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Common fixed point theorems for weakly compatible mappings satisfying contractive conditions of integral type

Abstract

Two common fixed theorems for weakly compatible mappings satisfying general contractive conditions of integral type in metric spaces are proved and an illustrative example is provided. The results obtained in this paper substantially extend and improve several previous results, particularly Theorem 2.1 of Branciari (Int. J. Math. Math. Sci. 29(9):531-536, 2002), Theorem 2 of Rhoades (Int. J. Math. Math. Sci. 2003(63):4007-4013, 2003) and Theorem 2 of Vijayaraju et al. (Int. J. Math. Math. Sci. 2005(15):2359-2364, 2005). A nontrivial example with uncountably many points is also provided to support the results presented herein.

MSC:54H25.

1 Introduction and preliminaries

In 2002, Branciari [1] introduced the notion of contractive mappings of integral type in metric spaces and proved the following fixed point theorem for the contractive mapping of integral type, which is a nice generalization of the Banach contraction principle.

Theorem 1.1 ([1])

Let (X,d) be a complete metric space, c(0,1), and let f:XX be a mapping such that

0 d ( f x , f y ) φ(s)dsc d ( x , y ) φ(s)ds,x,yX,

where φ:[0,+)[0,+) is a Lebesgue integrable mapping which is summable on each compact subset of [0,+) and such that for all ε>0,

0 ε φ(s)ds>0.

Then f has a unique fixed point aX such that for each xX, lim n f n x=a.

Afterward, the researchers [217] and others extended the result to more general contractive conditions of integral type. In particular, Rhoades [13] proved the following extension of Theorem 1.1.

Theorem 1.2 ([13])

Let (X,d) be a complete metric space, k[0,1), f:XX be a mapping such that

0 d ( f x , f y ) φ(s)dsk m ( x , y ) φ(s)ds,x,yX,

where

m(x,y)=max { d ( x , y ) , d ( x , f x ) , d ( y , f y ) , d ( x , f y ) + d ( y , f x ) 2 } ,

φ:[0,+)[0,+) is a Lebesgue integrable mapping which is summable on each compact subset of [0,+) and such that for all ε>0,

0 ε φ(s)ds>0.

Then f has a unique fixed point aX and, for each xX, lim n f n x=a.

Vijayaraju et al. [17] extended further Theorems 1.1 and 1.2 from a single mapping to a pair of mappings. Using a rational expression for a contractive condition of integral type, Vetro [16] extended also Theorem 1.1 and proved the following common fixed point theorem for weakly compatible mappings.

Theorem 1.3 ([16])

Let (X,d) be a metric space and let A, B, S and T be self-mappings of X with S(X)B(X) and T(X)A(X) such that

0 d ( S x , T y ) φ(s)dsα 0 m ( x , y ) φ(s)ds+β 0 M ( x , y ) φ(s)ds,x,yX,

where

m ( x , y ) = d ( B y , T y ) 1 + d ( A x , S x ) 1 + d ( A x , B y ) , M ( x , y ) = max { d ( A x , B y ) , d ( A x , S x ) , d ( B y , T y ) } ,

α>0, β>0, α+β<1 and φ:[0,+)[0,+) is a Lebesgue integrable mapping on each compact subset of [0,+) and such that for all ε>0,

0 ε φ(s)ds>0.

Suppose that one of A(X), B(X), S(X) and T(X) is a complete subset of X and the pairs {A,S} and {B,T} are weakly compatible. Then A, B, S and T have a unique common fixed point in X.

Motivated and inspired by the results in [117], in this paper we introduce more general contractive mappings of integral type, which include the contractive mappings of integral type in [1, 4, 13, 16, 17] as special cases, and we establish the existence and uniqueness of common fixed points for these contractive mappings of integral type with weak compatibility. Our results extend, improve and unify the corresponding results in [1, 4, 13, 16, 17]. A nontrivial example with uncountably many points is also provided to support the results presented herein.

Throughout this paper, we assume that R + =[0,+), R=(,+), N 0 ={0}N, where denotes the set of all positive integers and

Φ = {φ:φ: R + R + satisfies that φ is Lebesgue integrable, summable on each compact subset of R + and 0 ε φ(t)dt>0 for each ε>0},

Ψ = {ψ:ψ: R + R + is upper semi-continuous on R + {0}, ψ(0)=0 and ψ(t)<t for each t>0},

Ψ 1 = {ψ:ψ: R + R + is nondecreasing on R + , ψ(t)<t and n = 1 ψ n (t)<+ for each t>0}.

Recall that a pair of self-mappings f and g in a metric space (X,d) are said to be weakly compatible if for all tX the equality ft=gt implies fgt=gft.

Lemma 1.4 ([9])

Let φΦ and { r n } n N be a nonnegative sequence. Then

lim n 0 r n φ(t)dt=0

if and only if lim n r n =0.

2 Common fixed point theorems

Now we show two common fixed point theorems for four contractive mappings of integral type in metric spaces.

