Open Access

Fixed point theorems of contractive mappings of integral type

Fixed Point Theory and Applications20132013:300

https://doi.org/10.1186/1687-1812-2013-300

Received: 4 July 2013

Accepted: 22 October 2013

Published: 19 November 2013

Abstract

Three fixed point theorems for three general classes of contractive mappings of integral type in complete metric spaces are proved. Three examples are included.

MSC:54H25.

Keywords

contractive mappings of integral type fixed point theorems complete metric space

1 Introduction

Branciari [1] was the first to study the existence of fixed points for the contractive mapping of integral type. He established a nice integral version of the Banach contraction principle and proved the following fixed point theorem.

Theorem 1.1 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying
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 and φ Φ 1 . Then f has a unique fixed point a X such that lim n f n x = a for each x X .

Afterwards, many authors continued the study of Branciari and obtained many fixed point theorems for several classes of contractive mappings of integral type; see, e.g., [18] and the references therein. In particular, in 2011, Liu et al. [5] extended the result of Branciari [1] and deduced the following fixed point theorems.

Theorem 1.2 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying
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 φ Φ 1 and α : R + [ 0 , 1 ) is a function with
lim sup s t α ( s ) < 1 , t > 0 .

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

Theorem 1.3 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying
0 d ( f x , f y ) φ ( t ) d t α ( d ( x , y ) ) 0 d ( x , f x ) φ ( t ) d t + β ( d ( x , y ) ) 0 d ( y , f y ) φ ( t ) d t , x , y X ,
where φ Φ 1 and α , β : R + [ 0 , 1 ) are two functions with
α ( t ) + β ( t ) < 1 , t R + , lim sup s 0 + β ( s ) < 1 , lim sup s t + α ( s ) 1 β ( s ) < 1 , t > 0 .

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

In 2008, Dutta and Choudhuty [9] proved the following result.

Theorem 1.4 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying
ψ ( d ( f x , f y ) ) ψ ( d ( x , y ) ) φ ( d ( x , y ) ) , x , y X ,

where ψ , φ : R + R + are both continuous and monotone nondecreasing functions with ψ ( t ) = φ ( t ) = 0 if and only if t = 0 . Then f has a unique fixed point a X such that lim n f n x = a for each x X .

However, to the best of our knowledge, no one studied the following contractive mappings of integral type:
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) ψ ( 0 d ( x , y ) φ ( t ) d t ) ϕ ( 0 d ( x , y ) φ ( t ) d t ) , x , y X ,
(1.1)
where ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 ;
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) α ( d ( x , y ) ) ψ ( 0 d ( x , y ) φ ( t ) d t ) , x , y X ,
(1.2)
where ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 ;
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) + β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) , x , y X ,
(1.3)

where ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 and ( α , β ) Φ 6 .

It is clear that the above contractive mappings of integral type include these mappings in Theorems 1.1-1.4 as special cases. The purpose of this paper is to investigate the existence of fixed points for contractive mappings (1.1)-(1.3) of integral type. Under certain conditions, we prove the existence, uniqueness and iterative approximations of fixed points for contractive mappings (1.1)-(1.3) of integral type in complete metric spaces. Three examples with uncountably many points are constructed.

2 Preliminaries

Throughout this paper, we assume that R + = [ 0 , + ) , N 0 = N { 0 } , denotes the set of all positive integers, ( X , d ) is a metric space, f : X X is a self-mapping and
d n = d ( f n x , f n + 1 x ) , ( n , x ) N 0 × X ,

Φ 1 = { φ : φ : R + R + is Lebesgue integrable, summable on each compact subset of R + and 0 ε φ ( t ) d t > 0 for each ε > 0 };

Φ 2 = { φ : φ : R + R + satisfies that lim inf n φ ( a n ) > 0 lim inf n a n > 0 for each { a n } n N R + };

Φ 3 = { φ : φ : R + R + is nondecreasing continuous and φ ( t ) = 0 t = 0 };

Φ 4 = { φ : φ : R + R + satisfies that φ ( 0 ) = 0 };

Φ 5 = { φ : φ : R + [ 0 , 1 ) satisfies that lim sup s t φ ( s ) < 1 for each t > 0 };

Φ 6 = { ( α , β ) : α , β : R + [ 0 , 1 ) satisfy that lim sup s 0 + β ( s ) < 1 , lim sup s t + α ( s ) 1 β ( s ) < 1 and α ( t ) + β ( t ) < 1 for each t > 0 }.

The following lemmas play important roles in this paper.

