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Fixed point and coupled fixed point theorems on b-metric-like spaces

Abstract

We first introduce the concept of b-metric-like space which generalizes the notions of partial metric space, metric-like space and b-metric space. Then we establish the existence and uniqueness of fixed points in a b-metric-like space as well as in a partially ordered b-metric-like space. As an application, we derive some new fixed point and coupled fixed point results in partial metric spaces, metric-like spaces and b-metric spaces. Moreover, some examples and an application to integral equations are provided to illustrate the usability of the obtained results.

MSC:47H10, 54H25, 55M20.

1 Introduction

There exist many generalizations of the concept of metric spaces in the literature. In particular, Matthews [1] introduced the notion of partial metric space and proved that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification. After that, fixed point results in partial metric spaces have been studied by many authors [1, 2]. The concept of b-metric space was introduced and studied by Bakhtin [3] and Czerwik [4]. Since then several papers have dealt with fixed point theory for single-valued and multi-valued operators in b-metric spaces (see [58] and references therein). Recently, Amini-Harandi [9, 10] introduced the notion of metric-like space, which is an interesting generalization of partial metric space and dislocated metric space [1113]. In this paper, we first introduce a new generalization of metric-like space and partial metric space which is called a b-metric-like space. Then, we give some fixed point results in such spaces. Our fixed point theorems, even in the case of metric-like spaces and partial metric spaces, generalize and improve some well-known results in the literature. Moreover, some examples and an application to integral equations are provided to illustrate the usability of the obtained results.

2 b-Metric-like spaces

Matthews [1] introduced the concept of a partial metric space as follows.

Definition 2.1 A mapping p:X×X R + , where X is a nonempty set, is said to be a partial metric on X if for any x,y,zX the following four conditions hold true:

(P1) x=y if and only if p(x,x)=p(y,y)=p(x,y);

(P2) p(x,x)p(x,y);

(P3) p(x,y)=p(y,x);

(P4) p(x,z)p(x,y)+p(y,z)p(y,y).

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

Definition 2.2 [9]

A mapping σ:X×X R + , where X is a nonempty set, is said to be a metric-like on X if for any x,y,zX the following three conditions hold true:

(σ 1) σ(x,y)=0x=y;

(σ 2) σ(x,y)=σ(y,x);

(σ 3) σ(x,z)σ(x,y)+σ(y,z).

The pair (X,σ) is then called a metric-like space. A metric-like on X satisfies all of the conditions of a metric except that σ(x,x) may be positive for xX.

Every partial metric space is a metric-like space but not conversely in general (see [9, 10]).

The concept of b-metric space was introduced by Czerwik in [4]. Since then, several papers have been published on the fixed point theory of various classes of single-valued and multi-valued operators in b-metric spaces (see, e.g., [68]).

Definition 2.3 A b-metric on a nonempty set X is a function D:X×X[0,+) such that for all x,y,zX and a constant K1 the following three conditions hold true:

(1) if D(x,y)=0x=y;

(2) D(x,y)=D(y,x);

(3) D(x,y)K(D(x,z)+D(z,y)).

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

Definition 2.4 A b-metric-like on a nonempty set X is a function D:X×X[0,+) such that for all x,y,zX and a constant K1 the following three conditions hold true:

(1) if D(x,y)=0x=y;

(2) D(x,y)=D(y,x);

(3) D(x,y)K(D(x,z)+D(z,y)).

The pair (X,D) is called a b-metric-like space.

Example 2.5 Let X=[0,). Define the function D: X 2 [0,) by D(x,y)= ( x + y ) 2 . Then (X,D) is a b-metric-like space with constant K=2. Clearly, (X,D) is not a b-metric or metric-like space. Indeed, for all x,y,zX,

D ( x , y ) = ( x + y ) 2 ( x + z + z + y ) 2 = ( x + z ) 2 + ( z + y ) 2 + 2 ( x + z ) ( z + y ) 2 [ ( x + z ) 2 + ( z + y ) 2 ] = 2 ( D ( x , z ) + D ( z , y ) )

and so (3) holds. Clearly, (1) and (2) hold.

Similarly, we have the following example.

Example 2.6 Let X=[0,). Define the function D: X 2 [0,) by D(x,y)= ( max { x , y } ) 2 . Then (X,D) is a b-metric-like space with constant K=2. Clearly, (X,D) is not a b-metric or metric-like space.

Example 2.7 Let C b (X)={f:XR: sup x X |f(x)|<+}. The function D:X×X R + , defined by

D(f,g)= sup x X ( | f ( x ) | + | g ( x ) | ) 3 3 for all f,g C b (X),

is a b-metric-like with constant K= 4 3 , and so (X,D, 4 3 ) is a b-metric-like space.

For this, note that if a, b are two nonnegative real numbers, then

( a + b ) 3 4 ( a 3 + b 3 ) and a + b 3 a 3 + b 3 .

This implies that

D(f,g) 4 3 ( D ( f , h ) + D ( h , g ) ) for all f,g,h C b (X).

Let (X,D) be a b-metric-like space. Let xX and r>0, then the set

B(x,r)= { y X : | D ( x , y ) D ( x , x ) | < r }

is called an open ball with center at x and radius r>0.

Now we have the following definitions.

Definition 2.8 Let (X,D) be a b-metric-like space, and let { x n } be a sequence of points of X. A point xX is said to be the limit of the sequence { x n } if lim n + D(x, x n )=D(x,x), and we say that the sequence { x n } is convergent to x and denote it by x n x as n.

Definition 2.9 Let (X,D) be a b-metric-like space.

(S1) A sequence { x n } is called Cauchy if and only if lim m , n D( x n , x m ) exists and is finite.

(S2) A b-metric-like space (X,D) is said to be complete if and only if every Cauchy sequence { x n } in X converges to xX so that

lim m , n D( x n , x m )=D(x,x)= lim n D( x n ,x).

Proposition 2.10Let(X,D,K)be ab-metric-like space, and let{ x n }be a sequence inXsuch that lim n D( x n ,x)=0. Then

  1. (A)

    xis unique;

  2. (B)

    1 K D(x,y) lim n D( x n ,y)KD(x,y)for allyX.

Proof Let us prove (A).

Assume that there exists a yX such that lim n D( x n ,y)=0, then

0D(y,x)K ( lim n D ( x n , y ) + lim n D ( x n , x ) ) =0.

Hence, from (1) we have y=x.