Theorem 2.1 Let A, B, S and T be self-mappings of a metric space (X,d) such that

(C1) S(X)B(X) and T(X)A(X);

(C2) the pairs {A,S} and {B,T} are weakly compatible;

(C3) one of A(X), B(X), S(X) and T(X) is a complete subset of X and

0 d ( S x , T y ) φ(t)dtψ ( max { 0 m i ( x , y ) φ ( t ) d t : 1 i 4 } ) ,x,yX,
(2.1)

where (ψ,φ) is in Ψ×Φ and

m 1 ( x , y ) = d ( B y , T y ) 1 + d ( A x , S x ) 1 + d ( A x , B y ) , m 2 ( x , y ) = d ( A x , S x ) 1 + d ( B y , T y ) 1 + d ( A x , B y ) , m 3 ( x , y ) = d ( S x , B y ) d ( T y , A x ) 1 + d ( A x , B y ) , m 4 ( x , y ) = max { d ( A x , B y ) , d ( A x , S x ) , d ( B y , T y ) , m 4 ( x , y ) = 1 2 [ d ( S x , B y ) + d ( T y , A x ) ] } .
(2.2)

Then A, B, S and T have a unique common fixed point in X.

Proof Let x 0 X. It follows from (C1) that there exist two sequences { y n } n N and { x n } n N 0 in X satisfying

y 2 n + 1 =S x 2 n =B x 2 n + 1 and y 2 n + 2 =T x 2 n + 1 =A x 2 n + 2 ,n N 0 .
(2.3)

Put d n =d( y n , y n + 1 ) for each nN.

Firstly we show that A, B, S and T have at most a common fixed point in X. Suppose that u and v are two different common fixed points of A, B, S and T in X. It follows from (2.1), (2.2) and (ψ,φ)Ψ×Φ that

m 1 ( u , v ) = d ( B v , T v ) 1 + d ( A u , S u ) 1 + d ( A u , B v ) = 0 , m 2 ( u , v ) = d ( A u , S u ) 1 + d ( B v , T v ) 1 + d ( A u , B v ) = 0 , m 3 ( u , v ) = d ( S u , B v ) d ( T v , A u ) 1 + d ( A u , B v ) = d 2 ( u , v ) 1 + d ( u , v ) , m 4 ( u , v ) = max { d ( A u , B v ) , d ( A u , S u ) , d ( B v , T v ) , 1 2 [ d ( S u , B v ) + d ( T v , A u ) ] } = d ( u , v )

and

0 d ( u , v ) φ ( t ) d t = 0 d ( S u , T v ) φ ( t ) d t ψ ( max { 0 m i ( u , v ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 , 0 , 0 d 2 ( u , v ) 1 + d ( u , v ) φ ( t ) d t , 0 d ( u , v ) φ ( t ) d t } ) = ψ ( 0 d ( u , v ) φ ( t ) d t ) < 0 d ( u , v ) φ ( t ) d t ,

which is a contradiction. Hence A, B, S and T have at most a common fixed point in X.

Secondly we show that A, B, S and T have a common fixed point AaX if there exist a,bX satisfying

Aa=Sa=Bb=Tb.
(2.4)

Assume that (2.4) holds for some a,bX. Put c=Aa. Note that (C2) implies that

Sc=SAa=ASa=AcandBc=BTb=TBb=Tc.
(2.5)

Suppose that cTc. In view of (2.1), (2.2), (2.4), (2.5) and (ψ,φ)Ψ×Φ, we infer that

m 1 ( a , c ) = d ( B c , T c ) 1 + d ( A a , S a ) 1 + d ( A a , B c ) = 0 , m 2 ( a , c ) = d ( A a , S a ) 1 + d ( B c , T c ) 1 + d ( A a , B c ) = 0 , m 3 ( a , c ) = d ( S a , B c ) d ( T c , A a ) 1 + d ( A a , B c ) = d ( c , B c ) d ( T c , c ) 1 + d ( c , B c ) = d 2 ( c , T c ) 1 + d ( c , T c ) , m 4 ( a , c ) = max { d ( A a , B c ) , d ( A a , S a ) , d ( B c , T c ) , 1 2 [ d ( S a , B c ) + d ( T c , A a ) ] } = max { d ( c , B c ) , 0 , 0 , 1 2 [ d ( c , B c ) + d ( T c , c ) ] } = d ( c , T c )

and

0 d ( c , T c ) φ ( t ) d t = 0 d ( S a , T c ) φ ( t ) d t ψ ( max { 0 m i ( a , c ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 , 0 , 0 d 2 ( c , T c ) 1 + d ( c , T c ) φ ( t ) d t , 0 d ( c , T c ) φ ( t ) d t } ) = ψ ( 0 d ( c , T c ) φ ( t ) d t ) < 0 d ( c , T c ) φ ( t ) d t ,

which is impossible. Consequently, c=Tc=Bc. Similarly we conclude that c=Ac=Sc. That is, c is a common fixed point of A, B, S and T.

Thirdly we show that (2.4) holds for some a,bX. In order to prove (2.4), we have to consider three possible cases as follows.