Lemma 2.1 ([5])

Let φ Φ 1 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 ([5])

Let φ Φ 1 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 .

Lemma 2.3 Let φ Φ 2 . Then φ ( t ) > 0 if and only if t > 0 .

Proof Let t > 0 . Put a n = t for each n N . It is easy to see that t = lim inf n a n > 0 , which together with φ Φ 2 ensures that
φ ( t ) = lim inf n φ ( a n ) > 0 .
Conversely, suppose that φ ( t ) > 0 for some t R + . Set a n = t for each n N . It is clear that φ ( t ) = lim inf n φ ( a n ) > 0 , which together with φ Φ 2 guarantees that
t = lim inf n a n > 0 .

This completes the proof. □

3 Main results

In this section we show the existence, uniqueness and iterative approximations of fixed points for contractive mappings (1.1)-(1.3) of integral type, respectively.

Theorem 3.1 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying (1.1). Then f has a unique fixed point a X such that lim n f n x = a for each x X .

Proof Let x be an arbitrary point in X. Firstly, we show that
d n d n 1 , n N .
(3.1)
Suppose that (3.1) does not hold. It follows that there exists some n 0 N satisfying
d n 0 > d n 0 1 .
(3.2)
Note that (3.2) and φ Φ 1 imply that
0 d n 0 φ ( t ) d t > 0 .
(3.3)
Using (1.1), (3.2) and ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 , we conclude immediately that
ψ ( 0 d n 0 1 φ ( t ) d t ) ψ ( 0 d n 0 φ ( t ) d t ) = ψ ( 0 d ( f n 0 x , f n 0 + 1 x ) φ ( t ) d t ) ψ ( 0 d ( f n 0 1 x , f n 0 x ) φ ( t ) d t ) ϕ ( 0 d ( f n 0 1 x , f n 0 x ) φ ( t ) d t ) = ψ ( 0 d n 0 1 φ ( t ) d t ) ϕ ( 0 d n 0 1 φ ( t ) d t ) ψ ( 0 d n 0 1 φ ( t ) d t ) ,
which yields that
ψ ( 0 d n 0 φ ( t ) d t ) = ψ ( 0 d n 0 1 φ ( t ) d t )
(3.4)
and
ϕ ( 0 d n 0 1 φ ( t ) d t ) = 0 .
(3.5)
Combining (3.5) and Lemma 2.3, we get that
0 d n 0 1 φ ( t ) d t = 0 ,
which together with ψ Φ 3 and (3.4) means that
ψ ( 0 d n 0 φ ( t ) d t ) = ψ ( 0 d n 0 1 φ ( t ) d t ) = ψ ( 0 ) = 0 ,
that is,
0 d n 0 φ ( t ) d t = 0 ,

which contradicts (3.3). Hence (3.1) holds.

Secondly, we show that
lim n d n = 0 .
(3.6)
In view of (3.1), we deduce that the nonnegative sequence { d n } n N 0 is nonincreasing, which means that there exists a constant c with lim n d n = c 0 . Suppose that c > 0 . It follows from (1.1) that
ψ ( 0 d n φ ( t ) d t ) = ψ ( 0 d ( f n x , f n + 1 x ) φ ( t ) d t ) ψ ( 0 d ( f n x , f n 1 x ) φ ( t ) d t ) ϕ ( 0 d ( f n x , f n 1 x ) φ ( t ) d t ) = ψ ( 0 d n 1 φ ( t ) d t ) ϕ ( 0 d n 1 φ ( t ) d t ) , n N .
(3.7)
Taking upper limit in (3.7) and using Lemma 2.1 and ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 , we conclude that
ψ ( 0 c φ ( t ) d t ) = lim sup n ψ ( 0 d n φ ( t ) d t ) lim sup n [ ψ ( 0 d n 1 φ ( t ) d t ) ϕ ( 0 d n 1 φ ( t ) d t ) ] lim sup n ψ ( 0 d n 1 φ ( t ) d t ) lim inf n ϕ ( 0 d n 1 φ ( t ) d t ) = ψ ( 0 c φ ( t ) d t ) lim inf n ϕ ( 0 d n 1 φ ( t ) d t ) < ψ ( 0 c φ ( t ) d t ) ,

which is a contradiction. Hence c = 0 .

