-ψ-contractive mapping of integral type and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    if { x n } is a sequence in X such that α( x n , x n + 1 )1 for all n and x n xX as n, then there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) ,x)1 for all k.

Then T has a fixed point, that is, there exists zX such that Tz=z.

Notice that in the theorems above, the authors proved only the existence of a fixed point. To guarantee the uniqueness of the fixed point, they needed the following condition.

(U): For all x,yFix(T), there exists zX such that α(x,z)1 and α(y,z)1, where Fix(T) denotes the set of fixed points of T.

2 Main results

In this section, we present our main results. First, we introduce two classes of generalized α-ψ-contractive type mappings of integral type in the following way.

Definition 2.1 Let (X,d) be a metric space and T:XX be a given mapping. We say that T is a generalized α-ψ-contractive mapping of integral type I if there exist two functions α:X×X[0,+) and ψΨ such that for each x,yX,

α(x,y) 0 d ( T x , T y ) φ(t)dtψ ( 0 M ( x , y ) φ ( t ) d t ) ,
(3)

where φΦ and M(x,y)=max{d(x,y),d(x,Tx),d(y,Ty),[ d ( x , T y ) + d ( y , T x ) 2 ]}.

Definition 2.2 Let (X,d) be a metric space and T:XX be a given mapping. We say that T is a generalized α-ψ-contractive mapping of integral type II if there exist two functions α:X×X[0,+) and ψΨ such that for each x,yX,

α(x,y) 0 d ( T x , T y ) φ(t)dtψ ( 0 M ( x , y ) φ ( t ) d t ) ,
(4)

where φΦ and M(x,y)=max{d(x,y),[ d ( x , T x ) + d ( y , T y ) 2 ],[ d ( x , T y ) + d ( y , T x ) 2 ]}.

Remark 2.1 It is evident that if T:XX is an α-ψ-contractive mapping of integral type, then T is a generalized α-ψ-contractive mapping of integral types I and II.

The following is the first main result of this manuscript.

Theorem 2.1 Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is a generalized α-ψ-contractive mapping of integral type I and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    T is continuous.

Then T has a fixed point, that is, there exists zX such that Tz=z.

Proof Let x 0 be an arbitrary point of X such that α( x 0 ,T x 0 )1. We construct an iterative sequence { x n } in X in the following way:

x n + 1 =T x n for all n0.

If x n 0 = x n 0 + 1 for some n 0 , then, obviously, x = x n 0 is a fixed point of T and the proof is completed. Hence, from now on, we suppose that x n x n + 1 for all n. Due to the fact that T is α-admissible, we find that

α( x 0 , x 1 )=α( x 0 ,T x 0 )1α(T x 0 ,T x 1 )=α( x 1 , x 2 )1.

Iteratively, we obtain that

α( x n , x n + 1 )1
(5)

for all n0.

By applying inequality (3) with x= x n 1 and y= x n and using (5), we deduce that

0 d ( x n , x n + 1 ) φ ( t ) d t = 0 d ( T x n 1 , T x n ) φ ( t ) d t α ( x n 1 , x n ) 0 d ( T x n 1 , T x n ) φ ( t ) d t ψ ( 0 M ( x n 1 , x n ) φ ( t ) d t ) ,
(6)

where

M ( x n 1 , x n ) = max { d ( x n 1 , x n ) , d ( x n 1 , x n ) , d ( x n , x n + 1 ) , d ( x n 1 , x n + 1 ) + d ( x n , x n ) 2 } max { d ( x n 1 , x n ) , d ( x n , x n + 1 ) } .
(7)

By utilizing (7) and regarding the properties of the function ψ, we derive from (6) that

0 d ( x n , x n + 1 ) φ ( t ) d t = 0 d ( T x n 1 , T x n ) φ ( t ) d t α ( x n 1 , x n ) 0 d ( T x n 1 , T x n ) φ ( t ) d t ψ ( 0 max { d ( x n 1 , x n ) , d ( x n , x n + 1 ) } φ ( t ) d t ) ψ ( max { 0 d ( x n 1 , x n ) φ ( t ) d t , 0 d ( x n , x n + 1 ) φ ( t ) d t } ) ψ ( 0 d ( x n 1 , x n ) φ ( t ) d t ) .
(8)

Notice that the case

0 d ( x n , x n + 1 ) φ(t)dtψ ( 0 d ( x n , x n + 1 ) φ ( t ) d t ) < 0 d ( x n , x n + 1 ) φ(t)dt

is impossible due to the property ψ(t)<t for all t>0. By using mathematical induction, we get, for all nN,

0 d ( x n , x n + 1 ) φ(t)dt ψ n ( 0 d ( x 0 , x 1 ) φ ( t ) d t ) = ψ n (d),
(9)

where d= 0 d ( x 0 , x 1 ) φ(t)dt.