(B)

From (D3) we have

1 K D(x,y) lim n D( x n ,x) lim n D( x n ,y)K ( D ( x , y ) + lim n D ( x n , x ) ) ,

and so

1 K D(x,y) lim n D( x n ,y)KD(x,y)for all yX.

 □

Definition 2.11 Let (X,D) be a b-metric-like space, and let U be a subset of X. We say U is an open subset of X if for all xU there exists r>0 such that B(x,r)U. Also, VX is a closed subset of X if XV is an open subset of X.

Proposition 2.12Let(X,D,K)be ab-metric-like space, and letVbe a subset ofX. ThenVis closed if and only if for any sequence{ x n }inV, which converges tox, we havexV.

Proof At first, we suppose that V is closed. Let xV. By the above definition, XV is an open set. Then there is an r>0 such that B(x,r)XV. On the other hand, since x n x as n, then

lim n | D ( x n , x ) D ( x , x ) | =0.

Hence, there exists n 0 N such that for all n n 0 we have

| D ( x n , x ) D ( x , x ) | <r.

That is, for all n n 0 , { x n }B(x,r)XV, which is a contradiction. Since for all nN, { x n }V. Conversely, suppose that for any sequence { x n } in V which converges to x, we have xV. Let yV. Let us prove that there exists r 0 >0 such that B(y, r 0 )V=. Assume to the contrary that for all r>0, we have B(y,r)V. Then, for all nN, chose x n B(y,1/n)V. Therefore, |D( x n ,y)D(y,y)|<1/n for all nN. Hence, x n y as n. Our assumption on V implies yV, which is a contradiction. Then, for all yV, there exists r 0 >0 such that B(y, r 0 )V=. That is, V is closed. □

Lemma 2.13Let(X,D,K)be ab-metric-like space, and let { x k } k = 0 n X. Then

D( x n , x 0 )KD( x 0 , x 1 )++ K n 1 D( x n 2 , x n 1 )+ K n 1 D( x n 1 , x n ).

From Lemma 2.13, we deduce the following result.

Lemma 2.14Let{ y n }be a sequence in ab-metric-like space(X,D,K)such that

D( y n , y n + 1 )λD( y n 1 , y n )

for someλ, 0<λ<1/K, and eachnN. Then lim m , n D( y m , y n )=0.

Let (X,D,K) be a b-metric-like space. Define D s : X 2 [0,) by

D s (x,y)= | 2 D ( x , y ) D ( x , x ) D ( y , y ) | .

Clearly, D s (x,x)=0 for all xX.

3 Fixed point results for expansive mappings

The study of expansive mappings is a very interesting research area in fixed point theory (see, e.g., [1421]). In this section we prove some new fixed point results on expansive mappings in the setting of a b-metric-like space. Our results generalize and extend some old and recent fixed point results in the literature.

Theorem 3.1Let(X,D,K)be a completeb-metric-like space. Assume that the mappingT:XXis onto and satisfies

D(Tx,Ty) [ R + L min { D s ( x , T x ) , D s ( y , T y ) , D s ( x , T y ) , D s ( y , T x ) } ] D(x,y)
(3.1)

for allx,yX, whereR>K, L0. ThenThas a fixed point.

Proof Let x 0 X, since T is onto, then there exists x 1 X such that x 0 =T x 1 . By continuing this process, we get x n =T x n + 1 for all nN{0}. In case x n 0 = x n 0 + 1 for some n 0 N{0}, then it is clear that x n 0 is a fixed point of T. Now assume that x n x n + 1 for all n. From (3.1) with x= x n and y= x n + 1 we get

D ( T x n , T x n + 1 ) [ R + L min { D s ( x n , T x n ) , D s ( x n + 1 , T x n + 1 ) , D s ( x n , T x n + 1 ) , D s ( x n + 1 , T x n ) } ] D ( x n , x n + 1 ) ,

which implies

D ( x n 1 , x n ) [ R + L min { D s ( x n , x n 1 ) , D s ( x n + 1 , x n ) , D s ( x n , x n ) , D s ( x n + 1 , x n 1 ) } ] D ( x n , x n + 1 ) = R D ( x n , x n + 1 ) ,

and so

D( x n , x n + 1 )hD( x n 1 , x n ),where h= 1 R < 1 K .

Then by Lemma 2.14 we have lim m , n D( x n , x m )=0. Now, since lim m , n D( x n , x m )=0 exists (and is finite), so { x n } is a Cauchy sequence. Since (X,D,K) is a complete b-metric-like space, the sequence { x n } in X converges to zX so that

lim m , n D( x n ,z)=D(z,z)= lim m , n D( x n , x m )=0.

Since T is onto, there exists wX such that z=Tw. From (3.1) we have

D ( x n , z ) = D ( T x n + 1 , T w ) [ R + L min { D s ( x n + 1 , T x n + 1 ) , D s ( w , T w ) , D s ( x n + 1 , T w ) , D s ( w , T x n + 1 ) } ] D ( x n + 1 , w ) = [ R + L min { D s ( x n + 1 , x n ) , D s ( w , z ) , D s ( x n + 1 , z ) , D s ( w , x n ) } ] D ( x n + 1 , w ) .

Taking limit as n in the above inequality, we get

0= lim n D( x n ,z)R lim n D( x n + 1 ,w),

which implies lim n D( x n + 1 ,w)=0. By Proposition 2.10 (A), we get z=w. That is, z=Tz. □

If in Theorem 3.1 we take L=0, then we deduce the following corollary.

Corollary 3.2Let(X,D,K)be a completeb-metric-like space. Assume that the mappingT:XXis onto and satisfies

D(Tx,Ty)RD(x,y)

for allx,yX, whereR>K. ThenThas a fixed point.

Example 3.3 Let X=[0,) and let a b-metric-like D:X×X R + be defined by

D(x,y)= ( x + y ) 2 .

Clearly, (X,D,2) is a complete b-metric-like space. Let T:XX be defined by

Tx={ 6 x if  x [ 0 , 1 ) , 5 x + 1 if  x [ 1 , 2 ) , 4 x + 3 if  x [ 2 , ) .

Also, clearly, T is an onto mapping. Now, we consider following cases:

Let x,y[0,1), then

D(Tx,Ty)= ( 6 x + 6 y ) 2 =36 ( x + y ) 2 3 ( x + y ) 2 =3D(x,y).

