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Some common fixed point results in ordered partial b-metric spaces

Abstract

In this paper, we introduce a modified version of ordered partial b-metric spaces. We demonstrate a fundamental lemma for the convergence of sequences in such spaces. Using this lemma, we prove some fixed point and common fixed point results for (ψ,φ)-weakly contractive mappings in the setup of ordered partial b-metric spaces. Finally, examples are presented to verify the effectiveness and applicability of our main results.

MSC: 47H10, 54H25.

1 Introduction

Fixed points theorems in partially ordered metric spaces were firstly obtained in 2004 by Ran and Reurings [1], and then by Nieto and Lopez [2]. In this direction several authors obtained further results under weak contractive conditions (see, e.g., [38]).

The concept of b-metric space was introduced by Bakhtin [9] and extensively used by Czerwik in [10, 11]. After that, several interesting results about the existence of a fixed point for single-valued and multi-valued operators in (ordered) b-metric spaces have been obtained (see, e.g., [1226]).

Definition 1 [10]

Let X be a (nonempty) set and s1 be a given real number. A function d:X×X R + is a b-metric on X if, for all x,y,zX, the following conditions hold:

  • (b1) d(x,y)=0 if and only if x=y,

  • (b2) d(x,y)=d(y,x),

  • (b3) d(x,z)s[d(x,y)+d(y,z)].

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

On the other hand, Matthews [27] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. In partial metric spaces, self-distance of an arbitrary point need not be equal to zero. Several authors obtained many useful fixed point results in these spaces - we mention just [2833].

Definition 2 [27]

A partial metric on a nonempty set X is a mapping p:X×X R + such that for all x,y,zX:

  • (p1) x=y if and only if p(x,x)=p(x,y)=p(y,y),

  • (p2) p(x,x)p(x,y),

  • (p3) p(x,y)=p(y,x),

  • (p4) p(x,y)p(x,z)+p(z,y)p(z,z).

In this case, (X,p) is called a partial metric space.

It is clear that if p(x,y)=0, then from (p1) and (p2), x=y. But if x=y, p(x,y) may not be 0. A basic example of a partial metric space is the pair ( R + ,p), where p(x,y)=max{x,y} for all x,y R + .

Each partial metric p on a set X generates a T 0 topology τ p on X which has as a base the family of open p-balls { B p (x,ε):xX,ε>0}, where B p (x,ε)={yX:p(x,y)<p(x,x)+ε} for all xX and ε>0.

Definition 3 [27]

Let (X,p) be a partial metric space, and let { x n } be a sequence in X and xX. Then:

  1. (i)

    The sequence { x n } is said to converge to x with respect to τ p if lim n p( x n ,x)=p(x,x).

  2. (ii)

    The sequence { x n } is said to be Cauchy in (X,p) if lim n , m p( x n , x m ) exists and is finite.

  3. (iii)

    (X,p) is said to be complete if every Cauchy sequence { x n } in X converges, with respect to τ p , to a point xX such that lim n , m p( x n , x m )= lim n p( x n ,x)=p(x,x).

The following example shows that a convergent sequence { x n } in a partial metric space (X,p) may not be Cauchy. In particular, it shows that the limit may not be unique.

Example 1 [32]

Let X=[0,) and p(x,y)=max{x,y}. Let

x n ={ 0 , n = 2 k , 1 , n = 2 k + 1 .

Then, clearly, { x n } is a convergent sequence and for every x1, we have lim n p( x n ,x)=p(x,x). But lim n , m p( x n , x m ) does not exist, that is, { x n } is not a Cauchy sequence.

As a generalization and unification of partial metric and b-metric spaces, Shukla [34] introduced the concept of partial b-metric space as follows.

Definition 4 [34]

A partial b-metric on a nonempty set X is a mapping p b :X×X R + such that for all x,y,zX:

  • ( p b 1 ) x=y if and only if p b (x,x)= p b (x,y)= p b (y,y),

  • ( p b 2 ) p b (x,x) p b (x,y),

  • ( p b 3 ) p b (x,y)= p b (y,x),

  • ( p b 4 ) p b (x,y)s[ p b (x,z)+ p b (z,y)] p b (z,z).

A partial b-metric space is a pair (X, p b ) such that X is a nonempty set and p b is a partial b-metric on X. The number s1 is called the coefficient of (X, p b ).

In a partial b-metric space (X, p b ), if x,yX and p b (x,y)=0, then x=y, but the converse may not be true. It is clear that every partial metric space is a partial b-metric space with the coefficient s=1 and every b-metric space is a partial b-metric space with the same coefficient and zero self-distance. However, the converse of these facts need not hold.

Example 2 [34]

Let X= R + , q>1 be a constant and p b :X×X R + be defined by

p b (x,y)= [ max { x , y } ] q +|xy | q for all x,yX.

Then (X, p b ) is a partial b-metric space with the coefficient s= 2 q 1 >1, but it is neither a b-metric nor a partial metric space.

Note that in a partial b-metric space the limit of a convergent sequence may not be unique (see [[34], Example 2]).

Some more examples of partial b-metrics can be constructed with the help of the following propositions.

Proposition 1 [34]

Let X be a nonempty set, and let p be a partial metric and d be a b-metric with the coefficient s1 on X. Then the function p b :X×X R + , defined by p b (x,y)=p(x,y)+d(x,y) for all x,yX, is a partial b-metric on X with the coefficient s.

Proposition 2 [34]

Let (X,p) be a partial metric space and q1. Then (X, p b ) is a partial b-metric space with the coefficient s= 2 q 1 , where p b is defined by p b (x,y)= [ p ( x , y ) ] q .

Altering distance functions were introduced by Khan et al. in [35].

Definition 5 [35]

A function ψ:[0,)[0,) is called an altering distance function if the following properties are satisfied:

  1. 1.

    ψ is continuous and nondecreasing;

  2. 2.

    ψ(t)=0 if and only if t=0.

So far, many authors have studied fixed point theorems which are based on altering distance functions (see, e.g., [12, 28, 3641]).

In this paper, we introduce a modified version of ordered partial b-metric spaces. We demonstrate a fundamental lemma for the convergence of sequences in such spaces. Using this lemma, we prove some fixed point and common fixed point results for (ψ,φ)-weakly contractive mappings in the setup of ordered partial b-metric spaces. Finally, examples are presented to verify the effectiveness and applicability of our main results.

2 Definition and basic properties of partial b-metric spaces

In the following definition, we modify Definition 4 in order to obtain that each partial b-metric p b generates a b-metric d p b .

Definition 6 Let X be a (nonempty) set and s1 be a given real number. A function p b :X×X R + is a partial b-metric if, for all x,y,zX, the following conditions are satisfied:

  • ( p b 1 ) x=y p b (x,x)= p b (x,y)= p b (y,y),

  • ( p b 2 ) p b (x,x) p b (x,y),

  • ( p b 3 ) p b (x,y)= p b (y,x),

  • ( p b 4 ) p b (x,y)s( p b (x,z)+ p b (z,y) p b (z,z))+( 1 s 2 )( p b (x,x)+ p b (y,y)).

