1 Introduction and preliminaries

There exist many generalizations of the concept of metric spaces in the literature. In particular, Matthews [1] introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks, showing 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 were studied by many other authors [211]. In this paper, we first introduce a new generalization of a partial metric space which is called a metric-like space. Then, we give some fixed point results in such spaces. Our fixed point theorems, even in the case of partial metric spaces, generalize and improve some well-known results in the literature.

In the rest of this section, we recall some definitions and facts which will be used throughout the paper.

Definition 1.1 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. A sequence { x n } in a partial metric space (X,p) converges to a point xX if lim n p( x n ,x)=p(x,x). A sequence { x n } of elements of X is called p-Cauchy if the limit lim m , n p( x m , x n ) exists and is finite. The partial metric space (X,p) is called complete if for each p-Cauchy sequence { x n } n = 0 , there is some xX such that

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

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 + . For some other examples of partial metric spaces see [111] and references therein.

2 Main results

We first introduce the concept of a metric-like space.

Definition 2.1 A mapping σ:X×X R + , where X is a nonempty set, is said to be 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. Then a metric-like on X satisfies all of the conditions of a metric except that σ(x,x) may be positive for xX. Each metric-like σ on X generates a topology τ σ on X whose base is the family of open σ-balls

B σ (x,ε)= { y X : | σ ( x , y ) σ ( x , x ) | < ε } ,for all xX and ε>0.

Then a sequence { x n } in the metric-like space (X,σ) converges to a point xX if and only if lim n σ( x n ,x)=σ(x,x).

Let (X,σ) and (Y,τ) be metric-like spaces, and let f:XX be a continuous mapping. Then

lim n x n =x lim n f( x n )=f(x).

A sequence { x n } n = 0 of elements of X is called σ-Cauchy if the limit lim m , n σ( x m , x n ) exists and is finite. The metric-like space (X,σ) is called complete if for each σ-Cauchy sequence { x n } n = 0 , there is some xX such that

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

Every partial metric space is a metric-like space. Below we give another example of a metric-like space.

Example 2.2 Let X={0,1}, and let

σ(x,y)={ 2 if  x = y = 0 , 1 otherwise .

Then (X,σ) is a metric-like space, but since σ(0,0)σ(0,1), then (X,σ) is not a partial metric space.

Remark 2.3 Let X={0,1}, let σ(x,y)=1 for each x,yX, and let x n =1 for each nN. Then it is easy to see that x n 0 and x n 1, and so in metric-like spaces the limit of a convergent sequence is not necessarily unique.

Some slight modifications of the proof of Theorem 2.1 in [12] yield the following result which is a generalization of the well-known fixed point theorem of Ćirić [13].

Theorem 2.4 Let (X,σ) be a complete metric-like space, and let T:XX be a map such that

σ(Tx,Ty)ψ ( M ( x , y ) ) ,

for all x,yX, where

M(x,y)=max { σ ( x , y ) , σ ( x , T x ) , σ ( y , T y ) , σ ( x , T y ) , σ ( y , T x ) , σ ( x , x ) , σ ( y , y ) } ,

where ψ:[0,)[0,) is a nondecreasing function satisfying

ψ(t)<tfor all t>0, lim s t + ψ(s)<tfor all t>0and lim t ( t ψ ( t ) ) =.

Then T has a fixed point.

Proof Let x 0 X be arbitrary, and let x n + 1 =T x n for n{0,1,2,}. Denote

O( x 0 ,n)={T x 0 ,T x 1 ,,T x n }andO( x 0 )={T x 0 ,T x 1 ,,T x n ,}.

First we show that O( x 0 ) is a bounded set. We shall show that for each nN,

δ n ( x 0 )=diam ( O ( x 0 , n ) ) =σ(T x 0 ,T x k ),
(1)

where k=k(n){0,1,2,,n}. Suppose, to the contrary, that there are positive integers 1i(n)=ij=j(n) such that

δ n ( x 0 )=σ(T x i ,T x j )>0.

