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Some new common fixed point results through generalized altering distances on partial metric spaces

Abstract

We establish common fixed point results for two pairs of weakly compatible mappings on a partial metric space, satisfying a weak contractive condition involving generalized control functions. The presented theorems extend and unify various known fixed point results. Examples are given to show that our results are proper extensions of the known ones.

MSC:47H10, 54H25, 54H10.

1 Introduction

In [1], Matthews introduced the notion of a partial metric space as a part of the study of denotational semantics of dataflow networks. He showed that the Banach contraction mapping theorem can be generalized to the partial metric context for applications in program verification.

Subsequently, several authors (see, e.g., Altun and Erduran [2], Oltra et al. [3], Romaguera and Schellekens [4], Romaguera and Valero [5], Rus [6], Djukić et al. [7], Nashine et al. [8], Di Bari and Vetro [9], Paesano and Vetro [10], Shatanawi et al. [11], Shatanawi and Nashine [12], Aydi et al. [13]) derived fixed point theorems in partial metric spaces.

Altering distance functions (also called control functions) were introduced by Khan et al. [14]. Subsequently, they were used by many authors to obtain fixed point results, including those in partial metric spaces (e.g., Abdeljawad [15], Abdeljawad et al. [16, 17], Altun et al. [18], Ćirić et al. [19], Karapinar and Yüksel [20]). Generalized altering distance functions with several variables were used on metric spaces by Berinde [21], Choudhury [22] and Rao et al. [23].

In this paper, an attempt has been made to derive some common fixed point theorems for two pairs of weakly compatible mappings on partial metric spaces, satisfying a weak contractive condition involving generalized control functions. The presented theorems extend and unify various known fixed point results. Examples are given to show that our results are proper extensions of the known ones.

2 Preliminaries

The following definitions and details about partial metrics can be seen, e.g., in [1, 2428].

Definition 1 A partial metric on a nonempty set X is a function p:X×X R + such that for all x,y,zX:

(p1) x=yp(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).

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

It is clear that, if p(x,y)=0, then from (p1) and (p2), it follows that x=y. But p(x,x) 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,ε):xX,ε>0}, where B p (x,ε)={yX:p(x,y)<p(x,x)+ε} for all xX and ε>0. A sequence { x n } in (X,p) converges to a point xX, with respect to τ p , if lim n p(x, x n )=p(x,x). This will be denoted as x n x, n or lim n x n =x. If (X,p) is a partial metric space, and T:XX is a mapping, continuous at x 0 X (in τ p ) then, for each sequence { x n } in X, we have

x n x 0 T x n T x 0 .

Clearly, a limit of a sequence in a partial metric space need not be unique. Moreover, the function p(,) need not be continuous in the sense that x n x and y n y implies p( x n , y n )p(x,y).

Definition 2 Let (X,p) be a partial metric space. Then:

  1. 1

    A sequence { x n } in (X,p) is called a Cauchy sequence if lim n , m p( x n , x m ) exists (and is finite).

  2. 2

    The 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 n , m p( x n , x m ).

It is easy to see that every closed subset of a complete partial metric space is complete.

If p is a partial metric on X, then the function p s :X×X R + given by

p s (x,y)=2p(x,y)p(x,x)p(y,y)
(2.1)

is a metric on X. Furthermore, lim n p s ( x n ,x)=0 if and only if

p(x,x)= lim n p( x n ,x)= lim n , m p( x n , x m ).
(2.2)

Lemma 1 Let (X,p) be a partial metric space.

  1. (a)

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

  2. (b)

    The space (X,p) is complete if and only if the metric space (X, p s ) is complete.

Definition 3 ([22, 23])

A function ψ:[0,+ ) n [0,+) is said to be a generalized altering distance function if:

  1. 1

    ψ is continuous;

  2. 2

    ψ is increasing in each of its variables;

  3. 3

    ψ( t 1 ,, t n )=0 if and only if t 1 == t n =0.

