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Some fixed point theorems for rational Geraghty contractive mappings in ordered b-metric spaces

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Abstract

In this paper, new classes of rational Geraghty contractive mappings in the setup of b-metric spaces are introduced. Moreover, the existence of some fixed point for such mappings in ordered b-metric spaces are investigated. Also, some examples are provided to illustrate the results presented herein. Finally, an application of the main result is given.

MSC:47H10, 54H25.

1 Introduction

Using different forms of contractive conditions in various generalized metric spaces, there is a large number of extensions of the Banach contraction principle [1]. Some of such generalizations are obtained via rational contractive conditions. Recently, Azam et al. [2] established some fixed point results for a pair of rational contractive mappings in complex valued metric spaces. Also, in [3], Nashine et al. proved some common fixed point theorems for a pair of mappings satisfying certain rational contractions in the framework of complex valued metric spaces. In [4], the authors proved some unique fixed point results for an operator T satisfying certain rational contractive condition in a partially ordered metric space. In fact, their results generalize the main result of Jaggi [5].

Ran and Reurings started the studying of fixed point results on partially ordered sets in [6], where they gave many useful results in matrix equations. Recently, many researchers have focused on different contractive conditions in complete metric spaces endowed with a partial order and obtained many fixed point results in such spaces. For more details on fixed point results in ordered metric spaces we refer the reader to [7, 8] and [9].

Czerwik in [10] introduced the concept of a b-metric space. Since then, several papers dealt with fixed point theory for single-valued and multi-valued operators in b-metric spaces (see, e.g., [1116] and [17, 18]).

Definition 1 Let X be a (nonempty) set and s1 be a given real number. A function d:X×X R + is a b-metric if the following conditions are satisfied:

(b1) d(x,y)=0 iff x=y,

(b2) d(x,y)=d(y,x),

(b3) d(x,z)s[d(x,y)+d(y,z)]

for all x,y,zX.

In this case, the pair (X,d) is called a b-metric space.

Definition 2 [19]

Let (X,d) be a b-metric space.

  1. (a)

    A sequence { x n } in X is called b-convergent if and only if there exists xX such that d( x n ,x)0 as n.

  2. (b)

    { x n } in X is said to be b-Cauchy if and only if d( x n , x m )0, as n,m.

  3. (c)

    The b-metric space (X,d) is called b-complete if every b-Cauchy sequence in X is b-convergent.

The following example (corrected from [20]) illustrates that a b-metric need not be a continuous function.

Example 1 Let X=N{} and d:X×XR be defined by

d(m,n)={ 0 , if  m = n , | 1 m 1 n | , if one of  m , n  is even and the other is even or  , 5 , if one of  m , n  is odd and the other is odd  ( and  m n )  or  , 2 , otherwise .

Then d(m,p) 5 2 (d(m,n)+d(n,p)) for all m,n,pX. Thus, (X,d) is a b-metric space (with s=5/2). Let x n =2n for each nN. So d(2n,)= 1 2 n 0 as n that is, x n , but d( x n ,1)=25=d(,1) as n.

Lemma 1 [21]

Let (X,d) be a b-metric space with s1, and suppose that { x n } and { y n } are b-convergent to x and y, respectively. Then

1 s 2 d(x,y) lim inf n d( x n , y n ) lim sup n d( x n , y n ) s 2 d(x,y).

Moreover, for each zX, we have

1 s d(x,z) lim inf n d( x n ,z) lim sup n d( x n ,z)sd(x,z).

Let S denote the class of all real functions β:[0,+)[0,1) satisfying the condition

β( t n )1implies that t n 0,as n.

In order to generalize the Banach contraction principle, Geraghty proved the following.

Theorem 1 [22]

Let (X,d) be a complete metric space, and let f:XX be a self-map. Suppose that there exists βS such that

d(fx,fy)β ( d ( x , y ) ) d(x,y)

holds for all x,yX. Then f has a unique fixed point zX and for each xX the Picard sequence { f n x} converges to z.

Amini-Harandi and Emami [23] generalized the result of Geraghty to the framework of a partially ordered complete metric space as follows.

Theorem 2 Let (X,d,) be a complete partially ordered metric space. Let f:XX be an increasing self-map such that there exists x 0 X with x 0 f x 0 . Suppose that there exists βS such that

d(fx,fy)β ( d ( x , y ) ) d(x,y)

holds for all x,yX with yx. Assume that either f is continuous or X is such that if an increasing sequence { x n } in X converges to xX, then x n x for all n. Then f has a fixed point in X. Moreover, if for each x,yX there exists zX comparable with x and y, then the fixed point of f is unique.

In [24], some fixed point theorems for mappings satisfying Geraghty-type contractive conditions are proved in various generalized metric spaces. As in [24], we will consider the class of functions β:[0,)[0,1/s) such that

β( t n ) 1 s implies that t n 0,as n.

Theorem 3 [24]

Let s>1, and let (X,D,s) be a complete metric type space. Suppose that a mapping f:XX satisfies the condition

D(fx,fy)β ( D ( x , y ) ) D(x,y)

for all x,yX and some βF. Then f has a unique fixed point zX, and for each xX the Picard sequence { f n x} converges to z in (X,D,s).

Also, by unification of the recent results obtained by Zabihi and Razani [25] we have the following result.

Theorem 4 Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that (X,d) is a b-complete b-metric space (with parameter s>1). Let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose there exists βF such that

sd(fx,fy)β ( d ( x , y ) ) M(x,y)+LN(x,y)
(1.1)

for all comparable elements x,yX, where L0,

M(x,y)=max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( f x , f y ) }

and

N(x,y)=min { d ( x , f x ) , d ( x , f y ) , d ( y , f x ) , d ( y , f y ) } .

