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Integral type contractions in modular metric spaces

Abstract

We prove the existence and uniqueness of a common fixed point of compatible mappings of integral type in modular metric spaces.

MSC:47H09, 47H10, 46A80.

1 Introduction and preliminaries

The metric fixed point theory is very important and useful in mathematics. It can be applied in various branches of mathematics, variational inequalities optimization and approximation theory. In 1976, Jungck [1] proved a common fixed point theorem for commuting maps generalizing the Banach contraction mapping principle. This result was further generalized and extended in various ways by many authors. On the other hand, Sessa [2] defined weak commutativity as follows.

Let (X,d) be a metric space, the self-mappings f, g are said to be weakly commuting if d(fg(x),gf(x))d(g(x),f(x)) for all xX. Further, Jungck [3] introduced more generalized commutativity, the so-called compatibility, which is more general than weak commutativity. Let f, g be self-mappings of a metric space (X,d). The mappings f and g are said to be compatible if lim n d(fg( x n ),gf( x n ))=0, whenever { x n } n = 1 is a sequence in X such that lim n f( x n )= lim n g( x n )=z for some zX. Clearly, weakly commuting mappings are compatible, but neither implication is reversible. Let X=[0,1) with the usual metric. We define mappings f and g on X by

f(x):={ 2 3 if  0 x < 2 3 , 1 x 2 if  2 3 x < 1 andg(x):={ 2 3 if  0 x < 2 3 , 4 3 x if  2 3 x < 1 .

Let { x n } n = 1 be a sequence in X with lim n f( x n )= lim n g( x n )=z, then z= 2 3 and lim n fg( x n )= lim n gf( x n )= 2 3 . Thus the pair (f,g) is compatible on X. In [4] Branciari obtained a fixed point theorem for a single mapping satisfying an analogue of the Banach contraction principle for integral type inequality (see also [57]). Vijayaraju et al. [8] proved the existence of the unique common fixed point theorem for a pair of maps satisfying a general contractive condition of integral type. Recently, Razani and Moradi [9] proved the common fixed point theorem of integral type in modular spaces. The purpose of this paper is to generalize and improve Jungck’s fixed point theorem [3] and Branciari’s result [4] to compatible maps in metric modular spaces. The notions of a metric modular on an arbitrary set and the corresponding modular space, more general than a metric space, were introduced and studied recently by Chistyakov [10]. In the sequel, we recall some basic concepts about modular metric spaces.

Definition 1.1 A function ω:(0,)×X×X[0,] is said to be a metric modular on X if it satisfies the following three axioms:

  1. (i)

    given x,yX, ω λ (x,y)=0 for all λ>0 if and only if x=y;

  2. (ii)

    ω λ (x,y)= ω λ (y,x) for all λ>0 and x,yX;

  3. (iii)

    ω λ + μ (x,y) ω λ (x,z)+ ω μ (z,y) for all λ,μ>0 and x,y,zX.

If, instead of (i), we have only the condition (i)′ ω λ (x,x)=0 for all λ>0 and xX, then ω is said to be a (metric) pseudo-modular on X. The main property of a (pseudo)modular ω on a set X is the following: given x,yX, the function 0<λ ω λ (x,y)[0,] is non-increasing on (0,). In fact, if 0<μ<λ, then (iii), (i)′ and (ii) imply

ω λ (x,y) ω λ μ (x,x)+ ω μ (x,y)= ω μ (x,y)

for all x,yX. If follows that at each point λ>0 the right limit ω λ + 0 (x,y):= lim ε + 0 ω λ + ε (x,y) and the left limit ω λ 0 (x,y):= lim ε + 0 ω λ ε (x,y) exist in [0,] and the following two inequalities hold:

ω λ + 0 (x,y) ω λ (x,y) ω λ 0 (x,y)

for all x,yX. We know that if x 0 X, the set X ω ={xX: lim λ ω λ (x, x 0 )=0} is a metric space, called a modular space, whose metric is given by

d ω 0 (x,y)=inf { λ > 0 : ω λ ( x , y ) λ }

for all x,y X ω . We know that (see [10]) if X is a real linear space, ρ:X[0,] and

ω λ (x,y)=ρ ( x y λ )

for all λ>0 and x,yX, then ρ is modular on X if and only if ω is metric modular on X.