Letting n+ in (9) and taking the property of ψ on the account, we find that

0 d ( x n , x n + 1 ) φ(t)dt=0,
(10)

which, from (1), implies that

d( x n , x n + 1 )0as n.
(11)

We shall prove that { x n } is a Cauchy sequence. Suppose, on the contrary, that there exist an ϵ>0 and subsequences {m(p)} and {n(p)} such that m(p)<n(p)<m(p+1) with

d( x m ( p ) , x n ( p ) )ϵ,d( x m ( p ) , x n ( p ) 1 )<ϵ.
(12)

Due to the definition of M(x,y), we have that

M ( x m ( p ) 1 , x n ( p ) 1 ) = max { d ( x m ( p ) 1 , x n ( p ) 1 ) , d ( x m ( p ) 1 , x m ( p ) ) , d ( x n ( p ) 1 , x n ( p ) ) , d ( x m ( p ) 1 , x n ( p ) ) + d ( x n ( p ) 1 , x m ( p ) ) 2 } .
(13)

By elementary evaluation, (11), we find that

lim p 0 d ( x m ( p ) 1 , x m ( p ) ) φ(t)dt= lim p 0 d ( x n ( p ) 1 , x n ( p ) ) φ(t)dt=0.
(14)

In view of (11), (12) and the triangular inequality, we deduce that

d ( x m ( p ) 1 , x n ( p ) 1 ) d ( x m ( p ) 1 , x m ( p ) ) + d ( x m ( p ) , x n ( p ) 1 ) < ϵ + d ( x m ( p ) 1 , x m ( p ) ) .

Letting n in the inequality above, we conclude that

lim p 0 d ( x m ( p ) 1 , x n ( p ) 1 ) φ(t)dt 0 ϵ φ(t)dt.
(15)

Owing to the transitivity of α, we infer from (5) that

α( x m ( p ) 1 , x n ( p ) 1 )1.
(16)

Regarding inequality (3) and by using (16), we obtain

0 d ( x m ( p ) , x n ( p ) ) φ ( t ) d t = 0 d ( T x m ( p ) 1 , T x n ( p ) 1 ) φ ( t ) d t α ( x m ( p ) 1 , x n ( p ) 1 ) 0 d ( T x m ( p ) 1 , T x n ( p ) 1 ) φ ( t ) d t ψ ( 0 M ( x m ( p ) 1 , x n ( p ) 1 ) φ ( t ) d t ) .
(17)

In view of (12) and using the triangular inequality, we get

t ( m , n ) = d ( x m ( p ) 1 , x n ( p ) ) + d ( x n ( p ) 1 , x m ( p ) ) 2 d ( x m ( p ) 1 , x m ( p ) ) + 2 d ( x m ( p ) , x n ( p ) 1 ) + d ( x n ( p ) 1 , x n ( p ) ) 2 < d ( x m ( p ) 1 , x m ( p ) ) + d ( x n ( p ) 1 , x n ( p ) ) 2 + ϵ .
(18)

Therefore, using (11), we infer that

lim p 0 t ( m , n ) φ(t)dt 0 ϵ φ(t)dt.
(19)

Now, from (3), (12), (13), (14), (15), (16) and (19), it then follows that

0 ϵ φ ( t ) d t 0 d ( x m ( p ) , x n ( p ) ) φ ( t ) d t α ( x m ( p ) 1 , x n ( p ) 1 ) 0 d ( T x m ( p ) 1 , T x n ( p ) 1 ) φ ( t ) d t ψ ( 0 M ( x m ( p ) 1 , x n ( p ) 1 ) φ ( t ) d t ) ψ ( 0 ϵ φ ( t ) d t ) ,
(20)

which is a contradiction. This implies that { x n } is a Cauchy sequence in (X,d). Due to the completeness of (X,d), there exists zX such that x n z as n+. The continuity of T yields that T x n Tz as n+, that is, x n + 1 Tz as n+. By the uniqueness of the limit, we obtain z=Tz. Therefore, z is a fixed point of T. □

Theorem 2.2 Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is a generalized α-ψ-contractive mapping of integral type I and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    if { x n } is a sequence in X such that α( x n , x n + 1 )1 for all n and x n xX as n, then there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) ,x)1 for all k;

  4. (iv)

    ψ is continuous for all t>0.