If f is continuous, or, whenever { x n } is a nondecreasing sequence in X such that x n uX, one has x n u for all nN, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

The aim of this paper is to present some fixed point theorems for rational Geraghty contractive mappings in partially ordered b-metric spaces. Our results extend some existing results in the literature.

2 Main results

Let denotes the class of all functions β:[0,)[0, 1 s ) satisfying the following condition:

lim sup n β( t n )= 1 s implies that t n 0,as n.

Definition 3 Let (X,d,) be a b-metric space. A mapping f:XX is called a rational Geraghty contraction of type I if there exists βF such that

d(fx,fy)β ( M ( x , y ) ) M(x,y)
(2.1)

for all comparable elements x,yX, where

M(x,y)=max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( f x , f y ) } .

Theorem 5 Let (X,) be a partially ordered set and suppose there exists a b-metric d on X such that (X,d) is a b-complete b-metric space (with parameter s>1). Let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose f is a rational Geraghty contraction of type I. If

  1. (I)

    f is continuous, or,

  2. (II)

    whenever { x n } is a nondecreasing sequence in X such that x n uX, one has x n u for all nN,

then f has a fixed point.

Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Let x n = f n ( x 0 ) for all n0. Since x 0 f( x 0 ) and f is increasing, we obtain by induction that

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

We do the proof in the following steps.

Step I: We show that lim n d( x n , x n + 1 )=0. Since x n x n + 1 for each nN, then by (2.1)

d ( x n , x n + 1 ) = d ( f x n 1 , f x n ) β ( M ( x n 1 , x n ) ) M ( x n 1 , x n ) ,
(2.2)

where

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

If max{d( x n 1 , x n ),d( x n , x n + 1 )}=d( x n , x n + 1 ), then from (2.2),

d ( x n , x n + 1 ) β ( M ( x n , x n + 1 ) ) d ( x n , x n + 1 ) < 1 s d ( x n , x n + 1 ) < d ( x n , x n + 1 ) ,
(2.3)

which is a contradiction.

Hence, max{d( x n 1 , x n ),d( x n , x n + 1 )}=d( x n 1 , x n ), so from (2.2),

d( x n , x n + 1 )β ( M ( x n 1 , x n ) ) d( x n 1 , x n ).
(2.4)

Since {d( x n , x n + 1 )} is a decreasing sequence, then there exists r0 such that lim n d( x n , x n + 1 )=r. We prove r=0. Suppose on contrary that r>0. Then, letting n, from (2.4) we have

r lim n β ( M ( x n 1 , x n ) ) r,

which implies that 1 s 1 lim n β(M( x n 1 , x n )). Now, as βF we conclude that M( x n 1 , x n )0, which yields r=0, a contradiction. Hence, r=0. That is,

lim n d( x n 1 , x n )=0.
(2.5)

Step II: Now, we prove that the sequence { x n } is a b-Cauchy sequence. Suppose the contrary, i.e., { x n } is not a b-Cauchy sequence. Then there exists ε>0 for which we can find two subsequences { x m i } and { x n i } of { x n } such that n i is the smallest index for which

n i > m i >iandd( x m i , x n i )ε.
(2.6)

This means that

d( x m i , x n i 1 )<ε.
(2.7)

From (2.5) and using the triangular inequality, we get

εd( x m i , x n i )sd( x m i , x m i + 1 )+sd( x m i + 1 , x n i ).

By taking the upper limit as i, we get

ε s lim sup i d( x m i + 1 , x n i ).
(2.8)

The definition of M(x,y) and (2.8) imply

lim sup i M ( x m i , x n i 1 ) = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , f x m i ) d ( x n i 1 , f x n i 1 ) 1 + d ( x m i , x n i 1 ) , d ( x m i , f x m i ) d ( x n i 1 , f x n i 1 ) 1 + d ( f x m i , f x n i 1 ) } = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , x m i + 1 ) d ( x n i 1 , x n i ) 1 + d ( x m i , x n i 1 ) , d ( x m i , x m i + 1 ) d ( x n i 1 , x n i ) 1 + d ( x m i + 1 , x n i ) } ε .

Now, from (2.1) and the above inequalities, we have

ε s lim sup i d ( x m i + 1 , x n i ) lim sup i β ( M ( x m i , x n i 1 ) ) lim sup i M ( x m i , x n i 1 ) ε lim sup i β ( M ( x m i , x n i 1 ) ) ,

which implies that 1 s lim sup i β(M( x m i , x n i 1 )). Now, as βF we conclude that M( x m i , x n i 1 )0, which yields d( x m i , x n i 1 )0. Consequently,

d( x m i , x n i )sd( x m i , x n i 1 )+sd( x n i 1 , x n i )0,

which is a contradiction to (2.6). Therefore, { x n } is a b-Cauchy sequence. b-Completeness of X shows that { x n } b-converges to a point uX.

Step III: u is a fixed point of f.

First, let f be continuous, so we have

u= lim n x n + 1 = lim n f x n =fu.

Now, let (II) holds. Using the assumption on X we have x n u. Now, we show that u=fu. By Lemma 1

1 s d ( u , f u ) lim sup n d ( x n + 1 , f u ) lim sup n β ( M ( x n , u ) ) lim sup n M ( x n , u ) ,

where

lim n M ( x n , u ) = lim n max { d ( x n , u ) , d ( x n , f x n ) d ( u , f u ) 1 + d ( x n , u ) , d ( x n , f x n ) d ( u , f u ) 1 + d ( f x n , f u ) } = max { 0 , 0 } = 0 .

