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Some fixed point theorems for mappings satisfying contractive conditions of integral type

Abstract

Five fixed point theorems for mappings satisfying contractive conditions of integral type in complete metric spaces are proved. Two examples are added to illustrate the results obtained.

MSC:54H25.

1 Introduction and preliminaries

Rhoades [1] and Branciari [2] proved the following fixed point theorems for the weakly contraction mapping and contractive mapping of integral type, respectively, which are generalizations of the Banach fixed point theorem.

Theorem 1.1 ([1])

Let T be a mapping from a complete metric space (X,d) into itself satisfying

d(Tx,Ty)d(x,y)ψ ( d ( x , y ) ) ,x,yX,
(1.1)

where ψ: R + R + is continuous and nondecreasing such that ψ is positive on R + {0}, ψ(0)=0 and lim t + ψ(t)=+. Then T has a unique fixed point in X.

Theorem 1.2 ([2])

Let T be a mapping from a complete metric space (X,d) into itself satisfying

0 d ( T x , T y ) φ(t)dtc 0 d ( x , y ) φ(t)dt,x,yX,
(1.2)

where c(0,1) is a constant and φ Φ 1 . Then T has a unique fixed point aX such that lim n T n x=a for each xX.

Recently several years, the researchers in [314] and others continued the study of Rhoades and Branciari, proved some fixed point and common fixed point theorems for various generalized weakly contraction mappings and contractive mappings of integral type in complete metric spaces, Banach spaces, modular spaces and symmetric spaces. Suzuki [15] proved that contractive condition of integral type in complete metric spaces is a special case of Meir-Keeler type.

The objective of this article is both to introduce several mappings satisfying contractive conditions of integral type, one of which extends the mapping (1.1) and is different from the mapping (1.2), and to provide sufficient conditions which ensure the existence of fixed points and convergence of iterative methods for these mappings in complete metric spaces. Two nontrivial examples are given to explain the main results obtained.

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

Φ 1 = { φ : φ : R + R +  is Lebesgue integrable, summable on each Φ 1 = compact subset of  R +  and  0 ε φ ( t ) d t > 0  for each  ε > 0 } ; Φ 2 = { ψ : ψ : R + R +  is a lower semicontinuous function with  ψ ( 0 ) = 0 Φ 2 = and  ψ ( t ) > 0  for each  t > 0 } .

For a self mapping T in a metric space (X,d) and (x,y,n) X 2 × N 0 , define

x n = T n x , d n = d ( x n , x n + 1 ) ; 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 ) ] } ; N ( x , y ) = max { d ( x , T x ) , d ( y , T y ) } ; P ( x , y ) = max { d ( x , y ) , d ( x , T x ) , d ( y , T y ) } ; Q ( x , y ) = max { d ( x , T x ) , d ( y , T y ) , 1 2 [ d ( x , T y ) + d ( y , T x ) ] } .

Lemma 1.1 ([10])

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)dt= 0 a φ(t)dt.

Lemma 1.2 ([10])

Let φ Φ 1 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 Main results

Now we prove the existence, uniqueness, and iterative approximations of fixed points for the mappings (2.1), (2.8), and (2.19)(2.21), respectively.

Theorem 2.1 Let (φ,ψ) be in Φ 1 × Φ 2 and T be a mapping from a complete metric space (X,d) into itself satisfying

0 d ( T x , T y ) φ(t)dt 0 d ( x , y ) φ(t)dt 0 ψ ( d ( x , y ) ) φ(t)dt,x,yX.
(2.1)

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

Proof Let x be an arbitrary point in X. Suppose that there exists some n 0 N 0 with x n 0 = x n 0 + 1 . Clearly,

x n 0 =T x n 0 = T 2 x n 0 == T m x n 0 == lim n T n x n 0 ,

that is, x n 0 is a fixed point of T. Suppose that x n x n + 1 for each n N 0 . It follows from (2.1) and (φ,ψ) Φ 1 × Φ 2 that