Example 1.2 The following indexed objects ω are simple examples of (pseudo)modulars on a set X. Let λ>0 and x,yX, we have:

  1. (a)

    ω λ a (x,y)= if xy, ω λ a (x,y)=0 if x=y;

and if (X,d) is a (pseudo)metric space with (pseudo)metric d, then we also have:

  1. (b)

    ω λ b (x,y)= d ( x , y ) φ ( λ ) , where φ:(0,)(0,) is a nondecreasing function;

  2. (c)

    ω λ c (x,y)= if λd(x,y), and ω λ c (x,y)=0 if λ>d(x,y);

  3. (d)

    ω λ d (x,y)= if λ<d(x,y), and ω λ d (x,y)=0 if λd(x,y).

Definition 1.3 Let X ω be a modular metric space.

  1. (1)

    The sequence { x n } n = 1 in X ω is said to be convergent to x X ω if ω λ ( x n ,x)0 as n for all λ>0.

  2. (2)

    The sequence { x n } n = 1 in X ω is said to be Cauchy to x X ω if ω λ ( x n , x m )0 as m,n for all λ>0.

  3. (3)

    A subset C of X ω is said to be closed if the limit of a convergent sequence of C always belongs to C.

  4. (4)

    A subset C of X ω is said to be complete if any Cauchy sequence of C is a convergent sequence and its limit is in C.

  5. (5)

    A subset C of X ω is said to be bounded if for all λ>0, δ ω (C)=sup{ ω λ (x,y);x,yC}<.

2 A common fixed point theorem for contractive condition maps

Here, the existence of a common fixed point for ω-compatible mappings satisfying a contractive condition of integral type in modular metric spaces is studied. We recall the following definition.

Definition 2.1 Let X ω be a modular metric space induced by metric modular ω. Two self-mappings T, h of X ω are called ω-compatible if ω λ (Th x n ,hT x n )0, whenever { x n } n = 1 is a sequence in X ω such that h x n z and T x n z for some point z X ω and for λ>0.

Theorem 2.2 Let X ω be a complete modular metric space. Suppose that c,k,l R + , c>l and T,h: X ω X ω are two ω-compatible mappings such that T( X ω )h( X ω ) and

0 ω λ c ( T x , T y ) φ(t)dtk 0 ω λ l ( h x , h y ) φ(t)dt,
(2.1)

for some k(0,1) and for λ>0, where φ: R + R + is a Lebesgue integrable function which is summable, nonnegative and for all ε>0,

0 ε φ(t)dt>0.
(2.2)

If one of T or h is continuous, then there exists a unique common fixed point of T and h.

Proof Let x be an arbitrary point of X ω and generate inductively the sequence { T x n } n = 1 as follows: T x n =h x n + 1 for each n and x 1 =x, that is possible as T( X ω )h( X ω ). For each integer n1 and for all λ>0, (2.1) shows that

0 ω λ c ( T x n + 1 , T x n ) φ ( t ) d t k 0 ω λ l ( h x n + 1 , h x n ) φ ( t ) d t k 0 ω λ c ( T x n , T x n 1 ) φ ( t ) d t k 2 0 ω λ l ( h x n , h x n 1 ) φ ( t ) d t .

By the principle of mathematical induction, we can easily show that

0 ω λ c ( T x n + 1 , T x n ) φ(t)dt k n 0 ω λ l ( T x , x ) φ(t)dt,

which, upon taking the limit as n, yields

lim n 0 ω λ c ( T x n + 1 , T x n ) φ(t)dt0.

Hence (2.2) implies that

lim n ω λ c (T x n + 1 ,T x n )=0.

