1. Introduction
The compactness plays an essential role in the Schauder's fixed point theorem and however, there are some important problems where the operators are not compact. G. Darbo in 1955 [1], extended the Schauder theorem to noncompact operators . The main aim of their study is to define a new class of operators that map any bounded set to a compact set. The first measure of noncompactness was defined and studied by Kuratowski [2] in 1930.
Suppose (X, d) be a metric space the Kuratowski measure of noncompactness of a subset
A ⊂ X defined as
where diam (A) denotes the diameter of a set A ⊂ X namely
Also, in recent years measures of noncompactness have been used to define new geometrical properties of Banach spaces which are interesting for fixed point theory [3]. In this paper first, some essential concepts and results concerning the measure of noncompactness are called [4-7]. In the second section, a common fixed point for the maps T and S where T and S are self-map continuous, commuting continuous on a closed convex subset C of a Banach space E and also S is a linear map is shown. In the third section, we apply our result to obtain a coupled fixed point [8-11] . Finally by applying our results a solution of an integral equation is obtained [12-15].
Now, we recall some basic facts concerning measures of noncompactness. Suppose R denotes the set of real numbers and put
and let
be a Banach space. The symbol
will denote the closure and closed convex hull of a subset X of E , respectively. Moreover, let indicate the family of all nonempty and bounded subsets of E and indicate the family of all nonempty and relatively compact subsets.
We begin by recalling some needed definitions and results.
Definition 1.1 A mapping
is said to be a measure of noncompactness in E if it satisfies the following conditions:
The family
is nonempty and
.
2.
.
3.
.
4.
.
5.
for
.
6. If
is a sequence of closed sets from
such that
for
, and if
, then
Theorem 1.1 (Schauder [9]) Let C be a closed and convex subset of a Banach space E. Then every compact and continuous map has at least one fixed point.
In 1955, G. Darbo [1] used the measure of noncompactness to generalize Schauder's theorem to a wide class of operators, called k-set contractive operators, which satisfy the following condition
for some
. In 1967 Sadovskii generalized Darbo's theorem to set-condensing operators [16,17].
Definition 1.2 Let E1 and E2 be two Banach spaces and µ1 and µ2 be arbitrary measures of noncompactness on E1 and E2 respectively [5] . An operator T from E1 to E2 is called a (µ1, µ2) condensing operator if it is continuous and for every bounded noncompact set
the following inequality holds
The following lemmas and theorems from [16-18] are necessary for the main results.
Theorem 1.2 (Darbo's fixed point theorem) Let Ω be a nonempty, bounded, closed, and convex subset of a Banach space E and let
be a continuous mapping such that there exists a constant
with the property [18].
For any nonempty subset X of Ω Then T has a fixed point in the set Ω.
Lemma 1.3 For every nondecreasing and upper semicontinuous function
The following two conditions are equivalent:
i.
for any
.
ii.
for any
.
The following theorem is an extension of Darbo's fixed point theorem.
Theorem 1.4 [7] Let C be a nonempty, bounded, closed, and convex subset of a Banach space E and
be a continuous operator satisfying
for any subset X of C, where µ is an arbitrary measure of noncompactness and
is a nondecreasing and upper semicontinuous function such that
for all
. Then T has at least one fixed point.
2. Common fixed point
Theorem 2.1 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E.
and let
be continuous operators and S be a linear operator such
that and also
for each
, where µ is an arbitrary measure of noncompactness and
is a nondecreasing function such that
for each
and
. Then T, S have a common fixed point in C.
Proof. Set
And
in general, set
For n = 1,2, …
Then we have
for ever n = 1,2,3, …
Indeed it is clear that
and
.
So () holds for n = 1.
Assuming now that () is true for
.
Then
And
We obtain
Now if there exists an integer
such that
, then CN is relatively compact and since
, thus Schauder's fixed point theorem implies that T has a fixed point. So we assume that
for
. By assumptions we have
which implies that
is a positive decreasing sequence of real numbers thus, there is an
so that
as
We show that r = 0 . Suppose, in the contrary, that
. Then we have
.
.
.
By Lemma 1.3 and assumption with choose
, we have
for any t > 0.