Let x,y[1,2), then

D(Tx,Ty)= ( 5 x + 5 y + 2 ) 2 ( 5 x + 5 y ) 2 =25 ( x + y ) 2 3 ( x + y ) 2 =3D(x,y).

Let x,y[2,), then

D(Tx,Ty)= ( 4 x + 4 y + 6 ) 2 ( 4 x + 4 y ) 2 =16 ( x + y ) 2 3 ( x + y ) 2 =3D(x,y).

Let x[0,1) and y[1,2), then

D(Tx,Ty)= ( 6 x + 5 y + 1 ) 2 ( 5 x + 5 y ) 2 =25 ( x + y ) 2 3 ( x + y ) 2 =3D(x,y).

Let x[0,1) and y[2,), then

D(Tx,Ty)= ( 6 x + 4 y + 3 ) 2 ( 4 x + 4 y ) 2 =16 ( x + y ) 2 3 ( x + y ) 2 =3D(x,y).

Let x[1,2) and y[2,), then

D(Tx,Ty)= ( 5 x + 4 y + 4 ) 2 ( 4 x + 4 y ) 2 =16 ( x + y ) 2 3 ( x + y ) 2 =3D(x,y).

That is, D(Tx,Ty)RD(x,y) for all x,yX, where R=3>2=K. The conditions of Corollary 3.2 are satisfied and T has a fixed point x=0.

Let Ψ B L denote the class of those functions B:(0,)( L 2 ,) which satisfy the condition B( t n ) ( L 2 ) + t n 0, where L>0.

Theorem 3.4Let(X,D,K)be a completeb-metric-like space. Assume that the mappingT:XXis onto and satisfies

D(Tx,Ty)B ( D ( x , y ) ) D(x,y)
(3.2)

for allx,yX, whereB Ψ B K . ThenThas a fixed point.

Proof Let x 0 X, since T is onto, so there exists x 1 X such that x 0 =T x 1 . By continuing this process, we get x n =T x n + 1 for all nN{0}. In case x n 0 = x n 0 + 1 for some n 0 N{0}, then it is clear that x n 0 is a fixed point of T. Now assume that x n x n + 1 for all n. From (3.2) with x= x n and y= x n + 1 , we get

D ( x n 1 , x n ) = D ( T x n , T x n + 1 ) B ( D ( x n , x n + 1 ) ) D ( x n , x n + 1 ) K 2 D ( x n , x n + 1 ) D ( x n , x n + 1 ) .
(3.3)

Then the sequence {D( x n , x n + 1 )} is a decreasing sequence in R + and so there exists s0 such that lim n D( x n , x n + 1 )=s. Let us prove that s=0. Suppose to the contrary that s>0. By (3.3) we can deduce

K 2 D ( x n 1 , x n ) D ( x n , x n + 1 ) D ( x n 1 , x n ) D ( x n , x n + 1 ) B ( D ( x n , x n + 1 ) ) K 2 .

By taking limit as n in the above inequality, we have lim n B(D( x n , x n + 1 ))= K 2 . Hence,

s= lim n D( x n , x n + 1 )=0,

which is a contradiction. That is, s=0. We shall show that lim sup m , n D( x n , x m )=0. Suppose to the contrary that

lim sup m , n D( x n , x m )>0.

By (3.2) we have

D( x n , x m )=D(T x n + 1 ,T x m + 1 )B ( D ( x n + 1 , x m + 1 ) ) D( x n + 1 , x m + 1 ).

That is,

D ( x n , x m ) B ( D ( x n + 1 , x m + 1 ) ) D( x n + 1 , x m + 1 ).

Then by (3) we get

D ( x n , x m ) K D ( x n , x n + 1 ) + K 2 D ( x n + 1 , x m + 1 ) + K 2 D ( x m + 1 , x m ) K D ( x n , x n + 1 ) + K 2 D ( x n , x m ) B ( D ( x n + 1 , x m + 1 ) ) + K 2 D ( x m + 1 , x m ) .

Therefore,

D( x n , x m ) ( 1 K 2 B ( D ( x n + 1 , x m + 1 ) ) ) 1 ( K D ( x n , x n + 1 ) + K 2 D ( x m + 1 , x m ) ) .

By taking limit as m,n in the above inequality, since lim sup m , n D( x n , x m )>0 and s= lim n D( x n , x n + 1 )=0, then we obtain

lim sup m , n ( 1 K 2 B ( D ( x n + 1 , x m + 1 ) ) ) 1 =,

which implies

lim sup m , n B ( D ( x n + 1 , x m + 1 ) ) = ( K 2 ) + ,

and so

lim sup m , n D( x n + 1 , x m + 1 )=0,

which is a contradiction. Hence, lim sup m , n D( x n , x m )=0. Now, since lim m , n D( x n , x m )=0 exists (and is finite), so { x n } is a Cauchy sequence. Since (X,D,K) is a complete b-metric-like space, the sequence { x n } in X converges to zX so that

lim m , n D( x n ,z)=D(z,z)= lim m , n D( x n , x m )=0.

As T is onto, so there exists wX such that z=Tw. Let us prove that w=z. Suppose to the contrary that zw. Then by (3.2) we have

D( x n ,z)=D(T x n + 1 ,Tw)B ( D ( x n + 1 , w ) ) D( x n + 1 ,w).

By taking limit as n in the above inequality and applying Proposition 2.10(B), we have

0= lim n D( x n ,z) lim n B ( D ( x n + 1 , w ) ) lim n D( x n + 1 ,w) 1 K lim n B ( D ( x n + 1 , z ) ) D(z,w)

and hence

lim n B ( D ( x n + 1 , z ) ) =0,

which is a contradiction. Indeed, lim n B(D( x n + 1 ,z)) K 2 . Since B(t)> K 2 for all t[0,), therefore z=w. That is, z=Tw=Tz. □

Example 3.5 Let X=[0,) and D:X×X R + be defined by

D(x,y)= ( max { x , y } ) 2 .

Clearly, (X,D,2) is a complete b-metric-like space. Let T:XX be defined by

Tx=4x 1 + x 2 .

Also define B:(0,)(4,) by B(t)=4(1+t). At first we show that T is an onto mapping. For a given aX, we choose x 0 = 1 2 4 + a 2 2 . Then

T x 0 = ( 4 + a 2 2 ) ( 4 + a 2 + 2 ) =a.