We now show that { T x n } n = 1 is Cauchy. So, for all ε>0, there exists n 0 N such that ω λ c (T x n + 1 ,T x n )< ε c for all nN with n n 0 and λ>0. Without loss of generality, suppose m,nN and m>n. Observe that for λ c ( m n ) >0, there exists n λ m n N such that

ω λ c ( m n ) (T x n + 1 ,T x n )< ε c ( m n )

for all n n λ m n . We thus obtain

ω λ c ( T x n , T x m ) ω λ c ( m n ) ( T x n , T x n + 1 ) + ω λ c ( m n ) ( T x n + 1 , T x n + 2 ) + + ω λ c ( m n ) ( T x m 1 , T x m ) < ε c ( m n ) + ε c ( m n ) + + ε c ( m n ) = ε c

for all n,m n λ m n . This implies that { T x n } n = 1 is a Cauchy sequence. Since X ω is complete, there exists z X ω such that ω λ c (T x n ,z)0 as n. If T is continuous, then T 2 x n Tz and Th x n Tz. By the ω-compatibility of X ω , we have ω λ (hT x n ,Th x n )0 as n for λ>0. Moreover, hT x n Tz since ω λ (hT x n ,Tz) ω λ 2 (hT x n ,Th x n )+ ω λ 2 (Th x n ,Tz).

In the sequel, we prove that z is a common fixed point of T and h. By (2.1), we get

0 ω λ c ( T 2 x n , T x n ) φ(t)dtk 0 ω λ l ( h T x n , h x n ) φ(t)dt.

Taking the limit as n yields

0 ω λ c ( T z , z ) φ ( t ) d t k 0 ω λ l ( T z , z ) φ ( t ) d t k 0 ω λ c ( T z , z ) φ ( t ) d t ,

which implies that ω λ c (Tz,z)=0 for λ>0. Hence Tz=z. It follows from T( X ω )h( X ω ) that there exists a point z 1 such that z=Tz=h z 1 . By (2.1), we get

0 ω λ c ( T 2 x n , T z 1 ) φ(t)dtk 0 ω λ l ( h T x n , h z 1 ) φ(t)dt.

Taking the limit as n yields

0 ω λ c ( T z , T z 1 ) φ(t)dtk 0 ω λ l ( T z , h z 1 ) φ(t)dt,

and so

0 ω λ c ( z , T z 1 ) φ ( t ) d t k 0 ω λ l ( z , h z 1 ) φ ( t ) d t k 0 ω λ l ( z , z ) φ ( t ) d t .

Hence z=T z 1 =h z 1 and also hz=hT z 1 =Th z 1 =Tz=z (see [11]). In addition, if one considers h to be continuous (instead of T), then by a similar argument (as above), one can prove hz=Tz=z.

Finally, suppose that z and ω are two arbitrary common fixed points of T and h. Then we have

0 ω λ c ( z , ω ) φ ( t ) d t = 0 ω λ c ( T z , T ω ) φ ( t ) d t k 0 ω λ l ( h z , h ω ) φ ( t ) d t k 0 ω λ c ( z , ω ) φ ( t ) d t ,

which implies that ω λ c (z,ω)=0 for λ>0 and hence z=ω. □

The following theorem is another version of Theorem 2.2 when l=c, by adding the restriction that T,h:BB, where B is a closed and bounded subset of X ω .

Theorem 2.3 Let X ω be a complete modular metric space, and let B be a closed and bounded subset of X ω . Suppose that T,h: X ω X ω are two ω-compatible mappings such that T( X ω )h( X ω ) and

0 ω λ c ( T x , T y ) φ(t)dtk 0 ω λ c ( h x , h y ) φ(t)dt
(2.3)

for all x,yB and for λ>0, where c,k R + with k(0,1), and φ: R + R + is a Lebesgue integrable function which is summable, nonnegative and for all ε>0, 0 ε φ(t)dt>0. If one of T or h is continuous, then T and h have a unique common fixed point.