Then T has a fixed point, that is, there exists zX such that Tz=z.

Proof From the proof of Theorem 2.1, we infer that the sequence { x n } defined by x n + 1 =T x n for all n0 converges to zX. We obtain, from hypothesis (iii) and (3), that there exists a subsequence { x n ( k ) } of x n such that α( x n ( k ) ,z)1 for all k. Now, applying inequality (3), we get, for all k,

0 d ( x n ( k ) + 1 , T z ) φ ( t ) d t = 0 d ( T x n ( k ) , T z ) φ ( t ) d t α ( x n ( k ) , z ) 0 d ( T x n ( k ) , T z ) φ ( t ) d t ψ ( 0 M ( x n ( k ) , z ) φ ( t ) d t ) .
(21)

On the other hand, we have

M( x n ( k ) ,z)=max { d ( x n ( k ) , z ) , d ( x n ( k ) , x n ( k ) + 1 ) , d ( z , T z ) , d ( x n ( k ) , T z ) + d ( z , x n ( k ) + 1 ) 2 } .
(22)

Recall from the proof of Theorem 2.1 that the sequence { x n } converges to zX. Consequently, as k, the limit of the terms d( x n ( k ) ,z), d( x n ( k ) , x n ( k ) + 1 ), d(z, x n ( k ) + 1 ) tends to 0. Thus, by letting k in (22), we get that

lim k M( x n ( k ) ,z)=d(z,Tz).
(23)

Assume that d(z,Tz)>0. In view of (23) and for k large enough, we get M( x n ( k ) ,z)>0, which implies from (21) that

0 d ( x n ( k ) + 1 , T z ) φ(t)dtψ ( 0 M ( x n ( k ) , z ) φ ( t ) d t ) .
(24)

Letting k in (24) and by using (23), assumption (iv), together with the property of ψ(t)<t, we derive that

0 d ( z , T z ) φ(t)dtψ ( 0 d ( z , T z ) φ ( t ) d t ) < 0 d ( z , T z ) φ(t)dt,
(25)

which is a contradiction. Thus, we have d(z,Tz)=0, that is, z=Tz. □

One can easily deduce the following result from Theorem 2.1.

Theorem 2.3 Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is a generalized α-ψ-contractive mapping of integral type II and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    T is continuous.

Then T has a fixed point, that is, there exists zX such that Tz=z.

In the next theorem, we exclude the continuity hypothesis of T in Theorem 2.3.

Theorem 2.4 Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is a generalized α-ψ-contractive mapping of integral type II and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    if { x n } is a sequence in X such that α( x n , x n + 1 )1 for all n and x n xX as n, then there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) ,x)1 for all k.

Then T has a fixed point, that is, there exists zX such that Tz=z.

Proof From the proof of Theorem 2.3, we infer that the sequence { x n } defined by x n + 1 =T x n for all n0 converges to zX. We obtain, from hypothesis (iii) and (3), that there exists a subsequence { x n ( k ) } of x n such that α( x n ( k ) ,z)1 for all k. Now, applying inequality (4), we get, for all k,

0 d ( x n ( k ) + 1 , T z ) φ ( t ) d t = 0 d ( T x n ( k ) , T z ) φ ( t ) d t α ( x n ( k ) , z ) 0 d ( T x n ( k ) , T z ) φ ( t ) d t ψ ( 0 M ( x n ( k ) , z ) φ ( t ) d t ) .
(26)

On the other hand, we have

M ( x n ( k ) , z ) = max { d ( x n ( k ) , z ) , d ( x n ( k ) , x n ( k ) + 1 ) + d ( z , T z ) 2 , d ( x n ( k ) , T z ) + d ( z , x n ( k ) + 1 ) 2 } .
(27)

Letting k in the above equality, we get that

lim k M( x n ( k ) ,z)= d ( z , T z ) 2 .
(28)

Assume that d(z,Tz)>0. In view of (28) and for k large enough, we get M( x n ( k ) ,z)>0, which implies from (26) that

