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A discussion on generalized almost contractions via rational expressions in partially ordered metric spaces

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Abstract

The main purpose of this paper is to give some fixed point results for mappings involving generalized (ϕ,ψ)-contractions in partially ordered metric spaces. Our results generalize, extend, and unify several well-known comparable results in the literature (Jaggi in Indian J. Pure Appl. Math. 8(2):223-230, 1977, Harjani et al. in Nonlinear Anal. 71:3403-3410, 2009, Luong and Thuan in Fixed Point Theory Appl. 2011:46, 2011). The presented results are supported by three illustrative examples.

MSC: 46N40, 47H10, 54H25, 46T99.

1 Introduction and preliminaries

The Banach contraction mapping principle [1] is one of the pivotal results of analysis. It is widely considered as the source of metric fixed point theory. Also, its significance lies in its application in a vast number of branches of mathematics. Generalizations of this principle have been investigated heavily (see Jaggi [2], Harjani et al. [3], Luong and Thuan [4]). In particular, in 1977, Jaggi [2] proved the following theorem satisfying a contractive condition of a rational type.

Theorem 1 Let (X,d) be a complete metric space. Let T:XX be a continuous mapping such that

d(Tx,Ty)α d ( x , T x ) d ( y , T y ) d ( x , y ) +βd(x,y)
(1.1)

for all distinct points x,yX where α,β[0,1) with α+β<1. Then T has a unique fixed point.

Existence of fixed point in partially ordered sets has been recently studied in [353].

Recently, Harjani et al. [3] proved the ordered version of Theorem 1. Very recently, Luong and Thuan [4] generalized the results of [3] and proved the following.

Theorem 2 Let (X,) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a metric space. Let T:XX be a non-decreasing mapping such that

d(Tx,Ty)M(x,y)ψ ( M ( x , y ) )
(1.2)

for all distinct points x,yX with yx where ψ:[0,)[0,) is a lower semi-continuous function with the property that ψ(t)=0 if and only if t=0, and

M(x,y)=max { d ( x , T x ) d ( y , T y ) d ( x , y ) , d ( x , y ) } .
(1.3)

Also, assume either

  1. (i)

    T is continuous or

  2. (ii)

    if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Set Φ={ϕϕ:[0,)[0,) is continuous and non-decreasing with ϕ(t)=0 if andonly if t=0} and Ψ={ψψ:[0,)[0,) is lower semi continuous,ψ(t)>0 for allt>0, and ψ(0)=0}. For some work on the class of Φ or the class of Ψ, we refer the reader to [21, 51, 54].

In 2004, Berinde [55] introduced an almost contraction, a new class of contractive type mappings which exhibits totally different features more than the one of the particular results incorporated [1, 16, 39, 50], i.e., an almost contraction generally does not have a unique fixed point; see Example 1 in [55]. Thereafter, many authors presented several interesting and useful facts about almost contractions; see [42, 5659].

The purpose of this article is to generalize the above results for a mapping T:XX involving a generalized (ϕ,ψ)-almost contraction. Some examples are also presented to show that our results are effective.

2 Main result

Our essential result is given as follows.

Theorem 3 Let (X,) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:XX be a non-decreasing mapping which satisfies the inequality

ϕ ( d ( T x , T y ) ) ϕ ( M ( x , y ) ) ψ ( M ( x , y ) ) + L min { d ( x , T y ) , d ( y , T x ) , d ( x , T x ) , d ( y , T y ) }
(2.1)

for all distinct points x,yX with yx where ϕΦ, ψΨ, L0 and

M(x,y)=max { d ( x , T x ) d ( y , T y ) d ( x , y ) , d ( x , y ) } .

Also, assume either

  1. (i)

    T is continuous or

  2. (ii)

    if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Proof Let x 0 X such that x 0 T x 0 . We define a sequence { x n } in X as follows:

x n =T x n 1 for n1.
(2.2)

Since T is a non-decreasing mapping together with (2.2), we have x 2 =T x 1 . Inductively, we obtain

x 0 x 1 x 2 x n 1 x n x n + 1 .
(2.3)

Assume that there exists n 0 such that x n 0 = x n 0 + 1 . Since x n 0 = x n 0 + 1 =T x n 0 , then T has a fixed point. Suppose that x n x n + 1 for all nN. Thus, by (2.3) we have

x 0 < x 1 < x 2 << x n 1 < x n < x n + 1 <.
(2.4)