0 d n + 1 φ ( t ) d t = 0 d ( x n + 1 , x n + 2 ) φ ( t ) d t = 0 d ( T n + 1 x , T n + 2 x ) φ ( t ) d t 0 d ( T n x , T n + 1 x ) φ ( t ) d t 0 ψ ( d ( T n x , T n + 1 x ) ) φ ( t ) d t = 0 d n φ ( t ) d t 0 ψ ( d n ) φ ( t ) d t < 0 d n φ ( t ) d t , n N 0 ,

which yields

d n + 1 < d n ,n N 0 ,

which implies that there exists a constant c with lim n d n =c0. Suppose that c>0. Put lim inf n ψ( d n )=α. It is easy to see that there exists a subsequence { d n ( k ) } n N of { d n } n N 0 satisfying lim k ψ( d n ( k ) )=α. Since ψ is lower semicontinuous and ψ Φ 2 , it follows that αψ(c)>0. Using (2.1), Lemma 1.1 and (φ,ψ) Φ 1 × Φ 2 , we get

0 < 0 c φ ( t ) d t = lim sup k 0 d n ( k ) + 1 φ ( t ) d t = lim sup k 0 d ( x n ( k ) + 1 , x n ( k ) + 2 ) φ ( t ) d t = lim sup k 0 d ( T n ( k ) + 1 x , T n ( k ) + 2 x ) φ ( t ) d t lim sup k ( 0 d ( T n ( k ) x , T n ( k ) + 1 x ) φ ( t ) d t 0 ψ ( d ( T n ( k ) x , T n ( k ) + 1 x ) ) φ ( t ) d t ) lim sup k ( 0 d n ( k ) φ ( t ) d t 0 ψ ( d n ( k ) ) φ ( t ) d t ) lim sup k 0 d n ( k ) φ ( t ) d t lim inf k 0 ψ ( d n ( k ) ) φ ( t ) d t = 0 c φ ( t ) d t 0 α φ ( t ) d t 0 c φ ( t ) d t 0 ψ ( c ) φ ( t ) d t < 0 c φ ( t ) d t ,

which is impossible. Hence c=0 and

lim n d n =0.
(2.2)

Now we prove that { x n } n N 0 is a Cauchy sequence. If it is not a Cauchy sequence, then there exist a constant ε>0 and two subsequences { x m ( k ) } k N and { x n ( k ) } k N of { x n } n N 0 such that n(k) is minimal in the sense that n(k)>m(k)>k and d( x m ( k ) , x n ( k ) )>ε. It follows that d( x m ( k ) , x n ( k ) 1 )ε. Observe 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 m ( k ) 1 + d ( x m ( k ) 1 , x m ( k ) ) + d ( x m ( k ) , x n ( k ) 1 ) + d n ( k ) 1 2 d m ( k ) 1 + ε + d n ( k ) 1 , k N ,
(2.3)

and

| d ( x m ( k ) 1 , x n ( k ) 1 ) d ( x m ( k ) , x n ( k ) 1 ) | d m ( k ) 1 ,kN.
(2.4)

Letting k in (2.3) and (2.4) and using (2.2), we infer that

lim k d( x m ( k ) , x n ( k ) )= lim k d( x m ( k ) , x n ( k ) 1 )= lim k d( x m ( k ) 1 , x n ( k ) 1 )=ε.
(2.5)

Put

lim inf k ψ ( d ( x m ( k ) 1 , x n ( k ) 1 ) ) =β.

Clearly, there exists a subsequence { d ( x m ( k j ) 1 , x n ( k j ) 1 ) } j N of { d ( x m ( k ) 1 , x n ( k ) 1 ) } k N such that

lim j ψ ( d ( x m ( k j ) 1 , x n ( k j ) 1 ) ) =β.
(2.6)

Since ψ is lower semicontinuous, it follows from (2.5), (2.6), and ψ Φ 2 that βψ(ε)>0. By means of (2.1), (2.5), (2.6), Lemma 1.1, and φ Φ 1 , we deduce that