From our assumption, we have

Thus, from the above and the contractive condition on T, we have

δ n ( x 0 ) = σ ( T x i , T x j ) ψ ( max { σ ( x i , x j ) , σ ( x i , T x i ) , σ ( x j , T x j ) , σ ( x i , T x j ) , σ ( x j , T x i ) , σ ( x i , x i ) , σ ( x j , x j ) } ) ψ ( δ n ( x 0 ) ) < δ n ( x 0 ) ,

a contradiction. Thus, (1) holds. Since by the triangle inequality,

σ(T x 0 ,T x k )σ(T x 0 ,T x 1 )+σ(T x 1 ,T x k ),

then from (1)

δ n ( x 0 )σ(T x 0 ,T x 1 )+σ(T x 1 ,T x k ).
(2)

From our assumption on T, we have

σ(T x 1 ,T x k )ψ ( M ( x 1 , x k ) ) ψ ( δ n ( x 0 ) ) .

Now by (2),

δ n ( x 0 )σ(T x 0 ,T x 1 )+ψ ( δ n ( x 0 ) ) .

Hence,

(Iψ) ( δ n ( x 0 ) ) σ(T x 0 ,T x 1 ),
(3)

where I is the identity map. Since the sequence { δ n ( x 0 )} is nondecreasing, there exists lim n δ n ( x 0 ). Suppose that lim n δ n ( x 0 )=. Then from (3), we get

lim t ( t ψ ( t ) ) = lim n ( δ n ( x 0 ) ψ ( δ n ( x 0 ) ) ) σ(T x 0 ,T x 1 )<,

a contradiction. Therefore, lim n δ n ( x 0 )=δ( x 0 )<, that is,

δ( x 0 )=diam ( { T x 0 , T x 1 , , T x n , } ) <.
(4)

Now we show that { x n } is a σ-Cauchy sequence. Set

δ( x n )=diam ( { T x n , T x n + 1 , } ) .

Since δ( x n )δ( x 0 ), then by (4) we conclude that {δ( x n )} is a nonincreasing finite nonnegative number and so it converges to some δ0. We shall prove that δ=0. Let nN be arbitrary, and let r, s be any positive integers such that r,sn+1. Then T x r 1 ,T x s 1 {T x n ,T x n + 1 ,} and hence we conclude that M( x r , x s )δ( x n ). Then

σ(T x r ,T x s )ψ ( M ( x r , x s ) ) ψ ( δ ( x n ) ) .

Hence, we get

δ( x n + 1 )=sup { σ ( T x r , T x s ) : r , s n + 1 } ψ ( δ ( x n ) ) .

Hence, as δδ( x n ) for all n0, δψ(δ( x n )). Suppose that δ>0. Then we get

δ lim n ψ ( δ ( x n ) ) = lim s δ + ψ(s)<δ,

a contradiction. Therefore, δ=0. Thus, we have proved that

lim n diam ( { T x n , T x n + 1 , } ) =0.

Hence, from the triangle inequality, we conclude that { x n + 1 =T x n } is a σ-Cauchy sequence. By the completeness of X, there is some uX such that lim n T x n =u, that is,

lim n σ(T x n ,u)=σ(u,u)= lim m , n σ(T x n ,T x m )=0.

We show that Tu=u. Suppose, by way of contradiction, that σ(Tu,u)>0. Then we have

σ ( T u , u ) σ ( u , T x n + 1 ) + σ ( T u , T x n + 1 ) σ ( u , T x n + 1 ) + ψ ( M ( u , x n + 1 ) ) ,
(5)

where

From the triangle inequality, we have

| σ ( T u , T x n + 1 ) σ ( T u , u ) | σ(u,T x n + 1 )0as n.

Thus, lim n σ(Tu,T x n + 1 )=σ(Tu,u). Since lim n σ(u,T x n )=0, lim n σ(T x n ,Tu)=σ(Tu,u), for large enough n, we have

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

If M(u, x n + 1 )=σ(u,Tu), then from (5), we get

σ(Tu,u)σ(u,T x n + 1 )+ψ ( σ ( T u , u ) ) .

Letting n tend to infinity, we get

0<σ(Tu,u)ψ ( σ ( T u , u ) ) <σ(Tu,u),

a contradiction. If M(u, x n + 1 )=σ(T x n ,Tu), then we have

σ(T x n ,Tu)=M(u, x n + 1 )σ(Tu,u),

and so σ(T x n ,Tu)σ ( T u , u ) + . Then from (5) and our assumptions on ψ, we get σ(Tu,u)<σ(Tu,u), a contradiction. Thus, σ(Tu,u)=0 and so Tu=u. □

Example 2.5 Let ψ 1 (t)=kt for each t[0,), where k[0,1), and let ψ 2 (t)=tln(1+t) for each t[0,). Then ψ 1 and ψ 2 satisfy the conditions of Theorem 2.4.