The pair (X, p b ) is called a partial b-metric space.

Since s1, from ( p b 4 ) we have

p b (x,y)s ( p b ( x , z ) + p b ( z , y ) p b ( z , z ) ) s ( p b ( x , z ) + p b ( z , y ) ) p b (z,z).

Hence, a partial b-metric in the sense of Definition 6 is also a partial b-metric in the sense of Definition 4.

It should be noted that the class of partial b-metric spaces is larger than the class of partial metric spaces, since a partial b-metric is a partial metric when s=1. We present an example which shows that a partial b-metric on X (in the sense of Definition 6) might be neither a partial metric, nor a b-metric on X.

Example 3 Let (X,d) be a metric space and p b (x,y)=d ( x , y ) q +a, where q>1 and a0 are real numbers. We will show that p b is a partial b-metric with s= 2 q 1 .

Obviously, conditions ( p b 1 )-( p b 3 ) of Definition 6 are satisfied.

Since q>1, the convexity of the function f(x)= x q (x>0) implies that ( a + b ) q 2 q 1 ( a q + b q ) holds for a,b0. Thus, for each x,y,zX, we obtain

p b ( x , y ) = d ( x , y ) q + a ( d ( x , z ) + d ( z , y ) ) q + a 2 q 1 ( d ( x , z ) q + d ( z , y ) q ) + a = 2 q 1 ( d ( x , z ) q + a + d ( z , y ) q + a a ) + a 2 q 1 a = 2 q 1 ( p b ( x , z ) + p b ( z , y ) p b ( z , z ) ) + ( 1 2 q 1 2 ) ( p b ( x , x ) + p b ( y , y ) ) .

Hence, condition ( p b 4 ) of Definition 6 is fulfilled and p b is a partial b-metric on X.

Note that (X, p b ) is not necessarily a partial metric space. For example, if X=R is the set of real numbers, d(x,y)=|xy|, q=2 and a=3, then p b (x,y)= ( x y ) 2 +3 is a partial b-metric on X with s= 2 2 1 =2, but it is not a partial metric on X. Indeed, the ordinary (partial) triangle inequality does not hold. To see this, let x=2, y=5 and z= 5 2 . Then p b (2,5)=12, p b (2, 5 2 )= 13 4 and p b ( 5 2 ,5)= 37 4 , hence p b (2,5)=12 38 4 = p b (2, 5 2 )+ p b ( 5 2 ,5) p b ( 5 2 , 5 2 ).

Also, p b is not a b-metric since p b (x,x)0 for xX.

Proposition 3 Every partial b-metric p b defines a b-metric d p b , where

d p b (x,y)=2 p b (x,y) p b (x,x) p b (y,y)

for all x,yX.

Proof Let x,y,zX. Then we have

d p b ( x , y ) = 2 p b ( x , y ) p b ( x , x ) p b ( y , y ) 2 [ s ( p b ( x , z ) + p b ( z , y ) p b ( z , z ) ) + ( 1 s 2 ) ( p b ( x , x ) + p b ( y , y ) ) ] p b ( x , x ) p b ( y , y ) = 2 s p b ( x , z ) + 2 s p b ( z , y ) 2 s p b ( z , z ) + ( 1 s ) ( p b ( x , x ) + p b ( y , y ) ) p b ( x , x ) p b ( y , y ) = 2 s p b ( x , z ) + 2 s p b ( z , y ) 2 s p b ( z , z ) s p b ( x , x ) s p b ( y , y ) = s [ 2 p b ( x , z ) p b ( x , x ) p b ( z , z ) + 2 p b ( z , y ) p b ( z , z ) p b ( y , y ) ] = s [ d p b ( x , z ) + d p b ( z , y ) ] .

 □

Hence, the advantage of our definition of partial b-metric is that by using it we can define a dependent b-metric which we call the b-metric associated with p b . This allows us to readily transport many concepts and results from b-metric spaces into a partial b-metric space.

Now, we present some definitions and propositions in a partial b-metric space.

Definition 7 Let (X, p b ) be a partial b-metric space. Then, for xX and ϵ>0, the p b -ball with center x and radius ϵ is

B p b (x,ϵ)= { y X p b ( x , y ) < p b ( x , x ) + ϵ } .

For example, let (X, p b ) be the partial b-metric space from Example 3 (with X=R, q=2 and a=3). Then

B p b ( 1 , 4 ) = { y X p b ( 1 , y ) < p b ( 1 , 1 ) + 4 } = { y X ( y 1 ) 2 + 3 < 3 + 4 } = { y X ( y 1 ) 2 < 4 } = ( 1 , 3 ) .

Proposition 4 Let (X, p b ) be a partial b-metric space, xX and r>0. If y B p b (x,r), then there exists δ>0 such that B p b (y,δ) B p b (x,r).

Proof Let y B p b (x,r). If y=x, then we choose δ=r. Suppose that yx. Then we have p b (x,y)0. Now, we consider two cases.

Case 1. If p b (x,y)= p b (x,x), then for s=1 we choose δ=r. If s>1, then we consider the set

A= { n N | r 2 s n + 1 ( s 1 ) < p b ( x , x ) } .

By the Archimedean property, A is a nonempty set; then by the well ordering principle, A has the least element m. Since m1A, we have p b (x,x)r/(2 s m (s1)) and we choose δ=r/(2 s m + 1 ). Let z B p b (y,δ); by the property ( p b 4 ), we have

p b ( x , z ) s ( p b ( x , y ) + p b ( y , z ) p b ( y , y ) ) s ( p b ( x , x ) + δ ) p b ( x , x ) + r 2 s m + r 2 s m = p b ( x , x ) + r s m < p b ( x , x ) + r .

Hence, B p b (y,δ) B p b (x,r).

Case 2. If p b (x,y) p b (x,x), then from the property ( p b 2 ) we have p b (x,x)< p b (x,y) and for s=1 we consider the set

B= { n N | r 2 n + 3 < p b ( x , y ) p b ( x , x ) } .

Similarly, by the well ordering principle, there exists an element m such that p b (x,y) p b (x,x)r/( 2 m + 2 ), and we choose δ=r/( 2 m + 2 ). One can easily obtain that B p b (y,δ) B p b (x,r).

For s>1, we consider the set

C= { n N | r 2 s n + 2 < p b ( x , y ) 1 s p b ( x , x ) }

and by the well ordering principle, there exists an element m such that p b (x,y) 1 s p b (x,x) r 2 s m + 1 and we choose δ= r 2 s m + 1 . Let z B p b (y,δ). By the property ( p b 4 ), we have

p b ( x , z ) s ( p b ( x , y ) + p b ( y , z ) p b ( y , y ) ) s ( p b ( x , y ) + δ ) p b ( x , x ) + r 2 s m + r 2 s m = p b ( x , x ) + r s m < p b ( x , x ) + r .

