1 Introduction and preliminaries

Throughout this paper, by R + we denote the set of all nonnegative real numbers, while ℕ is the set of all natural numbers. In 1994, Mattews [1] introduced the following notion of partial metric spaces.

Definition 1 [1]

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

( p 1 ) x=y if and only if p(x,x)=p(x,y)=p(y,y);

( p 2 ) p(x,x)p(x,y);

( p 3 ) p(x,y)=p(y,x);

( p 4 ) p(x,y)p(x,z)+p(z,y)p(z,z).

A partial metric space is a pair (X,p) such that X is a nonempty set and p is a partial metric on X.

Remark 1 It is clear that if p(x,y)=0, then from ( p 1 ) and ( p 2 ), x=y. But if x=y, p(x,y) may not be 0.

Each partial metric p on X generates a T 0 topology τ p on X which has as a base the family of open p-balls { B p (x,γ):xX,γ>0}, where B p (x,γ)={yX:p(x,y)<p(x,x)+γ} for all xX and γ>0. If p is a partial metric on X, then the function d p :X×X R + given by

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

is a metric on X.

We recall some definitions of a partial metric space as follows.

Definition 2 [1]

Let (X,p) be a partial metric space. Then

  1. (1)

    a sequence { x n } in a partial metric space (X,p) converges to xX if and only if p(x,x)= lim n p(x, x n );

  2. (2)

    a sequence { x n } in a partial metric space (X,p) is called a Cauchy sequence if and only if lim m , n p( x m , x n ) exists (and is finite);

  3. (3)

    a partial metric space (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 p(x,x)= lim m , n p( x m , x n );

  4. (4)

    a subset A of a partial metric space (X,p) is closed if whenever { x n } is a sequence in A such that { x n } converges to some xX, then xA.

Remark 2 The limit in a partial metric space is not unique.

Lemma 1 [1, 2]

  1. (1)

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

  2. (2)

    a partial metric space (X,p) is complete if and only if the metric space (X, d p ) is complete. Furthermore, lim n d p ( x n ,x)=0 if and only if p(x,x)= lim n p( x n ,x)= lim n p( x n , x m ).

In recent years, fixed point theory has developed rapidly on partial metric spaces, see [210].

In this study, we also recall the Meir-Keeler-type contraction [11] and α-admissible one [12]. In 1969, Meir and Keeler [11] introduced the following notion of Meir-Keeler-type contraction in a metric space (X,d).

Definition 3 Let (X,d) be a metric space, f:XX. Then f is called a Meir-Keeler-type contraction whenever, for each η>0, there exists γ>0 such that

ηd(x,y)<η+γd(fx,fy)<η.

The following definition was introduced in [12].

Definition 4 Let f:XX be a self-map** of a set X and α:X×X R + . Then f is called α-admissible if

x,yX,α(x,y)1α(fx,fy)1.

The purpose of this paper is to study fixed point theorems for a map** satisfying the generalized Meir-Keeler-type ϕ-α-contractions in complete partial metric spaces. Our results generalize or improve many recent fixed point theorems in the literature.

2 Main results

In the article, we denote by Φ the class of functions ϕ: R + 4 R + satisfying the following conditions:

( ϕ 1 ) ϕ is an increasing and continuous function in each coordinate;

( ϕ 2 ) for t R + {0}, ϕ(t,t,t,t)t, ϕ(t,0,0,t)t, ϕ(0,0,t, t 2 )t; and ϕ( t 1 , t 2 , t 3 , t 4 )=0 iff t 1 = t 2 = t 3 = t 4 =0.

We now state the new notions of generalized Meir-Keeler-type ϕ-contractions and generalized Meir-Keeler-type ϕ-α-contractions in partial metric spaces as follows.

Definition 5 Let (X,p) be a partial metric space, f:XX and ϕΦ. Then f is called a generalized Meir-Keeler-type ϕ-contraction whenever, for each η>0, there exists δ>0 such that

η ϕ ( p ( x , y ) , p ( x , f x ) , p ( y , f y ) , 1 2 [ p ( x , f y ) + p ( y , f x ) ] ) < η + δ p ( f x , f y ) < η .