Case 1. There exists n 0 N satisfying d 2 n 0 =0. We claim that d 2 n 0 + 1 =0. Otherwise d 2 n 0 + 1 >0. Using (2.1)-(2.3) and (ψ,φ)Ψ×Φ, we deduce that

m 1 ( x 2 n 0 , x 2 n 0 + 1 ) = d ( B x 2 n 0 + 1 , T x 2 n 0 + 1 ) 1 + d ( A x 2 n 0 , S x 2 n 0 ) 1 + d ( A x 2 n 0 , B x 2 n 0 + 1 ) m 1 ( x 2 n 0 , x 2 n 0 + 1 ) = d ( y 2 n 0 + 1 , y 2 n 0 + 2 ) 1 + d ( y 2 n 0 , y 2 n 0 + 1 ) 1 + d ( y 2 n 0 , y 2 n 0 + 1 ) = d 2 n 0 + 1 , m 2 ( x 2 n 0 , x 2 n 0 + 1 ) = d ( A x 2 n 0 , S x 2 n 0 ) 1 + d ( B x 2 n 0 + 1 , T x 2 n 0 + 1 ) 1 + d ( A x 2 n 0 , B x 2 n 0 + 1 ) m 2 ( x 2 n 0 , x 2 n 0 + 1 ) = d ( y 2 n 0 , y 2 n 0 + 1 ) 1 + d ( y 2 n 0 + 1 , y 2 n 0 + 2 ) 1 + d ( y 2 n 0 , y 2 n 0 + 1 ) = 0 , m 3 ( x 2 n 0 , x 2 n 0 + 1 ) = d ( S x 2 n 0 , B x 2 n 0 + 1 ) d ( T x 2 n 0 + 1 , A x 2 n 0 ) 1 + d ( A x 2 n 0 , B x 2 n 0 + 1 ) m 3 ( x 2 n 0 , x 2 n 0 + 1 ) = d ( y 2 n 0 + 1 , y 2 n 0 + 1 ) d ( y 2 n 0 + 2 , y 2 n 0 ) 1 + d ( y 2 n 0 , y 2 n 0 + 1 ) = 0 , m 4 ( x 2 n 0 , x 2 n 0 + 1 ) = max { d ( A x 2 n 0 , B x 2 n 0 + 1 ) , d ( A x 2 n 0 , S x 2 n 0 ) , d ( B x 2 n 0 + 1 , T x 2 n 0 + 1 ) , m 4 ( x 2 n 0 , x 2 n 0 + 1 ) = 1 2 [ d ( S x 2 n 0 , B x 2 n 0 + 1 ) + d ( T x 2 n 0 + 1 , A x 2 n 0 ) ] } m 4 ( x 2 n 0 , x 2 n 0 + 1 ) = max { d ( y 2 n 0 , y 2 n 0 + 1 ) , d ( y 2 n 0 , y 2 n 0 + 1 ) , d ( y 2 n 0 + 1 , y 2 n 0 + 2 ) , m 4 ( x 2 n 0 , x 2 n 0 + 1 ) = 1 2 [ d ( y 2 n 0 + 1 , y 2 n 0 + 1 ) + d ( y 2 n 0 + 2 , y 2 n 0 ) ] } = max { d 2 n 0 , d 2 n 0 + 1 } = d 2 n 0 + 1

and

0 d 2 n 0 + 1 φ ( t ) d t = 0 d ( y 2 n 0 + 1 , y 2 n 0 + 2 ) φ ( t ) d t = 0 d ( S x 2 n 0 , T x 2 n 0 + 1 ) φ ( t ) d t ψ ( max { 0 m i ( x 2 n 0 , x 2 n 0 + 1 ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 d 2 n 0 + 1 φ ( t ) d t , 0 , 0 , 0 d 2 n 0 + 1 φ ( t ) d t } ) < 0 d 2 n 0 + 1 φ ( t ) d t ,

which is a contradiction. Hence d 2 n 0 + 1 =0. It follows that

A x 2 n 0 = y 2 n 0 = y 2 n 0 + 1 =S x 2 n 0 andB x 2 n 0 + 1 = y 2 n 0 + 1 = y 2 n 0 + 2 =T x 2 n 0 + 1 .

Put a= x 2 n 0 and b= x 2 n 0 + 1 . It is easy to see that (2.4) holds and y 2 n 0 is a common fixed point of A, B, S and T.

Case 2. There exists n 0 N satisfying d 2 n 0 1 =0. As in the proof of Case 1, we infer similarly that (2.4) holds for a= x 2 n 0 and b= x 2 n 0 1 , and y 2 n 0 1 is a common fixed point of A, B, S and T.