The set of generalized altering distance functions with n variables will be denoted by F n . If ψ F n , we will write Ψ(t)=ψ(t,t,,t) (obviously, this function belongs to F 1 ).

Simple examples of generalized altering distance functions with, say, four variables are:

Recall also the following notions. Let X be a nonempty set and T 1 , T 2 :XX be given self-maps on X. If w= T 1 x= T 2 x for some xX, then x is called a coincidence point of T 1 and T 2 , and w is called a point of coincidence of T 1 and T 2 . The pair { T 1 , T 2 } is said to be weakly compatible if T 1 T 2 t= T 2 T 1 t, whenever T 1 t= T 2 t for some t in X.

3 Results

3.1 Some auxiliary results

Assertions similar to the following lemma (see, e.g., [29]) were used (and proved) in the course of proofs of several fixed point results in various papers.

Lemma 2 Let (X,d) be a metric space and let { y n } be a sequence in X such that {d( y n + 1 , y n )} is nonincreasing and

lim n d( y n + 1 , y n )=0.

If { y 2 n } is not a Cauchy sequence, then there exist ε>0 and two sequences {m(k)} and {n(k)} of positive integers such that n(k)>m(k)>k and the following four sequences tend to ε when k:

d( y 2 m ( k ) , y 2 n ( k ) ),d( y 2 m ( k ) , y 2 n ( k ) + 1 ),d( y 2 m ( k ) 1 , y 2 n ( k ) ),d( y 2 m ( k ) 1 , y 2 n ( k ) + 1 ).

As a corollary (putting d= p s for a partial metric p), we obtain

Lemma 3 Let (X,p) be a partial metric space and let { y n } be a sequence in X such that {p( y n + 1 , y n )} is nonincreasing and

lim n p( y n + 1 , y n )=0.
(3.1)

If { y 2 n } is not a Cauchy sequence in (X,p), then there exist ε>0 and two sequences { m k } and {n(k)} of positive integers such that n(k)>m(k)>k and the following four sequences tend to ε when k:

(3.2)

3.2 Main results

Theorem 1 Let (X,p) be a complete partial metric space. Let T,S,I,J:XX be given mappings satisfying for every pair (x,y)X×X:

Ψ 1 ( p ( S x , T y ) ) ψ 1 ( p ( I x , J y ) , p ( I x , S x ) , p ( J y , T y ) , 1 2 [ p ( I x , T y ) + p ( J y , S x ) ] ) ψ 2 ( p ( I x , J y ) , p ( I x , S x ) , p ( J y , T y ) ) ,
(3.3)

where ψ 1 F 4 and ψ 2 F 3 are generalized altering distance functions, and Ψ 1 (t)= ψ 1 (t,t,t,t). Suppose that

  1. (i)

    TXIX and SXJX;

  2. (ii)

    one of the ranges IX, JX, TX and SX is a closed subset of (X,p).

Then

  1. (a)

    I and S have a coincidence point,

  2. (b)

    J and T have a coincidence point.

Moreover, if the pairs {I,S} and {J,T} are weakly compatible, then I, J, T and S have a unique common fixed point.

Proof Let x 0 be an arbitrary point in X. Since TXIX and SXJX, we can define sequences { x n } and { y n } in X by

y 2 n 1 =S x 2 n 2 =J x 2 n 1 , y 2 n =T x 2 n 1 =I x 2 n ,nN.
(3.4)

Without loss of the generality, we may assume that

p( y 2 n , y 2 n + 1 )>0,nN.
(3.5)

If not, then p( y 2 n , y 2 n + 1 )=0 and hence y 2 n = y 2 n + 1 , for some n. Taking x= x 2 n and y= x 2 n + 1 , from (3.4) and the considered contraction condition (3.3), we have