Definition 6 Let (X,p) be a partial metric space, f:XX, ϕΦ and α:X×X R + . Then f is called a generalized Meir-Keeler-type ϕ-α-contraction if the following conditions hold:

  1. (1)

    f is α-admissible;

  2. (2)

    for each η>0, there exists δ>0 such that

    η ϕ ( p ( x , y ) , p ( x , f x ) , p ( y , f y ) , 1 2 [ p ( x , f y ) + p ( y , f x ) ] ) < η + δ α ( x , x ) α ( y , y ) p ( f x , f y ) < η .
    (2.1)

Remark 3 Note that if f is a generalized Meir-Keeler-type ϕ-α-contraction, then we have that for all x,yX,

α ( x , x ) α ( y , y ) p ( f x , f y ) ϕ ( p ( x , y ) , p ( x , f x ) , p ( y , f y ) , 1 2 [ p ( x , f y ) + p ( y , f x ) ] ) .

Further, if ϕ(p(x,y),p(x,fx),p(y,fy), 1 2 [p(x,fy)+p(y,fx)])=0, then p(fx,fy)=0. On the other hand, if ϕ(p(x,y),p(x,fx),p(y,fy), 1 2 [p(x,fy)+p(y,fx)])>0, then α(x,x)α(y,y)p(fx,fy)<ϕ(p(x,y),p(x,fx),p(y,fy), 1 2 [p(x,fy)+p(y,fx)]).

We now state our main result for the generalized Meir-Keeler-type ϕ-α-contraction as follows.

Theorem 1 Let (X,p) be a complete partial metric space, and ϕΦ. If α:X×X R + satisfies the following conditions:

( α 1 ) there exists x 0 X such that α( x 0 , x 0 )1;

( α 2 ) if α( x n , x n )1 for all nN, then lim n α( x n , x n )1;

( α 3 ) α:X×X R + is a continuous function in each coordinate.

Suppose that f:XX is a generalized Meir-Keeler-type ϕ-α-contraction. Then f has a fixed point in X.

Proof Let x 0 and let x n + 1 =f x n = f n x 0 for n=0,1,2, . Since f is α-admissible and α( x 0 , x 0 )1, we have

α(f x 0 ,f x 0 )=α( x 1 , x 1 )1.

By continuing this process, we get

α( x n , x n )1for all nN{0}.
(2.2)

If there exists n 0 N such that x n 0 + 1 = x n 0 , then we finished the proof. Suppose that x n + 1 x n for any n=0,1,2, . By the definition of the function ϕ, we have ϕ(p( x n , x n + 1 ),p( x n ,f x n ),p( x n + 1 ,f x n + 1 ), 1 2 [p( x n ,f x n + 1 )+p( x n + 1 ,f x n )])>0 for all nN{0}.

Step 1. We shall prove that

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

By Remark 3 and ( p 4 ), using (2.2), we have

p ( x n + 1 , x n + 2 ) = p ( f x n , f x n + 1 ) α ( x n , x n ) α ( x n + 1 , x n + 1 ) p ( f x n , f x n + 1 ) < ϕ ( p ( x n , x n + 1 ) , p ( x n , f x n ) , p ( x n + 1 , f x n + 1 ) , 1 2 [ p ( x n , f x n + 1 ) + p ( x n + 1 , f x n ) ] ) = ϕ ( p ( x n , x n + 1 ) , p ( x n , x n + 1 ) , p ( x n + 1 , x n + 2 ) , 1 2 [ p ( x n , x n + 2 ) + p ( x n + 1 , x n + 1 ) ] ) ϕ ( p ( x n , x n + 1 ) , p ( x n , x n + 1 ) , p ( x n + 1 , x n + 2 ) , 1 2 [ p ( x n , x n + 1 ) + p ( x n + 1 , x n + 2 ) ] ) .
(2.3)

If p( x n , x n + 1 )p( x n + 1 , x n + 2 ), then

p ( x n + 1 , x n + 2 ) = p ( f x n , f x n + 1 ) < ϕ ( p ( x n + 1 , x n + 2 ) , p ( x n + 1 , x n + 2 ) , p ( x n + 1 , x n + 2 ) , p ( x n + 1 , x n + 2 ) ) p ( x n + 1 , x n + 2 ) ,

which implies a contradiction, and hence p( x n , x n + 1 )<p( x n 1 , x n ). From the argument above, we also have that for each nN,

p ( x n + 1 , x n + 2 ) = p ( f x n , f x n + 1 ) < ϕ ( p ( x n , x n + 1 ) , p ( x n , x n + 1 ) , p ( x n , x n + 1 ) , p ( x n , x n + 1 ) ) p ( x n , x n + 1 ) .
(2.4)