Case 3. y n y n + 1 for all nN. Now we claim that d 2 n d 2 n 1 for all nN. Suppose that d 2 n > d 2 n 1 for some nN. By virtue of (2.1), (2.2) and (ψ,φ)Ψ×Φ, we arrive at

m 1 ( x 2 n , x 2 n 1 ) = d ( B x 2 n 1 , T x 2 n 1 ) 1 + d ( A x 2 n , S x 2 n ) 1 + d ( A x 2 n , B x 2 n 1 ) m 1 ( x 2 n , x 2 n 1 ) = d ( y 2 n 1 , y 2 n ) 1 + d ( y 2 n , y 2 n + 1 ) 1 + d ( y 2 n , y 2 n 1 ) = d 2 n 1 1 + d 2 n 1 + d 2 n 1 < d 2 n , m 2 ( x 2 n , x 2 n 1 ) = d ( A x 2 n , S x 2 n ) 1 + d ( B x 2 n 1 , T x 2 n 1 ) 1 + d ( A x 2 n , B x 2 n 1 ) m 2 ( x 2 n , x 2 n 1 ) = d ( y 2 n , y 2 n + 1 ) 1 + d ( y 2 n 1 , y 2 n ) 1 + d ( y 2 n , y 2 n 1 ) = d 2 n , m 3 ( x 2 n , x 2 n 1 ) = d ( S x 2 n , B x 2 n 1 ) d ( T x 2 n 1 , A x 2 n ) 1 + d ( A x 2 n , B x 2 n 1 ) m 3 ( x 2 n , x 2 n 1 ) = d ( y 2 n + 1 , y 2 n 1 ) d ( y 2 n , y 2 n ) 1 + d ( y 2 n , y 2 n 1 ) = 0 , m 4 ( x 2 n , x 2 n 1 ) = max { d ( A x 2 n , B x 2 n 1 ) , d ( A x 2 n , S x 2 n ) , d ( B x 2 n 1 , T x 2 n 1 ) , m 4 ( x 2 n , x 2 n 1 ) = 1 2 [ d ( S x 2 n , B x 2 n 1 ) + d ( T x 2 n 1 , A x 2 n ) ] } m 4 ( x 2 n , x 2 n 1 ) = max { d ( y 2 n , y 2 n 1 ) , d ( y 2 n , y 2 n + 1 ) , d ( y 2 n 1 , y 2 n ) , m 4 ( x 2 n , x 2 n 1 ) = 1 2 [ d ( y 2 n + 1 , y 2 n 1 ) + d ( y 2 n , y 2 n ) ] } m 4 ( x 2 n , x 2 n 1 ) = max { d 2 n 1 , d 2 n } = d 2 n
(2.6)

and

0 d 2 n φ ( t ) d t = 0 d ( y 2 n + 1 , y 2 n ) φ ( t ) d t = 0 d ( S x 2 n , T x 2 n 1 ) φ ( t ) d t ψ ( max { 0 m i ( x 2 n , x 2 n 1 ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 d 2 n 1 1 + d 2 n 1 + d 2 n 1 φ ( t ) d t , 0 d 2 n φ ( t ) d t , 0 , 0 d 2 n φ ( t ) d t } ) = ψ ( 0 d 2 n φ ( t ) d t ) < 0 d 2 n φ ( t ) d t ,
(2.7)

which is absurd. Hence d 2 n d 2 n 1 for each nN. As in the proofs of (2.6) and (2.7), we infer similarly that d 2 n + 1 d 2 n for all nN. Consequently, { d n } n N is a nonincreasing positive sequence, which means that there exists a constant r0 with

lim n d n =r.
(2.8)

Suppose that r>0. Making use of (2.1), (2.2), (2.6), (2.8) and (ψ,φ)Ψ×Φ and Lemma 1.4, we get that

0 r φ ( t ) d t = lim sup n 0 d 2 n φ ( t ) d t = lim sup n 0 d ( y 2 n + 1 , y 2 n ) φ ( t ) d t = lim sup n 0 d ( S x 2 n , T x 2 n 1 ) φ ( t ) d t lim sup n ψ ( max { 0 m i ( x 2 n , x 2 n 1 ) φ ( t ) d t : 1 i 4 } ) ψ ( lim sup n max { 0 d 2 n 1 1 + d 2 n 1 + d 2 n 1 φ ( t ) d t , 0 d 2 n φ ( t ) d t , 0 , 0 max { d 2 n 1 , d 2 n } φ ( t ) d t } ) = ψ ( max { 0 r φ ( t ) d t , 0 r φ ( t ) d t , 0 , 0 r φ ( t ) d t } ) = ψ ( 0 r φ ( t ) d t ) < 0 r φ ( t ) d t ,

which is a contradiction. Hence r=0. That is,

lim n d n =0.
(2.9)

In order to prove that { y n } n N is a Cauchy sequence, by (2.9) we need only to prove that { y 2 n } n N is a Cauchy sequence. Suppose that { y 2 n } n N is not a Cauchy sequence. It follows that there exists ε>0 such that for each even integer 2k there are even integers 2m(k), 2n(k) with 2m(k)>2n(k)>2k and

d( y 2 n ( k ) , y 2 m ( k ) )ε.
(2.10)

For every even integer 2k, let 2m(k) be the least even integer exceeding 2n(k) satisfying (2.10). It follows that

d( y 2 n ( k ) , y 2 m ( k ) 2 )<ε,kN.
(2.11)

