Open Access

Fixed point theorems for mappings satisfying contractive conditions of integral type

Fixed Point Theory and Applications20132013:267

https://doi.org/10.1186/1687-1812-2013-267

Received: 19 June 2013

Accepted: 28 August 2013

Published: 7 November 2013

Abstract

Two results involving the existence, uniqueness and iterative approximations of fixed points for two contractive mappings of integral type are proved in complete metric spaces. Two nontrivial examples are included.

MSC:54H25.

Keywords

contractive mappings of integral typefixed pointcomplete metric space

1 Introduction

In recent years, there has been increasing interest in the study of fixed points and common fixed points of mappings satisfying contractive conditions of integral type, see, for example, [114] and the references cited therein. Branciari [4] introduced first the contractive mapping of integral type as follows:
0 d ( f x , f y ) φ ( t ) d t c 0 d ( x , y ) φ ( t ) d t , x , y X ,
where c ( 0 , 1 ) is a constant, φ Φ = { φ : φ : R + R + satisfies that φ is Lebesgue integrable, summable on each compact subset of R + and 0 ε φ ( t ) d t > 0 for each ε > 0 } and proved the existence of a fixed point for the mapping in complete metric spaces. Rhoades [10] and Liu et al. [8] extended Branciari’s result and obtained a few fixed point theorems for the contractive mappings of integral type below:
0 d ( f x , f y ) φ ( t ) d t c 0 M ( x , y ) φ ( t ) d t , x , y X
and
0 d ( f x , f y ) φ ( t ) d t α ( d ( x , y ) ) 0 d ( x , y ) φ ( t ) d t , x , y X ,

where c ( 0 , 1 ) is a constant, φ Φ and α : R + [ 0 , 1 ) is a function with lim sup s t α ( s ) < 1 , t > 0 . Mongkolkeha and Kumam [9] proved fixed point and common fixed point theorems for ρ-compatible mapping satisfying a generalized weak contraction of integral type in modular spaces. Sintunavarat and Kumam [11, 12] gave common fixed point theorems for single-valued and multi-valued mappings satisfying strict general contractive conditions of integral type.

Inspired and motivated by the results in [114], in this paper, we introduce two new classes of contractive mappings of integral type in complete metric spaces and study the existence, uniqueness and iterative approximations of fixed points for the mappings. The results obtained in this paper generalize and improve Theorem 2.1 in [4], Theorem 3.1 in [8] and Theorem 2 in [10]. Two nontrivial examples are constructed.

2 Preliminaries

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

Φ 1 = { ϕ : ϕ : R + R + is upper semi-continuous on R + { 0 } , ϕ ( 0 ) = 0 and ϕ ( t ) < t , t > 0 };

Φ 2 = { ϕ : ϕ : R + R + is right upper semi-continuous on R + { 0 } , ϕ ( 0 ) = 0 and ϕ ( t ) < t , t > 0 };

Φ 3 = { ϕ : ϕ : R + R + is continuous, ϕ ( 0 ) = 0 , ϕ ( t ) > 0 , t > 0 and lim n t n = 0 , for each sequence { t n } n N R + with lim n ϕ ( t n ) = 0 };

Φ 4 = { ϕ : ϕ : R + R + is strictly increasing, ϕ ( 0 ) = 0 , continuous at 0 and lim n t n = 0 for each sequence { t n } n N in R + with lim n ϕ ( t n ) = 0 };

Φ 5 = { ϕ : ϕ is in Φ 4 and is left continuous on R + { 0 } };
  1. (a1)

    ( φ , ϕ , ψ ) Φ × Φ 1 × Φ 3 ;

     
  2. (a2)

    ( φ , ϕ , ψ ) Φ × Φ 2 × Φ 4 ;

     
  3. (a3)

    ( φ , ϕ , ψ ) Φ × Φ 2 × Φ 5 .

     
Let T be a mapping from a metric space ( X , d ) into itself, and let ψ : R + R + be a function. Put
M ( x , y ) = max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) , 1 2 [ d ( x , T y ) + d ( y , T x ) ] } , x , y X ,

ψ ( s + ) and ψ ( s ) denote the right and left limits of the function ψ at s R + , respectively.

The following lemmas play important roles in this paper.

