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

• 908 Accesses

• 12 Citations

## 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 . Some of such generalizations are obtained via rational contractive conditions. Recently, Azam et al.  established some fixed point results for a pair of rational contractive mappings in complex valued metric spaces. Also, in , 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 , 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 .

Ran and Reurings started the studying of fixed point results on partially ordered sets in , 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 .

Czerwik in  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.,  and [17, 18]).

Definition 1 Let X be a (nonempty) set and $s≥1$ 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 $x,y,z∈X$.

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

Definition 2 

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 $x∈X$ 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 ) illustrates that a b-metric need not be a continuous function.

Example 1 Let $X=N∪{∞}$ and $d:X×X→R$ be defined by

Then $d(m,p)≤ 5 2 (d(m,n)+d(n,p))$ for all $m,n,p∈X$. Thus, $(X,d)$ is a b-metric space (with $s=5/2$). Let $x n =2n$ for each $n∈N$. So $d(2n,∞)= 1 2 n →0$ as $n→∞$ that is, $x n →∞$, but $d( x n ,1)=2↛5=d(∞,1)$ as $n→∞$.

Lemma 1 

Let $(X,d)$ be a b-metric space with $s≥1$, 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 $z∈X$, 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

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

Theorem 1 

Let $(X,d)$ be a complete metric space, and let $f:X→X$ be a self-map. Suppose that there exists $β∈S$ such that

$d(fx,fy)≤β ( d ( x , y ) ) d(x,y)$

holds for all $x,y∈X$. Then f has a unique fixed point $z∈X$ and for each $x∈X$ the Picard sequence ${ f n x}$ converges to z.

Amini-Harandi and Emami  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 $f:X→X$ be an increasing self-map such that there exists $x 0 ∈X$ with $x 0 ⪯f x 0$. Suppose that there exists $β∈S$ such that

$d(fx,fy)≤β ( d ( x , y ) ) d(x,y)$

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

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

Theorem 3 

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

$D(fx,fy)≤β ( D ( x , y ) ) D(x,y)$

for all $x,y∈X$ and some $β∈F$. Then f has a unique fixed point $z∈X$, and for each $x∈X$ 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  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 $f:X→X$ 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 $β∈F$ such that

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

for all comparable elements $x,y∈X$, where $L≥0$,

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

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 $x n →u∈X$, one has $x n ⪯u$ for all $n∈N$, 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:

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

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

for all comparable elements $x,y∈X$, 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 $f:X→X$ 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 $x n →u∈X$, one has $x n ⪯u$ for all $n∈N$,

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 $n≥0$. 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 $n∈N$, 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)

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 $r≥0$ 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 $β∈F$ 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

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

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 $β∈F$ 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 $u∈X$.

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 $u≠v$. 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 $f:X→X$ is called a rational Geraghty contraction of type II if there exists $β∈F$ such that

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

for all comparable elements $x,y∈X$, 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 $f:X→X$ 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 $x n →u∈X$, one has $x n ⪯u$ for all $n∈N$,

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 $n∈N$, 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 $r≥0$ 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 $β∈F$ we conclude that ${ x n }$ is a b-Cauchy sequence. b-Completeness of X shows that ${ x n }$ b-converges to a point $u∈X$.

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 $f:X→X$ is called a rational Geraghty contraction of type III if there exists $β∈F$ such that

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

for all comparable elements $x,y∈X$, 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 $f:X→X$ 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 $x n →u∈X$, one has $x n ⪯u$ for all $n∈N$,

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 $n∈N$, 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

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

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 $β∈F$ we conclude that ${ x n }$ is a b-Cauchy sequence. b-Completeness of X shows that ${ x n }$ b-converges to a point $u∈X$.

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 $0≤r< 1 s$, 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 $f:X→X$ 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 $x,y∈X$, 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 $x n →u∈X$ one has $x n ⪯u$ for all $n∈N$, 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 $f:X→X$ 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 $x,y∈X$, where $a,b,c≥0$ and $0≤a+b+c< 1 s$.