0 < 0 ε φ ( t ) d t = lim sup j 0 d ( x m ( k j ) , x n ( k j ) ) φ ( t ) d t = lim sup j 0 d ( T m ( k j ) x , T n ( k j ) x ) φ ( t ) d t lim sup j ( 0 d ( T m ( k j ) 1 x , T n ( k j ) 1 x ) φ ( t ) d t 0 ψ ( d ( T m ( k j ) 1 x , T n ( k j ) 1 x ) ) φ ( t ) d t ) = lim sup j ( 0 d ( x m ( k j ) 1 , x n ( k j ) 1 ) φ ( t ) d t 0 ψ ( d ( x m ( k j ) 1 , x n ( k j ) 1 ) ) φ ( t ) d t ) lim sup j 0 d ( x m ( k j ) 1 , x n ( k j ) 1 ) φ ( t ) d t lim inf j 0 ψ ( d ( x m ( k j ) 1 , x n ( k j ) 1 ) ) φ ( t ) d t = 0 ε φ ( t ) d t 0 β φ ( t ) d t 0 ε φ ( t ) d t 0 ψ ( ε ) φ ( t ) d t < 0 ε φ ( t ) d t ,

which is a contradiction. Thus { x n } n N 0 is a Cauchy sequence. Since (X,d) is complete, it follows that there exists aX such that

lim n T n x=a.
(2.7)

Next we prove that a is a fixed point of T. In view of (2.1), (2.7), and Lemma 1.2, we obtain

0 0 d ( T n + 1 x , T a ) φ ( t ) d t 0 d ( T n x , a ) φ ( t ) d t 0 ψ ( d ( T n x , a ) ) φ ( t ) d t 0 d ( T n x , a ) φ ( t ) d t 0 as  n ,

which implies that

lim n 0 d ( T n + 1 x , T a ) φ(t)dt=0,

which together with Lemma 1.2 gives

lim n d ( T n + 1 x , T a ) =0.

Consequently, we have

d(a,Ta)d ( a , T n + 1 x ) +d ( T n + 1 x , T a ) 0as n,

that is, a=Ta.

Lastly, we prove that a is a unique fixed point of T in X. Suppose that T has another fixed point bX{a}. It follows from (2.1), φ Φ 1 , and ψ(d(a,b))>0 that

0 < 0 d ( a , b ) φ ( t ) d t = 0 d ( T a , T b ) φ ( t ) d t 0 d ( a , b ) φ ( t ) d t 0 ψ ( d ( a , b ) ) φ ( t ) d t < 0 d ( a , b ) φ ( t ) d t ,

which is a contradiction. This completes the proof. □

Remark 2.1 In the case ϕ(t)=1 for all t R + , Theorem 2.1 reduces to Theorem 1.1. On the other hand, the example below demonstrates that Theorem 2.1 is different from Theorem 1.2.

Example 2.1 Let X= R + be endowed with the Euclidean metric d=||, T:XX and φ,ψ: R + R + be defined by

Tx= x 1 + x 2 ,xX

and

φ(t)=4 t 3 ,ψ(t)= t 2 1 + t 2 ,t R + .

Obviously, (φ,ψ) Φ 1 × Φ 2 . Let x,yX. It is clear that

( 1 x y ) 2 = 1 2 x y + x 2 y 2 1 + x 2 + y 2 + x 2 y 2 = ( 1 + x 2 ) ( 1 + y 2 ) , 1 + ( x y ) 2 = 1 + x 2 2 x y + y 2 1 + x 2 + y 2 + x 2 y 2 = ( 1 + x 2 ) ( 1 + y 2 ) ,

which imply that

( 1 x y ) 4 [ 1 + ( x y ) 2 ] 2 ( 1 + x 2 ) 4 ( 1 + y 2 ) 4 [ 1 + 2 ( x y ) 2 ] ( 1 + x 2 ) 4 ( 1 + y 2 ) 4 ,

which gives

0 d ( T x , T y ) φ ( t ) d t = ( x 1 + x 2 y 1 + y 2 ) 4 = ( x y ) 4 ( 1 x y ) 4 ( 1 + x 2 ) 4 ( 1 + y 2 ) 4 ( x y ) 4 [ 1 + 2 ( x y ) 2 ] [ 1 + ( x y ) 2 ] 2 = ( x y ) 4 ( x y ) 8 [ 1 + ( x y ) 2 ] 2 = 0 d ( x , y ) φ ( t ) d t 0 ψ ( d ( x , y ) ) φ ( t ) d t ,

that is, (2.1) holds. Thus the conditions of Theorem 2.1 are satisfied. It follows from Theorem 2.1 that T has a unique fixed point 0X and lim n T n x=0 for each xX.

