1. Introduction
For a metric space
, let
be the class of all nonempty closed and bounded subsets of
and
be the class of all nonempty compact subsets of
(it is well known that
). The mapping
defined by
is called the Pompeiu–Hausdorff metric induced by
d, where
is the distance from
p to
. For example, if we consider the set of real numbers with the usual metric
, then, for any two closed intervals
and
we have
.
In 1969, Nadler [
1] extended the Banach contraction principle as follows:
Theorem 1 ([
1])
. Let be a complete metric space and be a multi-valued mapping such thatfor all , where . Then, Υ has at least one fixed point. Recently, Wardowski [
2] gave a new generalization of Banach contraction to show the existence of the fixed point for such contraction by a more simple method of proof than the Banach’s one. After that, several authors studied different variations of Wardowski contraction for single-valued and multivalued mappings—for example, see [
3,
4,
5,
6,
7,
8]. On the other hand, Aydi et al. [
9] studied a common fixed point for generalized multi-valued contractions. In this paper, we introduce the concept of
-
-contraction for a pair of multi-valued mappings and prove the existence of common fixed point for such mappings. Our results generalize and improve many existing results in the literature (for instance, [
7,
9]). In addition, an illustrative example and an application to the system of Volterra-type integral inclusions are given.
2. Preliminaries
In the sequel, we recall some definitions and results which will be used in this article. Following [
2], denote by
the collection of all functions
satisfying the following conditions:
- (Ω1)
is strictly increasing,
- (Ω2)
For each sequence in , if and only if ,
- (Ω3)
There exists such that .
Definition 1 ([
2])
. Let be a metric space. A mapping is said to be an Ω-contraction if there exist and such that for all , It should be noted that any contraction is an
-contraction. To see this, suppose that
is a contraction on a metric space
with constant
that is,
, for all
. If
,
and we have nothing to prove. In the case where
, taking ln on both sides of the contraction, we get
for all
with
. Putting
and
in the above inequality, we have an
-contraction.
Example 1 ([
2])
. The functions defined by- (1)
,
- (2)
,
- (3)
,
- (4)
,
belong to Ξ.
Theorem 2 ([
2])
. Let be a complete metric space and be an Ω-contraction. Then, Υ has a unique fixed point μ in Λ and for any point the sequence converges to μ. In 2012, Samet et al. [
10] introduced the notion of
-admissible mapping as follows:
Let be a nonempty set. The selfmap on is called -admissible whenever there exists a map such that implies , for all . In addition, it is well known that is called -regular, if for any sequence in that and for all n, then for all n. In 2013, Mohammadi et al. introduced the notion of -admissiblity for multi-valued mappings as follows:
Definition 2 ([
11])
. Let Λ be a nonempty set and is the set of all nonempty subsets of Λ. A multi-valued mapping is called α-admissible, if there exists a function such that, for each and with then for all . 3. Main Results
Let denote the set of all the functions satisfying:
- ()
for all ;
- ()
for all
- ()
is nondecreasing and upper semi-continuous.
Example 2. The functions defined by
- (1)
where ,
- (2)
belong to Φ.
It is easy to see that any function satisfying () has the property that for all
Definition 3. Let Λ be a nonempty set. We say that a pair of multi-valued mappings is α-admissible, if there exists a function such that
- ()
for each and with then for all ,
- ()
for each and with then for all .
It is well known that a function is called symmetric if implies for all . We say that a pair of multi-valued mappings is symmetric -admissible if there exists a symmetric function such that is -admissible.
Definition 4. We say that a pair of mappings is α--contraction whenever there exist , and such thatfor all with and where Theorem 3. Let be a complete metric space and be two mappings such that is an α--contraction. Assume that the following assertions hold:
- (i)
There exists and such that ,
- (ii)
is a symmetric α-admissible pair.
Then, Υ and Γ have a common fixed point provided that one of the following holds:
- (C1)
Υ and Γ are continuous,
- (C2)
Ω is continuous and Λ is α-regular.
Proof. It is easy to check that, if , then and it is a common fixed point of and . Let be as in the assumption (i) that is, and be such that . We consider the following steps:
Step 1: If
, then
is a common fixed point of
and
. Thus, we may assume that
. Now, we have
Consider the following two cases:
- (Case a):
, that is,
. In this case, since
is symmetric
-admissible pair,
and
by
we have
. If
, then by
-
-contractivity of the pair
we have
which is a contradiction. Hence,
and so
is a common fixed point of
and
.