Ψ 1 ( p ( y 2 n + 1 , y 2 n + 2 ) ) = Ψ 1 ( p ( S x 2 n , T x 2 n + 1 ) ) ψ 1 ( p ( I x 2 n , J x 2 n + 1 ) , p ( I x 2 n , S x 2 n ) , p ( J x 2 n + 1 , T x 2 n + 1 ) , 1 2 [ p ( I x 2 n , T x 2 n + 1 ) + p ( J x 2 n + 1 , S x 2 n ) ] ) ψ 2 ( p ( I x 2 n , J x 2 n + 1 ) , p ( I x 2 n , S x 2 n ) , p ( J x 2 n + 1 , T x 2 n + 1 ) ) = ψ 1 ( p ( y 2 n , y 2 n + 1 ) , p ( y 2 n , y 2 n + 1 ) , p ( y 2 n + 1 , y 2 n + 2 ) , 1 2 [ p ( y 2 n , y 2 n + 2 ) + p ( y 2 n + 1 , y 2 n + 1 ) ] ) ψ 2 ( p ( y 2 n , y 2 n + 1 ) , p ( y 2 n , y 2 n + 1 ) , p ( y 2 n + 1 , y 2 n + 2 ) ) ψ 1 ( p ( y 2 n , y 2 n + 1 ) , p ( y 2 n , y 2 n + 1 ) , p ( y 2 n + 1 , y 2 n + 2 ) , 1 2 [ p ( y 2 n , y 2 n + 1 ) + p ( y 2 n + 1 , y 2 n + 2 ) ] ) ψ 2 ( p ( y 2 n , y 2 n + 1 ) , p ( y 2 n , y 2 n + 1 ) , p ( y 2 n + 1 , y 2 n + 2 ) ) ,
(3.6)

since

p( y 2 n , y 2 n + 2 )+p( y 2 n + 1 , y 2 n + 1 )p( y 2 n , y 2 n + 1 )+p( y 2 n + 1 , y 2 n + 2 ).

Suppose that p( y 2 n + 1 , y 2 n + 2 )>0. Using (3.6) together with p( y 2 n , y 2 n + 1 )=0 and the properties of the generalized altering distance functions ψ 1 , ψ 2 , we get

Ψ 1 ( p ( y 2 n + 1 , y 2 n + 2 ) ) ψ 1 ( 0 , 0 , p ( y 2 n + 1 , y 2 n + 2 ) , 1 2 p ( y 2 n + 1 , y 2 n + 2 ) ) ψ 2 ( 0 , 0 , p ( y 2 n + 1 , y 2 n + 2 ) ) , < ψ 1 ( 0 , 0 , p ( y 2 n + 1 , y 2 n + 2 ) , 1 2 p ( y 2 n + 1 , y 2 n + 2 ) ) Ψ 1 ( p ( y 2 n + 1 , y 2 n + 2 ) ) ,

which is a contradiction. It follows that p( y 2 n + 1 , y 2 n + 2 )=0 and hence y 2 n + 1 = y 2 n + 2 . Following similar arguments, we obtain y 2 n + 2 = y 2 n + 3 . Thus { y n } becomes an eventually constant sequence and y 2 n is a point of coincidence of I and S, while y 2 n + 1 is a point of coincidence of J and T.

Assume further that (3.5) holds. We claim that

lim n p( y n + 1 , y n + 2 )=0.
(3.7)

Suppose that, for some nN,

p( y 2 n + 1 , y 2 n + 2 )>p( y 2 n , y 2 n + 1 ).

Using this together with the properties of generalized altering distance functions ψ 1 , ψ 2 , we get from (3.6) that

This implies that

ψ 2 ( p ( y 2 n , y 2 n + 1 ) , p ( y 2 n , y 2 n + 1 ) , p ( y 2 n + 1 , y 2 n + 2 ) ) =0,

which yields that p( y 2 n + 1 , y 2 n )=0. Hence, we obtain a contradiction with (3.5). We deduce that

p( y 2 n + 1 , y 2 n + 2 )p( y 2 n , y 2 n + 1 ),nN.
(3.8)

By a similar reasoning, we obtain that

p( y 2 n + 2 , y 2 n + 3 )p( y 2 n + 1 , y 2 n + 2 ),nN.
(3.9)

Combining (3.8) and (3.9), we obtain

p( y n + 1 , y n + 2 )p( y n + 2 , y n + 3 ),nN.