Proof Let xB and m,nN. Let { x n } n = 1 be the sequence generated in the proof of Theorem 2.2. Then

0 ω λ c ( T x n + m , T x m ) φ ( t ) d t k 0 ω λ c ( h x n + m , h x m ) φ ( t ) d t = k 0 ω λ c ( T x n + m 1 , T x m 1 ) φ ( t ) d t k 2 0 ω λ c ( T x n + m 2 , T x m 2 ) φ ( t ) d t k m 0 ω λ c ( T x n , x ) φ ( t ) d t k m 0 δ ω ( B ) φ ( t ) d t

for λ>0. Since B is bounded,

lim n , m 0 ω λ c ( T x n + m , T x m ) φ(t)dt0,

which implies that lim n , m ω λ c (T x n + m ,T x m )=0. Therefore, { T x n } n = 1 is Cauchy. Since X ω is complete and B is closed, there exists zB such that lim n ω λ c (T x n ,z)=0. If T is continuous, then T 2 x n Tz and Th x n Tz. Then, by ω-compatibility of X ω , we have ω λ (hT x n ,Th x n )0 as n for λ>0. Moreover, hT x n Tz. Next, we prove that z is a fixed point of T. It follows from (2.3) that

0 ω λ c ( T 2 x n , T x n ) φ(t)dtk 0 ω λ c ( h T x n , h x n ) φ(t)dt.

Taking the limit as n yields

0 ω λ c ( T z , z ) φ(t)dtk 0 ω λ c ( T z , z ) φ(t)dt.

So ω λ c (Tz,z)=0 for λ>0 and hence Tz=z. Since T( X ω )h( X ω ), there exists a point z 1 such that z=Tz=h z 1 , and

0 ω λ c ( T 2 x n , T z 1 ) φ(t)dtk 0 ω λ c ( h T x n , h z 1 ) φ(t)dt.

Taking the limit as n yields

0 ω λ c ( z , T z 1 ) φ(t)dtk 0 ω λ c ( z , z ) φ(t)dt.

Hence z=T z 1 =h z 1 and also hz=hT z 1 =Th z 1 =Tz=z (see [11]). In addition, if one considers h to be continuous (instead of T), then by a similar argument (as above), one can prove hz=Tz=z.

Let z and ω be two arbitrary common fixed points of T and h. Then

0 ω λ c ( z , ω ) φ ( t ) d t = 0 ω λ c ( T z , T ω ) φ ( t ) d t k 0 ω λ c ( z , ω ) φ ( t ) d t ,

which implies that ω λ c (z,ω)=0 for λ>0 and hence z=ω. □

3 A common fixed point theorem for quasi-contraction maps

In this section, we prove Theorem 2.2 for a quasi-contraction map of integral type. To this end, we present the following definition.

Definition 3.1 Two self-mappings T,h: X ω X ω of a modular metric space X ω are (c,l,q)-generalized contractions of integral type if there exist 0<q<1 and c,l R + with c>l such that

0 ω λ c ( T x , T y ) φ(t)dtq 0 m ( x , y ) φ(t)dt,
(3.1)

where m(x,y)=max{ ω λ l (hx,hy), ω λ l (hx,Tx), ω λ l (hy,Ty), ω λ l ( h x , T y ) + ω λ l ( h y , T x ) 2 }, and φ: R + R + is a Lebesgue integrable function which is summable, nonnegative and for all ε>0, 0 ε φ(t)dt>0, λ>0 and x,y X ω .

Theorem 3.2 Let X ω be a complete modular metric space. Suppose that T and h are (c,l,q)-generalized contractions of integral type self-maps of X ω and T( X ω )h( X ω ). If one of T or h is continuous, then T and h have a unique common fixed point.

Proof Choose c>2l. Let x be an arbitrary point of X ω and generate inductively the sequence { T x n } n = 1 as follows: T x n =h x n + 1 and T( X ω )h( X ω ). We thus obtain

0 ω λ c ( T x n + 1 , T x n ) φ(t)dtq 0 m ( x n + 1 , x n ) φ(t)dt

for λ>0, where

m ( x n + 1 , x n ) = max { ω λ l ( h x n + 1 , h x n ) , ω λ l ( T x n , h x n ) , ω λ l ( h x n + 1 , T x n + 1 ) , ω λ l ( h x n + 1 , T x n ) + ω λ l ( h x n , T x n + 1 ) 2 } .