Thirdly, we show that { f n x } n N is a Cauchy sequence. Suppose that { f n x } n N 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 ( f m ( k ) x , f n ( k ) x ) > ε .
(3.8)
For each positive integer k, let m ( k ) denote the least integer exceeding n ( k ) and satisfying (3.8). It follows that
d ( f m ( k ) x , f n ( k ) x ) > ε and d ( f m ( k ) 1 x , f n ( k ) x ) ε , k N .
(3.9)
Note that
d ( f m ( k ) x , f n ( k ) x ) d ( f n ( k ) x , f m ( k ) 1 x ) + d m ( k ) 1 , k N ; | d ( f m ( k ) x , f n ( k ) + 1 x ) d ( f m ( k ) x , f n ( k ) x ) | d n ( k ) , k N ; | d ( f m ( k ) + 1 x , f n ( k ) + 1 x ) d ( f m ( k ) x , f n ( k ) + 1 x ) | d m ( k ) , k N ; | d ( f m ( k ) + 1 x , f n ( k ) + 1 x ) d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) | d n ( k ) + 1 , k N .
(3.10)
In light of (3.9) and (3.10), we get that
ε = lim k d ( f n ( k ) x , f m ( k ) x ) = lim k d ( f m ( k ) x , f n ( k ) + 1 x ) = lim k d ( f m ( k ) + 1 x , f n ( k ) + 1 x ) = lim k d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) .
(3.11)
In view of (1.1), we deduce that
ψ ( 0 d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) φ ( t ) d t ) ψ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) ϕ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) , k N .
(3.12)
Taking upper limit in (3.12) and using (3.11), ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 and Lemma 2.1, we deduce that
ψ ( 0 ε φ ( t ) d t ) = lim sup k ψ ( 0 d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) φ ( t ) d t ) lim sup k [ ψ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) ϕ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) ] lim sup k ψ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) lim inf k ϕ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) = ψ ( 0 ε φ ( t ) d t ) lim inf k ϕ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) < ψ ( 0 ε φ ( t ) d t ) ,

which is impossible. Thus { f n x } n N is a Cauchy sequence.

Since ( X , d ) is complete, it follows that there exists a point a X satisfying lim n f n x = a . By virtue of (1.1), we infer that
ψ ( 0 d ( f n + 1 x , f a ) φ ( t ) d t ) ψ ( 0 d ( f n x , a ) φ ( t ) d t ) ϕ ( 0 d ( f n x , a ) φ ( t ) d t ) , n N ,
which together with ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 and Lemmas 2.1 and 2.2 gives that
ψ ( 0 d ( a , f a ) φ ( t ) d t ) = lim sup n ψ ( 0 d ( f n + 1 x , f a ) φ ( t ) d t ) lim sup n [ ψ ( 0 d ( f n x , a ) φ ( t ) d t ) ϕ ( 0 d ( f n x , a ) φ ( t ) d t ) ] lim sup n ψ ( 0 d ( f n x , a ) φ ( t ) d t ) lim inf n ϕ ( 0 d ( f n x , a ) φ ( t ) d t ) = ψ ( 0 ) 0 = 0 ,
which together with ψ Φ 3 yields that
0 d ( a , f a ) φ ( t ) d t = 0 ,

that is, a = f a .

Finally, we show that a is a unique fixed point of f in X. Suppose that f has another fixed point b X { a } . It follows from (1.1) and ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 that
ψ ( 0 d ( a , b ) φ ( t ) d t ) = ψ ( 0 d ( f a , f b ) φ ( t ) d t ) ψ ( 0 d ( a , b ) φ ( t ) d t ) ϕ ( 0 d ( a , b ) φ ( t ) d t ) < ψ ( 0 d ( a , b ) φ ( t ) d t ) ,

which is a contradiction. This completes the proof. □

Theorem 3.2 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying (1.2). Then f has a unique fixed point a X such that lim n f n x = a for each x X .