Now we illustrate our previous result by the following example.

Example 2.6 Let X={0,1,2}. Define σ:X×X R + as follows:

σ ( 0 , 0 ) = 0 , σ ( 1 , 1 ) = 3 , σ ( 2 , 2 ) = 1 , σ ( 0 , 1 ) = σ ( 1 , 0 ) = 7 , σ ( 0 , 2 ) = σ ( 2 , 0 ) = 3 , σ ( 1 , 2 ) = σ ( 2 , 1 ) = 4 .

Then (X,σ) is a complete metric-like space. Note that σ is not a partial metric on X because

σ(0,1)σ(0,2)+σ(2,1)σ(2,2).

Define the map T:XX by

T0=0,T1=2,andT2=0.

Then

σ(Tx,Ty) 3 4 σ(x,y) 3 4 M(x,y),

for each x,yX. Then all the required hypotheses of Theorem 2.4 are satisfied. Then T has a unique fixed point.

Theorem 2.7 Let (X,σ) be a complete metric-like space, and let T:XX be a map such that

σ(Tx,Ty)σ(x,y)φ ( σ ( x , y ) ) ,

for all x,yX, where φ:[0,)[0,) is a nondecreasing continuous function such that φ(t)=0 if and only if t=0. Then T has a unique fixed point.

Proof Let x 0 X and define x n + 1 =T x n for n0. Then by our assumption,

σ( x n + 1 , x n + 2 )=σ(T x n ,T x n + 1 )σ( x n , x n + 1 )φ ( σ ( x n , x n + 1 ) ) ,
(6)

for each nN. Then {σ( x n , x n + 1 )} is a nonnegative nonincreasing sequence and hence possesses a limit r 0 0. Since φ is nondecreasing, then from (6), we get

σ( x n + 1 , x n + 2 )σ( x n , x n + 1 )φ( r 0 )

for each nN. Then r 0 r 0 φ( r 0 ) and so r 0 =0. Therefore,

lim n σ( x n , x n + 1 )=0.

Now, we show that { x n } is a Cauchy sequence. Fix ε>0 and choose N such that

σ( x n , x n + 1 )<min { ε 2 , φ ( ε 2 ) } for nN.

We show that if σ(x, x N )ε, then σ(Tx, x N )ε. To show the claim, let us assume first that σ(x, x N ) ε 2 . Then

σ ( T x , x N ) σ ( T x , T x N ) + σ ( T x N , x N ) σ ( x , x N ) φ ( σ ( x , x N ) ) + σ ( x N + 1 , x N ) < ε 2 + ε 2 = ε .

Now we assume that ε 2 <σ(x, x N )ε. Then φ(σ(x, x N ))φ( ε 2 ). Therefore, from the above, we have

σ ( T x , x N ) σ ( x , x N ) φ ( σ ( x , x N ) ) + σ ( x N + 1 , x N ) σ ( x , x N ) φ ( ε 2 ) + φ ( ε 2 ) = σ ( x , x N ) ε .

Since σ( x N + 1 , x N )ε, then from the above, we deduce that σ( x n , x N )ε for each nN. Since ε>0 is arbitrary, we get lim m , n σ( x m , x n )=0 and so { x n } is a Cauchy sequence. Since X is complete, there is some uX such that lim n x n =u, that is,

lim n σ( x n ,u)=σ(u,u)= lim m , n σ(T x n ,T x m )=0.
(7)

Since

σ( x n + 1 ,Tu)=σ(T x n ,Tu)σ( x n ,u)φ ( σ ( x n , u ) )
(8)

and φ is continuous, then from (7) and (8), we have

lim n σ( x n ,Tu)=0.
(9)

Since

σ(u,Tu)σ( x n ,u)+σ( x n ,Tu)

then by (7) and (9), we infer that σ(u,Tu)=0 and so Tu=u. To prove the uniqueness, let v be another fixed point of T, that is, Tv=v. Then

σ(u,v)=σ(Tu,Tv)σ(u,v)φ ( σ ( u , v ) ) ,

which gives φ(σ(u,v))=0 and so u=v. □

Example 2.8 Let X=[0,) and σ(x,y)=max{x,y}. Then (X,σ) is a complete metric-like space. Take φ(t)= t 1 + t for t[0,). Let Tx= x 2 1 + x for each xX. Take x,yX, without loss of generality, we may assume that yx. Then

σ(Tx,Ty)=Tx=xφ(x)=σ(x,y)φ ( σ ( x , y ) ) .