Therefore, from the above relations, we deduce that d(u,fu)=0, so u=fu.

Finally, suppose that the set of fixed point of f is well ordered. Assume to the contrary that u and v are two fixed points of f such that uv. Then by (2.1),

d(u,v)=d(fu,fv)β ( M ( u , v ) ) M(u,v)=β ( d ( u , v ) ) d(u,v)< 1 s d(u,v),
(2.9)

because

M(u,v)=max { d ( u , v ) , d ( u , u ) d ( v , v ) 1 + d ( u , v ) } =d(u,v).

So we get d(u,v)< 1 s d(u,v), a contradiction. Hence u=v, and f has a unique fixed point. Conversely, if f has a unique fixed point, then the set of fixed points of f is a singleton, and so it is well ordered. □

Definition 4 Let (X,d) be a b-metric space. A mapping f:XX is called a rational Geraghty contraction of type II if there exists βF such that

d(fx,fy)β ( M ( x , y ) ) M(x,y)
(2.10)

for all comparable elements x,yX, where

M ( x , y ) = max { d ( x , y ) , d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + s [ d ( x , f x ) + d ( y , f y ) ] , d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + d ( x , f y ) + d ( y , f x ) } .

Theorem 6 Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that (X,d) is a b-complete b-metric space. Let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose f is a rational Geraghty contractive mapping of type II. If

  1. (I)

    f is continuous, or,

  2. (II)

    whenever { x n } is a nondecreasing sequence in X such that x n uX, one has x n u for all nN,

then f has a fixed point.

Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Set x n = f n ( x 0 ). Since x 0 f( x 0 ) and f is increasing, we obtain by induction that

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

We do the proof in the following steps.

Step I: We show that lim n d( x n , x n + 1 )=0. Since x n x n + 1 for each nN, then by (2.10)

d ( x n , x n + 1 ) = d ( f x n 1 , f x n ) β ( M ( x n 1 , x n ) ) M ( x n 1 , x n ) β ( d ( x n 1 , x n ) ) d ( x n 1 , x n ) < 1 s d ( x n 1 , x n ) d ( x n 1 , x n ) ,
(2.11)

because

M ( x n 1 , x n ) = max { d ( x n 1 , x n ) , d ( x n 1 , f x n 1 ) d ( x n 1 , f x n ) + d ( x n , f x n ) d ( x n , f x n 1 ) 1 + s [ d ( x n 1 , f x n 1 ) + d ( x n , f x n ) ] , d ( x n 1 , f x n 1 ) d ( x n 1 , f x n ) + d ( x n , f x n ) d ( x n , f x n 1 ) 1 + d ( x n 1 , f x n ) + d ( x n , f x n 1 ) } = max { d ( x n 1 , x n ) , d ( x n 1 , x n ) d ( x n 1 , x n + 1 ) + d ( x n , x n + 1 ) d ( x n , x n ) 1 + s [ d ( x n 1 , x n ) + d ( x n , x n + 1 ) ] , d ( x n 1 , x n ) d ( x n 1 , x n + 1 ) + d ( x n , x n + 1 ) d ( x n , x n ) 1 + d ( x n 1 , x n + 1 ) + d ( x n , x n ) } = d ( x n 1 , x n ) .

Therefore, {d( x n , x n + 1 )} is decreasing. Then there exists r0 such that lim n d( x n , x n + 1 )=r. We will prove that r=0. Suppose to the contrary that r>0. Then, letting n, from (2.11)

1 s r lim n β ( d ( x n 1 , x n ) ) r,

which implies that d( x n 1 , x n )0. Hence, r=0, a contradiction. So,

lim n d( x n 1 , x n )=0
(2.12)

holds true.

Step II: Now, we prove that the sequence { x n } is a b-Cauchy sequence. Suppose the contrary, i.e., { x n } is not a b-Cauchy sequence. Then there exists ε>0 for which we can find two subsequences { x m i } and { x n i } of { x n } such that n i is the smallest index for which

n i > m i >iandd( x m i , x n i )ε.
(2.13)

This means that

d( x m i , x n i 1 )<ε.
(2.14)

As in the proof of Theorem 5, we have

ε s lim sup i d( x m i + 1 , x n i ).
(2.15)

From the definition of M(x,y) and the above limits,

lim sup i M ( x m i , x n i 1 ) = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , f x m i ) d ( x m i , f x n i 1 ) + d ( x n i 1 , f x n i 1 ) d ( x n i 1 , f x m i ) 1 + s [ d ( x m i , f x m i ) + d ( x n i 1 , f x n i 1 ) ] , d ( x m i , f x m i ) d ( x m i , f x n i 1 ) + d ( x n i 1 , f x n i 1 ) d ( x n i 1 , f x m i ) 1 + d ( x m i , f x n i 1 ) + d ( x n i 1 , f x m i ) } = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , x m i + 1 ) d ( x m i , x n i ) + d ( x n i 1 , x n i ) d ( x n i 1 , x m i + 1 ) 1 + s [ d ( x m i , x m i + 1 ) + d ( x n i 1 , x n i ) ] , d ( x m i , x m i + 1 ) d ( x m i , x n i ) + d ( x n i 1 , x n i ) d ( x n i 1 , x m i + 1 ) 1 + d ( x m i , x n i ) + d ( x n i 1 , x m i + 1 ) } ε .