Then, {p( y n + 1 , y n + 2 )} is a nonincreasing sequence of positive real numbers. This implies that there exists r0 such that

lim n p( y n + 1 , y n + 2 )=r.
(3.10)

By (3.6), we have

(3.11)

Letting n in (3.11) and using continuity of Ψ 1 and ψ 2 , we obtain

Ψ 1 (r) Ψ 1 (r) ψ 2 (r,r,r),

which implies that ψ 2 (r,r,r)=0, and thus r=0. Hence, (3.7) is proved.

Next, we claim that { y n } is a Cauchy sequence in the space (X,p) (and also in the metric space (X, p s ) by Lemma 1). For this it is sufficient to show that { y 2 n } is a Cauchy sequence. Suppose that this is not the case. Then, using Lemma 3 we get that there exist ε>0 and two sequences {m(i)} and {n(i)} of positive integers such that n(i)>m(i)>i and sequences (3.2) tend to ε when i. Applying condition (3.3) to elements x= x 2 m ( i ) and y= x 2 n ( i ) 1 , we get that

Ψ 1 ( p ( y 2 m ( i ) + 1 , y 2 n ( i ) ) ) = Ψ 1 ( p ( S x 2 m ( i ) , T x 2 n ( i ) 1 ) ) ψ 1 ( p ( y 2 m ( i ) , y 2 n ( i ) 1 ) , p ( y 2 m ( i ) , y 2 m ( i ) + 1 ) , p ( y 2 n ( i ) 1 , y 2 n ( i ) ) , 1 2 [ p ( y 2 m ( i ) , y 2 n ( i ) ) + p ( y 2 n ( i ) 1 , y 2 m ( i ) + 1 ) ] ) ψ 2 ( p ( y 2 m ( i ) , y 2 n ( i ) 1 ) , p ( y 2 m ( i ) , y 2 m ( i ) + 1 ) , p ( y 2 n ( i ) 1 , y 2 n ( i ) ) ) .

Passing to the limit as i in the last inequality (and using the continuity of the functions ψ 1 , ψ 2 ), we obtain

Ψ 1 (ε) ψ 1 (ε,0,0,ϵ) ψ 2 (ε,0,0) Ψ 1 (ε) ψ 2 (ε,0,0),

which implies that ψ 2 (ε,0,0)=0, that is a contradiction since ε>0. We deduce that { y n } is a Cauchy sequence.

Finally, we prove the existence of a common fixed point of the four mappings I, J, S and T.

Since (X,p) is complete, then from Lemma 1, (X, p s ) is a complete metric space. Therefore, the sequence { y n } p s -converges to some zX that is, lim n p s ( y n ,z)=0. From (2.2), we have

p(z,z)= lim n p( y n ,z)= lim m n p( y n , y m ).
(3.12)

Moreover, since { y n } is a Cauchy sequence in the metric space (X, p s ), then lim n , m p s ( y n , y m )=0. On the other hand, by (p2) and (3.7), we have p( y n , y n )p( y n , y n + 1 )0, n and hence

lim n p( y n , y n )=0.
(3.13)

Thus from the definition of p s and (3.13), we have lim m n p( y n , y m )=0. Therefore, from (3.12), we have

p(z,z)= lim n p( y n ,z)= lim m n p( y n , y m )=0.
(3.14)

This implies that

lim n p( y 2 n ,z)= lim n p( y 2 n + 1 ,z)=0.
(3.15)

Thus we have

lim n p(T x 2 n 1 ,z)= lim n p(I x 2 n ,z)=0

and

lim n p(S x 2 n ,z)= lim n p(J x 2 n + 1 ,z)=0.