In order to verify that Theorem 1.2 is useless in proving the existence of fixed points of T, we need to show that (1.2) does not hold. Otherwise, (1.2) holds, that is, there exists some constant c(0,1) satisfying

0 d ( T x , T y ) φ ( t ) d t = ( x 1 + x 2 y 1 + y 2 ) 4 = ( x y ) 4 ( 1 x y ) 4 ( 1 + x 2 ) 4 ( 1 + y 2 ) 4 c ( x y ) 4 = c 0 d ( x , y ) φ ( t ) d t , x , y X ,

which yields

( 1 x y ) 4 ( 1 + x 2 ) 4 ( 1 + y 2 ) 4 c,x,yX with xy,

which means that

1= lim ( x , y ) ( 0 + , 0 + ) x y ( 1 x y ) 4 ( 1 + x 2 ) 4 ( 1 + y 2 ) 4 c<1,

which is a contradiction.

Theorem 2.2 Let (φ,ψ) be in Φ 1 × Φ 2 and T be a mapping from a complete metric space (X,d) into itself satisfying

0 d ( T x , T y ) φ(t)dt 0 M ( x , y ) φ(t)dt 0 ψ ( M ( x , y ) ) φ(t)dt,x,yX.
(2.8)

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

Proof Let x be an arbitrary point in X. If x n 0 = x n 0 + 1 for some n 0 N 0 , then there is nothing to prove. Now suppose that x n x n + 1 for all n N 0 . Note that

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 n , d n + 1 , 1 2 d ( x n , x n + 2 ) } = max { d n , d n + 1 } , n N 0 ,
(2.9)

because

1 2 d( x n , x n + 2 ) 1 2 [ d ( x n , x n + 1 ) + d ( x n + 1 , x n + 2 ) ] max{ d n , d n + 1 },n N 0 .

Now we prove that

d n + 1 < d n ,n N 0 .
(2.10)

Or else there exists some n 0 N 0 such that d n 0 + 1 d n 0 . Making use of (2.8) and (2.9), we know that

0 d n 0 + 1 φ ( t ) d t = 0 d ( x n 0 + 1 , x n 0 + 2 ) φ ( t ) d t = 0 d ( T n 0 + 1 x , T n 0 + 2 x ) φ ( t ) d t 0 M ( T n 0 x , T n 0 + 1 x ) φ ( t ) d t 0 ψ ( M ( T n 0 x , T n 0 + 1 x ) ) φ ( t ) d t = 0 max { d n 0 , d n 0 + 1 } φ ( t ) d t 0 ψ ( max { d n 0 , d n 0 + 1 } ) φ ( t ) d t 0 d n 0 + 1 φ ( t ) d t 0 ψ ( d n 0 + 1 ) φ ( t ) d t < 0 d n 0 + 1 φ ( t ) d t ,

which is a contradiction. Note that (2.10) means that there exists a constant c with

lim n d n =c0.
(2.11)

Suppose that c>0. Set lim inf n ψ( d n )=γ. Obviously, there exists a subsequence { d n ( k ) } n N of { d n } n N 0 such that lim k ψ( d n ( k ) )=γ. Since ψ is lower semicontinuous, it follows from ψ Φ 2 that γψ(c)>0. On account of (2.8)(2.11), Lemma 1.1, and φ Φ 1 , we arrive at