Hence, B p b (y,δ) B p b (x,r). □

Thus, from the above proposition the family of all p b -balls

Δ= { B p b ( x , r ) x X , r > 0 }

is a base of a T 0 topology τ p b on X which we call the p b -metric topology.

The topological space (X, p b ) is T 0 , but need not be T 1 .

Definition 8 A sequence { x n } in a partial b-metric space (X, p b ) is said to be:

  1. (i)

    p b -convergent to a point xX if lim n p b (x, x n )= p b (x,x);

  2. (ii)

    a p b -Cauchy sequence if lim n , m p b ( x n , x m ) exists (and is finite).

  3. (iii)

    A partial b-metric space (X, p b ) is said to be p b -complete if every p b -Cauchy sequence { x n } in X p b -converges to a point xX such that lim n , m p b ( x n , x m )= lim n , m p b ( x n ,x)= p b (x,x).

The following lemma shows the relationship between the concepts of p b -convergence, p b -Cauchyness and p b -completeness in two spaces (X, p b ) and (X, d p b ) which we state and prove according to Lemma 2.2 of [31].

Lemma 1

  1. (1)

    A sequence { x n } is a p b -Cauchy sequence in a partial b-metric space (X, p b ) if and only if it is a b-Cauchy sequence in the b-metric space (X, d p b ).

  2. (2)

    A partial b-metric space (X, p b ) is p b -complete if and only if the b-metric space (X, d p b ) is b-complete. Moreover, lim n d p b (x, x n )=0 if and only if

    lim n p b (x, x n )= lim n , m p b ( x n , x m )= p b (x,x).

Proof First, we show that every p b -Cauchy sequence in (X, p b ) is a b-Cauchy sequence in (X, d p b ). Let { x n } be a p b -Cauchy sequence in (X, p b ). Then, there exists αR such that, for arbitrary ε>0, there is n ε N with

| p b ( x n , x m )α|< ε 4

for all n,m n ε . Hence,

| d p b ( x n , x m ) | = 2 p b ( x n , x m ) p b ( x n , x n ) p b ( x m , x m ) = | p b ( x n , x m ) α + α p b ( x n , x n ) + p b ( x m , x n ) α + α p b ( x m , x m ) | | p b ( x n , x m ) α | + | α p b ( x n , x n ) | + | p b ( x m , x n ) α | + | α p b ( x m , x m ) | < ε

for all n,m n ε . Hence, we conclude that { x n } is a b-Cauchy sequence in (X, d p b ).

Next, we prove that b-completeness of (X, d p b ) implies p b -completeness of (X, p b ). Indeed, if { x n } is a p b -Cauchy sequence in (X, p b ), then according to the above discussion, it is also a b-Cauchy sequence in (X, d p b ). Since the b-metric space (X, d p b ) is b-complete, we deduce that there exists yX such that lim n d p b (y, x n )=0. Hence,

lim n [ p b ( x n , y ) p b ( y , y ) + p b ( y , x n ) p b ( x n , x n ) ] =0,

therefore, lim n [ p b ( x n ,y) p b (y,y)]=0. Further, we have

lim n [ p b ( y , x n ) p b ( x n , x n ) ] =0.

Consequently,

lim n p b ( x n ,y)= p b (y,y)= lim n p b ( x n , x n ).

On the other hand,

lim n , m p b ( x n , x m ) lim n , m s p b ( x n , y ) + lim n , m s p b ( x m , y ) s p b ( y , y ) + ( 1 s 2 ) ( p b ( x n , x n ) + p b ( x m , x m ) ) = p b ( y , y ) .

Also, from ( p b 2 ),

p b (y,y) lim n , m p b ( x n ,y)= lim n , m p b ( x n , x n ) lim n , m p b ( x n , x m ).

Hence, we obtain that { x n } is a p b -convergent sequence in (X, p b ).

Now, we prove that every b-Cauchy sequence { x n } in (X, d p b ) is a p b -Cauchy sequence in (X, p b ). Let ε= 1 2 . Then there exists n 0 N such that d p b ( x n , x m )< 1 2 for all n,m n 0 . Since

p b ( x n , x n 0 ) p b ( x n 0 , x n 0 ) d p b ( x n , x n 0 )< 1 2 ,

hence

p b ( x n , x n ) p b ( x n , x n 0 ) d p b ( x n , x n 0 )+ p b ( x n 0 , x n 0 )< 1 2 + p b ( x n 0 , x n 0 ).

Consequently, the sequence { p b ( x n , x n )} is bounded in , and so there exists aR such that a subsequence { p b ( x n k , x n k )} of { p b ( x n , x n )} is convergent to a, i.e.,

lim k p b ( x n k , x n k )=a.

Now, we prove that { p b ( x n , x n )} is a Cauchy sequence in . Since { x n } is a b-Cauchy sequence in (X, d p b ) for given ε>0, there exists n ε N such that d p b ( x n , x m )<ε for all n,m n ε . Thus, for all n,m n ε ,

p b ( x n , x n ) p b ( x m , x m ) p b ( x n , x m ) p b ( x m , x m ) d p b ( x m , x n ) < ε .

Therefore, lim n p b ( x n , x n )=a.

On the other hand,

| p b ( x n , x m ) a | = | p b ( x n , x m ) p b ( x n , x n ) + p b ( x n , x n ) a | d p b ( x m , x n ) + | p b ( x n , x n ) a |

for all n,m n ε . Hence, lim n , m p b ( x n , x m )=a, and consequently, { x n } is a p b -Cauchy sequence in (X, p b ).

Conversely, let { x n } be a b-Cauchy sequence in (X, d p b ). Then { x n } is a p b -Cauchy sequence in (X, p b ), and so it is convergent to a point xX with

lim n p b (x, x n )= lim n , m p b ( x m , x n )= p b (x,x).

Then, for given ε>0, there exists n ε N such that

p b (x, x n ) p b (x,x)< ε 4

and

p b ( x n , x n ) p b (x,x) p b ( x m , x n ) p b (x,x)< ε 4 .

Therefore,

| d p b ( x n , x ) | = | p b ( x n , x ) p b ( x n , x n ) + p b ( x n , x ) p b ( x , x ) | | p b ( x n , x ) p b ( x , x ) | + | p b ( x , x ) p b ( x n , x n ) | + | p b ( x n , x ) p b ( x , x ) | < ε ,

whenever n n ε . Therefore, (X, d p b ) is complete.

Finally, let lim n d p b ( x n ,x)=0. So,

lim n [ p b ( x n , x ) p b ( x n , x n ) ] + lim n [ p b ( x n , x ) p b ( x , x ) ] =0.

On the other hand,

lim n , m [ p b ( x n , x m ) p b ( x , x ) ] lim n [ s p b ( x n , x ) + s p b ( x , x m ) s p b ( x , x ) + ( 1 s 2 ) ( p b ( x n , x n ) + p b ( x m , x m ) ) p b ( x , x ) ] = 0 .