Regarding (2.4), the condition (2.1) implies that

ϕ ( d ( x n , x n + 1 ) ) = ϕ ( d ( T x n 1 , T x n ) ) ϕ ( M ( x n 1 , x n ) ) ψ ( M ( x n 1 , x n ) ) + L min { d ( x n 1 , T x n ) , d ( T x n 1 , x n ) , d ( x n 1 , T x n 1 ) , d ( x n , T x n ) } ϕ ( M ( x n 1 , x n ) ) ψ ( M ( x n 1 , x n ) ) + L min { d ( x n 1 , x n + 1 ) , d ( x n , x n ) , d ( x n 1 , x n ) , d ( x n , x n + 1 ) } = ϕ ( M ( x n 1 , x n ) ) ψ ( M ( x n 1 , x n ) ) ,
(2.5)

where

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

Suppose that M( x n 1 , x n )=d( x n , x n + 1 ) for some n1. Then the inequality (2.5) turns into

ϕ ( d ( x n , x n + 1 ) ) ϕ ( d ( x n , x n + 1 ) ) ψ ( d ( x n , x n + 1 ) ) .

Regarding (2.4) and the property of ψ, this is a contradiction. Thus, M( x n 1 , x n )=d( x n 1 , x n ) for all n1. Therefore, the inequality (2.5) yields

ϕ ( d ( x n , x n + 1 ) ) ϕ ( d ( x n 1 , x n ) ) ψ ( d ( x n 1 , x n ) ) <ϕ ( d ( x n 1 , x n ) ) .
(2.6)

Since ϕ is non-decreasing, we have d( x n , x n + 1 )d( x n 1 , x n ). Consequently, {d( x n 1 , x n )} is a decreasing sequence of positive real numbers which is bounded below. So, there exists α0 such that lim n d( x n 1 , x n )=α. We claim that α=0. Suppose, to the contrary, that α>0. By taking the limit of the supremum in the relation ϕ(d( x n , x n + 1 ))ϕ(d( x n 1 , x n ))ψ(d( x n 1 , x n )), as n, we get

ϕ(α)ϕ(α)ψ(α)<ϕ(α),

which is a contradiction. Hence, we conclude that α=0, that is,

lim n d( x n 1 , x n )=0.
(2.7)

We prove that the sequence { x n } is Cauchy in X. Suppose, to the contrary, that { x n } is not a Cauchy sequence. So, there exists ε>0 such that

d( x m ( k ) , x n ( k ) )ε,
(2.8)

where { x m ( k ) } and { x n ( k ) } are subsequences of { x n } with

n(k)>m(k)k.
(2.9)

Moreover, n(k) is chosen to be the smallest integer satisfying (2.8). Thus, we have

d( x m ( k ) , x n ( k ) 1 )<ε.
(2.10)

By the triangle inequality, we get

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

Keeping (2.7) in mind and letting n in the above inequality, we get

lim n d( x m ( k ) , x n ( k ) )=ε.
(2.11)

Due to the triangle inequality, we have

d( x m ( k ) , x n ( k ) )d( x m ( k ) , x m ( k ) 1 )+d( x m ( k ) 1 , x n ( k ) 1 )+d( x n ( k ) 1 , x n ( k ) )
(2.12)

and

d( x m ( k ) 1 , x n ( k ) 1 )d( x m ( k ) 1 , x m ( k ) )+d( x m ( k ) , x n ( k ) )+d( x n ( k ) , x n ( k ) 1 ).
(2.13)

By using (2.7), (2.11), and letting n in (2.12) and (2.13), we get

lim n d( x m ( k ) 1 , x n ( k ) 1 )=ε.
(2.14)

Analogously, we derive

lim n d( x m ( k ) , x n ( k ) 1 )=εand lim n d( x m ( k ) 1 , x n ( k ) )=ε.
(2.15)

Since m(k)<n(k) we have x m ( k ) 1 < x n ( k ) 1 . By (2.1) we have

ϕ ( d ( x m ( k ) , x n ( k ) ) ) = ϕ ( d ( T x m ( k ) 1 , T x n ( k ) 1 ) ) ϕ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) + L min { d ( x n ( k ) 1 , T x m ( k ) 1 ) , d ( x m ( k ) 1 , T x n ( k ) 1 ) , d ( x m ( k ) 1 , T x m ( k ) 1 ) , d ( x n ( k ) 1 , T x n ( k ) 1 ) } ϕ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) + L min { d ( x n ( k ) 1 , x m ( k ) ) , d ( x m ( k ) 1 , x n ( k ) ) , d ( x m ( k ) 1 , x m ( k ) ) , d ( x n ( k ) 1 , x n ( k ) ) } ,
(2.16)

where

M ( x m ( k ) 1 , x n ( k ) 1 ) = max { d ( x m ( k ) 1 , T x m ( k ) 1 ) d ( x n ( k ) 1 , T x n ( k ) 1 ) d ( x m ( k ) 1 , x n ( k ) 1 ) , d ( x m ( k ) 1 , x n ( k ) 1 ) } = max { d ( x m ( k ) 1 , x m ( k ) ) d ( x n ( k ) 1 , x n ( k ) ) d ( x m ( k ) 1 , x n ( k ) 1 ) , d ( x m ( k ) 1 , x n ( k ) 1 ) } .
(2.17)