Now we can suppose, without loss of generality, that IX is a closed subset of the partial metric space (X,p). From (3.15), there exists uX such that z=Iu. We claim that Su=z. Suppose, to the contrary, that p(Su,z)>0. By (p4) we get

p ( z , S u ) p ( z , T x 2 n 1 ) + p ( S u , T x 2 n 1 ) p ( T x 2 n 1 , T x 2 n 1 ) p ( z , y 2 n ) + p ( S u , y 2 n ) .

It follows by (3.15) that

p(z,Su) lim sup n p(Su, y 2 n ).

Then, since Ψ 1 is increasing and continuous, we get that

Ψ 1 ( p ( z , S u ) ) lim sup n Ψ 1 ( p ( S u , y 2 n ) ) .
(3.16)

Now, from (3.3)

Ψ 1 ( p ( S u , y 2 n ) ) = Ψ 1 ( p ( S u , T x 2 n 1 ) ) ψ 1 ( p ( I u , J x 2 n 1 ) , p ( I u , S u ) , p ( J x 2 n 1 , T x 2 n 1 ) , 1 2 [ p ( I u , T x 2 n 1 ) + p ( S u , J x 2 n 1 ) ] ) ψ 2 ( p ( I u , J x 2 n 1 ) , p ( I u , S u ) , p ( J x 2 n 1 , T x 2 n 1 ) ) ψ 1 ( p ( z , y 2 n 1 ) , p ( z , S u ) , p ( y 2 n 1 , y 2 n ) , 1 2 [ p ( z , y 2 n ) + p ( S u , z ) + p ( z , y 2 n 1 ) p ( z , z ) ] ) ψ 2 ( p ( z , y 2 n 1 ) , p ( z , S u ) , p ( y 2 n 1 , y 2 n ) ) .
(3.17)

Passing to the upper limit as n in (3.17), we obtain using (3.14) and the continuity of ψ 1 , ψ 2 that

lim sup n Ψ 1 ( p ( S u , y 2 n ) ) ψ 1 ( 0 , p ( z , S u ) , 0 , 1 2 p ( S u , z ) ) ψ 2 ( 0 , p ( z , S u ) , 0 ) < Ψ 1 ( p ( z , S u ) ) .

Therefore, from (3.16) we have

Ψ 1 ( p ( z , S u ) ) < Ψ 1 ( p ( z , S u ) ) ,

which is a contradiction. Thus we deduce that

p(z,Su)=0andz=Su.
(3.18)

We get that Su=Iu=z, so u is a coincidence point of I and S.

From SXJX and (3.18), we have zJX. Hence we deduce that there exists vX such that z=Jv. We claim that Tv=z. Suppose, to the contrary, that p(Tv,z)>0. From (3.3), we have

Ψ 1 ( p ( z , T v ) ) = Ψ 1 ( p ( S u , T v ) ) ψ 1 ( p ( I u , J v ) , p ( I u , S u ) , p ( J v , T v ) , 1 2 [ p ( I u , T v ) + p ( S u , J v ) ] ) ψ 2 ( p ( I u , J v ) , p ( I u , S u ) , p ( J v , T v ) ) = ψ 1 ( p ( z , z ) , p ( z , z ) , p ( z , T v ) , 1 2 [ p ( z , T v ) + p ( z , z ) ] ) ψ 2 ( p ( z , z ) , p ( z , z ) , p ( z , T v ) ) < ψ 1 ( 0 , 0 , p ( z , T v ) , 1 2 p ( z , T v ) ) Ψ 1 ( p ( z , T v ) ) ,

which is a contradiction. Then, we deduce that

p(z,Tv)=0andz=Tv.
(3.19)

We get that Jv=Tv=z, so v is a coincidence point of J and S.

Since the pair {S,I} is weakly compatible, from (3.18), we have Sz=SIu=ISu=Iz. We claim that Sz=z. Suppose, to the contrary, that p(Sz,z)>0. Then we have

p(Sz,z)p(Sz, y 2 n )+p( y 2 n ,z)=p(Sz,T x 2 n 1 )+p( y 2 n ,z).