It follows from T x n =h x n + 1 that

m( x n + 1 , x n )=max { ω λ l ( h x n + 1 , h x n ) , ω λ l ( T x n , T x n + 1 ) , 0 + ω λ l ( h x n , T x n + 1 ) 2 } .

Moreover,

ω λ l ( h x n , T x n + 1 ) = ω λ l ( T x n 1 , T x n + 1 ) ω λ 2 l ( T x n 1 , T x n ) + ω λ 2 l ( T x n , T x n + 1 ) ω λ c ( T x n 1 , T x n ) + ω λ c ( T x n , T x n + 1 ) .

Then

m( x n + 1 , x n ) ω λ c (T x n ,T x n 1 )

and

0 ω λ c ( T x n + 1 , T x n ) φ(t)dtq 0 ω λ c ( T x n , T x n 1 ) φ(t)dt.

Continuing this process, we get

0 ω λ c ( T x n + 1 , T x n ) φ(t)dt q n 0 ω λ c ( T x , x ) φ(t)dt.

So lim n ω λ c (T x n ,T x n + 1 )=0 as n tends to infinity. Suppose l< c <2l. Since ω λ is a decreasing function, one may write ω λ c (T x n ,T x n + 1 ) ω λ c (T x n ,T x n + 1 ), whenever c <2lc. Taking the limit from both sides of this inequality shows that lim n ω λ c (T x n ,T x n + 1 )=0 for l< c <2l and λ>0. Thus we have lim n ω λ c (T x n ,T x n + 1 )=0 for any c>l. Now, we show that { T x n } n = 1 is Cauchy. Since lim n ω λ c (T x n ,T x n + 1 )=0 for λ>0, for ε>0, there exists n 0 N such that ω λ c (T x n + 1 ,T x n )< ε c for all nN with n n 0 and λ>0. Without loss of generality, suppose m,nN and m>n. Observe that for λ c ( m n ) >0, there exists n λ m n N such that

ω λ c ( m n ) (T x n + 1 ,T x n )< ε c ( m n )

for all n n λ m n . Now we have

ω λ c ( T x n , T x m ) ω λ c ( m n ) ( T x n , T x n + 1 ) + ω λ c ( m n ) ( T x n + 1 , T x n + 2 ) + + ω λ c ( m n ) ( T x m 1 , T x m ) < ε c ( m n ) + ε c ( m n ) + + ε c ( m n ) = ε c

for all n,m n λ m n . This implies { T x n } n = 1 is a Cauchy sequence. Since X ω is complete, there exists z X ω such that ω λ c (T x n ,z)0 as n. Next we prove that z is a fixed point of T. If T is continuous, then T 2 x n Tz and Th x n Tz. By the ω-compatibility of X ω , we have ω λ (hT x n ,Th x n )0 as n for λ>0. Moreover, hT x n Tz since ω λ (hT x n ,Tz) ω λ 2 (hT x n ,Th x n )+ ω λ 2 (Th x n ,Tz). Note that

0 ω λ c ( T x n , T 2 x n ) φ(t)dtq 0 m ( x n , T x n ) φ(t)dt,

where

m ( x n , T x n ) = max { ω λ l ( h x n , h T x n ) , ω λ l ( h x n , T x n ) , ω λ l ( h T x n , T T x n ) , ω λ l ( h x n , T T x n ) + ω λ l ( T x n , h T x n ) 2 } .

Taking the limit as n, we get

0 ω λ c ( z , T z ) φ(t)dtq 0 ω λ c ( z , T z ) φ(t)dt,

and so Tz=z. Since T( X ω )h( X ω ), there exists a point z 1 such that z=Tz=h z 1 . We have

0 ω λ c ( T 2 x n , T z 1 ) φ(t)dtq 0 m ( T x n , z 1 ) φ(t)dt

and

m ( T x n , z 1 ) = max { ω λ l ( h T x n , z ) , ω λ l ( h T x n , T 2 x n ) , ω λ l ( z , T z 1 ) , ω λ l ( h T x n , T z 1 ) + ω λ l ( z , T 2 x n ) 2 } .

