1. Introduction and Preliminaries
In 1968, Bryant relaxed the assumption of Banach contraction mapping principle by using an iteration of the mapping.
Theorem 1 ([
1])
. Let T be a self mapping on the complete metric space , and m a positive integer. Suppose that there exists such that T satisfies the inequalityfor all , where denotes the mth iterate of T. Then, there exists exactly one fixed point of T. After then a number of authors deepen the research by considering an iteration of the mapping, see e.g., [
2,
3,
4,
5,
6,
7]. We recollect some significant results in this direction. One of the pioneer report in this way was given Seghal [
4].
Theorem 2 ([
4])
. Let be a complete metric space, T a continuous self-mapping of which satisfies the condition that there exists a real number q, such that, for each there exists a positive integer such that, for each ,Then T has a unique fixed point in .
Guseman [
2] extended this result by removing the condition of continuity of
T and later, other extensions for a single mapping were discussed in several papers, see e.g., Iseki [
8], Matkowski [
3], Singh [
5] and the reference therein. One of the most interesting results for mappings which satisfy a general contractive conditions were announced by Singh.
Theorem 3 ([
5])
. Let be a complete metric space and be a mapping such that for all we can find a positive integer such thatwhere are nonnegative functions such thatThen T has a unique fixed point .
In this paper, we consider more general contractive condition in the setting of dislocated metric space. For sake of completeness, we shall recollect some basic notions and fundamental results.
Definition 1. For a nonempty set a dislocated metric is a function such that for all :
- ()
,
- ()
,
- ()
.
The space is said to be a dislocated metric space (DMS).
Example 1. Let and defined by . The pair forms a dislocated metric space.
It is obvious that any metric space is a dislocated metric space, but conversely this is not true.
Definition 2. Let be a DMS. A sequence in is called:
- (a)
convergent to a point if the following limit exists and is finite - (b)
Cauchy if the following limitexists and is finite.
Moreover, if , then is said that is a 0-Cauchy sequence.
Definition 3. The DMS is complete if for each Cauchy sequence in , there is some such that Particularly, if each 0-Cauchy sequence converges to a point the pair is said to be 0- complete-DMS.
Definition 4. Let be a DMS. A mapping is continuous if for any sequence in converging to , we have converges to .
Proposition 1 ([
9])
. Let be a DMS. For any we have the following- (i)
If then .
- (ii)
If then .
- (iii)
If is a sequence in such that then
Definition 5. By a comparison function we mean a function with the following properties:
- ()
φ is increasing;
- ()
, for .
We denote by Φ the class of the comparison function .
Next we list some basic properties of the comparison functions.
Proposition 2 ([
10,
11])
. If φ is a comparison function then:- (cfi)
each is a comparison function, for all ;
- (cfii)
φ is continuous at 0;
- (cfiii)
for all .
Definition 6 ([
10])
. A function is called a comparison function if:- (ccf1)
is monotone increasing;
- (ccf2)
, for all .
We denote by the family of comparison functions.
It can be shown that every comparison function is a comparison function.
Throughout this paper we denote by the collection of all comparison functions that satisfy the following condition
(ccf3) .
In the following we recall the concept of -admissible mappings. A function is said to be -admissible if
(A) ,
for all where is a given function. An -admissible map which satisfies the condition
(TA) and implies that ,
is said to be triangular -admissible.
Later, the notion of
-admissible mapping and triangular
-admissible mappings are refined by Popescu [
12], as follows:
Definition 7 ([
12])
. Let and We say that T is an orbital admissible mapping if for all we have(O) .
Every
admissible mapping is an
-orbital admissible mapping, for more details on admissible mapping, see e.g., [
13,
14,
15,
16,
17,
18,
19,
20,
21,
22,
23,
24].
Definition 8 ([
12])
. Let An α-orbital admissible function is said to be triangular α-orbital admissible if it satisfies(TO) and implies that , for all .
At the end of this section, we present two further concepts that will be essential in our next considerations.
A set is regular with respect to mapping if the following condition is satisfied:
- (R)
for any sequence in such that for all n and as we have , for all n.
A map is said to satisfy the condition (U) if
- (U)
for any fixed point v of we have for any , where is a positive integer.
2. Main Results
We are now prepared to establish the main result of this paper.
Theorem 4. Let be a complete DMS, a function , and . Suppose that for all we can find a positive integer such that for any Suppose also that:
- (i)
T is triangular α-orbital admissible;
- (ii)
there exists in such that ;
- (iii)
either T is continuous, or
- (iv)
the space is regular and α satisfies the condition (U).