Proof Let x be an arbitrary point in X. Suppose that (3.2) holds for some n 0 N . Using (1.2), (3.2) and ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 , we get that
ψ ( 0 d n 0 φ ( t ) d t ) > 0
and
ψ ( 0 d n 0 1 φ ( t ) d t ) ψ ( 0 d n 0 φ ( t ) d t ) = ψ ( 0 d ( f n 0 x , f n 0 + 1 x ) φ ( t ) d t ) α ( d ( f n 0 1 x , f n 0 x ) ) ψ ( 0 d ( f n 0 1 x , f n 0 x ) φ ( t ) d t ) = α ( d n 0 1 ) ψ ( 0 d n 0 1 φ ( t ) d t ) < ψ ( 0 d n 0 1 φ ( t ) d t ) ,
which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) is true. Notice that the nonnegative sequence { d n } n N 0 is nonincreasing, which implies that there exists a constant c 0 with lim n d n = c . Suppose that c > 0 . In light of (1.2), we infer that
ψ ( 0 d n φ ( t ) d t ) = ψ ( 0 d ( f n x , f n + 1 x ) φ ( t ) d t ) α ( d ( f n 1 x , f n x ) ) ψ ( 0 d ( f n 1 x , f n x ) φ ( t ) d t ) = α ( d n 1 ) ψ ( 0 d n 1 φ ( t ) d t ) , n N .
(3.13)
Taking upper limit in (3.13) and using Lemma 2.1 and ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 , we know that
ψ ( 0 c φ ( t ) d t ) = lim sup n ψ ( 0 d n φ ( t ) d t ) lim sup n [ α ( d n 1 ) ψ ( 0 d n 1 φ ( t ) d t ) ] lim sup n α ( d n 1 ) lim sup n ψ ( 0 d n 1 φ ( t ) d t ) < ψ ( 0 c φ ( t ) d t ) ,

which is a contradiction, and hence c = 0 , that is, (3.6) holds.

Now we show that { f n x } n N is a Cauchy sequence. Suppose that { f n x } n N is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist ε > 0 and { m ( k ) , n ( k ) : k N } N with m ( k ) > n ( k ) > k for each k N satisfying (3.8)-(3.11). By means of (1.2), (3.11), Lemma 2.1 and ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 , we get that
ψ ( 0 ε φ ( t ) d t ) = lim sup k ψ ( 0 d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) φ ( t ) d t ) = lim sup k [ α ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ψ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) ] lim sup k α ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) lim sup k ψ ( 0 d ( f m ( k ) x , f n ( k ) + 1 x ) φ ( t ) d t ) < ψ ( 0 ε φ ( t ) d t ) ,

which is a contradiction. Hence { f n x } n N is a Cauchy sequence.

It follows from completeness of ( X , d ) that there exists a X with lim n f n x = a . In view of (1.2), we have
ψ ( 0 d ( f n + 1 x , f a ) φ ( t ) d t ) α ( d ( f n x , a ) ) ψ ( 0 d ( f n x , a ) φ ( t ) d t ) , n N 0 .
(3.14)
Taking upper limit in (3.14) and making use of ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 and Lemmas 2.1 and 2.2, we get that
ψ ( 0 d ( a , f a ) φ ( t ) d t ) = lim sup n ψ ( 0 d ( f n + 1 x , f a ) φ ( t ) d t ) lim sup n [ α ( d ( f n x , a ) ) ψ ( 0 d ( f n x , a ) φ ( t ) d t ) ] lim sup n α ( d ( f n x , a ) ) lim sup n ψ ( 0 d ( f n x , a ) φ ( t ) d t ) = 0 ,
which means that
ψ ( 0 d ( a , f a ) φ ( t ) d t ) = 0 ,

that is, f a = a .

Next we prove that a is a unique fixed point of f in X. Suppose that f has another fixed point b X { a } . It follows from (1.2) and ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 that
ψ ( 0 d ( a , b ) φ ( t ) d t ) = ψ ( 0 d ( f a , f b ) φ ( t ) d t ) α ( d ( a , b ) ) ψ ( 0 d ( a , b ) φ ( t ) d t ) < ψ ( 0 d ( a , b ) φ ( t ) d t ) ,

which is a contradiction. This completes the proof. □

Theorem 3.3 Let f be a mapping from a complete metric space ( X , d ) into itself satisfying (1.3) and
ϕ ( t ) ψ ( t ) , t R + .
(3.15)

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

Proof Let x be an arbitrary point in X. If there exists n 0 N 0 satisfying d n 0 = 0 , it is clear that f n 0 x is a fixed point of f and lim n f n x = f n 0 x . Now we assume that d n 0 for all n N 0 . Suppose that (3.2) holds for some n 0 N . It follows from (1.3) that
ψ ( 0 d n 0 φ ( t ) d t ) = ψ ( 0 d ( f n 0 x , f n 0 + 1 x ) φ ( t ) d t ) α ( d ( f n 0 1 x , f n 0 x ) ) ϕ ( 0 d ( f n 0 1 x , f n 0 x ) φ ( t ) d t ) + β ( d ( f n 0 1 x , f n 0 x ) ) ψ ( 0 d ( f n 0 x , f n 0 + 1 x ) φ ( t ) d t ) = α ( d n 0 1 ) ϕ ( 0 d n 0 1 φ ( t ) d t ) + β ( d n 0 1 ) ψ ( 0 d n 0 φ ( t ) d t ) ,
which together with (3.2), (3.15), ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 and ( α , β ) Φ 6 implies that
0 < ψ ( 0 d n 0 1 φ ( t ) d t ) ψ ( 0 d n 0 φ ( t ) d t ) α ( d n 0 1 ) 1 β ( d n 0 1 ) ϕ ( 0 d n 0 1 φ ( t ) d t ) α ( d n 0 1 ) 1 β ( d n 0 1 ) ψ ( 0 d n 0 1 φ ( t ) d t ) < ψ ( 0 d n 0 1 φ ( t ) d t ) ,

which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) holds.