 □

Definition 9 Let (X, p b ) and ( X , p b ) be two partial b-metric spaces, and let f:(X, p b )( X , p b ) be a mapping. Then f is said to be p b -continuous at a point aX if for a given ε>0, there exists δ>0 such that xX and p b (a,x)<δ+ p b (a,a) imply that p b (f(a),f(x))<ε+ p b (f(a),f(a)). The mapping f is p b -continuous on X if it is p b -continuous at all aX.

Proposition 5 Let (X, p b ) and ( X , p b ) be two partial b-metric spaces. Then a mapping f:X X is p b -continuous at a point xX if and only if it is p b -sequentially continuous at x; that is, whenever { x n } is p b -convergent to x, {f( x n )} is p b -convergent to f(x).

Definition 10 A triple (X,, p b ) is called an ordered partial b-metric space if (X,) is a partially ordered set and p b is a partial b-metric on X.

3 Fixed point results in partial b-metric spaces

The following crucial lemma is useful in proving our main results.

Lemma 2 Let (X, p b ) be a partial b-metric space with the coefficient s>1 and suppose that { x n } and { y n } are convergent to x and y, respectively. Then we have

1 s 2 p b ( x , y ) 1 s p b ( x , x ) p b ( y , y ) lim inf n p b ( x n , y n ) lim sup n p b ( x n , y n ) s p b ( x , x ) + s 2 p b ( y , y ) + s 2 p b ( x , y ) .

In particular, if p b (x,y)=0, then we have lim n p b ( x n , y n )=0.

Moreover, for each zX, we have

1 s p b ( x , z ) p b ( x , x ) lim inf n p b ( x n , z ) lim sup n p b ( x n , z ) s p b ( x , z ) + s p b ( x , x ) .

In particular, if p b (x,x)=0, then we have

1 s p b (x,z) lim inf n p b ( x n ,z) lim sup n p b ( x n ,z)s p b (x,z).

Proof Using the triangle inequality in a partial b-metric space, it is easy to see that

p b (x,y)s p b (x, x n )+ s 2 p b ( x n , y n )+ s 2 p b ( y n ,y)

and

p b ( x n , y n )s p b ( x n ,x)+ s 2 p b (x,y)+ s 2 p b (y, y n ).

Taking the lower limit as n in the first inequality and the upper limit as n in the second inequality, we obtain the first desired result. If p b (x,y)=0, then by the triangle inequality we get p b (x,x)=0 and p b (y,y)=0. Therefore, we have lim n p b ( x n , y n )=0. Similarly, using again the triangle inequality, the other assertions follow. □

Let (X,, p b ) be an ordered partial b-metric space, and let f:XX be a mapping. Set

M s f (x,y)=max { p b ( x , y ) , p b ( x , f x ) , p b ( y , f y ) , p b ( x , f y ) + p b ( y , f x ) 2 s } .

Definition 11 Let (X, p b ) be an ordered partial b-metric space. We say that a mapping f:XX is a generalized ( ψ , φ ) s -weakly contractive mapping if there exist two altering distance functions ψ and φ such that

ψ ( s p b ( f x , f y ) ) ψ ( M s f ( x , y ) ) φ ( M s f ( x , y ) )
(3.1)

for all comparable x,yX.

First, we prove the following result.

Theorem 1 Let (X,, p b ) be a p b -complete ordered partial b-metric space. Let f:XX be a nondecreasing, with respect to , continuous mapping. Suppose that f is a generalized ( ψ , φ ) s -weakly contractive mapping. If there exists x 0 X such that x 0 f x 0 , then f has a fixed point.

Proof Let x 0 X be such that x 0 f x 0 . Then we define a sequence ( x n ) in X such that x n + 1 =f x n for all n0. Since x 0 f x 0 = x 1 and f is nondecreasing, we have x 1 =f x 0 x 2 =f x 1 . Again, as x 1 x 2 and f is nondecreasing, we have x 2 =f x 1 x 3 =f x 2 . By induction, we have

x 0 x 1 x n x n + 1 .

If x n = x n + 1 for some nN, then x n =f x n and hence x n is a fixed point of f. So, we may assume that x n x n + 1 for all nN. By (3.1), we have

ψ ( p b ( x n , x n + 1 ) ) ψ ( s p b ( x n , x n + 1 ) ) = ψ ( s p b ( f x n 1 , f x n ) ) ψ ( M s f ( x n 1 , x n ) ) φ ( M s f ( x n 1 , x n ) ) ,
(3.2)

where

M s f ( x n 1 , x n ) = max { p b ( x n 1 , x n ) , p b ( x n 1 , f x n 1 ) , p b ( x n , f x n ) , p b ( x n 1 , f x n ) + p b ( x n , f x n 1 ) 2 s } = max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) , p b ( x n 1 , x n + 1 ) + p b ( x n , x n ) 2 s } max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) , s p b ( x n 1 , x n ) + s p b ( x n , x n + 1 ) + ( 1 s ) p b ( x n , x n ) 2 s } = max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) } .

So, we have

M s f ( x n 1 , x n )=max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) } .
(3.3)

From (3.2), (3.3) we get

ψ ( p b ( x n , x n + 1 ) ) ψ ( max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) } ) φ ( max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) } ) .
(3.4)

If

max { p b ( x n 1 , x n ) , p b ( x n , x n + 1 ) } = p b ( x n , x n + 1 ),

then by (3.4) and properties of φ, we have

ψ ( p b ( x n , x n + 1 ) ) ψ ( p b ( x n , x n + 1 ) ) φ ( p b ( x n , x n + 1 ) ) < ψ ( p b ( x n , x n + 1 ) ) ,

which gives a contradiction. Thus,

ψ ( p b ( x n , x n + 1 ) ) ψ ( p b ( x n 1 , x n ) ) φ ( p b ( x n 1 , x n ) ) .
(3.5)

Therefore, { p b ( x n , x n + 1 ):nN{0}} is a nonincreasing sequence of positive numbers. So, there exists r0 such that

lim n p b ( x n , x n + 1 )=r.

Letting n in (3.5), we get

ψ(r)ψ(r)φ(r)ψ(r).

Therefore, φ(r)=0, and hence r=0. Thus, we have

lim n p b ( x n , x n + 1 )=0.
(3.6)

Next, we show that { x n } is a p b -Cauchy sequence in X. For this, we have to show that { x n } is a b-Cauchy sequence in (X, d p b ) (see Lemma 1). Suppose the contrary; that is, { 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 >i, d p b ( x m i , x n i )ε.
(3.7)

This means that

d p b ( x m i , x n i 1 )<ε.
(3.8)

From (3.7) and using the triangular inequality, we get

ε d p b ( x m i , x n i )s d p b ( x m i , x n i 1 )+s d p b ( x n i 1 , x n i ).
(3.9)

Taking the upper limit as i and using (3.8), we get

ε s lim inf i d p b ( x m i , x n i 1 ) lim sup i d p b ( x m i , x n i 1 )ε.
(3.10)

Also, from (3.9) and (3.10),

ε lim sup i d p b ( x m i , x n i )sε.