So, T is an onto mapping. Without loss of generality, we assume that xy. Now, since

Ty2y 1 + y 2 ,

so

( T y ) 2 4 ( 1 + y 2 ) y 2 ;

equivalently,

( max { T x , T y } ) 2 4 ( 1 + ( max { x , y } ) 2 ) ( max { x , y } ) 2

and hence

D(Tx,Ty)4 ( 1 + D ( x , y ) ) D(x,y).

That is,

D(Tx,Ty)B ( D ( x , y ) ) D(x,y).

The conditions of Theorem 3.4 hold and T has a fixed point (here, x=0 is a fixed point of T).

Note that b-metric-like spaces are a proper extension of partial metric, metric-like and b-metric spaces. Hence, we can deduce the following corollaries in the settings of partial metric, metric-like and b-metric spaces, respectively.

Corollary 3.6Let(X,p)be a complete partial metric space. Assume that the mappingT:XXis onto and satisfies

p(Tx,Ty)B ( p ( x , y ) ) p(x,y)

for allx,yX, whereB Ψ B 1 . ThenThas a fixed point.

Corollary 3.7Let(X,σ)be a complete metric-like space. Assume that the mappingT:XXis onto and satisfies

σ(Tx,Ty)B ( σ ( x , y ) ) σ(x,y)

for allx,yX, whereB Ψ B 1 . ThenThas a fixed point.

Corollary 3.8Let(X,d,K)be a completeb-metric space. Assume that the mappingT:XXis onto and satisfies

d(Tx,Ty)B ( d ( x , y ) ) d(x,y)
(3.4)

for allx,yX, whereB Ψ B K . ThenThas a fixed point.

4 Fixed point results in partially ordered b-metric-like spaces

In this section we prove certain new fixed point theorems in partially ordered b-metric-like spaces which generalize and extend corresponding results of Amini-Harandi [9, 10] and many others (see [22]).

Let Ψ L L denote the class of those functions L:(0,)(0, 1 L 2 ) which satisfy the condition L( t n ) ( 1 L 2 ) + t n 0, where L>0.

Theorem 4.1Let(X,D,K,)be a partially ordered completeb-metric-like space, and letT:XXbe a non-decreasing mapping such that

D(Tx,Ty)L ( M ( x , y ) ) M(x,y)+J ( N ( x , y ) ) N(x,y)
(4.1)

for allx,yXwithxy, whereL Ψ L K , J:[0,)[0,)is a bounded function and

M(x,y)=max { D ( x , y ) , D ( x , T x ) , D ( y , T y ) , D ( x , T y ) + D ( y , T x ) 6 K }

and

N(x,y)=min { D s ( x , T x ) , D s ( y , T y ) , D s ( x , T y ) , D s ( y , T x ) } .

Also, suppose that the following assertions hold:

  1. (i)

    there exists x 0 Xsuch that x 0 T x 0 ;

  2. (ii)

    for an increasing sequence{ x n }Xconverging toxX, we have x n xfor allnN;

thenThas a fixed point.

Proof Let x 0 T x 0 . If x 0 =T x 0 , then the result is proved. Hence we suppose that x 0 f x 0 . Define a sequence { x n } by x n = T n x 0 =T x n 1 for all nN. Since T is non-decreasing and x 0 T x 0 , then

x 0 x 1 x 2 ,
(4.2)

and hence { x n } is a non-decreasing sequence. If x n = x n + 1 =T x n for some nN, then the result is proved as x n is a fixed point of T. In what follows we will suppose that x n x n + 1 for all nN. From (4.1) and (4.2) we have

D ( x n , x n + 1 ) = D ( T x n 1 , T x n ) L ( M ( x n 1 , x n ) ) M ( x n 1 , x n ) + J ( N ( x n 1 , x n ) ) N ( x n 1 , x n ) ,

where

N ( x n 1 , x n ) = min { D s ( x n 1 , T x n 1 ) , D s ( x n , T x n ) , D s ( x n 1 , T x n ) , D s ( x n , T x n 1 ) } = min { D s ( x n 1 , x n ) , D s ( x n , x n + 1 ) , D s ( x n 1 , x n + 1 ) , D s ( x n , x n ) } = 0 .

Then

D( x n , x n + 1 )L ( M ( x n 1 , x n ) ) M( x n 1 , x n ).
(4.3)

On the other hand, from (3) we have

D( x n 1 , x n + 1 )K ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) )

and

D( x n , x n )2KD( x n , x n + 1 )2K ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) ) .

Then

D ( x n 1 , x n + 1 ) + D ( x n , x n ) 6 K 1 2 ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) )

and hence

M ( x n 1 , x n ) = max { D ( x n 1 , x n ) , D ( x n 1 , T x n 1 ) , D ( x n , T x n ) , D ( x n 1 , T x n ) + D ( x n , T x n 1 ) 6 K } = max { D ( x n 1 , x n ) , D ( x n , x n + 1 ) , D ( x n 1 , x n + 1 ) + D ( x n , x n ) 6 K } max { D ( x n 1 , x n ) , D ( x n , x n + 1 ) , 1 2 ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) ) } = max { D ( x n 1 , x n ) , D ( x n , x n + 1 ) } M ( x n 1 , x n ) .

That is,

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

Now by (4.3) we get

D( x n , x n + 1 )L ( max { D ( x n 1 , x n ) , 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 )L ( D ( x n , x n + 1 ) ) D( x n , x n + 1 )< 1 K 2 D( x n , x n + 1 )D( x n , x n + 1 ),

which is a contradiction. Hence,

D( x n , x n + 1 )L ( D ( x n 1 , x n ) ) D( x n 1 , x n )D( x n 1 , x n ),
(4.4)

and so the sequence {D( x n , x n + 1 )} is a decreasing sequence in R + . Then there exists s0 such that lim n D( x n , x n + 1 )=s. By (4.4) we can write

D ( x n , x n + 1 ) K 2 D ( x n 1 , x n ) D ( x n , x n + 1 ) D ( x n 1 , x n ) L ( D ( x n 1 , x n ) ) 1 K 2 .

Taking limit as n in the above inequality, we get

lim n L ( D ( x n 1 , x n ) ) = 1 K 2 ,

and so s= lim n D( x n 1 , x n )=0. Now we want to show that lim sup m , n D( x n , x m )=0. Suppose to the contrary that

lim sup m , n D( x n , x m )>0.