Taking the limit as n, we get

0 ω λ c ( z , T z 1 ) φ(t)dtq 0 ω λ c ( z , T z 1 ) φ(t)dt.

It follows that z=T z 1 =h z 1 and also hz=hT z 1 =Th z 1 =Tz=z (see [11]). Moreover, if h is continuous (instead of T), then by a similar argument (as above), we can prove hz=Tz=z. Let z and ω be two arbitrary fixed points of T and h. Then

m ( z , ω ) = max { ω λ l ( z , ω ) , 0 , 0 , ω λ l ( z , ω ) + ω λ l ( z , ω ) 2 } = ω λ l ( z , ω ) .

Therefore,

0 ω λ c ( z , ω ) φ(t)dtq 0 ω λ l ( z , ω ) φ(t)dt,

which implies that z=ω. □

4 Generalization

Here, we extend the results of the last section. We need a general contractive inequality of integral type. Let R + be a set of nonnegative real numbers and consider () ϕ: R + R + as a nondecreasing and right-continuous function such that ϕ(t)<t for any t>0.

To prove the next theorem, we need the following lemma [12].

Lemma 4.1 Let t>0. ϕ(t)<t if only if lim k ϕ k (t)=0, where ϕ k denotes the k-times repeated composition of ϕ with itself.

Next, we prove a modified version of Theorem 2.2.

Theorem 4.2 Let X ω be a complete modular metric space. Suppose that c,l R + , c>l and T,h: X ω X ω are two ω-compatible mappings such that T( X ω )h( X ω ) and

0 ω λ c ( T x , T y ) φ(t)dtϕ ( 0 ω λ l ( h x , h y ) φ ( t ) d t ) ,

where ϕ is a function satisfying the property (), and φ: R + R + is a Lebesgue integrable mapping which is summable, nonnegative and for all ε>0, 0 ε φ(t)dt>0 and λ>0. If one of T or h is continuous, then T and h have a unique common fixed point.

Proof Let x be an arbitrary point of X ω and generate inductively the sequence { T x n } n = 1 as follows: T x n =h x n + 1 and x 1 =x, that is possible as T( X ω )h( X ω ),

0 ω λ c ( T x n + 1 , T x n ) φ ( t ) d t ϕ ( 0 ω λ l ( h x n + 1 , h x n ) φ ( t ) d t ) ϕ ( 0 ω λ c ( T x n , T x n 1 ) φ ( t ) d t ) ϕ 2 ( 0 ω λ l ( h x n , h x n 1 ) φ ( t ) d t )

for each integer n1 and λ>0. By the principle of mathematical induction, we can easily see that

0 ω λ c ( T x n + 1 , T x n ) φ(t)dt ϕ n ( 0 ω λ l ( T x , x ) φ ( t ) d t ) .

Taking the limit as n, we obtain, by Lemma 4.1,

lim n 0 ω λ c ( T x n + 1 , T x n ) φ(t)dt=0.

Using the same method as in the proof of Theorem 2.2, we show that T and h have a unique common fixed point. □

Applying the method of proof of Theorem 3.2, we get the following result.

Theorem 4.3 Let X ω be a complete modular metric space. Suppose c,l R + , c>l and T,h: X ω X ω such that T( X ω )h( X ω )

0 ω λ c ( T x , T y ) φ(t)dtϕ ( 0 m ( x , y ) φ ( t ) d t ) ,

where m(x,y)=max{ ω λ l (hx,hy), ω λ l (hx,Tx), ω λ l (hy,Ty), ω λ l ( h x , T y ) + ω λ l ( h y , T x ) 2 }, ϕ is a function satisfying the property () and λ>0. If one of T or h is continuous, then there exists a unique common fixed point of h and T.