So r = 0 and hence
as
. Since
and
for all
, then from (6),
is a nonempty convex closed set, and
. Moreover, the set
is invariant under the operator T and belongs to ker µ. Thus, applying Schauder's fixed point theorem, T has a fixed point. Now, suppose that
. The set FT is closed by the continuity of T, by the assumption we have
then Sx is a fixed point of T for any
and
then
and have FT is compact.
Then by Schauder's fixed point theorem, we deduce that S has a fixed point and set
is closed by the continuity of S. Also, since
by Schauder's fixed point theorem, we have Tx is a fixed point of S for each
. Since
is a compact subset, then
are continuous self maps, now by Schauder's fixed point theorem we have a common fixed point in C.
Corollary 2.2 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E and let
be continuous operators and be a linear operator such that
and
for each
, where µ is an arbitrary measure of noncompactness and
. Then T, S have a common fixed point in C.
Proof. Let
in the Theorem 2.1 .
Corollary 2.3 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E and let
be continuous operators and S be a linear and condensing operator such that.
and
For each
, where µ is an arbitrary measure of noncompactness and
is a nondecreasing function such that
for each
and
. Then T, S have a common fixed point in C.
Proof. The result is followed by Definition 1.2 and Theorem 2.1.
Corollary 2.4 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E and let
be continuous operators and S be a linear operator such that T and S be two commuting map and
For each
, where µ is an arbitrary measure of noncompactness and
is a nondecreasing function such that
for each
and
. Then T, S have a common fixed point in C [19-23].
Proof. The proof is similar to the proof of Theorem 2.1.
Definition2.1 Let X be a Banach space. An operator (not necessarily linear)
is compact if the closure of F(Y) is compact whenever
is bounded.
Corollary 2.5 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E and let
be a linear and continuous operator such that T and S be two commuting map and
where
is a nondecreasing function such that
for each
and
. Assume that
is a compact, continuous operator. Define
and assume that
for all
. Then T, S have a common fixed point in C.
Proof. Let
be the Kuratowski measure of noncompactness defined by (1). Moreover, assume that X is a nonempty subset of C. As ϕ is non-decreasing, from (3), we have
so
By the definition of Kuratowski measure of noncompactness, for every
, there exist
such that
and
. As
and by assumption, ϕ is a non-decreasing function, from (4) we have
and
On the other hand, as G is compact, from (5) we obtain
Now, by Theorem 1.4, T has a fixed point in C, Now, suppose that
is closed by the continuity of T.
On the other hand, since S commuting with T, we see that Sx is a fixed point of T for any
.
Thus
and since
then
and have FT is compact.
Then by Schauder's fixed point theorem, we deduce that S has a fixed point and set
is closed by the continuity of S . Also, since S commutes with T, we have Tx is a fixed point of S for each
, therefore Fs is invariant by T or . Since Fs is convex closed and bounded and for any
we have
Then by Corollary 2.4, T and S have a common fixed point in D.
Corollary 2.6 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E and let
be continuous operators and S be a linear operator and G be a compact operator, define
and assume that
for all
, such that T and S be two commuting map. Then T, S have a common fixed point in C [2,24-27].
Proof. Since G is a compact operator, we have
and so
so T, S have a common fixed point in C.
Example 2.1 ([7]) Let C[a,b] denote the Banach space consisting of all real-valued functions, defined and continuous on [a,b]. The space C[a,b] is furnished with the standard norm
for every
.
A measure of noncompactness can be defined as follows. To this end let us fix a nonempty bounded subset X of C[a,b]. For
and
let us denote by
the modulus of continuity of the function x on the interval [a,b], i.e
and
It is easy to prove that ω0 is a measure of noncompactness and
In the following, we will show some examples of the results.
Example2.2 Let
a nonempty, bounded, closed, and convex subset of a Banach space
be continuous operators and S be a linear operator where
and also
For each
is holed.
Then consider
,
and
by
is a nondecreasing function such that
for
each and
.
Let µ be the same measure of noncompactness in Example 2.1. Then by Theorem 2.1, T, S have a common fixed point
in C.
Example 2.3 Let R be a Banach space and we define:
And
by
is a non-decreasing function such that
for each
and
. Clearly T, S are commuting maps and with corresponding Corollary 2.4, for any subset
, obviously we have
Then T, S have a common fixed point x = 0.