Note that

d ( y 2 n ( k ) , y 2 m ( k ) ) d ( y 2 n ( k ) , y 2 m ( k ) 2 ) + d 2 m ( k ) 2 + d 2 m ( k ) 1 , k N ; | d ( y 2 n ( k ) + 1 , y 2 m ( k ) ) d ( y 2 n ( k ) , y 2 m ( k ) ) | d 2 n ( k ) , k N ; | d ( y 2 n ( k ) , y 2 m ( k ) 1 ) d ( y 2 n ( k ) , y 2 m ( k ) ) | d 2 m ( k ) 1 , k N ; | d ( y 2 n ( k ) + 1 , y 2 m ( k ) 1 ) d ( y 2 n ( k ) + 1 , y 2 m ( k ) ) | d 2 m ( k ) 1 , k N .
(2.12)

In terms of (2.9)-(2.12), we know that

ε = lim k d ( y 2 n ( k ) , y 2 m ( k ) ) = lim k d ( y 2 n ( k ) + 1 , y 2 m ( k ) ) = lim k d ( y 2 n ( k ) , y 2 m ( k ) 1 ) = lim k d ( y 2 n ( k ) + 1 , y 2 m ( k ) 1 ) .
(2.13)

In light of (2.1), (2.2), (2.9), (2.13), (ψ,φ)Ψ×Φ and Lemma 1.4, we deduce that

m 1 ( x 2 n ( k ) , x 2 m ( k ) 1 ) = d ( B x 2 m ( k ) 1 , T x 2 m ( k ) 1 ) 1 + d ( A x 2 n ( k ) , S x 2 n ( k ) ) 1 + d ( A x 2 n ( k ) , B x 2 m ( k ) 1 ) = d ( y 2 m ( k ) 1 , y 2 m ( k ) ) 1 + d ( y 2 n ( k ) , y 2 n ( k ) + 1 ) 1 + d ( y 2 n ( k ) , y 2 m ( k ) 1 ) 0 as  k , m 2 ( x 2 n ( k ) , x 2 m ( k ) 1 ) = d ( A x 2 n ( k ) , S x 2 n ( k ) ) 1 + d ( B x 2 m ( k ) 1 , T x 2 m ( k ) 1 ) 1 + d ( A x 2 n ( k ) , B x 2 m ( k ) 1 ) = d ( y 2 n ( k ) , y 2 n ( k ) + 1 ) 1 + d ( y 2 m ( k ) 1 , y 2 m ( k ) ) 1 + d ( y 2 n ( k ) , y 2 m ( k ) 1 ) 0 as  k , m 3 ( x 2 n ( k ) , x 2 m ( k ) 1 ) = d ( S x 2 n ( k ) , B x 2 m ( k ) 1 ) d ( T x 2 m ( k ) 1 , A x 2 n ( k ) ) 1 + d ( A x 2 n ( k ) , B x 2 m ( k ) 1 ) = d ( y 2 n ( k ) + 1 , y 2 m ( k ) 1 ) d ( y 2 m ( k ) , y 2 n ( k ) ) 1 + d ( y 2 n ( k ) , y 2 m ( k ) 1 ) ε 2 1 + ε as  k , m 4 ( x 2 n ( k ) , x 2 m ( k ) 1 ) = max { d ( A x 2 n ( k ) , B x 2 m ( k ) 1 ) , d ( A x 2 n ( k ) , S x 2 n ( k ) ) , d ( B x 2 m ( k ) 1 , T x 2 m ( k ) 1 ) , 1 2 [ d ( S x 2 n ( k ) , B x 2 m ( k ) 1 ) + d ( T x 2 m ( k ) 1 , A x 2 n ( k ) ) ] } = max { d ( y 2 n ( k ) , y 2 m ( k ) 1 ) , d ( y 2 n ( k ) , y 2 n ( k ) + 1 ) , d ( y 2 m ( k ) 1 , y 2 m ( k ) ) , 1 2 [ d ( y 2 n ( k ) + 1 , y 2 m ( k ) 1 ) + d ( y 2 m ( k ) , y 2 n ( k ) ) ] } ε as  k

and

0 ε φ ( t ) d t = lim sup k 0 d ( y 2 n ( k ) + 1 , y 2 m ( k ) ) φ ( t ) d t = lim sup k 0 d ( S x 2 n ( k ) , T x 2 m ( k ) 1 ) φ ( t ) d t lim sup k ψ ( max { 0 m i ( x 2 n ( k ) , x 2 m ( k ) 1 ) φ ( t ) d t : 1 i 4 } ) ψ ( lim sup k max { 0 m i ( x 2 n ( k ) , x 2 m ( k ) 1 ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 , 0 , 0 ε 2 1 + ε φ ( t ) d t , 0 ε φ ( t ) d t } ) = ψ ( 0 ε φ ( t ) d t ) < 0 ε φ ( t ) d t ,

which is a contradiction. Therefore { y n } n N is a Cauchy sequence.