0 d ( x n ( k ) + 1 , T z ) φ ( t ) d t ψ ( 0 M ( x n ( k ) , z ) φ ( t ) d t ) < 0 M ( x n ( k ) , z ) φ ( t ) d t .
(29)

Letting k in (29) and using (28), we obtain that

0 d ( z , T z ) φ(t)dt 0 d ( z , T z ) 2 φ(t)dt,
(30)

which is a contradiction. Thus, we have d(z,Tz)=0, that is, z=Tz. □

Remark 2.2 Notice that in Theorem 2.2, the continuity of ψ is assumed as an extra condition. Despite Remark 2.1, Theorem 2.4 can be derived from Theorem 2.2 due to the additional assumption on ψ.

In order to ensure the uniqueness of a fixed point of a generalized α-ψ-contractive mapping of integral type II, we need an additional condition (U) defined in the previous section.

Theorem 2.5 If the condition (U) is added to the hypotheses of Theorem  2.1, then the fixed point u of T is unique.

Proof We shall show the uniqueness of a fixed point of T by reductio ad absurdum. Suppose, on the contrary, that v is another fixed point of T with vu. From the hypothesis (U), we obtain that there exists zX such that

α(u,z)1,α(v,z)1.
(31)

Using the α-admissible property of T, we get from (31) for all nN

α ( u , T n z ) 1,α ( v , T n z ) 1.
(32)

Consider the sequence { z n } in X by z n + 1 =T z n for all n0 and z 0 =z. From (32), for all n, we infer that

0 d ( u , z n + 1 ) φ ( t ) d t = 0 d ( T u , T z n ) φ ( t ) d t α ( u , z n ) 0 d ( T u , T z n ) φ ( t ) d t ψ ( 0 M ( u , z n ) φ ( t ) d t ) .
(33)

On the other hand, we have

M ( u , z n ) = max { d ( u , z n ) , d ( u , T u ) + d ( z n , T z n ) 2 , d ( u , T z n ) + d ( z n , T u ) 2 } = max { d ( u , z n ) , d ( u , T u ) , d ( z n , T z n ) , d ( u , T z n ) + d ( z n , T u ) 2 } = max { d ( u , z n ) , 0 , d ( z n , z n + 1 ) , d ( u , z n + 1 ) + d ( z n , u ) 2 } max { d ( u , z n ) , d ( z n , z n + 1 ) , d ( u , z n + 1 ) + d ( z n , u ) 2 } max { d ( u , z n ) , d ( z n , z n + 1 ) , d ( u , z n + 1 ) } .
(34)

Due to the monotone property of ψ and using the above inequality, we infer from (33) that

0 d ( u , z n + 1 ) φ ( t ) d t ψ ( 0 M ( u , z n ) φ ( t ) d t ) ψ ( 0 max { d ( u , z n ) , d ( z n , z n + 1 ) , d ( u , z n + 1 ) } φ ( t ) d t ) ψ ( max { 0 d ( u , z n ) φ ( t ) d t , 0 d ( z n , z n + 1 ) φ ( t ) d t , 0 d ( u , z n + 1 ) φ ( t ) d t } )
(35)

for all n. Let us examine the possibilities for the inequality above. For simplicity, let

P(u, z n )=max { 0 d ( u , z n ) φ ( t ) d t , 0 d ( z n , z n + 1 ) φ ( t ) d t , 0 d ( u , z n + 1 ) φ ( t ) d t } .

If P(u, z n )= 0 d ( u , z n + 1 ) φ(t)dt, then due to the properties of the function ψ, we get

0 d ( u , z n + 1 ) φ(t)dtψ ( 0 d ( u , z n + 1 ) φ ( t ) d t ) < 0 d ( u , z n + 1 ) φ(t)dt,

which is a contradiction. If P(u, z n )= 0 d ( u , z n ) φ(t)dt, then

0 d ( u , z n + 1 ) φ(t)dtψ ( 0 d ( u , z n ) φ ( t ) d t ) ,

thereby implying that

0 d ( u , z n + 1 ) φ(t)dt ψ n ( 0 d ( u , z 0 ) φ ( t ) d t )
(36)

for all n1. Letting n in the above inequality, we obtain that

lim n 0 d ( u , z n + 1 ) φ(t)dt=0,
(37)

which from (1) implies that

lim n d( z n ,u)=0.
(38)

Let us analyze the last case: P(u, z n )= 0 d ( z n , z n + 1 ) φ(t)dt. Regarding the properties of ϕ and the triangle inequality, we have

d( z n , z n + 1 )d( z n ,u)+d(u, z n + 1 )2max { d ( z n , u ) , d ( u , z n + 1 ) } .