Now, from (2.10) and the above inequalities, we have

ε s lim sup i d ( x m i + 1 , x n i ) lim sup i β ( M ( x m i , x n i 1 ) ) lim sup i M ( x m i , x n i 1 ) ε lim sup i β ( M ( x m i , x n i 1 ) ) ,

which implies that 1 s lim sup i β(M( x m i , x n i 1 )). Now, as βF we conclude that { x n } is a b-Cauchy sequence. b-Completeness of X shows that { x n } b-converges to a point uX.

Step III: u is a fixed point of f.

First, let f be continuous, so we have

u= lim n x n + 1 = lim n f x n =fu.

Now, let (II) hold. Using the assumption on X we have x n u. Now, we show that u=fu. By Lemma 1

1 s d ( u , f u ) lim sup n d ( x n + 1 , f u ) lim sup n β ( M ( x n , u ) ) lim sup n M ( x n , u ) = 0 ,

because

lim n M ( x n , u ) = lim n max { d ( x n , u ) , d ( x n , f x n ) d ( x n , f u ) + d ( u , f u ) d ( u , f x n ) 1 + s [ d ( x n , f x n ) + d ( u , f u ) ] , d ( x n , f x n ) d ( x n , f u ) + d ( u , f u ) d ( u , f x n ) 1 + d ( x n , f u ) + d ( x n , f u ) } = max { 0 , 0 } = 0 .

Therefore, d(u,fu)=0, so u=fu. □

Definition 5 Let (X,d) be a b-metric space. A mapping f:XX is called a rational Geraghty contraction of type III if there exists βF such that

d(fx,fy)β ( M ( x , y ) ) M(x,y)
(2.16)

for all comparable elements x,yX, where

M ( x , y ) = max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + s [ d ( x , y ) + d ( x , f y ) + d ( y , f x ) ] , d ( x , f y ) d ( x , y ) 1 + s d ( x , f x ) + s 3 [ d ( y , f x ) + d ( y , f y ) ] } .

Theorem 7 Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that (X,d) is a b-complete b-metric space. Let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose f is a rational Geraghty contractive mapping of type III. If

  1. (I)

    f is continuous, or,

  2. (II)

    whenever { x n } is a nondecreasing sequence in X such that x n uX, one has x n u for all nN,

then f has a fixed point.

Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Set x n = f n ( x 0 ).

Step I: We show that lim n d( x n , x n + 1 )=0. Since x n x n + 1 for each nN, then by (2.16)

d ( x n , x n + 1 ) = d ( f x n 1 , f x n ) β ( M ( x n 1 , x n ) ) M ( x n 1 , x n ) β ( d ( x n 1 , x n ) ) d ( x n 1 , x n ) < 1 s d ( x n 1 , x n ) d ( x n 1 , x n ) ,
(2.17)

because

M ( x n 1 , x n ) = max { d ( x n 1 , x n ) , d ( x n 1 , f x n 1 ) d ( x n , f x n ) 1 + s [ d ( x n 1 , x n ) + d ( x n 1 , f x n ) + d ( x n , f x n 1 ) ] , d ( x n 1 , f x n ) d ( x n 1 , x n ) 1 + s d ( x n 1 , f x n 1 ) + s 3 [ d ( x n , f x n 1 ) + d ( x n , f x n ) ] } = max { d ( x n 1 , x n ) , d ( x n 1 , x n ) d ( x n , x n + 1 ) 1 + s [ d ( x n 1 , x n ) + d ( x n 1 , x n + 1 ) + d ( x n , x n ) ] , d ( x n 1 , x n + 1 ) d ( x n 1 , x n ) 1 + s d ( x n 1 , x n ) + s 3 [ d ( x n , x n ) + d ( x n , x n + 1 ) ] } max { d ( x n 1 , x n ) , d ( x n 1 , x n ) s [ d ( x n , x n 1 ) + d ( x n 1 , x n + 1 ) ] s [ d ( x n 1 , x n ) + d ( x n 1 , x n + 1 ) + d ( x n , x n ) ] } = d ( x n 1 , x n ) .

Therefore, {d( x n , x n + 1 )} is decreasing. Similar to what we have done in Theorems 5 and 6, we have

lim n d( x n 1 , x n )=0.
(2.18)

Step II: Now, we prove that the sequence { x n } is a b-Cauchy sequence. Suppose the contrary, i.e., { x n } is not a b-Cauchy sequence. Then there exists ε>0 for which we can find two subsequences { x m i } and { x n i } of { x n } such that n i is the smallest index for which

n i > m i >iandd( x m i , x n i )ε.
(2.19)

This means that

d( x m i , x n i 1 )<ε.
(2.20)

From (2.18) and using the triangular inequality, we get

εd( x m i , x n i )sd( x m i , x m i + 1 )+sd( x m i + 1 , x n i ).

By taking the upper limit as i, we get

ε s lim sup i d( x m i + 1 , x n i ).
(2.21)

Using the triangular inequality, we have

d( x m i , x n i )sd( x m i , x n i 1 )+sd( x n i 1 , x n i ).

Taking the upper limit as i in the above inequality and using (2.20) we get

lim sup i d( x m i , x n i )εs.
(2.22)

Again, using the triangular inequality, we have

d( x m i , x n i )sd( x m i , x m i + 1 )+ s 2 d( x m i + 1 , x n i 1 )+ s 2 d( x n i 1 , x n i ).