Assume that A(X) is complete. Notice that { y 2 n } n N A(X), which implies that { y 2 n } n N converges to a point cA(X). Obviously lim n y n =c. Put a A 1 c. It follows that Aa=c. Suppose that Sac. In view of (2.1)-(2.3), (ψ,φ)Ψ×Φ, Lemma 1.4 and lim n y n =c, we infer that

m 1 ( a , x 2 n 1 ) = d ( B x 2 n 1 , T x 2 n 1 ) 1 + d ( A a , S a ) 1 + d ( A a , B x 2 n 1 ) = d ( y 2 n 1 , y 2 n ) 1 + d ( c , S a ) 1 + d ( c , y 2 n 1 ) 0 as  n , m 2 ( a , x 2 n 1 ) = d ( A a , S a ) 1 + d ( B x 2 n 1 , T x 2 n 1 ) 1 + d ( A a , B x 2 n 1 ) = d ( c , S a ) 1 + d ( y 2 n 1 , y 2 n ) 1 + d ( c , y 2 n 1 ) d ( c , S a ) as  n , m 3 ( a , x 2 n 1 ) = d ( S a , B x 2 n 1 ) d ( T x 2 n 1 , A a ) 1 + d ( A a , B x 2 n 1 ) = d ( S a , y 2 n 1 ) d ( y 2 n , c ) 1 + d ( c , y 2 n 1 ) 0 as  n , m 4 ( a , x 2 n 1 ) = max { d ( A a , B x 2 n 1 ) , d ( A a , S a ) , d ( B x 2 n 1 , T x 2 n 1 ) , 1 2 [ d ( S a , B x 2 n 1 ) + d ( T x 2 n 1 , A a ) ] } = max { d ( c , y 2 n 1 ) , d ( c , S a ) , d ( y 2 n 1 , y 2 n ) , 1 2 [ d ( S a , y 2 n 1 ) + d ( y 2 n , c ) ] } d ( c , S a ) as  n

and

0 d ( S a , c ) φ ( t ) d t = lim sup n 0 d ( S a , y 2 n ) φ ( t ) d t = lim sup n 0 d ( S a , T x 2 n 1 ) φ ( t ) d t lim sup n ψ ( max { 0 m i ( a , x 2 n 1 ) φ ( t ) d t : 1 i 4 } ) ψ ( lim sup n max { 0 m i ( a , x 2 n 1 ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 , 0 d ( c , S a ) φ ( t ) d t , 0 , 0 d ( c , S a ) φ ( t ) d t } ) = ψ ( 0 d ( c , S a ) φ ( t ) d t ) < 0 d ( c , S a ) φ ( t ) d t ,

which is a contradiction. Therefore, Sa=c, which together with (C1) means that cB(X). Put b B 1 c, that is, Bb=c. Suppose that cTb. By means of (2.1), (2.2) and (ψ,φ)Ψ×Φ, we get that

m 1 ( a , b ) = d ( B b , T b ) 1 + d ( A a , S a ) 1 + d ( A a , B b ) = d ( c , T b ) , m 2 ( a , b ) = d ( A a , S a ) 1 + d ( B b , T b ) 1 + d ( A a , B b ) = 0 , m 3 ( a , b ) = d ( S a , B b ) d ( T b , A a ) 1 + d ( A a , B b ) = 0 , m 4 ( a , b ) = max { d ( A a , B b ) , d ( A a , S a ) , d ( B b , T b ) , 1 2 [ d ( S a , B b ) + d ( T b , A a ) ] } = max { 0 , 0 , d ( c , T b ) , 1 2 [ 0 + d ( T b , c ) ] } = d ( c , T b )

and

0 d ( c , T b ) φ ( t ) d t = 0 d ( S a , T b ) φ ( t ) d t ψ ( max { 0 m i ( a , b ) φ ( t ) d t : 1 i 4 } ) = ψ ( max { 0 d ( c , T b ) φ ( t ) d t , 0 , 0 , 0 d ( c , T b ) φ ( t ) d t } ) = ψ ( 0 d ( c , T b ) φ ( t ) d t ) < 0 d ( c , T b ) φ ( t ) d t ,

which is impossible. That is, c=Tb. Hence (2.4) holds.

Assume that T(X) is complete. Notice that { y 2 n } n N T(X), which implies that { y 2 n } n N converges to a point cT(X). Obviously lim n y n =c. Put b 1 T 1 c. It follows that T b 1 =c. Observe that T(X)A(X), which implies that there exists aX with Aa=T b 1 =c. As in the proof of completeness of A(X), we infer that (2.4) holds. Similarly we conclude that (2.4) holds if one of B(X) and S(X) is complete. This completes the proof. □

As in the proof of Theorem 2.1 we have the following result and omit its proof.

Theorem 2.2 Let A, B, S and T be self-mappings of a metric space (X,d) satisfying (C1)-(C3) and

0 d ( S x , T y ) φ(t)dtψ ( 0 m 4 ( x , y ) φ ( t ) d t ) ,x,yX,
(2.14)

where (ψ,φ) is in Ψ×Φ and m 4 is defined by (2.2). Then A, B, S and T have a unique common fixed point in X.

Remark 2.3 Theorems 2.1 and 2.2 extend, improve and unify Theorem 2.1 in [1, 4], Theorem 2 in [13, 16, 17] and Corollary 3 in [17]. The following example reveals that Theorem 2.2 is both an indeed generalization of Theorem 2.1 in [1], Theorem 2 in [13, 17] and Corollary 3 in [17], and different from Theorems 3.1-3.3 in [9].