Letting n in (2.16) (and hence in (2.17)), and taking (2.7), (2.11), (2.14), and (2.15) into account, we obtain

ϕ(ε)ϕ ( max { 0 , ε } ) ψ ( max { 0 , ε } ) +Lmin{ε,ε,0,0}<ϕ(ε),
(2.18)

which is a contradiction. Thus, { x n } is a Cauchy sequence in X. Since X is a complete metric space, there exists zX such that lim n x n =z.

We will show that z is a fixed point of T. Assume that (i) holds. Then by the continuity of T, we have

z= lim n x n = lim n T x n 1 =T ( lim n x n 1 ) =Tz.

Suppose that (ii) holds. Since { x n } is a non-decreasing sequence and lim n x n =z then z=sup{ x n }. Hence, x n z for all nN. Since T is a non-decreasing mapping, we conclude that T x n Tz, or equivalently,

x n x n + 1 Tzfor all nN.
(2.19)

Then z=sup{ x n }, and we get zTz.

To this end, we construct a new sequence { y n } as follows:

y 0 =zand y n =T y n 1 for all n1.

Since zTz, we have y 0 T y 0 = y 1 . Hence we find that { y n } is a non-decreasing sequence. By repeating the discussion above, one can conclude that { y n } is Cauchy. Thus there exists yX such that lim n y n =y. By (ii), we have y=sup{ y n } and so we have y n y. From (2.19), we get

x n <z= y 0 Tz=T y 0 y n yfor all nN.
(2.20)

If z=y then the proof is finished. Suppose that zy. On account of (2.20), the expression (2.1) implies that

ϕ ( d ( x n + 1 , y n + 1 ) ) = ϕ ( d ( T x n , T y n ) ) ϕ ( M ( x n , y n ) ) ψ ( M ( x n , y n ) ) + L min { d ( x n , T y n ) , d ( y n , T x n ) , d ( x n , T x n ) , d ( y n , T y n ) } ϕ ( M ( x n , y n ) ) ψ ( M ( x n , y n ) ) + L min { d ( x n , y n + 1 ) , d ( y n , x n + 1 ) , d ( x n , x n + 1 ) , d ( y n , y n + 1 ) } ,
(2.21)

where

M ( x n , y n ) = max { d ( x n , T x n ) d ( y n , T y n ) d ( x n , y n ) , d ( x n , y n ) } = max { d ( x n , x n + 1 ) d ( y n , y n + 1 ) d ( x n , y n ) , d ( x n , y n ) } .
(2.22)

Letting n in (2.21) (and hence (2.22)), we obtain

ϕ ( d ( y , z ) ) ϕ ( d ( y , z ) ) ψ ( d ( y , z ) ) <ϕ ( d ( y , z ) )

which is a contradiction. So y=z and we have zTzz, then Tz=z. □

If we take L=0 in Theorem 3 we get the following result.

Theorem 4 Let (X,) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:XX be a non-decreasing mapping which satisfies the inequality

ϕ ( d ( T x , T y ) ) ϕ ( M ( x , y ) ) ψ ( M ( x , y ) )
(2.23)

for all distinct x,yX with yx where ϕΦ, ψΨ and

M(x,y)=max { d ( x , T x ) d ( y , T y ) d ( x , y ) , d ( x , y ) } .

Also, assume either

  1. (i)

    T is continuous or

  2. (ii)

    if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Other corollaries could be derived.

Corollary 5 Let (X,) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:XX be a non-decreasing mapping such that

d(Tx,Ty)M(x,y)ψ ( M ( x , y ) ) +Lmin { d ( x , T y ) , d ( y , T x ) , d ( x , T x ) , d ( y , T y ) }
(2.24)

for all distinct x,yX with yx where ψΨ, L0 and

M(x,y)=max { d ( x , T x ) d ( y , T y ) d ( x , y ) , d ( x , y ) } .