Taking the upper limit as i in the above inequality and using (2.20) we get

lim sup i d( x m i + 1 , x n i 1 ) ε s 2 .
(2.23)

From the definition of M(x,y) and the above limits,

lim sup i M ( x m i , x n i 1 ) = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , f x m i ) d ( x n i 1 , f x n i 1 ) 1 + s [ d ( x m i , x n i 1 ) + d ( x m i , f x n i 1 ) + d ( x n i 1 , f x m i ) ] , d ( x m i , f x n i 1 ) d ( x m i , x n i 1 ) 1 + s d ( x m i , f x m i ) + s 3 [ d ( x n i 1 , f x m i ) + d ( x n i 1 , f x n i 1 ) ] } = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , x m i + 1 ) d ( x n i 1 , x n i ) 1 + s [ d ( x m i , x n i 1 ) + d ( x m i , x n i ) + d ( x n i 1 , x m i + 1 ) ] , d ( x m i , x n i ) d ( x m i , x n i 1 ) 1 + s d ( x m i , x m i + 1 ) + s 3 [ d ( x n i 1 , x m i + 1 ) + d ( x n i 1 , x n i ) ] } ε .

Now, from (2.16) and the above inequalities, we have

ε s lim sup i d ( x m i + 1 , x n i ) lim sup i β ( M ( x m i , x n i 1 ) ) lim sup i M ( x m i , x n i 1 ) ε lim sup i β ( M ( x m i , x n i 1 ) ) ,

which implies that 1 s lim sup i β(M( x m i , x n i 1 )). Now, as βF we conclude that { x n } is a b-Cauchy sequence. b-Completeness of X shows that { x n } b-converges to a point uX.

Step III: u is a fixed point of f.

When f is continuous, the proof is straightforward.

Now, let (II) hold. By Lemma 1

1 s d ( u , f u ) lim sup n d ( x n + 1 , f u ) lim sup n β ( M ( x n , u ) ) lim sup n M ( x n , u ) ,

where

lim n M ( x n , u ) = lim n max { d ( x n , u ) , d ( x n , f x n ) d ( u , f u ) 1 + s [ d ( x n , u ) + d ( x n , f u ) + d ( u , f x n ) ] , d ( x n , f u ) d ( x n , u ) 1 + s d ( x n , f x n ) + s 3 [ d ( u , f u ) + d ( u , f x n ) ] } = max { 0 , 0 } = 0 .

Therefore, from the above relations, we deduce that d(u,fu)=0, so u=fu. □

If in the above theorems we take β(t)=r, where 0r< 1 s , then we have the following corollary.

Corollary 1 Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that (X,d) is a b-complete b-metric space, and let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose that

d(fx,fy)rM(x,y)

for all comparable elements x,yX, where

M(x,y)=max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( f x , f y ) }

or

M ( x , y ) = max { d ( x , y ) , d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + s [ d ( x , f x ) + d ( y , f y ) ] , d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + d ( x , f y ) + d ( y , f x ) } ,

or

M ( x , y ) = max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + s [ d ( x , y ) + d ( x , f y ) + d ( y , f x ) ] , d ( x , f y ) d ( x , y ) 1 + s d ( x , f x ) + s 3 [ d ( y , f x ) + d ( y , f y ) ] } .

If f is continuous, or, for any nondecreasing sequence { x n } in X such that x n uX one has x n u for all nN, then f has a fixed point.

Corollary 2 Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that (X,d) is a b-complete b-metric space, and let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose

d(fx,fy)ad(x,y)+b d ( x , f x ) d ( y , f y ) 1 + d ( x , y ) +c d ( x , f x ) d ( y , f y ) 1 + d ( f x , f y )

or

d ( f x , f y ) a d ( x , y ) + b d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + s [ d ( x , f x ) + d ( y , f y ) ] + c d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + d ( x , f y ) + d ( y , f x ) ,

or

d ( f x , f y ) a d ( x , y ) + b d ( x , f x ) d ( y , f y ) 1 + s [ d ( x , y ) + d ( x , f y ) + d ( y , f x ) ] + c d ( x , f y ) d ( x , y ) 1 + s d ( x , f x ) + s 3 [ d ( y , f x ) + d ( y , f y ) ]

for all comparable elements x,yX, where a,b,c0 and 0a+b+c< 1 s .

If f is continuous, or, for any nondecreasing sequence { x n } in X such that x n uX one has x n u for all nN, then f has a fixed point.

Corollary 3 Let (X,,d) be an ordered b-complete b-metric space, and let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f m ( x 0 ) and

d ( f m x , f m y ) β ( M ( x , y ) ) M(x,y)

for all comparable elements x,yX, where

M(x,y)=max { d ( x , y ) , d ( x , f m x ) d ( y , f m y ) 1 + d ( x , y ) , d ( x , f m x ) d ( y , f m y ) 1 + d ( f m x , f m y ) }

or

M ( x , y ) = max { d ( x , y ) , d ( x , f m x ) d ( x , f m y ) + d ( y , f m y ) d ( y , f m x ) 1 + s [ d ( x , f m x ) + d ( y , f m y ) ] , d ( x , f m x ) d ( x , f m y ) + d ( y , f m y ) d ( y , f m x ) 1 + d ( x , f m y ) + d ( y , f m x ) } ,

or

M ( x , y ) = max { d ( x , y ) , d ( x , f m x ) d ( y , f m y ) 1 + s [ d ( x , y ) + d ( x , f m y ) + d ( y , f m x ) ] , d ( x , f m y ) d ( x , y ) 1 + s d ( x , f m x ) + s 3 [ d ( y , f m x ) + d ( y , f m y ) ] }

for some positive integer m.