Next we show that lim n d n = 0 . Note that the nonnegative sequence { d n } n N is nonincreasing, which implies that there exists a constant c 0 with lim n d n = c . Suppose that c > 0 . It follows from (1.3) that
ψ ( 0 d n φ ( t ) d t ) = ψ ( 0 d ( f n x , f n + 1 x ) φ ( t ) d t ) α ( d ( f n 1 x , f n x ) ) ϕ ( 0 d ( f n 1 x , f n x ) φ ( t ) d t ) + β ( d ( f n 1 x , f n x ) ) ψ ( 0 d ( f n x , f n + 1 x ) φ ( t ) d t ) = α ( d n 1 ) ϕ ( 0 d n 1 φ ( t ) d t ) + β ( d n 1 ) ψ ( 0 d n φ ( t ) d t ) , n N ,
which means that
ψ ( 0 d n φ ( t ) d t ) α ( d n 1 ) 1 β ( d n 1 ) ϕ ( 0 d n 1 φ ( t ) d t ) , n N .
(3.16)
Taking upper limit in (3.16) and using (3.15), ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 , ( α , β ) Φ 6 and Lemma 2.1, we arrive at
ψ ( 0 c φ ( t ) d t ) = lim sup n ψ ( 0 d n φ ( t ) d t ) lim sup n [ α ( d n 1 ) 1 β ( d n 1 ) ϕ ( 0 d n 1 φ ( t ) d t ) ] lim sup n α ( d n 1 ) 1 β ( d n 1 ) lim sup n ψ ( 0 d n 1 φ ( t ) d t ) lim sup s c + α ( s ) 1 β ( s ) ψ ( 0 c φ ( t ) d t ) < ψ ( 0 c φ ( t ) d t ) ,

which is impossible. Therefore c = 0 , that is, lim n d n = 0 .

Next we show that { f n x } n N is a Cauchy sequence. Suppose that { f n x } n N is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist ε > 0 and { m ( k ) , n ( k ) : k N } N with m ( k ) > n ( k ) > k for each k N satisfying (3.8)-(3.11). By means of (3.12), we deduce that
ψ ( 0 d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) φ ( t ) d t ) α ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ϕ ( 0 d ( f m ( k ) x , f m ( k ) + 1 x ) φ ( t ) d t ) + β ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ψ ( 0 d ( f n ( k ) + 1 x , f n ( k ) + 2 x ) φ ( t ) d t ) = α ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ϕ ( 0 d m ( k ) φ ( t ) d t ) + β ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ψ ( 0 d n ( k ) φ ( t ) d t ) , k N .
(3.17)
Taking upper limit in (3.17) and making use of (1.3), (3.11), Lemma 2.1, ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 and ( α , β ) Φ 6 , we deduce that
0 < ψ ( 0 ε φ ( t ) d t ) = lim sup k ψ ( 0 d ( f m ( k ) + 1 x , f n ( k ) + 2 x ) φ ( t ) d t ) lim sup k [ α ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ϕ ( 0 d m ( k ) φ ( t ) d t ) + β ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) ψ ( 0 d n ( k ) φ ( t ) d t ) ] lim sup k α ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) lim sup k ψ ( 0 d m ( k ) φ ( t ) d t ) + lim sup k β ( d ( f m ( k ) x , f n ( k ) + 1 x ) ) lim sup k ψ ( 0 d n ( k ) φ ( t ) d t ) lim sup s ε α ( s ) ψ ( 0 0 φ ( t ) d t ) + lim sup s ε β ( s ) ψ ( 0 0 φ ( t ) d t ) = 0 ,

which is a contradiction. Hence { f n x } n N is a Cauchy sequence.