Example 2.4 Let X=R be endowed with the Euclidean metric d(x,y)=|xy| for all x,yX. Let S,T:XX be defined by

Sx={ 2 , x X { 1 } , 3 2 , x = 1 , Tx=2,xX.

Now we claim that Theorem 2 and Corollary 3 in [17] cannot be used to prove the existence of common fixed points of the mappings S and T in X, and Theorem 2 in [13], Theorem 2.1 in [1] and Theorems 3.1-3.3 in [9] are useless in proving the existence of fixed points of the mapping S in X.

Suppose that there exists (φ,ψ)Φ× Ψ 1 satisfying the condition of Theorem 2 in [17], that is,

0 d ( S x , T y ) φ(t)dtψ ( 0 M ( x , y ) φ ( t ) d t ) ,x,yX,
(2.15)

where

M ( x , y ) = max { d ( x , y ) , d ( x , S x ) , d ( y , T y ) , 1 2 [ d ( y , S x ) + d ( x , T y ) ] } .
(2.16)

Put x 0 =1 and y 0 = 3 2 . It follows from (2.15), (2.16) and (φ,ψ)Φ× Ψ 1 that

M ( x 0 , y 0 ) = max { d ( x 0 , y 0 ) , d ( x 0 , S x 0 ) , d ( y 0 , T y 0 ) , 1 2 [ d ( y 0 , S x 0 ) + d ( x 0 , T y 0 ) ] } = max { d ( 1 , 3 2 ) , d ( 1 , 3 2 ) , d ( 3 2 , 2 ) , 1 2 [ d ( 3 2 , 3 2 ) + d ( 1 , 2 ) ] } = 1 2

and

0 1 2 φ ( t ) d t = 0 d ( S x 0 , T y 0 ) φ ( t ) d t ψ ( 0 M ( x 0 , y 0 ) φ ( t ) d t ) = ψ ( 0 1 2 φ ( t ) d t ) < 0 1 2 φ ( t ) d t ,

which is impossible.

Suppose that there exist φΦ and k[0,1) satisfying the condition of Corollary 3 in [17], that is,

0 d ( S x , T y ) φ(t)dtk 0 n ( x , y ) φ(t)dtfor each distinct x,yX,
(2.17)

where

n(x,y)=max { d ( x , y ) , d ( y , T y ) [ 1 + d ( x , S x ) ] 1 + d ( x , y ) } .
(2.18)

Take x 0 =1 and y 0 = 3 2 . It follows from (2.17), (2.18), k[0,1) and φΦ that

n ( x 0 , y 0 ) = max { d ( x 0 , y 0 ) , d ( y 0 , T y 0 ) [ 1 + d ( x 0 , S x 0 ) ] 1 + d ( x 0 , y 0 ) } = max { d ( 1 , 3 2 ) , d ( 3 2 , 2 ) [ 1 + d ( 1 , 3 2 ) ] 1 + d ( 1 , 3 2 ) } = 1 2

and

0 1 2 φ ( t ) d t = 0 d ( S x 0 , T y 0 ) φ ( t ) d t k 0 n ( x 0 , y 0 ) φ ( t ) d t = k 0 1 2 φ ( t ) d t < 0 1 2 φ ( t ) d t ,

which is a contradiction.

Suppose that there exist φΦ and k[0,1) satisfying the condition of Theorem 2 in [13], that is,

0 d ( S x , S y ) φ ( t ) d t k 0 m ( x , y ) φ ( t ) d t , x , y X ,
(2.19)

where

m ( x , y ) = max { d ( x , y ) , d ( x , S x ) , d ( y , S y ) , 1 2 [ d ( y , S x ) + d ( x , S y ) ] } .
(2.20)

Put x 0 =1 and y 0 = 3 2 . It follows from (2.19), (2.20), k[0,1) and φΦ that

m ( x 0 , y 0 ) = max { d ( x 0 , y 0 ) , d ( x 0 , S x 0 ) , d ( y 0 , S y 0 ) , 1 2 [ d ( y 0 , S x 0 ) + d ( x 0 , S y 0 ) ] } = max { d ( 1 , 3 2 ) , d ( 1 , S 1 ) , d ( 3 2 , S 3 2 ) , 1 2 [ d ( 3 2 , S 1 ) + d ( 1 , S 3 2 ) ] } = max { 1 2 , 1 2 , 1 2 , 1 2 } = 1 2

and

0 1 2 φ(t)dt= 0 d ( S x 0 , S y 0 ) φ(t)dtk 0 m ( x 0 , y 0 ) φ(t)dt=k 0 1 2 φ(t)dt< 0 1 2 φ(t)dt,

which is absurd. Observe that Theorem 2 in [13] generalizes Theorem 2.1 in [1], hence Theorem 2.1 in [1] cannot be used to prove the existence of fixed points of S in X.