If f m is continuous, or, for any nondecreasing sequence { x n } in X such that x n uX one has x n u for all nN, then f has a fixed point.

Let Ψ be the family of all nondecreasing functions ψ:[0,)[0,) such that

lim n ψ n (t)=0

for all t>0.

Lemma 2 If ψΨ, then the following are satisfied.

  1. (a)

    ψ(t)<t for all t>0;

  2. (b)

    ψ(0)=0.

As an example ψ 1 (t)=kt, for all t0, where k[0,1), and ψ 2 (t)=ln(t+1), for all t0, are in Ψ.

Theorem 8 Let (X,) be a partially ordered set and suppose that there exists a b-metric d on X such that (X,d) is a b-complete b-metric space, and let f:XX be an increasing mapping with respect to such that there exists an element x 0 X with x 0 f( x 0 ). Suppose that

sd(fx,fy)ψ ( M ( x , y ) ) ,
(2.24)

where

M(x,y)=max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + d ( f x , f y ) }

for all comparable elements x,yX. If f is continuous, then f has a fixed point. Moreover, the set of fixed points of f is well ordered if and only if f has one and only one fixed point.

Proof Since x 0 f( x 0 ) and f is increasing, we obtain by induction that

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

Putting x n = f n ( x 0 ), we have

x 0 x 1 x 2 x n x n + 1 .

If there exists n 0 N such that x n 0 = x n 0 + 1 then x n 0 =f x n 0 and so we have nothing to prove. Hence, we assume that d( x n , x n + 1 )>0, for all nN.

In the following steps, we will complete the proof.

Step I: We will prove that

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

Using condition (2.24), we obtain

d( x n + 1 , x n )sd( x n + 1 , x n )=sd(f x n ,f x n 1 )ψ ( M ( x n , x n 1 ) ) ,

because

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

If max{d( x n 1 , x n ),d( x n , x n + 1 )}=d( x n , x n + 1 ), then

d ( x n , x n + 1 ) s d ( x n , x n + 1 ) = s d ( f x n 1 , x n ) ψ ( M ( x n 1 , x n ) ) < M ( x n 1 , x n ) d ( x n , x n + 1 ) ,
(2.25)

which is a contradiction. Hence, max{d( x n 1 , x n ),d( x n , x n + 1 )}=d( x n 1 , x n ), so from (2.25),

d ( x n , x n + 1 ) s d ( x n , x n + 1 ) = s d ( f x n 1 , x n ) ψ ( M ( x n 1 , x n ) ) < M ( x n 1 , x n ) d ( x n 1 , x n ) .
(2.26)

Hence,

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

By induction,

d ( x n + 1 , x n ) ψ ( d ( x n , x n 1 ) ) ψ 2 ( d ( x n 1 , x n 2 ) ) ψ n ( d ( x 1 , x 0 ) ) .
(2.27)

As ψΨ, we conclude that

lim n d( x n , x n + 1 )=0.
(2.28)

Step II: Now, we prove that the sequence { x n } is a b-Cauchy sequence. Suppose the contrary, i.e., { x n } is not a b-Cauchy sequence. Then there exists ε>0 for which we can find two subsequences { x m i } and { x n i } of { x n } such that n i is the smallest index for which

n i > m i >iandd( x m i , x n i )ε.
(2.29)

This means that

d( x m i , x n i 1 )<ε.
(2.30)

From (2.29) and using the triangular inequality, we get

εd( x m i , x n i )sd( x m i , x m i + 1 )+sd( x m i + 1 , x n i ).

Taking the upper limit as i, we get

ε s lim sup i d( x m i + 1 , x n i ).
(2.31)

From the definition of M(x,y) and the above limits,

lim sup i M ( x m i , x n i 1 ) = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , f x m i ) d ( x n i 1 , f x n i 1 ) 1 + d ( x m i , x n i 1 ) , d ( x m i , f x m i ) d ( x n i 1 , f x n i 1 ) 1 + d ( f x m i , f x n i 1 ) } = lim sup i max { d ( x m i , x n i 1 ) , d ( x m i , x m i + 1 ) d ( x n i 1 , x n i ) 1 + d ( x m i , x n i 1 ) , d ( x m i , x m i + 1 ) d ( x n i 1 , x n i ) 1 + d ( x m i + 1 , x n i ) } ε .

Now, from (2.24) and the above inequalities, we have

ε = s ε s s lim sup i d ( x m i + 1 , x n i ) lim sup i ψ ( M ( x m i , x n i 1 ) ) ψ ( ε ) < ε ,

which is a contradiction. Consequently, { x n } is a b-Cauchy sequence. b-Completeness of X shows that { x n } b-converges to a point uX.

Step III: Now we show that u is a fixed point of f,

u= lim n x n + 1 = lim n f x n =fu,

as f is continuous. □

Theorem 9 Under the same hypotheses as Theorem  8, without the continuity assumption of f, assume that whenever { x n } is a nondecreasing sequence in X such that x n uX, x n u for all nN. Then f has a fixed point.

Proof By repeating the proof of Theorem 8, we construct an increasing sequence { x n } in X such that x n uX. Using the assumption on X we have x n u. Now we show that u=fu. By (2.24) we have

d(fu, x n )=d(fu,f x n 1 )ψ ( M ( u , x n 1 ) ) ,
(2.32)

where

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

Letting n,

lim sup n M(u, x n 1 )=0.
(2.33)

Again, taking the upper limit as n in (2.32) and using Lemma 1 and (2.33),

1 s d ( f u , u ) lim sup n d ( f u , x n ) lim sup n ψ ( M ( u , x n 1 ) ) = 0 .