Further,

d p b ( x m i + 1 , x n i )s d p b ( x m i + 1 , x m i )+s d p b ( x m i , x n i ),

and hence

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

Finally,

d p b ( x m i + 1 , x n i 1 )s d p b ( x m i + 1 , x m i )+s d p b ( x m i , x n i 1 ),

and hence

lim sup i d p b ( x m i + 1 , x n i 1 )sε.

On the other hand, by the definition of d p b and (3.6),

lim sup i d p b ( x m i , x n i 1 )=2 lim sup i p b ( x m i , x n i 1 ).

Hence, by (3.10),

ε 2 s lim inf i p b ( x m i , x n i 1 ) lim sup i p b ( x m i , x n i 1 ) ε 2 .
(3.11)

Similarly,

lim sup i p b ( x m i , x n i ) s ε 2 ,
(3.12)
ε 2 s lim sup i p b ( x m i + 1 , x n i ),
(3.13)
lim sup i p b ( x m i + 1 , x n i 1 ) s ε 2 .
(3.14)

From (3.1), we have

ψ ( s p b ( x m i + 1 , x n i ) ) = ψ ( s p b ( f x m i , f x n i 1 ) ) ψ ( M s f ( x m i , x n i 1 ) ) φ ( M s f ( x m i , x n i 1 ) ) ,
(3.15)

where

M s f ( x m i , x n i 1 ) = max { p b ( x m i , x n i 1 ) , p b ( x m i , f x m i ) , p b ( x n i 1 , f x n i 1 ) , p b ( x m i , f x n i 1 ) + p b ( f x m i , x n i 1 ) 2 s } = max { p b ( x m i , x n i 1 ) , p b ( x m i , x m i + 1 ) , p b ( x n i 1 , x n i ) , p b ( x m i , x n i ) + p b ( x m i + 1 , x n i 1 ) 2 s } .
(3.16)

Taking the upper limit as i in (3.16) and using (3.6), (3.11), (3.12) and (3.14), we get

lim sup i M s f ( x m i , x n i 1 ) = max { lim sup i p b ( x m i , x n i 1 ) , 0 , 0 , lim sup i p b ( x m i , x n i ) + lim sup i p b ( x m i + 1 , x n i 1 ) 2 s } max { ε 2 , ε s + ε s 2 2 s } = ε 2 .
(3.17)

Now, taking the upper limit as i in (3.15) and using (3.13) and (3.17), we have

ψ ( s ε 2 s ) ψ ( s lim sup i p b ( x m i + 1 , x n i ) ) ψ ( lim sup i M s f ( x m i , x n i 1 ) ) lim inf i φ ( M s f ( x m i , x n i 1 ) ) ψ ( ε 2 ) φ ( lim inf i M s f ( x m i , x n i 1 ) ) ,

which further implies that

φ ( lim inf i M s f ( x m i , x n i 1 ) ) =0,

so lim inf i M s f ( x m i , x n i 1 )=0, and by (3.16) we get lim inf i d p b ( x m i , x n i 1 )=0, a contradiction with (3.11).

Thus, we have proved that { x n } is a b-Cauchy sequence in the b-metric space (X, d p b ). Since (X, p b ) is p b -complete, then from Lemma 1, (X, d p b ) is a b-complete b-metric space. Therefore, the sequence { x n } converges to some zX, that is, lim n d p b ( x n ,z)=0. Again, from Lemma 1,

lim n p b (z, x n )= lim n p b ( x n , x n )= p b (z,z).

On the other hand, thanks to (3.6) and condition ( p b 2 ), lim n p b ( x n , x n )=0, which yields that

lim n p b (z, x n )= lim n p b ( x n , x n )= p b (z,z)=0.

Using the triangular inequality, we get

p b (z,fz)s p b (z,f x n )+s p b (f x n ,fz).

Letting n and using the continuity of f, we get

p b (z,fz)s lim n p b (z,f x n )+s lim n p b (f x n ,fz)=s p b (fz,fz).
(3.18)

Note that from (3.1), we have

ψ ( s p b ( f z , f z ) ) ψ ( M s f ( z , z ) ) φ ( M s f ( z , z ) ) ,
(3.19)

where

M s f (z,z)=max { p b ( z , z ) , p b ( z , f z ) , p b ( z , f z ) , p b ( z , f z ) + p b ( z , f z ) 2 s } = p b (fz,z).

Hence, as ψ is nondecreasing, we have s p b (fz,fz) p b (fz,z). Thus, by (3.18) we obtain that s p b (fz,fz)= p b (fz,z). But then, using (3.19), we get that φ( M s f (z,z))=0.

Hence, we have p b (fz,z)=0 and fz=z. Thus, z is a fixed point of f. □

We will show now that the continuity of f in Theorem 1 is not necessary and can be replaced by another assumption.

Theorem 2 Under the hypotheses of Theorem  1, without the continuity assumption on f, assume that whenever { x n } is a nondecreasing sequence in X such that x n xX, one has x n x for all nN. Then f has a fixed point in X.

Proof Following similar arguments as those given in Theorem 1, we construct an increasing sequence { x n } in X such that x n z for some zX. Using the assumption on X, we have x n z for all nN. Now, we show that fz=z. By (3.1), we have

ψ ( s p b ( x n + 1 , f z ) ) = ψ ( s p b ( f x n , f z ) ) ψ ( M s f ( x n , z ) ) φ ( M s f ( x n , z ) ) ,
(3.20)

where

M s f ( x n , z ) = max { p b ( x n , z ) , p b ( x n , f x n ) , p b ( z , f z ) , p b ( x n , f z ) + p b ( f x n , z ) 2 s } = max { p b ( x n , z ) , p b ( x n , x n + 1 ) , p b ( z , f z ) , p b ( x n , f z ) + p b ( x n + 1 , z ) 2 s } .
(3.21)

Letting n in (3.21) and using Lemma 2, we get

p b ( z , f z ) 2 s 2 = min { p b ( z , f z ) , p b ( z , f z ) s 2 s } lim inf i M s f ( x n , z ) lim sup i M s f ( x n , z ) max { p b ( z , f z ) , s p b ( z , f z ) 2 s } = p b ( z , f z ) .
(3.22)

Again, taking the upper limit as n in (3.20) and using Lemma 2 and (3.22), we get

ψ ( p b ( z , f z ) ) = ψ ( s 1 s p b ( z , f z ) ) ψ ( s lim sup n p b ( x n + 1 , f z ) ) ψ ( lim sup n M s f ( x n , z ) ) lim inf n φ ( M s f ( x n , z ) ) ψ ( p b ( z , f z ) ) φ ( lim inf n M s f ( x n , z ) ) .