Notice that if d( z n ,u)d(u, z n + 1 ), then, as in the analysis of the first case, we get a contradiction. Hence,

d( z n , z n + 1 )d( z n ,u)+d(u, z n + 1 )2max { d ( z n , u ) , d ( u , z n + 1 ) } 2d( z n ,u),

and hence we easily deduce that

0 d ( u , z n + 1 ) φ ( t ) d t ψ ( 0 d ( z n , z n + 1 ) φ ( t ) d t ) ψ ( 0 d ( z n , u ) + d ( u , z n + 1 ) φ ( t ) d t ) ψ ( 0 2 max { d ( z n , u ) , d ( u , z n + 1 ) } φ ( t ) d t ) ψ ( 0 2 d ( z n , u ) φ ( t ) d t )

for each n. Consequently, we find that

0 d ( u , z n + 1 ) φ(t)dt ψ n ( 0 2 d ( u , z 0 ) φ ( t ) d t )
(39)

for all n1. Letting n in the above inequality, we obtain that

lim n 0 d ( u , z n + 1 ) φ(t)dt=0,
(40)

which from (1) implies that

lim n d( z n ,u)=0.
(41)

Similarly, we can show that

lim n d( z n ,v)=0.
(42)

From equations (41) and (42), we obtain that u=v. Therefore, we have proved that u is the unique fixed point of T. □

The following result can be easily deduced from Theorem 2.5 due to Remark 2.1.

Theorem 2.6 Adding the condition (U) to the hypotheses of Theorem  2.3 (resp. Theorem  2.4), one obtains that u is the unique fixed point of T.

3 Consequences

In this section, we shall list some existing results in the literature that can be deduced easily from our Theorem 2.6.

3.1 Standard fixed point theorems

Theorem 1.1 and Theorem 1.2 are immediate consequences of our main results Theorem 2.1 and Theorem 2.3 where M(x,y)=d(x,y).

Corollary 3.1 (see Karapınar and Samet [7])

Let (X,d) be a complete metric space and α:X×X[0,+) be a transitive mapping. Suppose that T:XX is a generalized α-ψ-contractive mapping and satisfies the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    T is continuous.

Then T has a fixed point, that is, there exists zX such that Tz=z.

Proof It is sufficient to take φ(t)=1 for all t0 in Theorem 2.3. □

If one replaces φ(t)=1 for all t0 in Theorem 1.1, the following fixed point theorem is observed.

Corollary 3.2 (see Samet et al. [1])

Let (X,d) be a complete metric space and T:XX be an α-ψ-contractive mapping satisfying the following conditions:

  1. (i)

    T is α-admissible;

  2. (ii)

    there exists x 0 X such that α( x 0 ,T x 0 )1;

  3. (iii)

    T is continuous.

Then T has a fixed point, that is, there exists x X such that T x = x .

If we take α(x,y) = 1 for all x,yX and ψ(t)=kt for k[0,1) in Theorem 1.1, we derive the following result.

Corollary 3.3 (see Branciari [10])

Let (X,d) be a complete metric space, k[0,1), and let T:XX be a mapping such that for each x,yX,

0 d ( T x , T y ) φ(t)dtk 0 d ( x , y ) φ(t)dt,

where φΦ. Then T has a unique fixed point aX such that for each xX, lim n + T n x=a.

The following corollary is concluded from Corollary 3.1 by taking α(x,y)=1 for all x,yX.

Corollary 3.4 (see Karapınar and Samet [7])

Let (X,d) be a complete metric space and T:XX be a given mapping. Suppose that there exists a function ψΨ such that

d(Tx,Ty)ψ ( M ( x , y ) )

for all x,yX. Then T has a unique fixed point.

By taking ψ(t)=λt for λ[0,1) in Corollary 3.4, we get the next result.

Corollary 3.5 (see Ćirić [15])

Let (X,d) be a complete metric space and T:XX be a given mapping. Suppose that there exists a constant λ(0,1) such that

d(Tx,Ty)λmax { d ( x , y ) , d ( x , T x ) + d ( y , T y ) 2 , d ( x , T y ) + d ( y , T x ) 2 }

for all x,yX. Then T has a unique fixed point.