Since the sequence {p( x n , x n + 1 )} is decreasing, it must converge to some η0, that is,

lim n p( x n , x n + 1 )=η.
(2.5)

It follows from (2.4) and (2.5) that

lim n ϕ ( p ( x n , x n + 1 ) , p ( x n , x n + 1 ) , p ( x n , x n + 1 ) , p ( x n , x n + 1 ) ) =η.
(2.6)

Notice that η=inf{p( x n , x n + 1 ):nN}. We claim that η=0. Suppose, to the contrary, that η>0. Since f is a generalized Meir-Keeler-type ϕ-contraction, corresponding to η use, and taking into account the above inequality (2.6), there exist δ>0 and a natural number k such that

η ϕ ( p ( x k , x k + 1 ) , p ( x k , x k + 1 ) , p ( x k , x k + 1 ) , p ( x k , x k + 1 ) ) < η + δ α ( x k , x k ) α ( x k + 1 , x k + 1 ) p ( f x k , f x k + 1 ) < η ,

which implies

p( x k + 1 , x k + 2 )=p(f x k ,f x k + 1 )α( x k , x k )α( x k + 1 , x k + 1 )p(f x k ,f x k + 1 )<η.

So, we get a contradiction since η=inf{p( x n , x n + 1 ):nN}. Thus we have that

lim n p( x n , x n + 1 )=0.
(2.7)

By ( p 2 ), we also have

lim n p( x n , x n )=0.
(2.8)

Since d p (x,y)=2p(x,y)p(x,x)p(y,y) for all x,yX, using (2.7) and (2.8), we obtain that

lim n d p ( x n , x n + 1 )=0.
(2.9)

Step 2. We show that { x n } is a Cauchy sequence in the partial metric space (X,p), that is, it is sufficient to show that { x n } is a Cauchy sequence in the metric space (X, d p ).

Suppose that the above statement is false. Then there exists ϵ>0 such that for any kN, there are n k , m k N with n k > m k k satisfying

d p ( x m k , x n k )ϵ.
(2.10)

Further, corresponding to m k k, we can choose n k in such a way that it is the smallest integer with n k > m k k and d( x 2 m k , x 2 n k )ϵ. Therefore

d p ( x m k , x n k 2 )<ϵ.
(2.11)

Now we have that for all kN,

ϵ d p ( x m k , x n k ) d p ( x m k , x n k 2 ) + d p ( x n k 2 , x n k 1 ) + d p ( x n k 1 , x n k ) < ϵ + d p ( x n k 2 , x n k 1 ) + d p ( x n k 1 , x n k ) .
(2.12)

Letting k in the above inequality and using (2.12), we get

lim n d p ( x m k , x n k )=ϵ.
(2.13)

On the other hand, we have

ϵ d p ( x m k , x n k ) d p ( x m k , x m k + 1 ) + d p ( x m k + 1 , x n k + 1 ) + d p ( x n k + 1 , x n k ) d p ( x m k , x m k + 1 ) + d p ( x m k + 1 , x m k ) + d p ( x m k , x n k ) + d p ( x n k , x n k + 1 ) + d p ( x n k + 1 , x n k ) .

Letting n, we obtain that

lim n d p ( x m k + 1 , x n k + 1 )=ϵ.
(2.14)

Since d p (x,y)=2p(x,y)p(x,x)p(y,y) and using (2.13) and (2.14), we have that

lim n p( x m k , x n k )= ϵ 2
(2.15)

and

lim n p( x m k + 1 , x n k + 1 )= ϵ 2
(2.16)

By Remark 3 and ( p 4 ), we have

p ( x m k + 1 , x n k + 1 ) = p ( f x m k , f x n k ) α ( x m k , x m k ) α ( x n k , x n k ) p ( f x m k , f x n k ) < ϕ ( p ( x m k , x n k ) , p ( x m k , f x m k ) , p ( x n k , f x n k ) , 1 2 [ p ( x m k , f x n k ) + p ( x n k , f x m k ) ] ) = ϕ ( p ( x m k , x n k ) , p ( x m k , x m k + 1 ) , p ( x n k , x n k + 1 ) , 1 2 [ p ( x m k , x n k + 1 ) + p ( x n k , x m k + 1 ) ] ) .
(2.17)