Again from (3.15) we get that

p(Sz,z) lim sup n p(Sz,T x 2 n 1 ).

Then, since Ψ 1 is increasing and continuous, we get

Ψ 1 ( p ( S z , z ) ) lim sup n Ψ 1 ( p ( S z , T x 2 n 1 ) ) .
(3.20)

Now, from (3.3)

Ψ 1 ( p ( S z , T x 2 n 1 ) ) ψ 1 ( p ( I z , J x 2 n 1 ) , p ( I z , S z ) , p ( J x 2 n 1 , T x 2 n 1 ) , 1 2 [ p ( I z , T x 2 n 1 ) + p ( S z , J x 2 n 1 ) ] ) ψ 2 ( p ( I z , J x 2 n 1 ) , p ( I z , S z ) , p ( J x 2 n 1 , T x 2 n 1 ) ) ψ 1 ( [ p ( S z , z ) + p ( z , y 2 n 1 ) p ( z , z ) ] , p ( S z , S z ) , p ( y 2 n 1 , y 2 n ) , 1 2 [ p ( S z , y 2 n ) + p ( S z , z ) + p ( z , y 2 n 1 ) p ( z , z ) ] ) ψ 2 ( p ( S z , y 2 n 1 ) , p ( S z , S z ) , p ( y 2 n 1 , y 2 n ) ) .

Passing to the upper limit as n, we obtain (since p(Sz,Sz)p(Sz,z))

lim sup n Ψ 1 ( p ( S z , T x 2 n 1 ) ) ψ 1 ( p ( S z , z ) , p ( S z , S z ) , 0 , p ( S z , z ) ) ψ 2 ( p ( S z , z ) , p ( S z , S z ) , 0 ) < Ψ 1 ( p ( S z , z ) ) .

Therefore, from (3.20) we have

Ψ 1 ( p ( S z , z ) ) < Ψ 1 ( p ( S z , z ) ) ,

a contradiction. This implies that

p(Sz,z)=0andz=Sz.

Hence, we have

Sz=z=Iz.
(3.21)

Since the pair {T,J} is weakly compatible, from (3.19), we have Tz=TJv=JTv=Jz. We claim that Tz=z. Suppose, to the contrary, that p(Tz,z)>0, then by (3.3), we have

Ψ 1 ( p ( z , T z ) ) = Ψ 1 ( p ( S z , T z ) ) ψ 1 ( p ( I z , J z ) , p ( I z , S z ) , p ( J z , T z ) , 1 2 [ p ( I z , T z ) + p ( S z , J z ) ] ) ψ 2 ( p ( I z , J z ) , p ( I z , S z ) , p ( J z , T z ) ) = ψ 1 ( p ( z , T z ) , 0 , p ( T z , T z ) , 1 2 [ p ( z , T z ) + p ( z , T z ) ] ) ψ 2 ( p ( z , T z ) , 0 , p ( T z , T z ) ) Ψ 1 ( p ( z , T z ) ) ψ 2 ( p ( z , T z ) , 0 , p ( T z , T z ) ) .

Therefore, ψ 2 (p(Tz,z),0,p(Tz,Tz))=0. Hence, we have p(z,Tz)=0 and

Tz=z=Jz.
(3.22)

Now, combining (3.21) and (3.22), we deduce

z=Iz=Sz=Tz=Jz,

so z is a common fixed point of the four mappings I, J, S and T.

We claim that there is a unique common fixed point of S, T, I and J. Assume on contrary that, Su=Tu=Iu=Ju=u and Sv=Tv=Iv=Jv=v with p(u,v)>0. By supposition, we can replace x by u and y by v in (3.3) to obtain

Ψ 1 ( p ( u , v ) ) = Ψ 1 ( p ( S u , T v ) ) ψ 1 ( p ( I u , J v ) , p ( I u , S u ) , p ( J v , T v ) , 1 2 [ p ( I u , T v ) + p ( S u , J v ) ] ) ψ 2 ( p ( I u , J v ) , p ( I u , S u ) , p ( J v , T v ) ) = ψ 1 ( p ( u , v ) , 0 , 0 , p ( u , v ) ) ψ 2 ( p ( u , v ) , 0 , 0 ) < Ψ 1 ( p ( u , v ) ) ,

a contradiction. Hence p(u,v)=0, that is, u=v. We conclude that S, T, I and J have only one common fixed point in X. The proof is complete. □

It is easy to state the corollary of Theorem 1 involving a contraction of integral type.