Lemma 2.1 [8]

Let φ Φ and { r n } n N be a nonnegative sequence with lim n r n = a . Then
lim n 0 r n φ ( t ) d t = 0 a φ ( t ) d t .

Lemma 2.2 [8]

Let φ Φ and { r n } n N be a nonnegative sequence. Then
lim n 0 r n φ ( t ) d t = 0

if and only if lim n r n = 0 .

3 Fixed point theorems and examples

In this section, we prove two fixed point theorems for two classes of contractive mappings of integral type and display two examples as applications of the theorems.

Theorem 3.1 Let ( X , d ) be a complete metric space, and let T : X X be a mapping satisfying
0 ψ ( d ( T x , T y ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( x , y ) ) φ ( t ) d t ) , x , y X ,
(3.1)

where φ, ϕ and ψ satisfy (a1) or (a2). Then T has a unique fixed point a X and lim n T n x 0 = a for each x 0 X .

Proof Let x 0 be an arbitrary point in X. Put x n = T x n 1 for each n N . Assume that x n 0 = x n 0 1 for some n 0 N . It is easy to see that x n 0 1 is a fixed point of T, and there is nothing to prove. Assume that x n x n 1 for all n N . From (3.1) and one of (a1) and (a2), we obtain that
0 ψ ( d ( x n + 1 , x n ) ) φ ( t ) d t = 0 ψ ( d ( T x n , T x n 1 ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( x n , x n 1 ) ) φ ( t ) d t ) < 0 ψ ( d ( x n , x n 1 ) ) φ ( t ) d t , n N ,
(3.2)
which implies that
0 < ψ ( d ( x n + 1 , x n ) ) < ψ ( d ( x n , x n 1 ) ) , n N .
(3.3)
Note that (3.3) yields that the sequence { ψ ( d ( x n , x n 1 ) ) } n N is positive and strictly decreasing. Thus, there exists a constant c 0 with
lim n ψ ( d ( x n , x n 1 ) ) = c .
(3.4)
Suppose that c > 0 . Taking upper limit in (3.2) and using (3.4), Lemma 2.1 and one of (a1) and (a2), we conclude that
0 c φ ( t ) d t = lim sup n 0 ψ ( d ( x n + 1 , x n ) ) φ ( t ) d t lim sup n ϕ ( 0 ψ ( d ( x n , x n 1 ) ) φ ( t ) d t ) ϕ ( 0 c φ ( t ) d t ) < 0 c φ ( t ) d t ,
which is absurd, and hence c = 0 , that is,
lim n ψ ( d ( x n , x n 1 ) ) = 0 ,
which together with one of (a1) and (a2) guarantees that
lim n d ( x n , x n 1 ) = 0 .
(3.5)
Now, we show that { x n } n N 0 is a Cauchy sequence. Suppose that { x n } n N 0 is not a Cauchy sequence, which means that there is a constant ε > 0 such that for each positive integer k, there are positive integers m ( k ) and n ( k ) with m ( k ) > n ( k ) > k satisfying
d ( x m ( k ) , x n ( k ) ) > ε .
For each positive integer k, let m ( k ) denote the least integer exceeding n ( k ) and satisfying the inequality above. It follows that
d ( x m ( k ) , x n ( k ) ) > ε and d ( x m ( k ) 1 , x n ( k ) ) ε , k N .
(3.6)
Note that
ε < d ( x m ( k ) , x n ( k ) ) d ( x m ( k ) , x m ( k ) 1 ) + d ( x m ( k ) 1 , x n ( k ) 1 ) + d ( x n ( k ) 1 , x n ( k ) ) d ( x m ( k ) , x m ( k ) 1 ) + d ( x m ( k ) 1 , x n ( k ) ) + 2 d ( x n ( k ) , x n ( k ) 1 ) , k N .
(3.7)
Letting k in (3.7) and using (3.5) and (3.6), we conclude that
lim k d ( x m ( k ) , x n ( k ) ) = lim k d ( x m ( k ) 1 , x n ( k ) 1 ) = ε .
(3.8)
In view of (3.1), we deduce that
0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t = 0 ψ ( d ( T x m ( k ) 1 , T x n ( k ) 1 ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) , k N .
(3.9)
Assume that (a1) holds. Taking upper limit in (3.9) and using (3.8) and Lemma 2.1, we get that
0 ψ ( ε ) φ ( t ) d t = lim sup k 0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t lim sup k ϕ ( 0 ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) ϕ ( 0 ψ ( ε ) φ ( t ) d t ) < 0 ψ ( ε ) φ ( t ) d t ,

which is a contradiction.