So we get d(fu,u)=0, i.e., fu=u. □

Remark 1 In Theorems 8 and 9, we can replace M(x,y) by the following:

M ( x , y ) = max { d ( x , y ) , d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + s [ d ( x , f x ) + d ( y , f y ) ] , d ( x , f x ) d ( x , f y ) + d ( y , f y ) d ( y , f x ) 1 + d ( x , f y ) + d ( y , f x ) }

or

M ( x , y ) = max { d ( x , y ) , d ( x , f x ) d ( y , f y ) 1 + s [ d ( x , y ) + d ( x , f y ) + d ( y , f x ) ] , d ( x , f y ) d ( x , y ) 1 + s d ( x , f x ) + s 3 [ d ( y , f x ) + d ( y , f y ) ] } .

Example 2 Let X={0,1,3} and define the partial order on X by

:= { ( 0 , 0 ) , ( 1 , 1 ) , ( 3 , 3 ) , ( 0 , 3 ) , ( 3 , 1 ) , ( 0 , 1 ) } .

Consider the function f:XX given as

f= ( 0 1 3 3 1 1 ) ,

which is increasing with respect to . Let x 0 =0. Hence, f( x 0 )=3, so x 0 f x 0 . Define first the b-metric d on X by d(0,1)=6, d(0,3)=9, d(1,3)= 1 2 , and d(x,x)=0. Then (X,d) is a b-complete b-metric space with s= 18 13 . Let βF is given by

β(t)= 13 18 e t 9 ,t0

and β(0)[0, 13 18 ). Then

d(f0,f3)=d(3,1)= 1 2 β ( M ( 0 , 3 ) ) M(0,3)=9β(9).

This is because

M ( 0 , 3 ) = max { d ( 0 , 3 ) , d ( 0 , f 0 ) d ( 3 , f 3 ) 1 + d ( f 0 , f 3 ) , d ( 0 , f 0 ) d ( 3 , f 3 ) 1 + d ( 0 , 3 ) } = max { d ( 0 , 3 ) , d ( 0 , 3 ) d ( 3 , 1 ) 1 + d ( 3 , 1 ) , d ( 0 , 3 ) d ( 3 , 1 ) 1 + d ( 0 , 3 ) } = 9 .

Also,

d(f0,f1)=d(3,1)= 1 2 β ( M ( 0 , 1 ) ) M(0,1)=6β(6),

because

M ( 0 , 1 ) = max { d ( 0 , 1 ) , d ( 0 , f 0 ) d ( 1 , f 1 ) 1 + d ( f 0 , f 1 ) , d ( 0 , f 0 ) d ( 1 , f 1 ) 1 + d ( 0 , 1 ) } = max { d ( 0 , 1 ) , d ( 0 , 3 ) d ( 1 , 1 ) 1 + d ( 3 , 1 ) , d ( 0 , 3 ) d ( 1 , 1 ) 1 + d ( 0 , 1 ) } = 6 .

Also,

d(f1,f3)=d(1,1)=0β ( M ( 1 , 3 ) ) M(1,3).

Hence, f satisfies all the assumptions of Theorem 5 and thus it has a fixed point (which is u=1).

Example 3 Let X=[0,1] be equipped with the usual order and b-complete b-metric given by d(x,y)=|xy | 2 with s=2. Consider the mapping f:XX defined by f(x)= 1 16 x 2 e x 2 and the function β given by β(t)= 1 4 . It is easy to see that f is an increasing function and 0f(0)=0. For all comparable elements x,yX, by the mean value theorem, we have

d ( f x , f y ) = | 1 16 x 2 e x 2 1 16 y 2 e y 2 | 2 1 8 | x 2 e x 2 y 2 e y 2 | 2 1 8 | x y | 2 1 4 d ( x , y ) = β ( d ( x , y ) ) d ( x , y ) β ( M ( x , y ) ) M ( x , y ) .

So, from Theorem 5, f has a fixed point.

Example 4 Let X=[0,1] be equipped with the usual order and b-complete b-metric d be given by d(x,y)=|xy | 2 with s=2. Consider the mapping f:XX defined by f(x)= 1 4 ln( x 2 +1) and the function ψΨ given by ψ(t)= 1 4 t, t0. It is easy to see that f is increasing and 0f(0)=0. For all comparable elements x,yX, using the mean value problem, we have

d ( f x , f y ) = | 1 4 ln ( x 2 + 1 ) 1 4 ln ( y 2 + 1 ) | 2 1 4 | x y | 2 = 1 4 d ( x , y ) = ψ ( d ( x , y ) ) ψ ( M ( x , y ) ) ,

so, using Theorem 8, f has a fixed point.

3 Application

In this section, we present an application where Theorem 8 can be applied. This application is inspired by [9] (also, see [26] and [27]).

Let X=C([0,T]) be the set of all real continuous functions on [0,T]. We first endow X with the b-metric

d(u,v)= max t [ 0 , T ] ( | u ( t ) v ( t ) | ) p

for all u,vX where p>1. Clearly, (X,d) is a complete b-metric space with parameter s= 2 p 1 . Secondly, C([0,T]) can also be equipped with a partial order given by

xyiffx(t)y(t)for all t[0,T].

Moreover, as in [9] it is proved that (C([0,T]),) is regular, that is, whenever { x n } in X is an increasing sequence such that x n x as n, we have x n x for all nN{0}.

Consider the first-order periodic boundary value problem

{ x ( t ) = f ( t , x ( t ) ) , x ( 0 ) = x ( T ) ,
(3.1)

where tI=[0,T] with T>0 and f:[0,T]×RR is a continuous function.