Also, assume either

  1. (i)

    T is continuous or

  2. (ii)

    if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Proof Take ϕ(t)=t in Theorem 3. □

Corollary 6 Let (X,) be a partially ordered set. Suppose there exists a metric d X such that (X,d) is a complete metric space. Let T:XX be a non-decreasing mapping such that

d(Tx,Ty)kM(x,y)+Lmin { d ( x , T y ) , d ( y , T x ) , d ( x , T x ) , d ( y , T y ) } ,
(2.25)

for all distinct x,yX with yx where L0 and

M(x,y)=max { d ( x , T x ) d ( y , T y ) d ( x , y ) , d ( x , y ) } .

Also, assume either

  1. (i)

    T is continuous or

  2. (ii)

    if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Proof Take ψ(t)=(1k)ψ(t) for all t[0,) in Corollary 5. □

Corollary 7 Let (X,) be a partially ordered set. Suppose there exists a metric d such that (X,d) is a complete metric space. Let T:XX be a non-decreasing mapping such that

d(Tx,Ty)α d ( x , T x ) d ( y , T y ) d ( x , y ) +βd(x,y)
(2.26)

for all distinct x,yX with yx where α,β[0,1) with α+β<1. Also, assume either

  1. (i)

    T is continuous or

  2. (ii)

    if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }.

If there exists x 0 X such that x 0 T x 0 , then T has a fixed point.

Proof Take L=0 and k=α+β for all t[0,) in Corollary 6. Indeed,

d ( T x , T y ) α d ( x , T x ) d ( y , T y ) d ( x , y ) + β d ( x , y ) ( α + β ) max { d ( x , T x ) d ( y , T y ) d ( x , y ) , d ( x , y ) } .
(2.27)

 □

Theorem 8 In addition to the hypotheses of Theorem  3, assume that

for every x,yX there exists zX that is comparable to x and y,
(2.28)

then T has a unique fixed point.

Proof Suppose, to the contrary, that x and y are fixed points of T where xy. By (2.28), there exists a point zX which is comparable with x and y. Without loss of generality, we choose zx. We construct a sequence { z n } as follows:

z 0 =zand z n =T z n 1 for all n1.
(2.29)

Since T is non-decreasing, zx implies TzTx=x. By induction, we get z n x.

If x= z N 0 for some N 0 1 then z n =T z n 1 =Tx=x for all n N 0 1. So lim n z n =x. Analogously, we get lim n z n =y, which completes the proof.

Consider the other case, that is, x z n for all n=0,1,2, . Then, by (2.1), we observe that

ϕ ( d ( x , z n ) ) = ϕ ( d ( T x , T z n 1 ) ) ϕ ( M ( x , z n 1 ) ) ψ ( M ( x , z n 1 ) ) + L min { d ( x , T x ) , d ( z n 1 , T z n 1 ) , d ( x , T z n 1 ) , d ( z n 1 , T z n 1 ) } = ϕ ( M ( x , z n 1 ) ) ψ ( M ( x , z n 1 ) )
(2.30)

for all distinct x,yX with yx where ϕΦ, ψΨ and

M ( x , z n 1 ) = max { d ( x , T x ) d ( z n 1 , T z n 1 ) d ( x , z n 1 ) , d ( x , z n 1 ) } = max { d ( x , x ) d ( z n 1 , z n ) d ( x , z n 1 ) , d ( x , z n 1 ) } = d ( x , z n 1 ) .
(2.31)

Thus,

ϕ ( d ( x , z n ) ) ϕ ( d ( x , z n 1 ) ) ψ ( d ( x , z n 1 ) ) <ϕ ( d ( x , z n ) ) ,

which is a contradiction. This ends the proof. □

Remark

  • Corollary 5 is a generalization of Theorem 2.1 of Luong and Thuan [4].

  • Corollary 7 (with L=0) corresponds to Theorem 2.2 and Theorem 2.3 of Harjani, López and Sadarangani [3].

  • Theorem 2.28 generalizes Theorem 2.4 of Luong and Thuan [4].

Now, we give some examples illustrating our results.

Example 9 Let X={4,5,6} be endowed with the usual metric d(x,y)=|xy| for all x,yX, and :={(4,4),(5,5),(6,6),(6,4)}. Consider the mapping

T= ( 4 5 6 4 6 4 ) .

We define the functions ϕ,ψ:[0,+)[0,+) by ϕ(t)=2t and ψ(t)= 3 2 t. Now, we will check that all the hypotheses required by Theorem 4 (Theorem 3 with L=0) are satisfied.

First, X has the property: if { x n } is a non-decreasing sequence in X such that x n x, then x=sup{ x n }. Indeed, let { z n } be a non-decreasing sequence in X with respect to such that z n zX as n+. We have z n z n + 1 for all nN.

  • If z 0 =4, then z 0 =4 z 1 . From the definition of , we have z 1 =4. By induction, we get z n =4 for all nN and z=4. Then z n z for all nN and z=sup{ z n }.