Assume that (a2) holds. In view of (3.8), there exists K N satisfying
d ( x m ( k ) 1 , x n ( k ) 1 ) > ε 2 , k K .
It follows from the inequality above and ψ Φ 4 that
ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) > ψ ( ε 2 ) > 0 , k K ,
which together with (3.9) and ϕ Φ 2 gives that
0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) < 0 ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t , k K ,
which ensures that
ψ ( d ( x m ( k ) , x n ( k ) ) ) < ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) , k K ,
that is,
d ( x m ( k ) , x n ( k ) ) < d ( x m ( k ) 1 , x n ( k ) 1 ) , k K ,
which together with (3.6) and (3.8) implies that
lim k d ( x m ( k ) , x n ( k ) ) = lim k d ( x m ( k ) 1 , x n ( k ) 1 ) = ε .
Taking upper limit in (3.9) and using Lemma 2.1, (a2) and the equations above, we conclude that
0 ψ ( ε + ) φ ( t ) d t = lim sup k 0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t lim sup k ϕ ( 0 ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) ϕ ( 0 ψ ( ε + ) φ ( t ) d t ) < 0 ψ ( ε + ) φ ( t ) d t ,

which is a contradiction.

Thus, { x n } n N 0 is a Cauchy sequence. Since ( X , d ) is a complete metric space, there exists a point a X such that lim n x n = a . By (3.1), Lemma 2.2 and one of ψ Φ 3 and ψ Φ 4 , we arrive at
0 0 ψ ( d ( x n + 1 , T a ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( x n , a ) ) φ ( t ) d t ) 0 ψ ( d ( x n , a ) ) φ ( t ) d t 0 as  n ,
that is,
lim n 0 ψ ( d ( x n + 1 , T a ) ) φ ( t ) d t = 0 ,
which together with Lemma 2.2 means that
lim n ψ ( d ( x n + 1 , T a ) ) = 0 .
(3.10)
Note that (3.10) and one of ψ Φ 3 and ψ Φ 4 ensure that lim n d ( x n + 1 , T a ) = 0 . Consequently, we conclude immediately that
d ( a , T a ) d ( a , x n + 1 ) + d ( x n + 1 , T a ) 0 as  n ,

which gives that a = T a .

Next, we show that a is a unique fixed point T in X. Suppose that T has another fixed point b X { a } . It follows from (3.1) and one of (a1) and (a2) that
0 < 0 ψ ( d ( a , b ) ) φ ( t ) d t = 0 ψ ( d ( T a , T b ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( a , b ) ) φ ( t ) d t ) < 0 ψ ( d ( a , b ) ) φ ( t ) d t ,

which is a contradiction. This completes the proof. □

Remark 3.2 Theorem 3.1 generalizes Theorem 2.1 in [4] and Theorem 3.1 in [8]. The example below is an application of Theorem 3.1.

Example 3.3 Let X = [ 0 , 1 ] { 3 , 5 } be endowed with the Euclidean metric d = | | . Define T : X X and φ , ϕ , ψ : R + R + by
T x = { x 2 , x [ 0 , 1 ] , 0 , x = 3 , 1 , x = 5 , φ ( t ) = { 1 , t [ 0 , 1 ] , e t , t ( 1 , + ) , ϕ ( t ) = { 2 3 t , t [ 0 , 3 4 ] , 2 t 1 , t ( 3 4 , 1 ) , t 2 1 + t , t [ 1 , + ) , ψ ( t ) = { 1 2 t , t [ 0 , 1 ] , t 2 , t ( 1 , + ) .

It is easy to see that (a2) holds. Put x , y X with x < y . To verify (3.1), we need to consider four possible cases as follows.