If f is continuous, or, for any nondecreasing sequence ${ x n }$ in X such that $x n →u∈X$ one has $x n ⪯u$ for all $n∈N$, then f has a fixed point.

Corollary 3 Let $(X,⪯,d)$ be an ordered b-complete b-metric space, and let $f:X→X$ 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 $x,y∈X$, 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 $x n →u∈X$ one has $x n ⪯u$ for all $n∈N$, 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 $ψ∈Ψ$, then the following are satisfied.

1. (a)

$ψ(t) for all $t>0$;

2. (b)

$ψ(0)=0$.

As an example $ψ 1 (t)=kt$, for all $t≥0$, where $k∈[0,1)$, and $ψ 2 (t)=ln(t+1)$, for all $t≥0$, 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 $f:X→X$ 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 $x,y∈X$. 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 $n∈N$.

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 $ψ∈Ψ$, 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

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

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 $u∈X$.

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 $x n →u∈X$, $x n ⪯u$ for all $n∈N$. 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 $x n →u∈X$. 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 $f:X→X$ 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 $β∈F$ is given by

$β(t)= 13 18 e − t 9 ,t≥0$

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,

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

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 $d(x,y)=|x−y | 2$ with $s=2$. Consider the mapping $f:X→X$ 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 $0≤f(0)=0$. For all comparable elements $x,y∈X$, 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 $d(x,y)=|x−y | 2$ with $s=2$. Consider the mapping $f:X→X$ defined by $f(x)= 1 4 ln( x 2 +1)$ and the function $ψ∈Ψ$ given by $ψ(t)= 1 4 t$, $t≥0$. It is easy to see that f is increasing and $0≤f(0)=0$. For all comparable elements $x,y∈X$, 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  (also, see  and ).

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 $u,v∈X$ 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

Moreover, as in  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 $n∈N∪{0}$.

Consider the first-order periodic boundary value problem

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

where $t∈I=[0,T]$ with $T>0$ and $f:[0,T]×R→R$ 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 $t∈I=[0,T]$.

Assume that there exists $λ>0$ such that for all $x,y∈X$ 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 $t∈I$.

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 . Note that if $u∈C[0,T]$ is a fixed point of S then $u∈ C 1 [0,T]$ is a solution of (3.1).

Let $x,y∈X$. 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  it was shown that $α⪯S(α)$.

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 $x≠y∈X$.

## References

1. 1.

Banach S: Sur les opérations dans les ensembles abstraits et leur application aux équations intégrales. Fundam. Math. 1922, 3: 133-181.

2. 2.

Azam A, Fisher B, Khan M: Common fixed point theorems in complex valued metric spaces. Numer. Funct. Anal. Optim. 2011, 32: 243-253. 10.1080/01630563.2011.533046

3. 3.

Nashine HK, Imdad M, Hasan M: Common fixed point theorems under rational contractions in complex valued metric spaces. J. Nonlinear Sci. Appl. 2014, 7: 42-50.

4. 4.

Arshad M, Karapınar E, Ahmad J: Some unique fixed point theorems for rational contractions in partially ordered metric spaces. J. Inequal. Appl. 2013:Article ID 248

5. 5.

Jaggi DS: Some unique fixed point theorems. Indian J. Pure Appl. Math. 1977,8(2):223-230.

6. 6.

Ran ACM, Reurings MCB: A fixed point theorem in partially ordered sets and some application to matrix equations. Proc. Am. Math. Soc. 2004, 132: 1435-1443. 10.1090/S0002-9939-03-07220-4

7. 7.

Abbas M, Parvaneh V, Razani A: Periodic points of T -Ćirić generalized contraction mappings in ordered metric spaces. Georgian Math. J. 2012, 19: 597-610.

8. 8.

Nieto JJ, Rodríguez-López R: Contractive mapping theorems in partially ordered sets and applications to ordinary differential equations. Order 2005, 22: 223-239. 10.1007/s11083-005-9018-5

9. 9.