Proof The proof is similar to the proof of Theorem 3.2. □

Now we provide examples to validate and illustrate Theorems 2.2 and 3.2.

Example 4.4 Let X ω ={ 1 n :nN}{0} and ω λ (x,y):= | x y | λ for λ>0. Define the mapping T,h: X ω X ω by

T(x):={ 1 n + 4 if  x = 1 n , 0 otherwise andh(x):={ 1 n + 1 if  x = 1 n , 0 otherwise .

Then all the hypotheses of Theorem 2.2 are satisfied with φ(t)=2t for t>0 and c=2, l=1, k= 1 4 .

Example 4.5 Let X ω ={ 1 n :nN}{0} and ω λ (x,y):= | x y | λ for λ>0. Define the mapping T,h: X ω X ω by

T(x):={ 1 n + 1 if  x = 1 n , 0 otherwise andh(x):={ 1 n if  x = 1 n , 0 otherwise .

Then all the hypotheses of Theorem 3.2 are satisfied with φ(t)= t 1 t 2 (1logt) for t>0 and c=l=λ=1, k= 1 2 . See [4] for details.

5 Application

In the section, we assume that R=(,+), R + =(0,+), denotes the set of all positive integers, ‘opt’ stands for ‘sup’ or ‘inf’, Y is a Banach space and X ω is an ω-complete space. Suppose that S X ω , DY and B(S) denotes the complete space of all bounded real-valued functions on S with the norm

g=sup { | g ( x ) | : x S } ( g B ( S ) )

and Φ={φ;φ: R + R + } such that φ is Lebesgue integrable, summable on each compact subset of R + and 0 ε φ(t)dt>0 for each ε>0. We prove the solvability of the functional equations

f(x)= opt y D { u ( x , y ) + H ( x , y , f ( T ( x , y ) ) ) }
(5.1)

for all xS in B(S). First, we recall the following lemma [13].

Lemma 5.1 ([13])

Let E be a set, and let p,q:ER be mappings. If opt y E p(y) and opt y E q(y) are bounded, then

| opt y E p(t) opt y E q(t)| sup y E | p ( y ) q ( y ) | .

Theorem 5.2 Let u:S×DR, T:S×DS, H:S×D×RR, ϕΦ. Suppose that u and H are bounded such that

0 c | H ( x , y , g ( T ( x , y ) ) ) H ( x , y , h ( T ( x , y ) ) ) | λ φ(t)dtk 0 l g h λ φ(t)dt
(5.2)

for all (x,y,g,h,λ)S×D×B(S)×B(S)× R + , and some 0<k<1 and c>l>0. Then the functional equation (5.1) has a unique solution wB(S) and { A n z } n N converges to w for each zB(S), where the mapping A is defined by

Az(x)= opt y D { u ( x , y ) + H ( x , y , z ( T ( x , y ) ) ) } (xS).
(5.3)

Proof By boundedness of u and H, there exists M>0 such that

sup { | u ( x , y ) | + | H ( x , y , z ( T ( x , y ) ) ) | : ( x , y , t ) S × D × R } M.
(5.4)

It is easy to show that A is a self-mapping in B(S) by (5.3), (5.4) and Lemma 5.1. Using [[14], Theorem 12.34] and ϕΦ, we conclude that for each ε>0, there exists δ>0 such that

C φ(t)dt<ε,C[0,2M] with m(C)δ,
(5.5)

where m(C) denotes the Lebesgue measure of C.

Let xS, h,gB(S). Suppose that opt y D = inf y D . Clearly, for c>l, (5.3) implies that there exist y,zD satisfying

Ag(x)>u(x,y)+H ( x , y , g ( T ( x , y ) ) ) δ c ;
(5.6)
Ah(x)>u(x,z)+H ( x , z , h ( T ( x , z ) ) ) δ c ;
(5.7)
Ag(x)u(x,z)+H ( x , z , g ( T ( x , z ) ) ) ;
(5.8)
Ah(x)u(x,y)+H ( x , y , h ( T ( x , y ) ) ) .
(5.9)

Put H 1 =H(x,y,g(T(x,y))), H 2 =H(x,y,h(T(x,y))), H 3 =H(x,z,g(T(x,z))), H 4 =H(x,z,h(T(x,z))).