Case 1. Let x , y [ 0 , 1 ] . It follows that
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( 1 2 | x y | ) φ ( t ) d t = 1 4 | x y | 1 3 | x y | = ϕ ( 0 1 2 | x y | φ ( t ) d t ) = ϕ ( 0 ψ ( d ( x , y ) ) φ ( t ) d t ) ;
Case 2. Let x [ 0 , 1 ] and y = 3 . Note that | y x | 2 and e | y x | 2 + 1 e > e . It follows that
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( x 2 ) φ ( t ) d t = 1 4 x 1 4 < ( e | y x | 2 + 1 e ) 4 1 + ( e | y x | 2 + 1 e ) 2 = ϕ ( e | y x | 2 + 1 e ) = ϕ ( 0 1 φ ( t ) d t + 1 | y x | 2 φ ( t ) d t ) = ϕ ( 0 ψ ( d ( x , y ) ) φ ( t ) d t ) ;
Case 3. Let x [ 0 , 1 ] and y = 5 . Notice that | y x | 4 . It follows that
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( | 1 x 2 | ) φ ( t ) d t = 1 2 x 4 1 2 < ( e | y x | 2 + 1 e ) 4 1 + ( e | y x | 2 + 1 e ) 2 = ϕ ( e | y x | 2 + 1 e ) = ϕ ( 0 1 φ ( t ) d t + 1 | y x | 2 φ ( t ) d t ) = ϕ ( 0 ψ ( d ( x , y ) ) φ ( t ) d t ) ;
Case 4. Let x = 3 and y = 5 . It follows that
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( 1 ) φ ( t ) d t = 0 1 2 φ ( t ) d t = 1 2 < ( e 4 + 1 e ) 2 1 + e 4 + 1 e = ϕ ( e 4 + 1 e ) = ϕ ( 1 + 1 4 φ ( t ) d t ) = ϕ ( 0 1 φ ( t ) d t + 1 4 φ ( t ) d t ) = ϕ ( 0 ψ ( d ( x , y ) ) φ ( t ) d t ) .

That is, (3.1) holds. Thus, Theorem 3.1 implies that T has a unique fixed point 0 X and lim n T n x 0 = 0 for each x 0 X .

Theorem 3.4 Let ( X , d ) be a complete metric space, and let T : X X be a mapping satisfying
0 ψ ( d ( T x , T y ) ) φ ( t ) d t ϕ ( 0 ψ ( M ( x , y ) ) φ ( t ) d t ) , x , y X ,
(3.11)

where φ, ϕ and ψ satisfy (a1) or (a3). Then T has a unique fixed point a X and lim n T n x 0 = a for each x 0 X .

Proof Let x 0 be an arbitrary point in X. Put x n = T x n 1 for each n N . Assume that x n 0 = x n 0 1 for some n 0 N . It is easy to see that x n 0 1 is a fixed point of T, and there is nothing to prove. Assume that x n x n 1 for all n N . From (3.11) and one of (a1) and (a3), we obtain that
0 ψ ( d ( x n + 1 , x n ) ) φ ( t ) d t = 0 ψ ( d ( T x n , T x n 1 ) ) φ ( t ) d t ϕ ( 0 ψ ( M ( x n , x n 1 ) ) φ ( t ) d t ) < 0 ψ ( M ( x n , x n 1 ) ) φ ( t ) d t , n N ,
(3.12)
where
M ( x n , x n 1 ) = max { d ( x n , x n 1 ) , d ( x n , T x n ) , d ( x n 1 , T x n 1 ) , 1 2 [ d ( x n , T x n 1 ) + d ( x n 1 , T x n ) ] } = max { d ( x n , x n 1 ) , d ( x n , x n + 1 ) , d ( x n 1 , x n ) , 1 2 [ d ( x n , x n ) + d ( x n 1 , x n + 1 ) ] } = max { d ( x n , x n 1 ) , d ( x n , x n + 1 ) } , n N .
(3.13)
Suppose that d ( x n 0 , x n 0 1 ) d ( x n 0 , x n 0 + 1 ) for some n 0 N . It follows from (3.12) and (3.13) that
0 ψ ( d ( x n 0 + 1 , x n 0 ) ) φ ( t ) d t < 0 ψ ( M ( x n 0 , x n 0 1 ) ) φ ( t ) d t = 0 ψ ( d ( x n 0 + 1 , x n 0 ) ) φ ( t ) d t ,
which is a contradiction. Consequently, we deduce that
d ( x n , x n + 1 ) < d ( x n , x n 1 ) and M ( x n , x n 1 ) = d ( x n , x n 1 ) , n N .
(3.14)
In view of (3.12), (3.14) and one of (a1) and (a3), we get that
0 ψ ( d ( x n + 1 , x n ) ) φ ( t ) d t ϕ ( 0 ψ ( M ( x n , x n 1 ) ) φ ( t ) d t ) = ϕ ( 0 ψ ( d ( x n , x n 1 ) ) φ ( t ) d t ) < 0 ψ ( d ( x n , x n 1 ) ) φ ( t ) d t , n N ,
(3.15)
which implies that
ψ ( d ( x n + 1 , x n ) ) < ψ ( d ( x n , x n 1 ) ) , n N ,
(3.16)