- (Case b):
. In this case, we have
. Since
and the pair
is
-
-contraction, we have
In the case
, we have
, which contradicts with (
). Hence,
and then we have
On the other hand, since
is compact, there exists
such that
Substituting in (
6), we get
Note that, since
is symmetric
-admissible pair, we have
.
Step 2: If
, then
is a common fixed point of
and
. Thus, we may assume that
. Now, we have
Consider two cases:
- (Case c):
that is,
. In this case, since
is symmetric
-admissible pair,
and
by
we have
. If
, then, by
-
-contractivity of the pair
, we have
which is a contradiction. Hence,
and so
is a common fixed point of
and
.
- (Case d):
. In this case, we have
. Since
and the pair
is
-
-contraction, we have
In the case
, we have
, which contradicts with (
). Hence,
, and so
On the other hand, since
is compact, there exists
such that
. Substituting in (
9), we get
Substituting (
7) in (
10), we get
Continuing this process, either we find a common fixed point of
and
or we can construct a sequence
in
such that
for all
and
for all
.
Put
. Then, from (
12), we have
as
. Thus,
. From
,
. Then, for any
, we have
Taking the limit on both sides of the above inequality, we obtain
. In addition, from (
), there exists
such that
. Now, we have
Taking the limit on both sides of the above inequality, we obtain
, and so
. Therefore, there exists
such that
for all
. Now, for any
with
, we have
From the above and from the convergence of the series
we receive that
is a Cauchy sequence. From the completeness of
there exists
such that
.
Suppose that the condition
is satisfied. Then,
and
Thus,
is a common fixed point of
and
.
Now, suppose that holds. Since is -regular, we have . Then, we consider two cases:
- (i)
There exists such that for all one has Then, . Since and is closed we get .
- (ii)
There exists a subsequence
of
such that
In this case, suppose, on the contrary, that
. Then,
Taking the limit on both sides of the above inequality, we obtain , a contradiction. Thus, and so .
A similar technique can be used to show that . □
Taking in Theorem 3, we obtain the following result.
Corollary 1. Let be a complete metric space and be two mappings satisfyingfor all with and , where andAssume that the following assertions hold: - (i)
There exists and such that ,
- (ii)
is a symmetric α-admissible pair.
Then, Υ and Γ have a common fixed point provided that one of (C1) and (C2) holds.
Taking in Corollary 1, we obtain the following result.
Corollary 2. Let be a complete metric space and be two mappings satisfyingfor all with and , whereAssume that the following assertions hold: - (i)
There exists and such that ,
- (ii)
is a symmetric α-admissible pair.
Then, Υ and Γ have a common fixed point provided that one of (C1) and (C2) holds.
Example 3. Let and Define byandDefine a function by if and , otherwise. Then, for any with and we have the following cases: - Case 1:
and . Then,and . Hence, we have - Case 2:
and . Then,and . Hence, we have - Case 3:
and . Then,and . Hence, we have - Case 4:
and . Then,and . Hence, we have - Case 5:
. Then,andHence, we have - Case 6:
. Then,andHence, we haveOn the other hand, it is easy to see that is a symmetric α-admissible pair. In addition, if we take , then and . Thus, by Corollary 2, Υ and Γ have a common fixed point. Here, 0 is a common fixed point of Υ and Γ. Note that Υ and Γ are not a generalized contraction. SinceTheorem 2.2 in [
9]
can not apply to this example.
Defining by , for all in Theorem 3, we have the following result.
Theorem 4. Let be a complete metric space and be two mappings satisfyingfor all with where and . If Υ, Γ or Ω be continuous, then Υ and Γ have a common fixed point. Taking in Theorem 4, we obtain the following corollary.
Corollary 3. Let be a complete metric space and be two mappings satisfyingfor all with where , andIf Υ, Γ or Ω is continuous, then Υ and Γ have a common fixed point. Taking in the Corollary 3, we obtain the following corollary.
Corollary 4. Let be a complete metric space and be two mappings satisfyingfor all with , whereIf Υ, Γ are continuous, then Υ and Γ have a common fixed point.