A lower solution for (3.1) is a function α C 1 [0,T] such that

{ α ( t ) f ( t , α ( t ) ) , α ( 0 ) α ( T ) ,
(3.2)

where tI=[0,T].

Assume that there exists λ>0 such that for all x,yX we have

|f ( t , x ( t ) ) +λx(t)f ( t , y ( t ) ) λy(t)| λ 2 p 1 ln ( | x ( t ) y ( t ) | p + 1 ) p .
(3.3)

Then the existence of a lower solution for (3.1) provides the existence of an unique solution of (3.1).

Problem (3.1) can be rewritten as

{ x ( t ) + λ x ( t ) = f ( t , x ( t ) ) + λ x ( t ) , x ( 0 ) = x ( T ) .

Consider

{ x ( t ) + λ x ( t ) = δ ( t ) = F ( t , x ( t ) ) , x ( 0 ) = x ( T ) ,

where tI.

Using the variation of parameters formula, we get

x(t)=x(0) e λ t + 0 t e λ ( t s ) δ(s)ds,
(3.4)

which yields

x(T)=x(0) e λ T + 0 T e λ ( T s ) δ(s)ds.

Since x(0)=x(T), we get

x(0) [ 1 e λ T ] = e λ T 0 T e λ ( s ) δ(s)ds

or

x(0)= 1 e λ T 1 0 T e λ s δ(s)ds.

Substituting the value of x(0) in (3.4) we arrive at

x(t)= 0 T G(t,s)δ(s)ds,

where

G(t,s)= { e λ ( T + s t ) e λ T 1 , 0 s t T , e λ ( s t ) e λ T 1 , 0 t s T .

Now define the operator S:C[0,T]C[0,T] by

Sx(t)= 0 T G(t,s)F ( s , x ( s ) ) ds.

The mapping S is nondecreasing [26]. Note that if uC[0,T] is a fixed point of S then u C 1 [0,T] is a solution of (3.1).

Let x,yX. Then we have

2 p 1 | S x ( t ) S y ( t ) | = 2 p 1 | 0 T G ( t , s ) F ( s , x ( s ) ) d s 0 T G ( t , s ) F ( s , y ( s ) ) d s | 2 p 1 0 T | G ( t , s ) | [ | F ( s , x ( s ) ) F ( s , y ( s ) ) | ] d s 2 p 1 0 T | G ( t , s ) | λ 2 p 1 ln ( | x ( t ) y ( t ) | p + 1 ) p d s λ ln ( d ( x , y ) + 1 ) p [ 0 t e λ ( T + s t ) e λ T 1 d s + t T e λ ( s t ) e λ T 1 d s ] = λ ln ( d ( x , y ) + 1 ) p [ 1 λ ( e λ T 1 ) ( e λ ( T + s t ) | 0 t + e λ ( s t ) | t T ) ] = λ ln ( d ( x , y ) + 1 ) p [ 1 λ ( e λ T 1 ) ( e λ T e λ ( T t ) + e λ ( T t ) 1 ) ] = ln ( d ( x , y ) + 1 ) p ln ( M ( x , y ) + 1 ) p ,

or, equivalently,

2 p 1 ( | S x ( t ) S y ( t ) | ) p ln ( M ( x , y ) + 1 ) ,

which shows that

2 p 1 d(Sx,Sy)ln ( M ( x , y ) + 1 ) ,

where

M(x,y)=max { d ( x , y ) , d ( x , S x ) d ( y , S y ) 1 + d ( x , y ) , d ( x , S x ) d ( y , S y ) 1 + d ( S x , S y ) }

or

M ( x , y ) = max { d ( x , y ) , d ( x , S x ) d ( x , S y ) + d ( y , S y ) d ( y , S x ) 1 + 2 p 1 [ d ( x , S x ) + d ( y , S y ) ] , d ( x , S x ) d ( x , S y ) + d ( y , S y ) d ( y , S x ) 1 + d ( x , S y ) + d ( y , S x ) } ,

or

M ( x , y ) = max { d ( x , y ) , d ( x , S x ) d ( y , S y ) 1 + 2 p 1 [ d ( x , y ) + d ( x , S y ) + d ( y , S x ) ] , d ( x , S y ) d ( x , y ) 1 + 2 p 1 d ( x , S x ) + 2 3 p 3 [ d ( y , S x ) + d ( y , S y ) ] } .

Finally, let α be a lower solution for (3.1). In [26] it was shown that αS(α).

Hence, the hypotheses of Theorem 8 are satisfied with ψ(t)=ln(t+1). Therefore, there exists a fixed point x ˆ C[0,T] such that S x ˆ = x ˆ .

Remark 2 In the above theorem, we can replace (3.3) by the following inequality:

|f ( t , x ( t ) ) +λx(t)f ( t , y ( t ) ) λy(t)| λ 2 p 2 1 p e M ( x , y ) M ( x , y ) p
(3.5)

for all xyX.

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Acknowledgements

The authors are grateful to the referees for valuable remarks that helped them to improve the exposition in the paper.

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Correspondence to Abdolrahman Razani.

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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 paper. All authors read and approved the final manuscript.

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Shahkoohi, R.J., Razani, A. Some fixed point theorems for rational Geraghty contractive mappings in ordered b-metric spaces. J Inequal Appl 2014, 373 (2014) doi:10.1186/1029-242X-2014-373

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Keywords

  • fixed point
  • complete metric space
  • b-metric space
  • contractive mappings