which means that there exists a constant c 0 with lim n ψ ( d ( x n , x n 1 ) ) = c .

Now we show that c = 0 . Otherwise, c > 0 . Taking upper limit in (3.15) and using Lemma 2.1 and one of (a1) and (a3), we conclude that
0 c φ ( t ) d t = lim sup n 0 ψ ( d ( x n + 1 , x n ) ) φ ( t ) d t lim sup n ϕ ( 0 ψ ( d ( x n , x n 1 ) ) φ ( t ) d t ) ϕ ( 0 c φ ( t ) d t ) < 0 c φ ( t ) d t ,
which is impossible. Hence c = 0 , that is,
lim n ψ ( d ( x n , x n 1 ) ) = 0 ,
which together with one of (a1) and (a3) gives that
lim n d ( x n , x n 1 ) = 0 .
(3.17)
Next, we claim that { x n } n N 0 is a Cauchy sequence. Suppose that { x n } n N 0 is not a Cauchy sequence, which means that there is a constant ε > 0 such that for each positive integer k, there are positive integers m ( k ) and n ( k ) with m ( k ) > n ( k ) > k such that
d ( x m ( k ) , x n ( k ) ) > ε .
For each positive integer k, let m ( k ) denote the least integer exceeding n ( k ) and satisfying the inequality above. Obviously, (3.6)-(3.8) hold. Note that
M ( x m ( k ) 1 , x n ( k ) 1 ) = max { d ( x m ( k ) 1 , x n ( k ) 1 ) , d ( x m ( k ) 1 , T x m ( k ) 1 ) , d ( x n ( k ) 1 , T x n ( k ) 1 ) , 1 2 [ d ( x m ( k ) 1 , T x n ( k ) 1 ) + d ( x n ( k ) 1 , T x m ( k ) 1 ) ] } = max { d ( x m ( k ) 1 , x n ( k ) 1 ) , d ( x m ( k ) 1 , x m ( k ) ) , d ( x n ( k ) 1 , x n ( k ) ) , 1 2 [ d ( x m ( k ) 1 , x n ( k ) ) + d ( x n ( k ) 1 , x m ( k ) ) ] } max { d ( x m ( k ) 1 , x n ( k ) 1 ) , d ( x m ( k ) 1 , x m ( k ) ) , d ( x n ( k ) 1 , x n ( k ) ) , 1 2 [ d ( x m ( k ) 1 , x n ( k ) 1 ) + 2 d ( x n ( k ) 1 , x n ( k ) ) + d ( x n ( k ) , x m ( k ) ) ] } , k N .
(3.18)
Combining (3.8), (3.17) and (3.18), we infer that
lim k M ( x m ( k ) 1 , x n ( k ) 1 ) = ε .
(3.19)
In light of (3.11), we deduce that
0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t = 0 ψ ( d ( T x m ( k ) 1 , T x n ( k ) 1 ) ) φ ( t ) d t ϕ ( 0 ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) , k N .
(3.20)
Assume that (a1) holds. Taking upper limit in (3.20) and using (3.8), (3.19) and Lemma 2.1, we get that
0 ψ ( ε ) φ ( t ) d t = lim sup k 0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t lim sup k ϕ ( 0 ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) ϕ ( 0 ψ ( ε ) φ ( t ) d t ) < 0 ψ ( ε ) φ ( t ) d t ,

which is a contradiction.