0 < 0 c φ ( t ) d t = lim sup k 0 d n ( k ) + 1 φ ( t ) d t = lim sup k 0 d ( x n ( k ) + 1 , x n ( k ) + 2 ) φ ( t ) d t = lim sup k 0 d ( T n ( k ) + 1 x , T n ( k ) + 2 x ) φ ( t ) d t lim sup k ( 0 M ( T n ( k ) x , T n ( k ) + 1 x ) φ ( t ) d t 0 ψ ( M ( T n ( k ) x , T n ( k ) + 1 x ) ) φ ( t ) d t ) lim sup k ( 0 M ( x n ( k ) , x n ( k ) + 1 ) φ ( t ) d t 0 ψ ( M ( x n ( k ) , x n ( k ) + 1 ) ) φ ( t ) d t ) = lim sup k ( 0 d n k φ ( t ) d t 0 ψ ( d n k ) φ ( t ) d t ) lim sup k 0 d n k φ ( t ) d t lim inf k 0 ψ ( d n k ) φ ( t ) d t = 0 c φ ( t ) d t 0 γ φ ( t ) d t 0 c φ ( t ) d t 0 ψ ( c ) φ ( t ) d t < 0 c φ ( t ) d t ,

which is absurd. Hence c=0 and (2.2) holds. Suppose that { x n } n N 0 is not a Cauchy sequence. It follows that there exist a constant ε>0 and two subsequences { x m ( k ) } k N and { x n ( k ) } k N of { x n } n N 0 such that n(k) is minimal in the sense that n(k)>m(k)>k and d( x m ( k ) , x n ( k ) )>ε. It follows that (2.5) holds. Observe that (2.2) and (2.5) ensure that

| d ( x m ( k ) 1 , x n ( k ) ) d ( x m ( k ) , x n ( k ) ) | d m ( k ) 1 0as k
(2.12)

and

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 m ( k ) 1 , d n ( k ) 1 , = 1 2 [ d ( x m ( k ) 1 , x n ( k ) ) + d ( x n ( k ) 1 , x m ( k ) ) ] } max { ε , 0 , 0 , ε } = ε as  k .
(2.13)

Put

lim inf j ψ ( M ( x m ( k ) 1 , x n ( k ) 1 ) ) =λ.

Clearly, there exists a subsequence { M ( x m ( k j ) 1 , x n ( k j ) 1 ) } j N of { M ( x m ( k ) 1 , x n ( k ) 1 ) } k N such that

lim j ψ ( M ( x m ( k j ) 1 , x n ( k j ) 1 ) ) =λψ(ε).
(2.14)

Combining (2.5), (2.8), (2.12)(2.14), Lemma 1.1, and φ Φ 1 , we get

0 < 0 ε φ ( t ) d t = lim sup j 0 d ( x m ( k j ) , x n ( k j ) ) φ ( t ) d t = lim sup j 0 d ( T m ( k j ) x , T n ( k j ) x ) φ ( t ) d t lim sup j ( 0 M ( T m ( k j ) 1 x , T n ( k j ) 1 x ) φ ( t ) d t 0 ψ ( M ( T m ( k j ) 1 x , T n ( k j ) 1 x ) ) φ ( t ) d t ) = lim sup j ( 0 M ( x m ( k j ) 1 , x n ( k j ) 1 ) φ ( t ) d t 0 ψ ( M ( x m ( k j ) 1 , x n ( k j ) 1 ) ) φ ( t ) d t ) lim sup j 0 M ( x m ( k j ) 1 , x n ( k j ) 1 ) φ ( t ) d t lim inf j 0 ψ ( M ( x m ( k j ) 1 , x n ( k j ) 1 ) ) φ ( t ) d t = 0 ε φ ( t ) d t 0 λ φ ( t ) d t 0 ε φ ( t ) d t 0 ψ ( ε ) φ ( t ) d t < 0 ε φ ( t ) d t ,

which is a contradiction. Hence { x n } n N 0 is a Cauchy sequence. Completeness of (X,d) ensures that there exists aX satisfying (2.7). Suppose that d(a,Ta)>0. Let

M =max { d ( a , T a ) , d ( T a , T 2 a ) , 1 2 [ d ( a , T 2 a ) + d ( a , T a ) ] } .
(2.15)