Completeness of ( X , d ) implies that there exists a point a X such that lim n f n x = a . In view of (1.3), ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 , ( α , β ) Φ 6 and Lemma 2.1, we infer that
ψ ( 0 d ( a , f a ) φ ( t ) d t ) = lim sup n ψ ( 0 d ( f n + 1 x , f a ) φ ( t ) d t ) lim sup n [ α ( d ( f n x , a ) ) ϕ ( 0 d ( f n x , f n + 1 x ) φ ( t ) d t ) + β ( d ( f n x , a ) ) ψ ( 0 d ( a , f a ) φ ( t ) d t ) ] lim sup n α ( d ( f n x , a ) ) lim sup n ψ ( 0 d n φ ( t ) d t ) + lim sup n β ( d ( f n x , a ) ) ψ ( 0 d ( a , f a ) φ ( t ) d t ) lim sup s 0 + β ( s ) ψ ( 0 d ( a , f a ) φ ( t ) d t ) ,
which together with ( α , β ) Φ 6 yields that
ψ ( 0 d ( a , f a ) φ ( t ) d t ) = 0 ,

which gives that d ( f a , a ) = 0 , that is, f a = a .

Finally, we prove that a is a unique fixed point of f in X. Suppose that f has another fixed point b X { a } . It follows from (1.3) and ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 and ( α , β ) Φ 6 that
0 ψ ( 0 d ( f a , f b ) φ ( t ) d t ) α ( d ( a , b ) ) ϕ ( 0 d ( a , f a ) φ ( t ) d t ) + β ( d ( a , b ) ) ψ ( 0 d ( b , f b ) φ ( t ) d t ) = 0 ,

which is a contradiction. This completes the proof. □

4 Three examples

Now we construct three examples to explain Theorems 3.1-3.3.

Example 4.1 Let X = [ 0 , 1 2 ] { 1 } { 3 } be endowed with the Euclidean metric d = | | . Assume that f : X X and φ , ϕ , ψ : R + R + are defined by
f ( x ) = { x 2 , x [ 0 , 1 2 ] , 0 , x = 1 , 1 , x = 3 , φ ( t ) = { t 2 , t [ 0 , 1 ] , 1 , t ( 1 , + ) , ϕ ( t ) = { t 2 4 , t [ 0 , 1 ] , t 2 8 , t ( 1 , + ) , ψ ( t ) = { t , t [ 0 , 1 ] , t 2 + 1 2 , t ( 1 , + ) .

Clearly, ( X , d ) is a complete metric and ( φ , ϕ , ψ ) Φ 1 × Φ 2 × Φ 3 . Let x , y X with x < y . In order to verify (1.1), we have to consider the following four cases.

Case 1. Let x , y [ 0 , 1 2 ] . Note that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 1 2 | x y | φ ( t ) d t ) = ψ ( | x y | 2 16 ) = | x y | 2 16 | x y | 2 4 | x y | 4 16 = ψ ( | x y | 2 4 ) ϕ ( | x y | 2 4 ) = ψ ( 0 | x y | φ ( t ) d t ) ϕ ( 0 | x y | φ ( t ) d t ) = ψ ( 0 d ( x , y ) φ ( t ) d t ) ϕ ( 0 d ( x , y ) φ ( t ) d t ) .
Case 2. Let x [ 0 , 1 2 ] and y = 1 . It follows that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 x 2 φ ( t ) d t ) = ψ ( x 2 16 ) = x 2 16 ( 1 x ) 2 4 ( 1 x ) 4 16 = ψ ( ( 1 x ) 2 4 ) ϕ ( ( 1 x ) 2 4 ) = ψ ( 0 | x 1 | φ ( t ) d t ) ϕ ( 0 | x 1 | φ ( t ) d t ) = ψ ( 0 d ( x , y ) φ ( t ) d t ) ϕ ( 0 d ( x , y ) φ ( t ) d t ) .
Case 3. Let x [ 0 , 1 2 ] and y = 3 . It follows that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 2 x 2 φ ( t ) d t ) = ψ ( ( 2 x ) 2 16 ) = ( 2 x ) 2 16 < 1 2 1 2 [ ( 9 4 x ) 2 + 1 ] 1 8 ( 9 4 x ) 2 = ψ ( 9 4 x ) ϕ ( 9 4 x ) = ψ ( 0 1 φ ( t ) d t + 1 3 x φ ( t ) d t ) ϕ ( 0 1 φ ( t ) d t + 1 3 x φ ( t ) d t ) = ψ ( 0 3 x φ ( t ) d t ) ϕ ( 0 3 x φ ( t ) d t ) = ψ ( 0 d ( x , y ) φ ( t ) d t ) ϕ ( 0 d ( x , y ) φ ( t ) d t ) .
Case 4. Let x = 1 and y = 3 . Note that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 1 φ ( t ) d t ) = ψ ( 1 4 ) = 1 4 < 139 128 = 1 2 ( 25 16 + 1 ) 1 8 25 16 = ψ ( 5 4 ) ϕ ( 5 4 ) = ψ ( 0 1 φ ( t ) d t + 1 2 φ ( t ) d t ) ϕ ( 0 1 φ ( t ) d t + 1 2 φ ( t ) d t ) = ψ ( 0 2 φ ( t ) d t ) ϕ ( 0 2 φ ( t ) d t ) = ψ ( 0 d ( x , y ) φ ( t ) d t ) ϕ ( 0 d ( x , y ) φ ( t ) d t ) .