Suppose that there exists φΦ satisfying the condition of Theorem 3.1 in [9], that is,

0 d ( S x , S y ) φ(t)dtα ( d ( x , y ) ) 0 d ( x , y ) φ(t)dt,x,yX,
(2.21)

where

α: R + [0,1)and lim sup s t α(s)<1,t>0.
(2.22)

Put x 0 =1 and y 0 = 3 2 . It follows from (2.21), (2.22) and φΦ that

0 1 2 φ ( t ) d t = 0 d ( S x 0 , S y 0 ) φ ( t ) d t α ( d ( x 0 , y 0 ) ) 0 d ( x 0 , y 0 ) φ ( t ) d t = α ( 1 2 ) 0 1 2 φ ( t ) d t < 0 1 2 φ ( t ) d t ,

which is impossible.

Suppose that there exists φΦ satisfying the condition of Theorem 3.2 in [9], that is,

0 d ( S x , S y ) φ ( t ) d t α ( d ( x , y ) ) 0 d ( x , S x ) φ ( t ) d t + β ( d ( x , y ) ) 0 d ( y , S y ) φ ( t ) d t , x , y X ,
(2.23)

where

α , β : R + [ 0 , 1 )  satisfy that  α ( t ) + β ( t ) < 1 , t R + , lim sup s 0 + β ( s ) < 1 and lim sup s t + α ( s ) 1 β ( s ) < 1 , t > 0 .
(2.24)

Put x 0 =1 and y 0 = 3 2 . It follows from (2.23), (2.24) and φΦ that

0 1 2 φ ( t ) d t = 0 d ( S x 0 , S y 0 ) φ ( t ) d t α ( d ( x 0 , y 0 ) ) 0 d ( x 0 , S x 0 ) φ ( t ) d t + β ( d ( x 0 , y 0 ) ) 0 d ( y 0 , S y 0 ) φ ( t ) d t = [ α ( 1 2 ) + β ( 1 2 ) ] 0 1 2 φ ( t ) d t < 0 1 2 φ ( t ) d t ,

which is a contradiction.

Suppose that there exists φΦ satisfying the condition of Theorem 3.3 in [9], that is,

0 d ( S x , S y ) φ ( t ) d t γ ( d ( x , y ) ) ( 0 d ( x , S x ) φ ( t ) d t + 0 d ( y , S y ) φ ( t ) d t ) , x , y X ,
(2.25)

where

γ : R + [ 0 , 1 2 ) and lim sup s t + γ ( s ) 1 γ ( s ) < 1 , t > 0 .
(2.26)

Put x 0 =1 and y 0 = 3 2 . It follows from (2.25), (2.26) and φΦ that

0 1 2 φ ( t ) d t = 0 d ( S x 0 , S y 0 ) φ ( t ) d t γ ( d ( x 0 , y 0 ) ) ( 0 d ( x 0 , S x 0 ) φ ( t ) d t + 0 d ( y 0 , S y 0 ) φ ( t ) d t ) = 2 γ ( 1 2 ) 0 1 2 φ ( t ) d t < 0 1 2 φ ( t ) d t ,

which is impossible.

Next we prove, by using Theorem 2.2, that the mappings A, B, S and T have a unique common fixed point in X, where A,B:XX are defined by

Ax= 1 2 x 2 andBx= 1 4 x 3 ,xX.

Define two functions ψ,φ: R + R + by

φ(t)=6 t 5 +3 t 2 +t,t[0,+)andψ(t)={ t 2 , t [ 0 , 1 2 ] , t 4 + t , t ( 1 2 , + ) .

It is easy to see that (C1), (C2) and (C3) hold. Let x,yX. In order to verify (2.14), we have to consider two possible cases as follows.

Case 1. xX{1}. It is clear that

0 d ( S x , T y ) φ ( t ) d t = 0 ψ ( 0 m 4 ( x , y ) φ ( t ) d t ) .

Case 2. x=1. Note that ψ is nondecreasing on ( 1 2 ,+),

m 4 ( 1 , y ) = max { d ( A 1 , B y ) , d ( A 1 , S 1 ) , d ( B y , T y ) , 1 2 [ d ( S 1 , B y ) + d ( T y , A 1 ) ] } = max { | 1 2 y 3 4 | , | 1 2 3 2 | , | y 3 4 2 | , 1 2 ( | 3 2 y 3 4 | + | 2 1 2 | ) } 1

and

0 d ( S 1 , T y ) φ ( t ) d t = 0 1 2 ( 6 t 5 + 3 t 2 + t ) d t = 17 64 < 5 13 = 5 2 4 + 5 2 = ψ ( 5 2 ) = ψ ( 0 1 ( 6 t 5 + 3 t 2 + t ) d t ) ψ ( 0 m 4 ( 1 , y ) φ ( t ) d t ) .

Hence (2.14) holds. That is, the conditions of Theorem 2.2 are satisfied. Consequently, Theorem 2.2 implies that A, B, S and T have a unique common fixed point 2X.

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Acknowledgements

This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380) and Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2013R1A1A2057665).

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Correspondence to Shin Min Kang.

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Keywords

  • contractive mappings of integral type
  • weakly compatible mappings
  • common fixed point