Note that (2.2) and (2.7) yield

lim n M ( x n + 1 , T a ) = lim n max { d ( x n + 1 , T a ) , d ( x n + 1 , T x n + 1 ) , d ( T a , T 2 a ) , 1 2 [ d ( x n + 1 , T 2 a ) + d ( T a , T x n + 1 ) ] } = lim n max { d ( x n + 1 , T a ) , d n + 1 , d ( T a , T 2 a ) , 1 2 [ d ( x n + 1 , T 2 a ) + d ( T a , x n + 2 ) ] } = max { d ( a , T a ) , 0 , d ( T a , T 2 a ) , 1 2 [ d ( a , T 2 a ) + d ( T a , a ) ] } = M
(2.16)

and

lim n M ( x n , a ) = lim n 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 ) ] } = lim n max { d ( x n , a ) , d n , d ( a , T a ) , 1 2 [ d ( x n , T a ) + d ( a , x n + 1 ) ] } = max { 0 , 0 , d ( a , T a ) , 1 2 d ( a , T a ) } = d ( a , T a ) .
(2.17)

Put lim inf n ψ(M( x n ,a))=η. Clearly, there exists a subsequence { M ( x n ( j ) , a ) } j N of { M ( x n , a ) } n N such that

lim j ψ ( M ( x n ( j ) , a ) ) =ηψ ( d ( a , T a ) ) .
(2.18)

In virtue of (2.8), (2.16)(2.18), and Lemma 1.1, we conclude that

0 < 0 M φ ( t ) d t = lim sup j 0 M ( x n ( j ) + 1 , T a ) φ ( t ) d t = lim sup j 0 M ( T n ( j ) + 1 x , T a ) φ ( t ) d t lim sup j ( 0 M ( T n ( j ) x , a ) φ ( t ) d t 0 ψ ( M ( T n ( j ) x , a ) ) φ ( t ) d t ) = lim sup j ( 0 M ( x n ( j ) , a ) φ ( t ) d t 0 ψ ( M ( x n ( j ) , a ) ) φ ( t ) d t ) lim sup j 0 M ( x n ( j ) , a ) φ ( t ) d t lim inf j 0 ψ ( M ( x n ( j ) , a ) ) φ ( t ) d t = 0 d ( a , T a ) φ ( t ) d t 0 η φ ( t ) d t 0 d ( a , T a ) φ ( t ) d t 0 ψ ( d ( a , T a ) ) φ ( t ) d t < 0 d ( a , T a ) φ ( t ) d t ,

which together with (2.15) means that

d(a,Ta) M <d(a,Ta),

which is impossible. Consequently, a=Ta is a fixed point of T in X. Suppose that T has another fixed point bX{a}. Notice that

M ( a , b ) = max { d ( a , b ) , d ( a , T a ) , d ( b , T b ) , 1 2 [ d ( a , T b ) + d ( b , T a ) ] } = max { d ( a , b ) , 0 , 0 , d ( a , b ) } = d ( a , b ) ,

which together with φ Φ 1 , (2.8), and ψ(d(a,b))>0 means that

0 < 0 d ( a , b ) φ ( t ) d t = 0 d ( T a , T b ) φ ( t ) d t 0 M ( a , b ) φ ( t ) d t 0 ψ ( M ( a , b ) ) φ ( t ) d t = 0 d ( a , b ) φ ( t ) d t 0 ψ ( d ( a , b ) ) φ ( t ) d t < 0 d ( a , b ) φ ( t ) d t ,

which is a contradiction. Consequently, T possesses a unique fixed point aX. This completes the proof. □

Remark 2.2 The below example is an application of Theorem 2.2.

Example 2.2 Let X=[0,2]{6} be endowed with the Euclidean metric d=||, T:XX and φ,ψ: R + R + be defined by

Tx={ x 2 , x [ 0 , 2 ] , 2 , x = 6

and

φ(t)=2t,t R + ,ψ(t)={ 3 4 t , t [ 0 , 2 ] , 1 + 1 1 + t , t ( 2 , + ) .