Assume that (a3) holds. Note that (3.19) implies that there exists K N with
M ( x m ( k ) 1 , x n ( k ) 1 ) > ε 2 , k K .
(3.21)
By virtue of (a3) and (3.21), we deduce that
ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) > ψ ( ε 2 ) > 0 , k K .
(3.22)
In terms of (3.20), (3.22) and (a3), we get that
0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t ϕ ( 0 ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) < 0 ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t , k K ,
which yields that
ψ ( d ( x m ( k ) , x n ( k ) ) ) < ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) , k K ,
that is,
d ( x m ( k ) , x n ( k ) ) < M ( x m ( k ) 1 , x n ( k ) 1 ) , k K ,
which together with (3.6), (3.8) and (3.19) implies that
lim k d ( x m ( k ) , x n ( k ) ) = lim k M ( x m ( k ) 1 , x n ( k ) 1 ) = ε .
(3.23)
Taking upper limit in (3.20) and using (3.23), (a3) and Lemma 2.1, we conclude that
0 ψ ( ε + ) φ ( t ) d t = lim sup k 0 ψ ( d ( x m ( k ) , x n ( k ) ) ) φ ( t ) d t lim sup k ϕ ( 0 ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) φ ( t ) d t ) ϕ ( 0 ψ ( ε + ) φ ( t ) d t ) < 0 ψ ( ε + ) φ ( t ) d t ,

which is a contradiction.

Thus, { x n } n N 0 is a Cauchy sequence. It follows from completeness of ( X , d ) that there exists a point a X with lim n x n = a .

Next, we show that a is a fixed point of T in X. Suppose that a T a . Notice that
M ( x n , a ) = max { d ( x n , a ) , d ( x n , T x n ) , d ( a , T a ) , 1 2 [ d ( x n , T a ) + d ( a , T x n ) ] } = max { d ( x n , a ) , d ( x n , x n + 1 ) , d ( a , T a ) , 1 2 [ d ( x n , T a ) + d ( a , x n + 1 ) ] } d ( a , T a ) as  n ,
which guarantees that there exists K 1 N satisfying
M ( x n , a ) = d ( a , T a ) , n K 1 .
(3.24)
Let (a1) hold. In light of (3.11), (3.24) and Lemma 2.1, we infer that
0 ψ ( d ( a , T a ) ) φ ( t ) d t = lim sup n 0 ψ ( d ( x n + 1 , T a ) ) φ ( t ) d t = lim sup n 0 ψ ( d ( T x n , T a ) ) φ ( t ) d t lim sup n ϕ ( 0 ψ ( M ( x n , a ) ) φ ( t ) d t ) = ϕ ( 0 ψ ( d ( a , T a ) ) φ ( t ) d t ) < 0 ψ ( d ( a , T a ) ) φ ( t ) d t ,

which is a contradiction.

Let (a3) hold. In view of (3.11) and (3.24), we deduce that
0 ψ ( d ( x n + 1 , T a ) ) φ ( t ) d t = 0 ψ ( d ( T x n , T a ) ) φ ( t ) d t ϕ ( 0 ψ ( M ( x n , a ) ) φ ( t ) d t ) = ϕ ( 0 ψ ( d ( a , T a ) ) φ ( t ) d t ) < 0 ψ ( d ( a , T a ) ) φ ( t ) d t , n K 1 ,
(3.25)
which yields that
ψ ( d ( x n + 1 , T a ) ) < ψ ( d ( a , T a ) ) , n K 1 ,
that is,
d ( x n + 1 , T a ) < d ( a , T a ) , n K 1 ,
which together with (3.25) and ( φ , ϕ , ψ ) Φ × Φ 2 × Φ 5 means that
0 ψ ( d ( a , T a ) ) φ ( t ) d t = 0 ψ ( d ( a , T a ) ) φ ( t ) d t = lim sup n 0 ψ ( d ( x n + 1 , T a ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( a , T a ) ) φ ( t ) d t ) < 0 ψ ( d ( a , T a ) ) φ ( t ) d t ,

which is a contradiction.

Hence T has a fixed point a X . Finally, we show that a is a unique fixed point of T in X. Suppose that T has another fixed point b X { a } . It follows from (3.11) and one of (a1) and (a3) that
0 < 0 ψ ( d ( a , b ) ) φ ( t ) d t = 0 ψ ( d ( T a , T b ) ) φ ( t ) d t ϕ ( 0 ψ ( d ( a , b ) ) φ ( t ) d t ) < 0 ψ ( d ( a , b ) ) φ ( t ) d t ,

which is a contradiction. This completes the proof. □

Remark 3.5 Theorem 3.4 extends Theorem 2 in [10]. The following example is an application of Theorem 3.4.