From (5.6) and (5.9), it follows that

c ( A g ( x ) A h ( x ) ) > c ( H ( x , y , g ( T ( x , y ) ) ) H ( x , y , h ( T ( x , y ) ) ) ) δ > l ( H ( x , y , g ( T ( x , y ) ) ) H ( x , y , h ( T ( x , y ) ) ) ) δ max { l | H ( x , y , g ( T ( x , y ) ) ) H ( x , y , h ( T ( x , y ) ) ) | , l | H ( x , z , g ( T ( x , z ) ) ) H ( x , z , h ( T ( x , z ) ) ) | } δ = max { l | H 1 H 2 | , l | H 3 H 4 | } δ .

Similarly, from (5.7) and (5.8), we get

c ( A h ( x ) A g ( x ) ) > c ( H ( x , z , h ( T ( x , z ) ) ) H ( x , z , g ( T ( x , z ) ) ) ) δ > l ( H ( x , z , h ( T ( x , z ) ) ) H ( x , z , g ( T ( x , z ) ) ) ) δ max { l | H ( x , y , g ( T ( x , y ) ) ) H ( x , y , h ( T ( x , y ) ) ) | , l | H ( x , z , g ( T ( x , z ) ) ) H ( x , z , h ( T ( x , z ) ) ) | } δ = max { l | H 1 H 2 | , l | H 3 H 4 | } δ .

So

c ( A g ( x ) A h ( x ) ) <max { l | H 1 H 2 | , l | H 3 H 4 | } +δ.

Then

c | A g ( x ) A h ( x ) | λ <max { l | H 1 H 2 | λ , l | H 3 H 4 | λ } + δ λ
(5.10)

for each λ>0.

Similarly, we infer that (5.10) holds also for opt y D = sup y D . Combining (5.2), (5.5) and (5.10) yields

0 c | A g ( x ) A h ( x ) | λ φ ( t ) d t 0 max { l | H 1 H 2 | λ , l | H 3 H 4 | λ } + δ λ φ ( t ) d t = max { 0 l | H 1 H 2 | λ + δ λ φ ( t ) d t , 0 l | H 3 H 4 | λ + δ λ φ ( t ) d t } = max { 0 l | H 1 H 2 | λ φ ( t ) d t + l | H 1 H 2 | λ l | H 1 H 2 | λ + δ λ φ ( t ) d t , 0 l | H 3 H 4 | λ φ ( t ) d t + l | H 3 H 4 | λ l | H 3 H 4 | λ + δ λ φ ( t ) d t } max { 0 l | H 1 H 2 | λ φ ( t ) d t , 0 l | H 3 H 4 | λ φ ( t ) d t } + max { l | H 1 H 2 | λ l | H 1 H 2 | λ + δ λ φ ( t ) d t , l | H 3 H 4 | λ l | H 3 H 4 | λ + δ λ φ ( t ) d t } k 0 l g h λ φ ( t ) d t + ε ,

which means that

0 c A g A h λ φ(t)dtk 0 l g h λ φ(t)dt+ε

for each λ>0. Letting ε 0 + in the above inequality, we deduce that

0 c A g A h λ φ(t)dtk 0 l g h λ φ(t)dt.

Thus Theorem 5.2 follows from Theorem 2.2. This completes the proof. □

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The authors are grateful to the two anonymous reviewers for their valuable comments and suggestions.

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All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

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Azadifar, B., Sadeghi, G., Saadati, R. et al. Integral type contractions in modular metric spaces. J Inequal Appl 2013, 483 (2013). https://doi.org/10.1186/1029-242X-2013-483

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Keywords

  • modular metric space
  • compatible maps
  • contractive inequality of integral type
  • fixed point