At first,

lim sup m , n N ( x n , x m ) = lim sup m , n min { D s ( x n , x n + 1 ) , D s ( x m , x m + 1 ) , D s ( x n , x m + 1 ) , D s ( x m , x n + 1 ) } = 0
(4.5)

and

lim sup m , n M ( x n , x m ) = lim sup m , n max { D ( x n , x m ) , D ( x n , x n + 1 ) , D ( x m , x m + 1 ) , D ( x n , x m + 1 ) + D ( x m , x n + 1 ) 6 K } lim sup m , n max { D ( x n , x m ) , D ( x n , x n + 1 ) , D ( x m , x m + 1 ) , K ( D ( x n , x m ) + D ( x m , x m + 1 ) ) + K ( D ( x m , x n ) + D ( x n , x n + 1 ) ) 6 K } = lim sup m , n D ( x n , x m ) lim sup m , n M ( x n , x m ) .

That is,

lim sup m , n M( x n , x m )= lim sup m , n D( x n , x m ).
(4.6)

Now, by (4.1) we have

lim sup m , n D ( x n + 1 , x m + 1 ) = lim sup m , n D ( T x n , T x m ) lim sup m , n L ( M ( x n , x m ) ) lim sup m , n M ( x n , x m ) + lim sup m , n J ( N ( x n , x m ) ) lim sup m , n N ( x n , x m ) ,

and so from (4.5) and (4.6) we get

lim sup m , n D( x n + 1 , x m + 1 ) lim sup m , n L ( M ( x n , x m ) ) lim sup m , n D( x n , x m ).
(4.7)

By (3) we have

D( x n , x m )KD( x n , x n + 1 )+ K 2 D( x n + 1 , x m + 1 )+ K 2 D( x m + 1 , x m ).

Taking limitsup as n in the above inequality, we have

1 K 2 lim sup m , n D( x n , x m ) lim sup m , n D( x n + 1 , x m + 1 ).

Then by (4.7) we deduce

1 K 2 lim sup m , n D( x n , x m ) lim sup m , n L ( M ( x n , x m ) ) lim sup m , n D( x n , x m ).

Now, since lim sup m , n D( x n , x m )>0, then

1 K 2 lim sup m , n L ( M ( x n , x m ) ) .

On the other hand, since lim sup m , n L(M( x n , x m )) 1 K 2 , hence

lim sup m , n L ( M ( x n , x m ) ) = 1 K 2 .

This implies that

lim sup m , n D( x n , x m )= lim sup m , n M( x n , x m )=0,

which is contradiction. Thus, lim sup m , n D( x n , x m )=0. Now, since lim m , n D( x n , x m )=0 exists (and is finite), so { x n } is a Cauchy sequence. As (X,D,K) is a complete b-metric-like space, the sequence { x n } in X converges to zX so that

lim m , n D( x n ,z)=D(z,z)= lim m , n D( x n , x m )=0.

From (ii) and (4.1), with x= x n and y=z, we obtain

D( x n + 1 ,Tz)=D(T x n ,Tz)L ( M ( x n , z ) ) M( x n ,z)+J ( N ( x n , z ) ) N( x n ,z).
(4.8)

On the other hand,

lim n N( x n ,z)= lim n min { D s ( x n , x n + 1 ) , D s ( z , T z ) , D s ( x n , T z ) , D s ( z , x n + 1 ) } =0

and

lim n M ( x n , z ) = lim n max { D ( x n , z ) , D ( x n , x n + 1 ) , D ( z , T z ) , D ( x n , T z ) + D ( z , x n + 1 ) 6 K } D ( z , T z ) (by applying Proposition 2.10(B)) .

Then lim n M( x n ,z)=D(z,Tz). Again, by using Proposition 2.10(B) and (4.8), we have

1 K 2 D(z,Tz) 1 K D(z,Tz) lim n D( x n + 1 ,Tz) lim n L ( M ( x n , z ) ) D(z,Tz).

Now, if D(z,Tz)>0, then lim n L(M( x n ,z))= 1 K 2 . This implies

D(z,Tz)= lim n M( x n ,z)=0,

which is a contradiction. Hence, D(z,Tz)=0. That is, z=Tz. □

Example 4.2 Let X=[0,) and D:X×X R + be defined by

D(x,y)= ( max { x , y } ) 2 .

Clearly, (X,D,2) is a complete b-metric-like space. Let T:XX be defined by

Tx={ x 5 1 + x 2 if  x [ 0 , 1 ) , x 4 2 1 + x if  x ( 1 , ) .

Also, define L:(0,)(0, 1 4 ) by L(t)= 1 4 ( 1 + t ) . Let xyxy. At first we assume that xy. Let y[0,1), then Ty y 2 1 + y 2 . Also, let y[1,), then Ty y 2 1 + y 2 . That is, for all yX, we have

Ty y 2 1 + y 2 ,

which implies

( T y ) 2 y 2 4 ( 1 + y 2 ) ;

equivalently,

( max { T x , T y } ) 2 ( max { x , y } ) 2 4 ( 1 + ( max { x , y } ) 2 ) ,

and so

D(Tx,Ty) D ( x , y ) 4 ( 1 + D ( x , y ) ) M ( x , y ) 4 ( 1 + M ( x , y ) ) =L ( M ( x , y ) ) M(x,y).

Then the conditions of Theorem 4.1 hold and T has a fixed point.

Also we have the following corollaries.

Corollary 4.3Let(X,p,)be a partially ordered complete partial metric space, and letT:XXbe a non-decreasing mapping such that

p(Tx,Ty)L ( M ( x , y ) ) M(x,y)+J ( N ( x , y ) ) N(x,y)

for allx,yXwithxy, whereL Ψ L 1 , J:[0,)[0,)is a bounded function and

M(x,y)=max { p ( x , y ) , p ( x , T x ) , p ( y , T y ) , p ( x , T y ) + p ( y , T x ) 6 }

and

N(x,y)=min { p s ( x , T x ) , p s ( y , T y ) , p s ( x , T y ) , p s ( y , T x ) } .

Also suppose that the following assertions hold:

  1. (i)

    there exists x 0 Xsuch that x 0 T x 0 ;

  2. (ii)

    for an increasing sequence{ x n }Xconverging toxX, we have x n xfor allnN;

thenThas a fixed point.

Corollary 4.4Let(X,d,K,)be a partially ordered completeb-metric space, and letT:XXbe a non-decreasing mapping such that

d(Tx,Ty)L ( M ( x , y ) ) M(x,y)+J ( N ( x , y ) ) N(x,y)
(4.9)

for allx,yXwithxy, whereL Ψ L K , J:[0,)[0,)is a bounded function and

M(x,y)=max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 4 K }

and

N(x,y)=2min { d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) } .