Corollary 1 Let T, S, I and J as well as ψ 1 , ψ 2 satisfy the conditions of Theorem 1, except that condition (3.3) is replaced by the following: there exists a positive Lebesgue integrable function u on R + such that 0 ε u(t)dt>0 for each ε>0 and that

0 Ψ 1 ( p ( S x , T y ) ) u ( t ) d t 0 ψ 1 ( p ( I x , J y ) , p ( I x , S x ) , p ( J y , T y ) , 1 2 [ p ( I x , T y ) + p ( J y , S x ) ] ) u ( t ) d t 0 ψ 2 ( ( p ( I x , J y ) , p ( I x , S x ) , p ( J y , T y ) ) u ( t ) d t ,

for all x,yX. Then, S, T, I and J have a unique common fixed point.

If in Theorem 1 I=J is the identity mapping on X, then we have the following consequence:

Theorem 2 Let (X,p) be a complete partial metric space. Let T,S:XX be given mappings satisfying for every pair (x,y)X×X

Ψ 1 ( p ( S x , T y ) ) ψ 1 ( p ( x , y ) , p ( x , S x ) , p ( y , T y ) , 1 2 [ p ( x , T y ) + p ( y , S x ) ] ) ψ 2 ( p ( x , y ) , p ( x , S x ) , p ( y , T y ) ) ,
(3.23)

where ψ 1 F 4 and ψ 2 F 3 are altering distance functions, and Ψ 1 (t)= ψ 1 (t,t,t,t). Then T and S have a unique common fixed point.

Remark 1 Several corollaries of Theorems 1 and 2 could be derived for particular choices of ψ 1 and ψ 2 . We state some of them.

Putting ψ 1 ( t 1 , t 2 , t 3 , t 4 )=ψ(max{ t 1 , t 2 , t 3 , t 4 }) and ψ 2 ( t 1 , t 2 , t 3 )=ϕ(max{ t 1 , t 2 , t 3 }) for ψ,ϕ F 1 , [15], Theorem 9] is obtained.

It is clear from the proof of Theorem 1 that condition (3.3), resp. (3.23), can be replaced by

(3.24)

resp.

(3.25)

where ψ 2 F 4 . Hence, Theorem 1 can be considered an extension of [23], Theorem 2.1] to the frame of partial metric spaces (since semi-compatible mappings are weakly compatible).

Putting ψ 1 ( t 1 , t 2 , t 3 , t 4 )=max{ t 1 , t 2 , t 3 , t 4 } and ψ 2 =(1r) ψ 1 , with 0r<1, in Theorem 2 (with condition (3.25)), we obtain [20], Theorem 8]. The same substitution in Theorem 1 (with (3.24)) gives an improvement of [20], Theorem 12] (since only weak compatibility and not commutativity of the respective mappings is assumed).

Putting ψ 1 ( t 1 , t 2 , t 3 , t 4 )=max{ t 1 , t 2 , t 3 , t 4 } and ψ 2 =φ ψ 1 for φ F 1 in Theorem 2 (with condition (3.25)), [16], Theorem 5] is obtained.

Of course, several known results from the frame of standard metric spaces (see, e.g., [30] and [31]) are also special cases of these theorems. For example, the following corollary can be obtained as a consequence of Theorem 2, which is a generalization and extension of [31], Corollary 3.2].