Clearly, (φ,ψ) Φ 1 × Φ 2 . For x,yX with yx, we consider the following four cases.

Case 1. Let x,y[0,2] with y x 2 . It is easy to verify that

M(x,y)=max { | x y | , x 2 , y 2 , 1 2 ( | x y 2 | + | y x 2 | ) } =xy2,

which yields

0 d ( T x , T y ) φ ( t ) d t = 1 4 ( x y ) 2 ( x y ) 2 3 16 ( x y ) 2 = 0 M ( x , y ) φ ( t ) d t 0 ψ ( M ( x , y ) ) φ ( t ) d t .

Case 2. Let x,y[0,2] with x 2 <yx. It is clear that

M(x,y)=max { | x y | , x 2 , y 2 , 1 2 ( | x y 2 | + | y x 2 | ) } = x 2 1,

which gives

0 d ( T x , T y ) φ ( t ) d t = 1 4 ( x y ) 2 = 1 4 x 2 ( 1 2 x y 1 4 y 2 ) = 1 4 x 2 ( 1 4 ( y x ) 2 + 1 4 x 2 ) 1 4 x 2 3 16 x 2 = 0 M ( x , y ) φ ( t ) d t 0 ψ ( M ( x , y ) ) φ ( t ) d t .

Case 3. Let y[0,2] and x=6. Obviously, we have

M(x,y)=max { 6 y , 4 , y 2 , 1 2 ( 6 y 2 + 2 y ) } =6y4,

which implies that

0 d ( T x , T y ) φ ( t ) d t = ( 2 1 2 y ) 2 4 < 4 2 ( 1 + 1 1 + 4 ) ( 6 y ) 2 ( 1 + 1 1 + 6 y ) = 0 M ( x , y ) φ ( t ) d t 0 ψ ( M ( x , y ) ) φ ( t ) d t .

Case 4. Let x=y=6. It follows that

M(x,y)=max { 0 , 6 2 , 6 2 , 1 2 ( 6 2 + 6 2 ) } =4,

which means that

0 d ( T x , T y ) φ ( t ) d t = 0 < 4 2 ( 1 + 1 1 + 4 ) = 0 M ( x , y ) φ ( t ) d t 0 ψ ( M ( x , y ) ) φ ( t ) d t .

That is, (2.8) holds. Thus the conditions of Theorem 2.2 are satisfied. It follows from Theorem 2.2 that T has a unique fixed point 0X and lim n T n x=0 for every xX.

Similar to the proofs of Theorems 2.1 and 2.2, we have the following results and we omit their proofs.

Theorem 2.3 Let (φ,ψ) be in Φ 1 × Φ 2 and T be a mapping from a complete metric space (X,d) into itself satisfying

0 d ( T x , T y ) φ(t)dt 0 N ( x , y ) φ(t)dt 0 ψ ( N ( x , y ) ) φ(t)dt,x,yX.
(2.19)

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

Theorem 2.4 Let (φ,ψ) be in Φ 1 × Φ 2 and T be a mapping from a complete metric space (X,d) into itself satisfying

0 d ( T x , T y ) φ(t)dt 0 P ( x , y ) φ(t)dt 0 ψ ( P ( x , y ) ) φ(t)dt,x,yX.
(2.20)

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

Theorem 2.5 Let (φ,ψ) be in Φ 1 × Φ 2 and T be a mapping from a complete metric space (X,d) into itself satisfying

0 d ( T x , T y ) φ(t)dt 0 Q ( x , y ) φ(t)dt 0 ψ ( Q ( x , y ) ) φ(t)dt,x,yX.
(2.21)

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

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Acknowledgements

The authors wish to express their gratitude to the referees for giving valuable comments. 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 Jeong Sheok Ume.

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Liu, Z., Wu, H., Ume, J.S. et al. Some fixed point theorems for mappings satisfying contractive conditions of integral type. Fixed Point Theory Appl 2014, 69 (2014). https://doi.org/10.1186/1687-1812-2014-69

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Keywords

  • fixed point
  • contractive conditions of integral type
  • complete metric space