Also suppose that the following assertions hold:

  1. (i)

    there exists x 0 Xsuch that x 0 T x 0 ;

  2. (ii)

    for an increasing sequence{ x n }Xconverging toxX, we have x n xfor allnN;

thenThas a fixed point.

Remark 4.5 By utilizing the technique of Amini-Harandi [10] and Samet et al. [23], we can obtain corresponding coupled fixed point results from our Theorem 4.1 and Corollaries 4.3 and 4.4 on the basis of the following simple lemma. For more detailed literature on coupled fixed theory, we refer to [2428].

Lemma 4.6 [23] (A coupled fixed point is a fixed point)

LetF:X×XXbe a given mapping. Define the mappingT:X×XX×Xby

T(x,y)= ( F ( x , y ) , F ( y , x ) )

for all(x,y)X×X. Then(x,y)is a coupled fixed point ofFif and only if(x,y)is a fixed point ofT.

5 Fixed point results for cyclic Edelstein-Suzuki contraction

In 1962, Edelstein [29] proved an important version of the Banach contraction principle. In 2009, Suzuki [30] improved the results of Banach and Edelstein (see also [31, 32]). In recent years, cyclic contraction and cyclic contractive type mapping have appeared in several works (see [3338]). In this section we first prove the following result, which generalizes corresponding results of Edelstein [29], Suzuki [30] and Kirk et al. [33] to the setting of b-metric-like spaces.

Theorem 5.1Let(X,D,K)be a completeb-metric-like space, and let { A j } j = 1 m be a family of nonempty closed subsets ofXwithY= j = 1 m A j . LetT:YYbe a map satisfying

T( A j ) A j + 1 ,j=1,2,,m, where  A m + 1 = A 1 .
(5.1)

Assume that

1 2 K D ( x , T x ) D ( x , y ) D ( T x , T y ) α ( K + 1 ) K D ( x , y ) + β [ D ( x , T x ) + D ( y , T y ) ] D ( T x , T y ) + γ [ D ( x , T y ) + D ( y , T x ) 3 K ] + δ [ D ( x , x ) + D ( y , y ) 4 K ]
(5.2)

for allx A i andy A i + 1 , whereα,β,γ,δ0andα+β+γ+δ< 1 K + 1 . ThenThas a fixed point in j = 1 m A j .

Proof Let x 0 A 1 and define a sequence { x n } in the following way:

x n =T x n 1 ,n=1,2,3,.
(5.3)

We have x 0 A 1 , x 1 =T x 0 A 2 , x 2 =T x 1 A 3 , … . If x n 0 + 1 = x n 0 for some n 0 N, then, clearly, the fixed point of the map T is x n 0 . Hence, we assume that x n x n + 1 for all nN. Clearly, 1 2 K D( x n 1 ,T x n 1 )D( x n 1 , x n ). Now, from (5.2) we have

D ( T x n 1 , T x n ) α ( K + 1 ) K D ( x n 1 , x n ) + β ( D ( x n 1 , T x n 1 ) + D ( x n , T x n ) ) + γ ( D ( x n 1 , T x n ) + D ( x n , T x n 1 ) 3 K ) + δ ( D ( x n 1 , x n 1 ) + D ( x n , x n ) 4 K ) ,

which implies

D ( x n , x n + 1 ) α ( K + 1 ) K D ( x n 1 , x n ) + β ( D ( x n 1 , x n ) + D ( x n , x n + 1 ) ) + γ ( D ( x n 1 , x n + 1 ) + D ( x n , x n ) 3 K ) + δ ( D ( x n 1 , x n 1 ) + D ( x n , x n ) 4 K ) .
(5.4)

From (3) we have

D( x n 1 , x n + 1 )K ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) )

and

D( x n , x n )2KD( x n , x n + 1 )2K ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) ) ;

and so

D ( x n 1 , x n + 1 ) + D ( x n , x n ) 3 K D( x n , x n 1 )+D( x n , x n + 1 ).
(5.5)

Also,

D( x n 1 , x n 1 )2KD( x n , x n 1 )2K ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) ) .

Then

D ( x n 1 , x n 1 ) + D ( x n , x n ) 4 K D( x n , x n 1 )+D( x n , x n + 1 ).
(5.6)

Hence, by (5.4), (5.5) and (5.6) we get

D( x n , x n + 1 ) α ( K + 1 ) K D( x n 1 , x n )+(β+γ+δ) ( D ( x n , x n 1 ) + D ( x n , x n + 1 ) ) ,

and then

D( x n , x n + 1 )hD( x n , x n 1 ),

where

h= [ α ( K + 1 ) K + β + γ + δ ] 1 ( β + γ + δ ) .

Now since (K+1)(α+β+γ+δ)<1, then

α K + 1 K +β+γ+δ+ 1 K (β+γ+δ)< 1 K ,

which implies

α K + 1 K +β+γ+δ+< 1 K [ 1 ( β + γ + δ ) ] .

Then by Lemma 2.14 we have lim m , n D( x n , x m )=0. Now, since lim m , n D( x n , x m )=0 exists (and is finite), so { x n } is a Cauchy sequence. Since (X,D,K) is a complete b-metric-like space, the sequence { x n } in X converges to zX so that

lim m , n D( x n ,z)=D(z,z)= lim m , n D( x n , x m )=0.

It is easy to see that z j = 1 m A j . Since x 0 A 1 , so the subsequence { x m ( n 1 ) } n = 1 A 1 , the subsequence { x m ( n 1 ) + 1 } n = 1 A 2 and, continuing in this way, the subsequence { x m n 1 } n = 1 A m . All the m subsequences are convergent in the closed sets A j , and hence they all converge to the same limit z j = 1 m A j . Suppose that there exists n 0 N such that the following inequalities hold:

1 2 K D( x n 0 , x n 0 + 1 )>D( x n 0 ,z)and 1 2 K D( x n 0 + 1 , x n 0 + 2 )>D( x n 0 + 1 ,z).