Therefore, φ( lim inf n M s f ( x n ,z))0, equivalently, lim inf n M s f ( x n ,z)=0. Thus, from (3.22) we get z=fz, and hence z is a fixed point of f. □

Corollary 1 Let (X,, p b ) be a p b -complete ordered partial b-metric space. Let f:XX be a continuous mapping, nondecreasing with respect to . Suppose that there exists k[0,1) such that

p b (fx,fy) k s max { p b ( x , y ) , p b ( x , f x ) , p b ( y , f y ) , p b ( x , f y ) + p b ( y , f x ) 2 s }

for all comparable elements x,yX. If there exists x 0 X such that x 0 f x 0 , then f has a fixed point.

Proof Follows from Theorem 1 by taking ψ(t)=t and φ(t)=(1k)t, for all t[0,+). □

Corollary 2 Under the hypotheses of Corollary  1, without the continuity assumption on f, for any nondecreasing sequence { x n } in X such that x n xX, let us have x n x for all nN. Then f has a fixed point in X.

Now, in order to support the usability of our results, we present the following example.

Example 4 Let X=[0,+) be equipped with the partial order defined by

xyx=y ( x , y [ 0 , 1 ] x y ) ,

and with the partial b-metric p b given by p b (x,y)= [ max { x , y } ] 2 (with s=2). Consider the mapping f:XX given by

fx={ x 2 1 + x , x [ 0 , 1 ] , x 2 , x > 1 .

Then f is continuous and increasing, and 0f0. Take altering distance functions

ψ(t)=t,φ(t)={ t t 1 + t , 0 t 1 , t 2 , t > 1 .

In order to check the contractive condition (3.1) of Theorem 1, without loss of generality, we may take x,yX such that yx. Consider the following two possible cases.

Case 1. 0yx1. Then

p b (fx,fy)= [ max { x 2 1 + x , y 2 1 + y } ] 2 = x 2 2 ( 1 + x )

and

M s f (x,y)=max { x 2 , x 2 , y 2 , x 2 + max 2 { y , x 2 1 + x } 2 s } = x 2 .

Thus, (3.1) reduces to

ψ ( 2 x 2 2 ( 1 + x ) ) = x 2 1 + x x 2 x 3 1 + x =ψ ( x 2 ) φ ( x 2 ) .

Case 2. x=y1. Then p b (fx,fy)= x 2 4 and M s f (x,y)= x 2 , so (3.1) reduces to

ψ ( 2 x 2 4 ) = x 2 2 x 2 x 2 2 =ψ ( x 2 ) φ ( x 2 ) .

Hence, all the conditions of Theorem 1 are satisfied and f has a fixed point (which is z=0).

4 Common fixed point results in partial b-metric spaces

Let (X,, p b ) be an ordered partial b-metric space with the coefficient s1, and let f,g:XX be two mappings. Set

M s f , g (x,y)=max { p b ( x , y ) , p b ( x , f x ) , p b ( y , g y ) , p b ( x , g y ) + p b ( y , f x ) 2 s } .

Now, we present the following definition.

Definition 12 Let (X,, p b ) be an ordered partial b-metric space, and let ψ and φ be altering distance functions. We say that a pair (f,g) of self-mappings f,g:XX is a generalized ( ψ , φ ) s -contraction pair if

ψ ( s 2 p b ( f x , g y ) ) ψ ( M s f , g ( x , y ) ) φ ( M s f , g ( x , y ) )
(4.1)

for all comparable x,yX.

Definition 13 [42]

Let (X,) be a partially ordered set. Then two mappings f,g:XX are said to be weakly increasing if fxgfx and gxfgx for all xX.

Theorem 3 Let (X,, p b ) be a p b -complete ordered partial b-metric space with the coefficient s1, and let f,g:XX be two weakly increasing mappings with respect to . Suppose that (f,g) is a generalized ( ψ , φ ) s -contraction pair for some altering distance functions ψ and φ. If f and g are continuous, then f and g have a common fixed point.

Proof Let us divide the proof into two parts as follows.

First part. We prove that uX is a fixed point of f if and only if it is a fixed point of g. Suppose that u is a fixed point of f, that is, fu=u. As uu, by (4.1), we have

ψ ( s 2 p b ( u , g u ) ) = ψ ( s 2 p b ( f u , g u ) ) ψ ( max { p b ( u , u ) , p b ( u , f u ) , p b ( u , g u ) , 1 2 s ( p b ( u , g u ) + p b ( u , f u ) ) } ) φ ( max { p b ( u , u ) , p b ( u , f u ) , p b ( u , g u ) , 1 2 s ( p b ( u , g u ) + p b ( u , f u ) ) } ) ψ ( p b ( u , g u ) ) φ ( max { p b ( u , u ) , p b ( u , f u ) , p b ( u , g u ) , 1 2 s ( p b ( u , g u ) + p b ( u , f u ) ) } ) ψ ( s 2 p b ( u , g u ) ) φ ( max { p b ( u , u ) , p b ( u , f u ) , p b ( u , g u ) , 1 2 s ( p b ( u , g u ) + p b ( u , f u ) ) } ) .

Therefore, p b (u,gu)=0 and hence gu=u. Similarly, we can show that if u is a fixed point of g, then u is a fixed point of f.

Second part (construction of a sequence by iterative technique).

Let x 0 X. We construct a sequence { x n } in X such that x 2 n + 1 =f x 2 n and x 2 n + 2 =g x 2 n + 1 for all nonnegative integers n. As f and g are weakly increasing with respect to , we have

x 1 = f x 0 g f x 0 = x 2 = g x 1 f g x 1 = x 3 x 2 n + 1 = f x 2 n g f x 2 n = x 2 n + 2 .

If x 2 n = x 2 n + 1 for some nN, then x 2 n =f x 2 n . Thus x 2 n is a fixed point of f. By the first part, we conclude that x 2 n is also a fixed point of g.

If x 2 n + 1 = x 2 n + 2 for some nN, then x 2 n + 1 =g x 2 n + 1 . Thus, x 2 n + 1 is a fixed point of g. By the first part, we conclude that x 2 n + 1 is also a fixed point of f. Therefore, we assume that x n x n + 1 for all nN. Now, we complete the proof in the following steps.

Step 1: We will prove that

lim n p b ( x n , x n + 1 )=0.

As x 2 n + 1 and x 2 n + 2 are comparable, by (4.1), we have

ψ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) ψ ( s 2 p b ( x 2 n + 1 , x 2 n + 2 ) ) = ψ ( s 2 p b ( f x 2 n , g x 2 n + 1 ) ) ψ ( M s f , g ( x 2 n , x 2 n + 1 ) ) φ ( M s f , g ( x 2 n , x 2 n + 1 ) ) ,

where

M s f , g ( x 2 n , x 2 n + 1 ) = max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n , f x 2 n ) , p b ( x 2 n + 1 , g x 2 n + 1 ) , p b ( f x 2 n , x 2 n + 1 ) + p b ( x 2 n , g x 2 n + 1 ) 2 s } = max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) , p b ( x 2 n + 1 , x 2 n + 1 ) + p b ( x 2 n , x 2 n + 2 ) 2 s } max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) , s p b ( x 2 n , x 2 n + 1 ) + s p b ( x 2 n + 1 , x 2 n + 2 ) 2 s } = max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) } .