That is, (1.1) holds. Thus Theorem 3.1 guarantees that f has a unique fixed point 0 X such that lim n f n x = 0 for each x X .

Example 4.2 Let X = [ 0 , 1 ] [ 4 , 5 ] be endowed with the Euclidean metric d = | | . Assume that f : X X and φ , ψ : R + R + and α : R + [ 0 , 1 ) are defined by
f ( x ) = { x 2 4 , x [ 0 , 1 ] , x 2 26 , x [ 4 , 5 ] , φ ( t ) = { 4 t 3 , t [ 0 , 1 ] , 2 t , t [ 4 , 5 ] , ψ ( t ) = t 1 2 , t R + , α ( t ) = { 1 3 + t 2 2 , t [ 0 , 1 ] , 1 2 t , t ( 1 , 3 ) , 1 t , t ( 3 , + ) .

Obviously, ( φ , ψ , α ) Φ 1 × Φ 3 × Φ 5 . Put x , y X with x < y . In order to verify (1.2), we have to consider three possible cases as follows.

Case 1. Let x , y [ 0 , 1 ] . It is clear that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ( 0 y 2 x 2 4 4 t 3 d t ) 1 2 = ( x + y ) 2 16 | x y | 2 1 4 | x y | 2 ( 1 3 + 1 2 | x y | 2 ) | x y | 2 = α ( d ( x , y ) ) ψ ( 0 d ( x , y ) φ ( t ) d t ) .
Case 2. Let x , y [ 4 , 5 ] . It follows that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ( 0 y 2 x 2 26 4 t 3 d t ) 1 2 = ( x + y 26 ) 2 | x y | 2 25 169 | x y | 2 ( 1 3 + 1 2 | x y | 2 ) | x y | 2 = α ( d ( x , y ) ) ψ ( 0 d ( x , y ) φ ( t ) d t ) .
Case 3. Let x [ 0 , 1 ] and y [ 4 , 5 ] . It follows that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ( 0 y 2 26 x 2 4 4 t 3 d t ) 1 2 = ( y 2 26 x 2 4 ) 2 ( 25 26 ) 2 < 1 < | x y | = α ( | x y | ) | x y | = α ( | x y | ) ( 0 1 4 t 3 d t + 1 | x y | 2 t d t ) 1 2 = α ( d ( x , y ) ) ( 0 | x y | φ ( t ) d t ) 1 2 = α ( d ( x , y ) ) ψ ( 0 d ( x , y ) φ ( t ) d t ) .

That is, (1.2) holds. Consequently, the conditions of Theorem 3.2 are satisfied. It follows from Theorem 3.2 that f has a unique fixed point 0 X such that lim n f n x = 0 for each x X .

Example 4.3 Let X = [ 1 2 , 1 ] [ 3 2 , 2 ] be endowed with the Euclidean metric d = | | . Assume that f : X X , φ , ϕ , ψ : R + R + and α , β : R + [ 0 , 1 ) are defined by
f ( x ) = { 1 , x [ 1 2 , 1 ] , x 2 , x [ 3 2 , 2 ] , ϕ ( t ) = { 0 , t [ 0 , 9 16 ) 32 t 2 9 , t [ 9 16 , + ) , φ ( t ) = 2 t , ψ ( t ) = 4 t 2 , α ( t ) = t ( 1 2 + t ) 2 , β ( t ) = t 2 ( 1 2 + t ) 2 , t R + .

It is easy to see that ( φ , ψ , ϕ ) Φ 1 × Φ 3 × Φ 4 , ( α , β ) Φ 6 and (3.15) holds. In order to verify (1.3), we have to consider the five possible cases below.