3. Common coupled fixed point
[8] An element
is called a coupled fixed point of the operator
if
and
Definition 3.2 The operators
is called commuting operator if
for all
Theorem 3.1 [7] Supposebe
measures of noncompactness on, Banach spaces
respectively. Moreover assume that the function
is convex and
if and only if
for
. Then
defines a measure of noncompactness on
where Xi denotes the natural projections of X into Ei for
Remark 3.1 [4] Let µ be a measure of noncompactness on a Banach space E considering
and
for
then conditions of Theorem 3.1 are satisfied. Therefore,
And
Define measures of noncompactness in the space
where
denote the natural
projections of X into E.
Theorem 3.2 Let C be a nonempty, bounded, closed, and convex subset of a Banach space E
and let
be continuous operators and S be a linear operator such that
and
for each
, where µ is an arbitrary measure of noncompactness and
is a non-decreasing function such that
for each
and
. Then T, S have a common coupled fixed point in C.
Proof. First note that, Remark 3.1 implies that
is a measure of noncompactness in the space
where
denote the natural projection of X. Now consider the map.
defined by the formula
Which is continuous on
. We claim that
satisfies all the conditions of Theorem 2.1. To prove this, let
be a nonempty subset. Then, by 2S0 and (6) we have
Then
and taking
we get
Since,
is also a measure of noncompactness, therefore, all the conditions of Theorem 2.1 are satisfied and G has a coupled fixed point.
4. Application
Let
be the space of Lebesgue integrable functions on the measurable subset R+ of R with the standard norm
Now, we define a measure of noncompactness in the space.
.
For
, let X be a nonempty, bounded, compact, and measurable subset of
, set
Where means(D) denotes the Lebesgue measure of the subset D and
Then we define
Where μ is a measure of noncompactness in
.
Our purpose is the study of the equation below:
under the following hypotheses.
i.
and there is a constant
such that
ii. The function
belongs to the space
, defined by
for
and
for
and
.
Therefore, we can see that for any
and for all
the following condition is satisfied:
iii.
is a non-decreasing function such that
for each
and
.
iv. The linear continuous operator K is defined by
Maps Qr into Qr. (Let E be an arbitrary Banach space with norm
and the zero element 0 and Br be a closed ball in E centered at 0 and of radius r, and also suppose Qr be the subset of Br consisting of all functions that are a.e. positive and nonincreasing on R+, which is a compact, bounded, closed, and convex subset of
.
Then we can prove the following result.
Theorem 4.1 Let the assumptions i), ii), iii), and iv) be satisfied. Then the equation (7) has at least one solution
such that
Proof.
Step 1: We consider the following operators
and
Thus the equation (7) becomes
Next, we consider
Step 2: For any
we have
hence, for
, we have
if we take
, then
. This implies that G maps the ball Br into itself, where
Step 3: For any consider and be arbitrary, let be with , then we have
When ε tends to zero and from definition
that is a defined measure in the
, we get
.
Step 4: For any
and
we have
.
Now with take
of the above inequality, we get
Where d is a defined measure in the
. Now by step 3 and step 4 we deduce that.
Step 5: Take
then
is a.e. positive and nonincreasing on R+ and consequently
is also of the same type in virtue of the assumptions (i), (ii), (iii) and (iv) we deduce that
is also a.e. positive and nonincreasing on R+. This fact, together with the assertion
gives that G is a self-mapping of the set Qr. For this reason that K is a linear and bounded operator, therefore K is continuous , and obviously F is a continuous operator then G is a continuous operator. Then K and G are continuous from Qr into Qr.
Step 6: We will show that
. We have
if and only if
therefore, if the function g satisfies
,
then g is a fixed point of K. Hence,
is a fixed point of K. Thus
Therefore,
, i.e. K and F are commuting maps. For every
we have
, so G and K are commuting maps Thus without the loss of generalities, in Corollary 2.4 enough that, we put
and
. Then K and G have at least one common fixed point, which is a solution of the equation (7) and satisfies
. Moreover,
is a common solution of the equations
and
Conclusion
This paper examines the existence of a fixed point in various cases based on the measures of incompressibility, which is a very important technique in existence proof.