Hence, we have

ψ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) ψ ( max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) } ) φ ( max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) } ) .
(4.2)

If

max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) } = p b ( x 2 n + 1 , x 2 n + 2 ),

then (4.2) becomes

ψ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) ψ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) φ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) < ψ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) ,

which gives a contradiction. Hence,

max { p b ( x 2 n , x 2 n + 1 ) , p b ( x 2 n + 1 , x 2 n + 2 ) } = p b ( x 2 n , x 2 n + 1 ),

and (4.2) becomes

ψ ( p b ( x 2 n + 1 , x 2 n + 2 ) ) ψ ( p b ( x 2 n , x 2 n + 1 ) ) φ ( p b ( x 2 n , x 2 n + 1 ) ) ψ ( p b ( x 2 n , x 2 n + 1 ) ) .
(4.3)

Similarly, we can show that

ψ ( p b ( x 2 n + 1 , x 2 n ) ) ψ ( p b ( x 2 n 1 , x 2 n ) ) φ ( p b ( x 2 n 1 , x 2 n ) ) ψ ( p b ( x 2 n 1 , x 2 n ) ) .
(4.4)

By (4.3) and (4.4), we get that { p b ( x n , x n + 1 ):nN} is a nonincreasing sequence of positive numbers. Hence, there is r0 such that

lim n p b ( x n , x n + 1 )=r.

Letting n in (4.3), we get

ψ(r)ψ(r)φ(r)ψ(r),

which implies that φ(r)=0 and hence r=0. So, we have

lim n p b ( x n , x n ) lim n p b ( x n , x n + 1 )=0.
(4.5)

Step 2. We will prove that { x n } is a p b -Cauchy sequence. Because of (4.5), it is sufficient to show that { x 2 n } is a p b -Cauchy sequence. By Lemma 1, we should show that { x 2 n } is b-Cauchy in (X, d p b ). Suppose the contrary, i.e., that { x 2 n } is not a b-Cauchy sequence in (X, d p b ). Then there exists ε>0 for which we can find two subsequences { x 2 m i } and { x 2 n i } of { x 2 n } such that n i is the smallest index for which

n i > m i >i, d p b ( x 2 m i , x 2 n i )ε.
(4.6)

This means that

d p b ( x 2 m i , x 2 n i 2 )<ε.
(4.7)

From (4.6) and using the triangular inequality, we get

ε d p b ( x 2 m i , x 2 n i )s d p b ( x 2 m i , x 2 m i + 1 )+s d p b ( x 2 m i + 1 , x 2 n i ).

Using (4.5) and taking the upper limit as i, we get

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

On the other hand, we have

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

Using (4.5), (4.7) and taking the upper limit as i, we get

lim sup i d p b ( x 2 m i , x 2 n i 1 )εs.
(4.8)

Again, using the triangular inequality, we have

d p b ( x 2 m i , x 2 n i ) s d p b ( x 2 m i , x 2 n i 2 ) + s d p b ( x 2 n i 2 , x 2 n i ) s d p b ( x 2 m i , x 2 n i 2 ) + s 2 d p b ( x 2 n i 2 , x 2 n i 1 ) + s 2 d p b ( x 2 n i 1 , x 2 n i )

and

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

Taking the upper limit as i in the above inequalities and using (4.5), (4.7) and (4.8), we get

lim sup i d p b ( x 2 m i , x 2 n i )εs

and

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

From the definition of d p b and (4.5), we have the following relations:

ε 2 s lim sup i p b ( x 2 m i + 1 , x 2 n i ),
(4.9)
ε 2 s lim inf i p b ( x 2 m i , x 2 n i 1 ) lim sup i p b ( x 2 m i , x 2 n i 1 ) s ε 2 ,
(4.10)
lim sup i p b ( x 2 m i , x 2 n i ) s ε 2 ,
(4.11)
lim sup i p b ( x 2 m i + 1 , x 2 n i 1 ) s 2 ε 2 .
(4.12)

Since x 2 m i and x 2 n i 1 are comparable, using (4.1) we have

ψ ( s 2 p b ( x 2 m i + 1 , x 2 n i ) ) = ψ ( s 2 p b ( f x 2 m i , g x 2 n i 1 ) ) ψ ( M s f , g ( x 2 m i , x 2 n i 1 ) ) φ ( M s f , g ( x 2 m i , x 2 n i 1 ) ) ,
(4.13)

where

M s f , g ( x 2 m i , x 2 n i 1 ) = max { p b ( x 2 m i , x 2 n i 1 ) , p b ( x 2 m i , x 2 m i + 1 ) , p b ( x 2 n i 1 , x 2 n i ) , p b ( x 2 m i , x 2 n i ) + p b ( x 2 m i + 1 , x 2 n i 1 ) 2 s } .
(4.14)

Taking the upper limit in (4.14) and using (4.5) and (4.10)-(4.12), we get

lim sup i M s f , g ( x 2 m i , x 2 n i 1 ) = max { lim sup i p b ( x 2 m i , x 2 n i 1 ) , 0 , 0 , lim sup i p b ( x 2 m i , x 2 n i ) + lim sup i p b ( x 2 m i + 1 , x 2 n i 1 ) 2 s } max { s ε 2 , ε s + ε s 2 2 2 s } = s ε 2 .
(4.15)

Now, taking the upper limit as i in (4.13) and using (4.9) and (4.15), we have

ψ ( s ε 2 ) = ψ ( s 2 ε 2 s ) ψ ( s 2 lim sup i p b ( x 2 m i + 1 , x 2 n i ) ) ψ ( lim sup i M s f , g ( x 2 m i , x 2 n i 1 ) ) φ ( lim inf i M s f , g ( x 2 m i , x 2 n i 1 ) ) ψ ( s ε 2 ) φ ( lim inf i M s f , g ( x 2 m i , x 2 n i 1 ) ) ,

which implies that φ( lim inf i M s f , g ( x 2 m i , x 2 n i 1 ))=0. By (4.14), it follows that

lim inf i p b ( x 2 m i , x 2 n i 1 )=0,

which is in contradiction with (4.10). Thus, we have proved that { x n } is a b-Cauchy sequence in the metric space (X, d p b ). Since (X, p b ) is p b -complete, then from Lemma 1, (X, d p b ) is a b-complete b-metric space. Therefore, the sequence { x n } converges to some zX, that is, lim n d p b ( x n ,z)=0. Again, from Lemma 1,

lim n p b (z, x n )= lim n p b ( x n , x n )= p b (z,z).

On the other hand, from (4.5) we get that

lim n p b (z, x n )= lim n p b ( x n , x n )= p b (z,z)=0.

Step 3 (Existence of a common fixed point). Using the triangular inequality, we get

p b ( z , f z ) s p b ( z , f x 2 n ) + s p b ( f x 2 n , f z ) , p b ( z , g z ) s p b ( z , g x 2 n + 1 ) + s p b ( g x 2 n + 1 , g z ) .