Then T satisfies the hypothesis of Theorem 2.7 and so T has a fixed point (x=0 is the unique fixed point of T). Now since lim t φ(t)=1<, we cannot invoke Theorem 2.1 of [9] to show the existence of fixed point of T.

The following corollary improves Theorem 1 in [2].

Corollary 2.9 Let (X,p) be a complete partial metric space, and let T:XX be a map such that

p(Tx,Ty)ψ ( max { p ( x , y ) , p ( x , T x ) , p ( y , T y ) , p ( x , T y ) , p ( y , T x ) , p ( x , x ) , p ( y , y ) } ) ,

for all x,yX, where ψ:[0,)[0,) is a nondecreasing function satisfying

ψ(t)<tfor all t>0, lim s t + ψ(s)<tfor all t>0and lim t ( t ψ ( t ) ) =.

Then T has a unique fixed point.

Proof The existence of a fixed point follows immediately from Theorem 2.4. To prove the uniqueness, let us suppose that x and y are fixed points of T. Then from our assumption on T, we get

p(x,x)=p(Tx,Ty)ψ ( max { p ( x , y ) , p ( x , x ) , p ( y , y ) } ) =ψ ( p ( x , y ) ) .

Thus, p(x,y)=0 and x=y. □

The following corollary improves Corollary 1 and Theorem 2 in [2] and the main fixed point result of Matthews [1].

Corollary 2.10 Let (X,p) be a complete partial metric space, and let T:XX be a map such that

p(Tx,Ty)λmax { p ( x , y ) , p ( x , T x ) , p ( y , T y ) , p ( x , T y ) , p ( y , T x ) , p ( x , x ) , p ( y , y ) } ,

for all x,yX, where λ[0,1). Then T has a unique fixed point.

Proof Let ψ(t)=λt for each t[0,) and apply Corollary 2.9. □

Now, we present the following version of Rakotch’s fixed point theorem [14] in metric-like spaces.

Theorem 2.11 Let (X,σ) be a complete metric-like space, and let T:XX be a mapping satisfying

σ(Tx,Ty)α ( σ ( x , y ) ) σ(x,y),

for each x,yX with xy, where α:[0,)[0,1) is nonincreasing. Then T has a unique fixed point.

Proof Fix xX and let x n = T n x for each nN. Following the lines of the proof of the Theorem 3.6 in [15], we get that

lim m , n σ ( T n x , T m x ) =0,

and so { x n } is a σ-Cauchy sequence. Since (X,σ) is complete, then there exists x 0 X such that

lim n σ ( T n x , x 0 ) =σ( x 0 , x 0 )= lim m , n σ ( T n x , T m x ) =0.

From our assumption, we have

σ ( T n x , T x 0 ) α ( σ ( T n 1 x , x 0 ) ) σ ( T n 1 x , x 0 ) ,

which yields lim n σ( T n x,T x 0 )=0. Also, notice that σ(T x 0 ,T x 0 )σ( x 0 , x 0 )=0 and hence σ(T x 0 ,T x 0 )=0. Thus,

lim n σ ( T n x , T x 0 ) =σ(T x 0 ,T x 0 )= lim m , n σ ( T n x , T m x ) =0.

By the triangle inequality, we have

σ( x 0 ,T x 0 )σ ( T n x , x 0 ) +σ ( T n x , T x 0 ) 0as n,

and so σ( x 0 ,T x 0 )=0, that is, T x 0 = x 0 . The uniqueness easily follows from our contractive condition on T. □

The following corollary is another new extension of Matthews’s fixed point result [1].

Corollary 2.12 Let (X,p) be a complete partial metric space, and let T:XX be a mapping satisfying

p(Tx,Ty)α ( p ( x , y ) ) p(x,y)

for each x,yX with xy, where α:[0,)[0,1) is nonincreasing. Then T has a unique fixed point.