Nieto JJ, Rodríguez-López R: Existence and uniqueness of fixed points in partially ordered sets and applications to ordinary differential equations. Acta Math. Sin. Engl. Ser. 2007, 23: 2205-2212. 10.1007/s10114-005-0769-0

10. 10.

Czerwik S: Contraction mappings in b -metric spaces. Acta Math. Inform. Univ. Ostrav. 1993, 1: 5-11.

11. 11.

Aydi H, Bota M, Karapınar E, Mitrović S: A fixed point theorem for set-valued quasicontractions in b -metric spaces. Fixed Point Theory Appl. 2012:Article ID 88

12. 12.

Jovanović M, Kadelburg Z, Radenović S: Common fixed point results in metric-type spaces. Abstr. Appl. Anal. 2010: Article ID 978121 10.1155/2010/978121

13. 13.

Khamsi MA: Remarks on cone metric spaces and fixed point theorems of contractive mappings. Fixed Point Theory Appl. 2010:Article ID 315398

14. 14.

Olatinwo MO: Some results on multi-valued weakly Jungck mappings in b -metric space. Cent. Eur. J. Math. 2008, 6: 610-621. 10.2478/s11533-008-0047-3

15. 15.

Pacurar M: Sequences of almost contractions and fixed points in b -metric spaces. An. Univ. Vest. Timiş., Ser. Mat.-Inform. 2010, XLVIII: 125-137.

16. 16.

Parvaneh V, Roshan JR, Radenović S: Existence of tripled coincidence points in ordered b -metric spaces and an application to a system of integral equations. Fixed Point Theory Appl. 2013:Article ID 130

17. 17.

Roshan JR, Parvaneh V, Shobkolaei N, Sedghi S, Shatanawi W: Common fixed points of almost generalized $( ψ , φ ) s$ -contractive mappings in ordered b -metric spaces. Fixed Point Theory Appl. 2013.2013: Article ID 159

18. 18.

Mustafa Z, Roshan JR, Parvaneh V, Kadelburg Z: Some common fixed point results in ordered partial b -metric spaces. J. Inequal. Appl. 2013: Article ID 562

19. 19.

Boriceanu M: Strict fixed point theorems for multivalued operators in b -metric spaces. Int. J. Mod. Math. 2009,4(3):285-301.

20. 20.

Hussain N, Ðorić D, Kadelburg Z, Radenović S: Suzuki-type fixed point results in metric type spaces. Fixed Point Theory Appl.2012: Article ID 126

21. 21.

Aghajani A, Abbas M, Roshan JR: Common fixed point of generalized weak contractive mappings in partially ordered b -metric spaces. Math. Slovaca 2014, 4: 941-960.

22. 22.

Geraghty M: On contractive mappings. Proc. Am. Math. Soc. 1973, 40: 604-608. 10.1090/S0002-9939-1973-0334176-5

23. 23.

Amini-Harandi A, Emami H: A fixed point theorem for contraction type maps in partially ordered metric spaces and application to ordinary differential equations. Nonlinear Anal. TMA 2010,72(5):2238-2242. 10.1016/j.na.2009.10.023

24. 24.

Dukić D, Kadelburg Z, Radenović S: Fixed points of Geraghty-type mappings in various generalized metric spaces. Abstr. Appl. Anal. 2011: Article ID 561245 10.1155/2011/561245

25. 25.

Zabihi F, Razani A: Fixed point theorems for hybrid rational Geraghty contractive mappings in ordered b -metric spaces. J. Appl. Math. 2014: Article ID 929821 10.1155/2014/929821

26. 26.

Harjani J, Sadarangani K: Fixed point theorems for weakly contractive mappings in partially ordered sets. Nonlinear Anal. 2009, 71: 3403-3410. 10.1016/j.na.2009.01.240

27. 27.

Hussain N, Parvaneh V, Roshan JR: Fixed point results for G - α -contractive maps with application to boundary value problems. Sci. World J. 2014: Article ID 585964

## Acknowledgements

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

## Author information

Correspondence to Abdolrahman Razani.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

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

## Rights and permissions 