Corollary 2 Let (X,p) be a complete partial metric space. Let T,S:XX be given mappings satisfying for every pair (x,y)X×X

where m and n are positive integers, ψ 1 F 4 and ψ 2 F 3 are altering distance functions, and Ψ 1 (t)= ψ 1 (t,t,t,t). Then T and S have a unique common fixed point.

Remark 2 However, it is not possible to use ψ 1 , ψ 2 F 5 in Theorems 1 and 2, as the following example, adapted from [23], Example 2.3], shows.

Example 1 Let X={1,2,3,4} and p:X×XX be given by p(x,x)= 1 2 for xX, p(1,2)=p(3,4)=2, p(1,3)=p(2,4)=1, p(1,4)=p(2,3)= 3 2 and p(y,x)=p(x,y) for x,yX. Then (X,p) is a (complete) partial metric space. Consider the mappings S,T:XX defined by

S=( 1 2 3 4 2 1 1 2 ),T=( 1 2 3 4 4 3 4 3 ),

and the functions ψ 1 , ψ 2 F 5 given as ψ 1 ( t 1 , t 2 , t 3 , t 4 , t 5 )=max{ t 1 , t 2 , t 3 , t 4 , t 5 } and ψ 2 = 1 4 ψ 1 . It is easy to check that

p ( S x , T y ) 3 4 max { p ( x , y ) , p ( x , S x ) , p ( y , T y ) , p ( x , T y ) , p ( y , S x ) } = ψ 1 ( ) ψ 2 ( )

holds for all x,yX. However, these mappings have no common fixed points; hence, condition (3.23) (or (3.25)) of Theorem 2 cannot be replaced by the respective condition with 5 variables. At the same time, condition (3.25) is not satisfied since, for x=3, y=1, p(Sx,Ty)=p(1,4)= 3 2 and

max { p ( x , y ) , p ( x , S x ) , p ( y , T y ) , 1 2 [ p ( x , T y ) + p ( y , S x ) ] } =max { 1 , 1 , 3 2 , 1 2 ( 2 + 1 2 ) } = 3 2 ,

hence,

Ψ 1 ( p ( S x , T y ) ) = 3 2 > ψ 1 () ψ 2 ()

whatever ψ 2 F 4 is chosen.

This example also shows (as in [23], Remark 2.4]) the importance of the second generalized altering distance function ψ 2 in Theorems 1 and 2.

The next example shows that Theorems 1 and 2 are proper extensions of the respective results in standard metric spaces.

Example 2 Let X=[0,1] be endowed with the partial metric p(x,y)=max{x,y}. Consider the mappings S,T:XX defined by

Sx=Tx= x 2 1 + x ,

and the functions ψ 1 , ψ 2 F 4 , given by

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

Take arbitrary elements, say yx, from X. Then

p(Sx,Ty)=max { x 2 1 + x , y 2 1 + y } = x 2 1 + x ,and Ψ 1 ( p ( S x , T y ) ) = x 2 1 + x .

On the other hand,

and

Hence, condition (3.25) is satisfied, as well as other conditions of Theorem 2. Mappings S,T have a common fixed point z=0.

On the other hand, consider the same problem in the standard metric d(x,y)= p s (x,y)=|xy| and take x=1 and y= 1 2 . Then

d(Sx,Ty)=| 1 2 1 6 |= 1 3

and

and hence

Thus, condition (3.25) for p=d does not hold and the existence of a common fixed point of these mappings cannot be derived from [23], Theorem 2.1].

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Acknowledgements

The first and second author would like to acknowledge the financial support received from University Kebangsaan Malaysia under the research grant OUP-UKM-FST-2012. The fourth and fifth author are thankful to the Ministry of Science and Technological Development of Serbia.

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Ahmad, A.G.B., Fadail, Z.M., Nashine, H.K. et al. Some new common fixed point results through generalized altering distances on partial metric spaces. Fixed Point Theory Appl 2012, 120 (2012). https://doi.org/10.1186/1687-1812-2012-120

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Keywords

  • partial metric space
  • generalized altering distance
  • coincidence point
  • common fixed point
  • weakly compatible mappings