  • If z 0 =5, then z 0 =5 z 1 . From the definition of , we have z 1 =5. By induction, we get z n =5 for all nN and z=5. Then z n z for all nN and z=sup{ z n }.

  • If z 0 =6, then z 0 =6 z 1 . From the definition of , we have z 1 {6,4}. By induction, we get z n {6,4} for all nN. Suppose that there exists p1 such that z p =4. From the definition of , we get z n = z p =4 for all np. Thus, we have z=4 and z n z for all nN. Now, suppose that z n =6 for all nN. In this case, we get z=6 and z n z for all nN and z=sup{ z n }.

Thus, we proved that in all cases, we have z=sup{ z n }.

Let x,yX such that xy and xy, so we have only x=6 and y=4. In particular

d(T6,T4)=0andM(6,4)=2,

so (2.23) holds easily. On the other hand, it is obvious that T is a non-decreasing mapping with respect to and there exists x 0 =6 such that x 0 T x 0 . All the hypotheses of Theorem 4 are verified and u=4 is a fixed point of T.

Note that Theorem 1 is not applicable. Indeed, taking x=4 and y=5

d(T4,T5)=2>β=α d ( 4 , T 4 ) d ( 5 , T 5 ) d ( 4 , 5 ) +βd(4,5),

for any α,β0 such that α+β<1. Also, we could not apply Theorem 2 in this example. Indeed, for x=6 and y=4 (that is, xy and xy), we have

0=d(T6,T4)>M(T6,T4)ψ ( M ( T 6 , T 4 ) ) =1.

Example 10 Let X=[0,) be endowed with the Euclidean metric and the order given as follows:

xy(x=y) or (x,y1,xy).

Define T:XX by Tx=x if 0x<1 and Tx=0 if x1. Define the functions ϕ,ψ:[0,+)[0,+) by ϕ(t)=4t and ψ(t)=3t.

Take xy and xy. It means that 1x<y. In particular, d(Tx,Ty)=0 and M(x,y)=yx. This implies that (2.23) holds. It is easy that X satisfies the property: if { x n } is a non-decreasing sequence in X such that x n x, then x n x for all nN. Also, the other conditions of Theorem 4 are satisfied and u=0 is a fixed point of T.

Notice that we cannot apply Theorem 1 (since T is not continuous) nor Theorem 2 to this example. Indeed, letting xy and xy (that is, 1x<y), we have

d(Tx,Ty)=0>M(x,y)ψ ( M ( x , y ) ) =2(yx).

Example 11 Let X={(0,1),(1,0),(1,1)} R 2 with the Euclidean distance d 2 . (X, d 2 ) is, obviously, a complete metric space. Moreover, we consider the order ≤ in X given by R={(x,x),xX}{((0,1),(1,1))}. We also consider T:XX given by T((0,1))=(0,1), T((1,0))=(1,0) and T((1,1))=(0,1). Take ϕ(t)=3t and ψ(t)=2t. Obviously, T is a continuous and non-decreasing mapping since (0,1)(1,1) and T(0,1)=(0,1)T(1,1)=(0,1). Let xy and xy, then necessarily x=(0,1) and y=(1,1). Then

d 2 (Tx,Ty)= d 2 ( ( 0 , 1 ) , ( 0 , 1 ) ) =0andM(x,y)= 2 ,

so (2.23) holds. Also, (0,1)T((0,1)), therefore all conditions in Theorem 4 hold and there are two fixed points which are (0,1) and (1,0). The non-uniqueness follows from the fact that the partial order ≤ is not total.

Note that Theorem 1 is not applicable. Indeed, taking x=(0,1) and y=(1,0)

d 2 (Tx,Ty)= 2 >(α+β) 2 =α d 2 ( x , T x ) d 2 ( y , T y ) d 2 ( x , y ) +β d 2 (x,y),

for any α,β0 such that α+β<1. Also, we could not apply Theorem 2 in this example. Indeed, for x=(0,1) and y=(1,1) we have

0= d 2 (Tx,Ty)> 2 2 2 =M(x,y)ψ ( M ( x , y ) ) .

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Acknowledgements

The authors express their gratitude to the referees for constructive and useful remarks and suggestions.

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Correspondence to Erdal Karapınar.

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The authors declare that they have no competing interests.

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All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

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Mustafa, Z., Karapınar, E. & Aydi, H. A discussion on generalized almost contractions via rational expressions in partially ordered metric spaces. J Inequal Appl 2014, 219 (2014) doi:10.1186/1029-242X-2014-219

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Keywords

  • ordered set
  • metric space
  • fixed point