Example 3.6 Let X = [ 0 , 1 ] { 2 n : n N } be endowed with the Euclidean metric d = | | . Define T : X X and φ , ϕ , ψ : R + R + by
T x = { x 2 , x [ 0 , 1 ] , 1 x , x { 2 n : n N } , φ ( t ) = 2 t , t R + , ϕ ( t ) = { t 2 , t [ 0 , 1 ] , t 2 1 + t , t ( 1 , + ) , ψ ( t ) = { t 2 , t [ 0 , 1 ] , t 2 , t ( 1 , + ) .

Clearly, (a1) holds, ϕ and ψ are strictly increasing in R + . Put x , y X with x < y . To prove (3.11), we need to consider three possible cases as follows.

Case 1. Let x , y [ 0 , 1 ] . It follows that
M ( x , y ) = max { | x y | , | x 1 2 x | , | y 1 2 y | , 1 2 ( | x 1 2 y | + | y 1 2 x | ) } ( 0 , 1 ]
and
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( 1 2 | x y | ) φ ( t ) d t = 1 16 ( x y ) 2 1 8 ( x y ) 2 = ϕ ( 1 4 ( x y ) 2 ) = ϕ ( 0 ψ ( d ( x , y ) ) φ ( t ) d t ) ϕ ( 0 ψ ( M ( x , y ) ) φ ( t ) d t ) .
Case 2. Let x , y { 2 n : n N } . Notice that
M ( x , y ) = max { | x y | , | x 1 x | , | y 1 y | , 1 2 ( | x 1 y | + | y 1 x | ) } > 1 .
Suppose that M ( x , y ) ( 1 , 4 ] . It follows that
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( | 1 x 1 y | ) φ ( t ) d t = 0 1 2 | 1 x 1 y | φ ( t ) d t = 1 4 ( 1 x 1 y ) 2 < 1 16 < 1 8 M ( x , y ) = ϕ ( 1 4 M ( x , y ) ) = ϕ ( 0 1 2 M ( x , y ) φ ( t ) d t ) = ϕ ( 0 ψ ( M ( x , y ) ) φ ( t ) d t ) .
Suppose that M ( x , y ) > 4 . It follows that
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( | 1 x 1 y | ) φ ( t ) d t = 0 1 2 | 1 x 1 y | φ ( t ) d t = 1 4 ( 1 x 1 y ) 2 < 1 16 < M 2 ( x , y ) 16 + 4 M ( x , y ) = 1 16 M 2 ( x , y ) 1 + 1 4 M ( x , y ) = ϕ ( 1 4 M ( x , y ) ) = ϕ ( 0 1 2 M ( x , y ) φ ( t ) d t ) = ϕ ( 0 ψ ( M ( x , y ) ) φ ( t ) d t ) .
Case 3. Let x [ 0 , 1 ] and y { 2 n : n N } . It follows that
M ( x , y ) = max { | x y | , | x x 2 | , | y 1 y | , 1 2 ( | x 1 y | + | y x 2 | ) } | y 1 y | > 1
and
0 ψ ( d ( T x , T y ) ) φ ( t ) d t = 0 ψ ( | x 2 1 y | ) φ ( t ) d t = 0 1 2 | x 2 1 y | φ ( t ) d t = 1 4 ( x 2 1 y ) 2 1 16 < max { 1 8 M ( x , y ) , M 2 ( x , y ) 16 + 4 M ( x , y ) } = ϕ ( 1 4 M ( x , y ) ) = ϕ ( 0 ψ ( M ( x , y ) ) φ ( t ) d t ) ,

that is, (3.11) holds. Thus, Theorem 3.4 implies that T has a unique fixed point 0 X and lim n T n x 0 = 0 for each x 0 X .

Declarations

Acknowledgements

The authors would like to thank the referees for useful comments and suggestions. This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380).

Authors’ Affiliations

(1)
Department of Mathematics, Liaoning Normal University
(2)
Department of Mathematics and RINS, Gyeongsang National University

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© Liu et al.; licensee Springer. 2013

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