Then

D ( x n 0 , x n 0 + 1 ) K ( D ( x n 0 , z ) + D ( T x n 0 , z ) ) < 1 2 D ( x n 0 , x n 0 + 1 ) + 1 2 D ( x n 0 + 1 , x n 0 + 2 ) < 1 2 D ( x n 0 , x n 0 + 1 ) + 1 2 D ( x n 0 , x n 0 + 1 ) = D ( x n 0 , x n 0 + 1 ) ,

which is a contradiction. Hence, for every nN, we have

1 2 K D( x n , x n + 1 )D( x n ,z)or 1 2 K D( x n + 1 , x n + 2 )D( x n + 1 ,z),

and so by (5.2) we have

D ( x n + 1 , T z ) α ( K + 1 ) K D ( x n , z ) + β ( D ( x n , x n + 1 ) + D ( z , T z ) ) + γ ( D ( x n , T z ) + D ( z , x n + 1 ) 3 K ) + δ ( D ( x n , x n ) + D ( z , z ) 4 K )
(5.7)

or

D ( x n + 2 , T z ) α ( K + 1 ) K D ( x n + 1 , z ) + β ( D ( x n + 1 , x n + 2 ) + D ( z , T z ) ) + γ ( D ( x n + 1 , T z ) + D ( z , x n + 2 ) 3 K ) + δ ( D ( x n + 1 , x n + 1 ) + D ( z , z ) 4 K ) .
(5.8)

Assume that (5.7) holds. Then, by taking limit as n in (5.7), we get

lim n D( x n + 1 ,Tz)βD(z,Tz)+ γ 3 K lim n D( x n ,Tz),

and hence by Proposition 2.10(B) we have

1 K D(z,Tz)βD(z,Tz)+ γ 3 D(z,Tz).

Therefore,

( 1 K β γ 3 ) D(z,Tz)0.

On the other hand, α,β,γ,δ0 and α+β+γ+δ< 1 K + 1 < 1 K . Then β+ γ 3 β+γ< 1 K . That is, 1 K β γ 3 >0. Hence, D(z,Tz)=0, i.e., z=Tz. If (5.8) holds, then by a similar method, we can deduce that z=Tz. □

If in the above theorem we take A i =X for all m, then we deduce the following corollary.

Corollary 5.2Let(X,D,K)be a completeb-metric-like space, and letTbe a self-mapping onX. Assume that

1 2 K D ( x , T x ) D ( x , y ) D ( T x , T y ) α ( K + 1 ) K D ( x , y ) + β ( D ( x , T x ) + D ( y , T y ) ) D ( T x , T y ) + γ ( D ( x , T y ) + D ( y , T x ) 3 K ) + δ ( D ( x , x ) + D ( y , y ) 4 K )

for allx,yX, whereα,β,γ,δ0andα+β+γ+δ< 1 K + 1 . ThenThas a fixed point.

If in Theorem 5.1 we take α ( K + 1 ) K =β= γ 3 K = δ 4 K =R, then we deduce the following corollary.

Corollary 5.3Let(X,D,K)be a completeb-metric-like space, and let { A j } j = 1 m be a family of nonempty closed subsets ofXwithY= j = 1 m A j . LetT:YYbe a map satisfying

T( A j ) A j + 1 ,j=1,2,,m, where  A m + 1 = A 1 .

Assume that

1 2 K D ( x , T x ) D ( x , y ) D ( T x , T y ) R [ D ( x , y ) + D ( x , T x ) + D ( y , T y ) + D ( x , T y ) D ( T x , T y ) + D ( y , T x ) + D ( x , x ) + D ( y , y ) ]

for allx A i andy A i + 1 , where0R< 1 ( K + 1 ) ( 7 K + 1 ) + K . ThenThas a fixed point in j = 1 m A j .

If in Corollary 5.2 we take α ( K + 1 ) K =β= γ 3 K = δ 4 K =R, then we deduce the following corollary.

Corollary 5.4Let(X,D,K)be a completeb-metric-like space, and letTbe a self-mapping onX. Assume that

1 2 K D ( x , T x ) D ( x , y ) D ( T x , T y ) R [ D ( x , y ) + D ( x , T x ) + D ( y , T y ) + D ( x , T y ) D ( T x , T y ) + D ( y , T x ) + D ( x , x ) + D ( y , y ) ]

for allx,yX, where0R< 1 ( K + 1 ) ( 7 K + 1 ) + K . ThenThas a fixed point.

Corollary 5.5Let(X,σ)be a complete metric-like space, mN, let A 1 , A 2 ,, A m be nonempty closed subsets ofXandY= i = 1 m A i . Suppose thatT:YYis an operator such that

  1. (i)

    Y= i = 1 m A i is a cyclic representation ofXwith respect toT;

  2. (ii)

    Assume that there exists 0R< 1 17 such that

    1 2 0 σ ( x , T x ) ρ(t)dt 0 σ ( x , y ) ρ(t)dt 0 σ ( T x , T y ) ρ(t)dtR 0 M ( x , y ) ρ(t)dt,

where

M(x,y)=σ(x,y)+σ(x,Tx)+σ(y,Ty)+σ(x,Ty)+σ(y,Tx)+σ(x,x)+σ(y,y)

for anyx A i , y A i + 1 , i=1,2,,m, where A m + 1 = A 1 , andρ:[0,)[0,)is a Lebesgue-integrable mapping satisfying 0 ε ρ(t)dt>0forε>0. ThenThas a fixed point.

Corollary 5.6Let(X,σ)be a complete metric-like space, and letT:XXbe a mapping such that for anyx,yXthere exists0R< 1 17 such that

1 2 0 σ ( x , T x ) ρ(t)dt 0 σ ( x , y ) ρ(t)dt 0 σ ( T x , T y ) ρ(t)dtR 0 M ( x , y ) ρ(t)dt,

where

M(x,y)=σ(x,y)+σ(x,Tx)+σ(y,Ty)+σ(x,Ty)+σ(y,Tx)+σ(x,x)+σ(y,y)

andρ:[0,)[0,)is a Lebesgue-integrable mapping satisfying 0 ε ρ(t)dtforε>0. ThenThas fixed point.

6 Application to the existence of solutions of integral equations

Motivated by the work in [3941], we study the existence of solutions for nonlinear integral equations using the results proved in the previous section.

Consider the integral equation

u(t)= 0 T G(t,s)f ( s , u ( s ) ) dsfor all t[0,T],
(6.1)

where T>0, f:[0,T]×RR and G:[0,T]×[0,T][0,) are continuous functions.

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

D (u,v)= sup t [ 0 , T ] ( | u ( t ) | + | v ( t ) | ) 2 for all u,vX.

Clearly, (X, D ,2) is a complete b-metric-like space.