Letting n and using the continuity of f and g, we get

p b ( z , f z ) s lim n p b ( z , f x 2 n ) + s lim n p b ( f x 2 n , f z ) = s p b ( f z , f z ) , p b ( z , g z ) s lim n p b ( z , g x 2 n + 1 ) + s lim n p b ( g x 2 n + 1 , g z ) = s p b ( g z , g z ) .

Therefore,

max { p b ( z , f z ) , p b ( z , g z ) } max { s p b ( f z , f z ) , s p b ( g z , g z ) } s 2 p b (gz,fz).
(4.16)

From (4.1), we have

ψ ( s 2 p b ( f z , g z ) ) ψ ( M s f , g ( z , z ) ) φ ( M s f , g ( z , z ) ) ,
(4.17)

where

M s f , g ( z , z ) = max { p b ( z , z ) , p b ( z , f z ) , p b ( z , g z ) , p b ( z , g z ) + p b ( z , f z ) 2 s } = max { p b ( z , f z ) , p b ( z , g z ) } .

As ψ is nondecreasing, we have s 2 p b (fz,gz)max{ p b (z,fz), p b (z,gz)}. Hence, by (4.16) we obtain that s 2 p b (fz,gz)=max{ p b (z,fz), p b (z,gz)}. But then, using (4.17), we get that φ( M s f , g (z,z))=0. Thus, we have fz=gz=z and z is a common fixed point of f and g. □

The continuity of functions f and g in Theorem 3 can be replaced by another condition.

Theorem 4 Under the hypotheses of Theorem  3, without the continuity assumption on the functions f and g, for any nondecreasing sequence { x n } in X such that x n xX, let us have x n x for all nN. Then f and g have a common fixed point in X.

Proof Reviewing the proof of Theorem 3, we construct an increasing sequence { x n } in X such that x n z for some zX. Using the given assumption on X, we have x n z for all nN. Now, we show that fz=gz=z. By (4.1), we have

ψ ( s 2 p b ( x 2 n + 1 , g z ) ) = ψ ( s 2 p b ( f x 2 n , g z ) ) ψ ( M s f , g ( x 2 n , z ) ) φ ( M s f , g ( x 2 n , z ) ) ,
(4.18)

where

M s f , g ( x 2 n , z ) = max { p b ( x 2 n , z ) , p b ( x 2 n , f x 2 n ) , p b ( z , g z ) , p b ( x 2 n , g z ) + p b ( f x 2 n , z ) 2 s } = max { p b ( x 2 n , z ) , p b ( x 2 n , x 2 n + 1 ) , p b ( z , g z ) , p b ( x 2 n , g z ) + p b ( x 2 n + 1 , z ) 2 s } .
(4.19)

Letting n in (4.19) and using Lemma 2, we get

p b ( z , g z ) s 2 max { p b ( z , g z ) , p b ( z , g z ) s 2 s } lim inf n M s f , g ( x 2 n , z ) lim sup n M s f , g ( x 2 n , z ) max { p b ( z , g z ) , s p b ( z , g z ) 2 s } = p b ( z , g z ) .
(4.20)

Again, taking the upper limit as n in (4.18) and using Lemma 2 and (4.20), we get

ψ ( p b ( z , g z ) ) = ψ ( s 2 1 s 2 p b ( z , g z ) ) ψ ( s 2 lim sup n d ( x 2 n + 1 , g z ) ) ψ ( lim sup n M s f , g ( x 2 n , z ) ) φ ( lim inf n M s f , g ( x 2 n , z ) ) ψ ( p b ( z , g z ) ) φ ( lim inf n M s f , g ( x 2 n , z ) ) .

Therefore, φ( lim inf n M s f , g ( x 2 n ,z))0, equivalently, lim inf n M s f , g ( x 2 n ,z)=0. Thus, from (4.20) we get z=gz and hence z is a fixed point of g. On the other hand, similar to the first part of the proof of Theorem 3, we can show that fz=z. Hence, z is a common fixed point of f and g. □

Also, we have the following results.

Corollary 3 Let (X,, p b ) be a p b -complete ordered partial b-metric space with the coefficient s1, and let f,g:XX be two weakly increasing mappings with respect to . Suppose that there exists k[0,1) such that

p b (fx,gy) k s 2 max { p b ( x , y ) , p b ( x , f x ) , p b ( y , g y ) , p b ( x , g y ) + p b ( f x , y ) 2 s }

for all comparable elements x,yX. If f and g are continuous, then f and g have a common fixed point.

Corollary 4 Under the hypotheses of Corollary  3, without the continuity assumption on the functions f and g, assume that whenever { x n } is a nondecreasing sequence in X such that x n xX, then x n x for all nN. Then f and g have a common fixed point in X.

Remark 1 Recall that a subset W of a partially ordered set X is said to be well ordered if every two elements of W are comparable. Note that in Theorems 1 and 2, it can be proved in a standard way that f has a unique fixed point provided that the fixed points of f are comparable. Similarly, in Theorems 3 and 4, the set of common fixed points of f and g is well ordered if and only if f and g have one and only one common fixed point.

The usability of these results is demonstrated by the following example.

Example 5 Let X={0,1,2,3,4} be equipped with the following partial order :

:= { ( 0 , 0 ) , ( 1 , 1 ) , ( 1 , 2 ) , ( 2 , 2 ) , ( 3 , 3 ) , ( 4 , 2 ) , ( 4 , 4 ) } .

Define a partial b-metric p b :X×X R + by

p b (x,y)={ 0 if  x = y , ( x + y ) 2 if  x y .

It is easy to see that (X, p b ) is a p b -complete partial b-metric space, with s=49/25.

Define self-maps f and g by

f= ( 0 1 2 3 4 0 2 2 1 2 ) ,g= ( 0 1 2 3 4 0 2 2 1 1 ) .

We see that f and g are weakly increasing mappings with respect to and that f and g are continuous.

Define ψ,φ:[0,)[0,) by ψ(t)= t and φ(t)= t 300 . In order to check that (f,g) is a generalized ( ψ , φ ) s -contractive pair, only the case x=2, y=4 is nontrivial (when x and y are comparable and the left-hand side of condition (4.1) is positive). Then

ψ ( s 2 p b ( f 2 , g 4 ) ) = s 2 3 2 = 147 25 = 36 36 300 =ψ ( M s f , g ( 2 , 4 ) ) φ ( M s f , g ( 2 , 4 ) ) .

Thus, all the conditions of Theorem 3 are satisfied and hence f and g have common fixed points. Indeed, 0 and 2 are two common fixed points of f and g. Note that the ordered set ({0,2},) is not well ordered.

Note that if the same example is considered in the space without order, then the contractive condition is not satisfied. For example,

ψ ( s 2 p b ( f 1 , g 4 ) ) = s 2 3 2 = 147 25 > 59 12 = 25 25 300 = ψ ( M s f , g ( 1 , 4 ) ) φ ( M s f , g ( 1 , 4 ) ) .

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