Corollary 3.6 (see Hardy and Rogers [16])

Let (X,d) be a complete metric space and T:XX be a given mapping. Suppose that there exist constants A,B,C0 with (A+2B+2C)(0,1) such that

d(Tx,Ty)Ad(x,y)+B [ d ( x , T x ) + d ( y , T y ) ] +C [ d ( x , T y ) + d ( y , T x ) ]

for all x,yX. Then T has a unique fixed point.

For the proof of the above corollary, it is sufficient to chose λ=max{A,B,C} in Corollary 3.5.

The next two results are obvious consequences of Corollary 3.5.

Corollary 3.7 (see Kannan [17])

Let (X,d) be a complete metric space and T:XX be a given mapping. Suppose that there exists a constant λ(0,1/2) such that

d(Tx,Ty)λ [ d ( x , T x ) + d ( y , T y ) ]

for all x,yX. Then T has a unique fixed point.

Corollary 3.8 (see Chatterjea [18])

Let (X,d) be a complete metric space and T:XX be a given mapping. Suppose that there exists a constant λ(0,1/2) such that

d(Tx,Ty)λ [ d ( x , T y ) + d ( y , T x ) ]

for all x,yX. Then T has a unique fixed point.

By taking y=Tx in Corollary 3.3, we obtain the following corollary.

Corollary 3.9 (Rhoades and Abbas [19])

Let T be a self-map of a complete metric space (X,d) satisfying

0 d ( T x , T 2 x ) φ(t)dtk 0 d ( x , T x ) φ(t)dt

for all xX and k[0,1), where φΦ. Then T has a unique fixed point aX.

Corollary 3.10 (Berinde [20])

Let (X,d) be a complete metric space and T:XX be a given mapping. Suppose that there exists a function ψΨ such that

d(Tx,Ty)ψ ( d ( x , y ) )

for all x,yX. Then T has a unique fixed point.

Proof Let α(x,y)=1 for all x,yX and φ(t)=1 for all t0 in Theorem 1.1. Then all the conditions of Theorem 1.1 are satisfied and the proof is completed. □

It is evident that we have the celebrated result of Banach.

Corollary 3.11 (Banach [2])

Let (X,d) be a complete metric space and T:XX be a given mapping satisfying

d(Tx,Ty)kd(x,y)for all x,yX,

where k[0,1). Then T has a unique fixed point.

3.2 Fixed point theorems on ordered metric spaces

Recently, there have been so many interesting developments in the field of existence of a fixed point in partially ordered sets. This idea was initiated by Ran and Reurings [21] where they extended the Banach contraction principle in partially ordered sets with some application to a matrix equation. Later, many remarkable results have been obtained in this direction (see, for example, [2229] and the references cited therein). In this section, we will establish various fixed point results on a metric space endowed with a partial order. For this, we require the following concepts.

Definition 3.1 Let (X,) be a partially ordered set and T:XX be a given mapping. We say that T is nondecreasing with respect to if

x,yX,xyTxTy.

Definition 3.2 Let (X,) be a partially ordered set. A sequence { x n }X is said to be nondecreasing with respect to if x n x n + 1 for all n.

Definition 3.3 [7]

Let (X,) be a partially ordered set and d be a metric on X. We say that (X,,d) is regular if for every nondecreasing sequence { x n }X such that x n xX as n, there exists a subsequence { x n ( k ) } of { x n } such that x n ( k ) x for all k.

Now, we have the following result.

Corollary 3.12 Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exist functions ψΨ and φΦ such that for all x,yX with xy, we have

0 d ( T x , T y ) φ(t)dtψ ( 0 M ( x , y ) φ ( t ) d t ) ,
(43)

where M(x,y)=max{d(x,y),[ d ( x , T x ) + d ( y , T y ) 2 ],[ d ( x , T y ) + d ( y , T x ) 2 ]}. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Proof Consider the mapping α:X×X[0,) by

α(x,y)= { 1 if  x y , 0 otherwise .

Clearly, α is transitive. In view of the definition of α, we infer that T is an α-ψ-contractive mapping of integral type, that is,

α(x,y) 0 d ( T x , T y ) φ(t)dtψ ( 0 M ( x , y ) φ ( t ) d t )
(44)

for all x,yX. From condition (i), we have α( x 0 ,T x 0 )1. Now, we proceed to show that T is α-admissible. For this, let α(x,y)1 for all x,yX. Moreover, owing to the monotone property of T, we have, for all x,yX,

α(x,y)1xyTxTyα(Tx,Ty)1.