Let (α,β) X 2 , ( α 0 , β 0 ) R 2 be such that

α 0 α(t)β(t) β 0 for all t[0,T].
(6.2)

Assume that for all t[0,T], we have

α(t) 0 T G(t,s)f ( s , β ( s ) ) ds
(6.3)

and

β(t) 0 T G(t,s)f ( s , α ( s ) ) ds.
(6.4)

Let, for all s[0,T], f(s,) be a decreasing function, that is,

x,yR,xyf(s,x)f(s,y).
(6.5)

Assume that

sup t [ 0 , T ] 0 T G(t,s)ds1.
(6.6)

Also, suppose that for all s[0,T], for all x,yR with (x β 0 and y α 0 ) or (x α 0 and y β 0 ),

| f ( s , x ) | + | f ( s , y ) | ( 3 α 2 ( x + y ) 2 + β ( ( x + T x ) 2 + ( y + T y ) 2 ) + γ ( ( x + T y ) 2 + ( y + T x ) 2 6 ) + δ ( ( 2 x ) 2 + ( 2 y ) 2 8 ) ) 1 2 ,
(6.7)

where α,β,γ,δ0 and α+β+γ+δ< 1 3 .

Theorem 6.1Under assumptions (6.2)-(6.7), integral equation (6.1) has a solution in{uC([0,T]):αu(t)β for all t[0,T]}.

Proof Define the closed subsets of X, A 1 and A 2 by

A 1 ={uX:uβ}

and

A 2 ={uX:uα}.

Also define the mapping T:XX by

Tu(t)= 0 T G(t,s)f ( s , u ( s ) ) dsfor all t[0,T].

Let us prove that

T( A 1 ) A 2 andT( A 2 ) A 1 .
(6.8)

Suppose that u A 1 , that is,

u(s)β(s)for all s[0,T].

Applying condition (6.5), since G(t,s)0 for all t,s[0,T], we obtain that

G(t,s)f ( s , u ( s ) ) G(t,s)f ( s , β ( s ) ) for all t,s[0,T].

The above inequality with condition (6.3) imply that

0 T G(t,s)f ( s , u ( s ) ) ds 0 T G(t,s)f ( s , β ( s ) ) dsα(t)

for all t[0,T]. Then we have Tu A 2 .

Similarly, let u A 2 , that is,

u(s)α(s)for all s[0,T].

Using condition (6.5), since G(t,s)0 for all t,s[0,T], we obtain that

G(t,s)f ( s , u ( s ) ) G(t,s)f ( s , α ( s ) ) for all t,s[0,T].

The above inequality with condition (6.4) imply that

0 T G(t,s)f ( s , u ( s ) ) ds 0 T G(t,s)f ( s , α ( s ) ) dsβ(t)

for all t[0,T]. Then we have Tu A 1 . Also, we deduce that (6.8) holds.

Now, let (u,v) A 1 × A 2 , that is, for all t[0,T],

u(t)β(t),v(t)α(t).

This implies from condition (6.2) that for all t[0,T],

u(t) β 0 ,v(t) α 0 .

Now, by conditions (6.6) and (6.7), we have, for all t[0,T],

( | T x | + | T y | ) 2 = ( | 0 T G ( t , s ) f ( s , x ( s ) ) d s | + | 0 T G ( t , s ) f ( s , y ( s ) ) d s | ) 2 ( 0 T G ( t , s ) | f ( s , x ( s ) ) | d s + 0 T G ( t , s ) | f ( s , y ( s ) ) | d s ) 2 = ( 0 T G ( t , s ) ( | f ( s , x ( s ) ) | + | f ( s , y ( s ) ) | ) d s ) 2 ( 0 T G ( t , s ) ( 3 α 2 ( x + y ) 2 + β ( ( x + T x ) 2 + ( y + T y ) 2 ) + γ ( ( x + T y ) 2 + ( y + T x ) 2 6 ) + δ ( ( 2 x ) 2 + ( 2 y ) 2 8 ) ) 1 2 d s ) 2 ( 0 T G ( t , s ) ( 3 α 2 ( | x | + | y | ) 2 + β ( ( | x | + | T x | ) 2 + ( | y | + | T y | ) 2 ) + γ ( ( | x | + | T y | ) 2 + ( | y | + | T x | ) 2 6 ) + δ ( ( 2 | x | ) 2 + ( 2 | y | ) 2 8 ) ) 1 2 d s ) 2 ( 0 T G ( t , s ) ( 3 α 2 D ( x , y ) + β ( D ( x , T x ) + D ( y , T y ) ) + γ ( D ( x , T y ) + D ( y , T x ) 6 ) + δ ( D ( x , x ) + D ( y , y ) 8 ) ) 1 2 d s ) 2 = 3 α 2 D ( x , y ) + β ( D ( x , T x ) + D ( y , T y ) ) + γ ( D ( x , T y ) + D ( y , T x ) 6 ) + δ ( D ( x , x ) + D ( y , y ) 8 ) × ( 0 T G ( t , s ) d s ) 2 3 α 2 D ( x , y ) + β ( D ( x , T x ) + D ( y , T y ) ) + γ ( D ( x , T y ) + D ( y , T x ) 6 ) + δ ( D ( x , x ) + D ( y , y ) 8 ) ,

which implies

D ( T x , T y ) 3 α 2 D ( x , y ) + β ( D ( x , T x ) + D ( y , T y ) ) + γ ( D ( x , T y ) + D ( y , T x ) 6 ) + δ ( D ( x , x ) + D ( y , y ) 8 ) .

By a similar method, we can show that the above inequality holds if (u,v) A 2 × A 1 .

Now, all the conditions of Theorem 5.1 hold and T has a fixed point z in

A 1 A 2 = { u C ( [ 0 , T ] ) : α u ( t ) β  for all  t [ 0 , T ] } .

That is, z A 1 A 2 is the solution to (6.1). □

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Acknowledgements

This research was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant No. (390-130-1433). The first and second authors acknowledge with thanks DSR, KAU for financial support. The authors would like to express their thanks to the referees for their helpful comments and suggestions.

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Correspondence to Nawab Hussain.

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Alghamdi, M.A., Hussain, N. & Salimi, P. Fixed point and coupled fixed point theorems on b-metric-like spaces. J Inequal Appl 2013, 402 (2013). https://doi.org/10.1186/1029-242X-2013-402

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Keywords

  • partial metric space
  • metric-like space
  • b-metric space
  • b-metric-like space
  • fixed point
  • integral equation