Since

p( x m k , x n k + 1 )p( x m k , x m k + 1 )+p( x m k + 1 , x n k + 1 )p( x m k + 1 , x m k + 1 )
(2.18)

and

p( x n k , x m k + 1 )p( x n k , x n k + 1 )+p( x n k + 1 , x m k + 1 )p( x n k + 1 , x n k + 1 ).
(2.19)

Taking into account the above inequalities (2.8), (2.17), (2.18) and (2.19), letting k, we have

ϵ 2 <ϕ ( ϵ 2 , 0 , 0 , ϵ 2 ) ϵ 2 ,

which implies a contradiction. Thus, { x n } is a Cauchy sequence in the metric space (X, d p ).

Step 3. We show that f has a fixed point ν in i = 1 m A i .

Since (X,p) is complete, then from Lemma 1, we have that (X, d p ) is complete. Thus, there exists νX such that

lim n d p ( x n ,ν)=0.

Moreover, it follows from Lemma 1 that

p(ν,ν)= lim n p( x n ,ν)= lim n , m p( x n , x m ).
(2.20)

On the other hand, since the sequence { x n } is a Cauchy sequence in the metric space (X, d p ), we also have

lim n d p ( x n , x m )=0.

Since d p (x,y)=2p(x,y)p(x,x)p(y,y), we can deduce that

lim n p( x n , x m )=0.
(2.21)

Using (2.20) and (2.21), we have

p(ν,ν)= lim n p( x n ,ν)= lim n p( x n k ,ν)=0.

Again, by Remark 3, ( p 4 ), and the conditions of the map** α, we have

p ( x n + 1 , f ν ) = p ( f x n , f ν ) α ( x n , x n ) α ( ν , ν ) p ( f x n , f ν ) < ϕ ( p ( x n , ν ) , p ( x n , f x n ) , p ( ν , f ν ) , 1 2 [ p ( x n , f ν ) + p ( ν , f x n ) ] ) = ϕ ( p ( x n , ν ) , p ( x n , x n + 1 ) , p ( ν , f ν ) , 1 2 [ p ( x n , f ν ) + p ( ν , x n + 1 ) ] ) .
(2.22)

Letting n in (2.22), we get

p(ν,fν)<ϕ ( 0 , 0 , p ( ν , f ν ) , 1 2 p ( ν , f ν ) ) p(ν,fν),

a contradiction. So, we have p(ν,fν)=0, that is, fν=ν. □

We give the following example to illustrate Theorem 2.

Example 1 Let X=[0,1]. We define the partial metric p on X by

p(x,y)=max{x,y}.

Let α:[0,1]×[0,1] R + be defined as

α(x,y)=1+x+y,

let f:XX be defined as

f(x)= 1 16 x 2 ,

and, let ϕ: R + 4 R + denote

ψ( t 1 , t 2 , t 3 , t 4 )= 1 2 max { t 1 , t 2 , t 3 , 1 2 t 4 } .

Then f is α-admissible.

Without loss of generality, we assume that x>y and verify the inequality (2.1). For all x,y[0,1] with x>y, we have

α ( x , x ) α ( y , y ) p ( f x , f y ) 1 16 x 2 , p ( x , y ) = x , p ( x , f x ) = x , p ( y , f y ) = y and 1 2 [ p ( x , f y ) + p ( y , f x ) ] = 1 2 [ max { x , y 2 } + max { y , x 2 } ] 1 2 [ max { x , y } + max { y , x } ] < x ,

and hence ϕ(p(x,y),p(x,fx),p(y,fy), 1 2 [p(x,fy)+p(y,fx)])= 1 2 x. Therefore, all the conditions of Theorem 1 are satisfied, and we obtained that 0 is a fixed point of f.

If we let

α(x,y)=1for x,yX,

then it is easy to get the following theorem.

Theorem 2 Let (X,p) be a complete partial metric space and ϕΦ. Suppose that f:XX is a generalized Meir-Keeler-type ϕ-contraction. Then f has a fixed point in X.