Thus, T is α-admissible. Now, if T is continuous, we obtain the existence of a fixed point from Theorem 2.3. Now, assume that (X,,d) is regular. Suppose that { x n } is a sequence in X such that α( x n , x n + 1 )1 for all n and x n xX as n. Due to the fact that the space (X,,d) is regular, there exists a subsequence { x n ( k ) } of { x n } such that x n ( k ) x for all k. Owing to the definition of α, we get that α( x n ( k ) ,x)1 for all k. In this case, we get the existence of a fixed point from Theorem 2.4. Now, we have to show the uniqueness of the fixed point. For this, let x,yX. By hypothesis, there exists zX such that xz and yz, which implies from the definition of α that α(x,z)1 and α(y,z)1. Therefore, we obtain the uniqueness of the fixed point from Theorem 2.6. □

We can now easily derive the following results from Corollary 3.12.

Corollary 3.13 (Shahi et al. [9])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a function ψΨ such that for all x,yX with xy, we have

0 d ( T x , T y ) φ(t)dtψ ( 0 d ( x , y ) φ ( t ) d t ) ,

where φΦ. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Corollary 3.14 (Karapınar and Samet [7])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a function ψΨ such that

d(Tx,Ty)ψ ( M ( x , y ) )

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Proof By taking φ(t)=1 for all t0 in Corollary 3.12, we get the proof of this corollary. □

Corollary 3.15 (Karapınar and Samet [7])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a function ψΨ such that

d(Tx,Ty)ψ ( d ( x , y ) )

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Proof By taking φ(t)=1 for all t0 in Corollary 3.13, we get the proof of this corollary. □

Corollary 3.16 (Shahi et al. [9])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a function ψΨ such that for all x,yX with xy, we have

0 d ( T x , T y ) φ(t)dtk 0 d ( x , y ) φ(t)dt,

where φΦ. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Proof By taking ψ(t)=kt for all t0 and some k[0,1) in Corollary 3.13, we get the proof of this corollary. □

Corollary 3.17 (Ran and Reurings [21], Nieto and Rodriguez-Lopez [29])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a constant k(0,1) such that

d(Tx,Ty)kd(x,y)

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Proof Taking φ(t)=1 for all t0 in Corollary 3.16, we get the proof of this corollary. □

Corollary 3.18 (see Karapınar and Samet [7])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a constant λ(0,1) such that

d(Tx,Ty)λmax { d ( x , y ) , d ( x , T x ) + d ( y , T y ) 2 , d ( x , T y ) + d ( y , T x ) 2 }

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Corollary 3.19 (see Karapınar and Samet [7])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exist constants A,B,C0 with (A+2B+2C)(0,1) such that

d(Tx,Ty)Ad(x,y)+B [ d ( x , T x ) + d ( y , T y ) ] +C [ d ( x , T y ) + d ( y , T x ) ]

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Corollary 3.20 (see Karapınar and Samet [7])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a constant λ(0,1/2) such that

d(Tx,Ty)λ [ d ( x , T x ) + d ( y , T y ) ]

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

Corollary 3.21 (see Karapınar and Samet [7])

Let (X,) be a partially ordered set and d be a metric on X such that (X,d) is complete. Let T:XX be a nondecreasing mapping with respect to . Suppose that there exists a constant λ(0,1/2) such that

d(Tx,Ty)λ [ d ( x , T y ) + d ( y , T x ) ]

for all x,yX with xy. Suppose also that the following conditions hold:

  1. (i)

    there exists x 0 X such that x 0 T x 0 ;

  2. (ii)

    T is continuous or (X,,d) is regular.

Then T has a fixed point. Moreover, if for all x,yX there exists zX such that xz and yz, we have uniqueness of the fixed point.

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Acknowledgements

The authors are grateful to the reviewers for their careful reviews and useful comments. The first author was supported by the Research Center, College of Science, King Saud University.

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Karapınar, E., Shahi, P. & Tas, K. Generalized α-ψ-contractive type mappings of integral type and related fixed point theorems. J Inequal Appl 2014, 160 (2014). https://doi.org/10.1186/1029-242X-2014-160

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