Case 1. Let x , y [ 3 2 , 2 ] with x y . Note that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 | x y | 2 φ ( t ) d t ) = ψ ( ( x y ) 2 4 ) = ( x y ) 4 4 x y 2 x y 2 x 4 ( 1 2 + x y ) 2 x y ( 1 2 + x y ) 2 32 9 ( x 2 4 ) 2 = α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) d + β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) .
Case 2. Let x , y [ 3 2 , 2 ] with y > x . Note that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 | y x | 2 φ ( t ) d t ) = ( y x ) 4 4 ( y x ) 2 4 y 4 ( 1 2 + y x ) 2 = ( y x ) 2 ( 1 2 + y x ) 2 y 4 4 = β ( d ( x , y ) ) ψ ( y 2 4 ) = β ( d ( x , y ) ) ψ ( 0 y 2 φ ( t ) d t ) = β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) + β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) .
Case 3. Let x [ 3 2 , 2 ] and y [ 1 2 , 1 ] . It follows that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 x 2 2 φ ( t ) d t ) = ψ ( ( x 2 ) 2 4 ) = ( x 2 ) 4 4 1 64 < 27 64 = 3 8 2 9 81 16 x y ( 1 2 + x y ) 2 2 9 x 4 = α ( d ( x , y ) ) ϕ ( x 2 4 ) = α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) + β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) .
Case 4. Let x [ 1 2 , 1 ] and y [ 3 2 , 2 ] . Note that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = ψ ( 0 | y 2 | 2 2 t d t ) = ( y 2 ) 4 4 1 64 < 1 4 81 64 ( y x ) 2 ( 1 2 + y x ) 2 y 4 4 = β ( d ( x , y ) ) ψ ( y 2 4 ) = β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) + β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) .
Case 5. Let x , y [ 1 2 , 1 ] . Notice that f x = f y = 1 . It follows that
ψ ( 0 d ( f x , f y ) φ ( t ) d t ) = 0 α ( d ( x , y ) ) ϕ ( 0 d ( x , f x ) φ ( t ) d t ) + β ( d ( x , y ) ) ψ ( 0 d ( y , f y ) φ ( t ) d t ) .

That is, (1.3) holds. Thus all the conditions of Theorem 3.3 are satisfied. It follows from Theorem 3.3 that f has a unique fixed point 1 X such that lim n f n x = 1 for each x X .

Declarations

Acknowledgements

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

References

  1. Branciari A: A fixed point theorem for mappings satisfying a general contractive condition of integral type. Int. J. Math. Math. Sci. 2002, 29: 531–536. 10.1155/S0161171202007524MathSciNetView ArticleGoogle Scholar
  2. Altun I, Türkoǧlu D, Rhoades BE: Fixed points of weakly compatible maps satisfying a general contractive of integral type. Fixed Point Theory Appl. 2007., 2007: Article ID 17301 10.1155/2007/17301Google Scholar
  3. Djoudi A, Merghadi F: Common fixed point theorems for maps under a contractive condition of integral type. J. Math. Anal. Appl. 2008, 341: 953–960. 10.1016/j.jmaa.2007.10.064MathSciNetView ArticleGoogle Scholar
  4. Jachymski J: Remarks on contractive conditions of integral type. Nonlinear Anal. 2009, 71: 1073–1081. 10.1016/j.na.2008.11.046MathSciNetView ArticleGoogle Scholar
  5. Liu Z, Li X, Kang SM, Cho SY: Fixed point theorems for mappings satisfying contractive conditions of integral type and applications. Fixed Point Theory Appl. 2011., 2011: Article ID 64 10.1186/1687-1812-2011-64Google Scholar
  6. Rhoades BE: Two fixed point theorems for mappings satisfying a general contractive condition of integral type. Int. J. Math. Math. Sci. 2003, 63: 4007–4013. 10.1155/S0161171203208024MathSciNetView ArticleGoogle Scholar
  7. Suzuki T: Meir-Keeler contractions of integral type are still Meir-Keeler contractions. Int. J. Math. Math. Sci. 2007., 2007: Article ID 39281 10.1155/2007/39281Google Scholar
  8. Vijayaraju P, Rhoades BE, Mohanraj R: A fixed point theorem for a pair of maps satisfying a general contractive condition of integral type. Int. J. Math. Math. Sci. 2005, 15: 2359–2364. 10.1155/IJMMS.2005.2359MathSciNetView ArticleGoogle Scholar
  9. Dutta PN, Choudhury BS: A generalization of contraction principle in metric spaces. Fixed Point Theory Appl. 2008., 2008: Article ID 406368 10.1155/2008/406368Google Scholar

Copyright

© Liu et al.; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.