Then the function T has exactly one fixed point.
Proof. Consider the initial value
and define a sequence
as follows:
If we denote for any , then we can write .
Now, T is α-orbital admissible and
. Thus, from condition (O), we have
and so forth
Taking into account (TO) and (
8) we easily infer that
Recursively, we can conclude that
for all
.
In the initial inequality (
6) letting
,
and using (
9) we can find a positive integer,
such that
Since
, the condition
is satisfied and applying
we obtain
Let
such that
Then from (
11) together with
we get that
Using the same arguments, we can find a positive integer
such that
where
is chosen such that
Very easily we can see from (
12), (
13) and taking into account
, that
Since ψ is monotone increasing, by continuing this process, we find that
for
. On one hand the inequality (
15) shows us, taking into account
from Definition 5 that
On the other hand using triangle inequality, for
, we have
We should focus our attention on the set
. More precisely, we will show that this set is bounded. In order to prove that, we mention first that by hypothesis there exists a positive integer
such that (
6) holds. Let
, where
and
. According to
, we can find
such that
It is clear then that
and we will show that
for all
. We suppose the contrary, that there exists
such that
. Note that (according to (
6), (
9) and triangle inequality)
Since
ψ is increasing, from (
19) we get
which contradicts (
18). This contradiction shows that our assumption was false. Thus, for all
We have thus demonstrated that the set
is bounded, and also, varying
, the set
is bounded. Hence,
With this observation, we return to (
17) and we get
The series
is convergent due to
and its sequence of partial sums, denoted by
, is convergent at
S. Then
as
, and, therefore
is a 0-Cauchy sequence. By completeness of
, there is some point
such that
From the continuity of
T it easily follows that
and by the uniqueness of the limit, we get
.
We claim now that
is a fixed point of
T under the hypothesis (iv). The first step in our proof is to show that
which means that
is a fixed point of
. Firstly we claim that
. Taking
and
in (
6), there exists
such that for any
we have
Let
such that
Using triangle inequality, we have
and
Then, from (
25) it follows that
Repeating this process and keeping in mind the properties
,
we find that
Suppose now that
. Then
. Since
as
and the space
is regular, by triangle inequality we have
Letting
in the above inequality, and taking (
24) respectively (
27) into account, we find that
which implies that
. Hence,
.
Let
another point such that
and
. Since
T satisfies (
6) and the function
α satisfies the condition (U) we get
But the above inequality is possible only if , that is . This is a contradiction.
From the uniqueness of the fixed point we can conclude that
is a fixed point for
T. Indeed,
shows that
is also fixed point of
. But,
has a unique fixed point
. Hence,
. □
Remark 1. Denoting by , we have from (22) Letting in the above inequality we obtain Example 2. Let and with for any , defined as follows(It is easy to see that the pair is a DMS but not a metric space.) Let be defined by For and we havefor any . Thus, T does not satisfy the Banach contraction condition. We show that the function T satisfies all the assumptions of Theorem 4 with for any and , . Since for all the assumptions , and are fulfilled. We discuss the following possible cases: - 1.
If and then for we have and inequality (6) holds. - 2.
If then and also (6) holds. - 3.
If then . Then, for we have . Thus, the condition (6) is satisfied. - 4.
If then and . Then, for we have and . Thus, - 5.
If then and . Then, for we have and . Thus,
On the other hand, we can note that: For this reason, there exists such that for any the condition (6) is satisfied (since ). The conclusion is that T satisfies all the assumptions of Theorem 4. Therefore T has exactly one fixed point, .
Taking, in Theorem 4,
we get the following result:
Corollary 1. Let be a complete DMS, a function and . Suppose that for all we can find a positive integer such that Then the function T has exactly one fixed point.
Corollary 2. Let be a DMS, a function and . Suppose that for all we can find a positive integer such that Then the function T has exactly one fixed point.
If we take
,
in Corollary 1 respectively in Corollary 2 we find the following consequences:
Corollary 3. Let be a complete DMS, a function . Suppose that for all we can find a positive integer such that Then the function T has exactly one fixed point.
Corollary 4. Let be a complete DMS, a function . Suppose that for all we can find a positive integer such that Then the function T has exactly one fixed point.
Corollary 5. Let be a complete DMS, a function . Suppose that for all we can find a positive integer such thatwhere and . Then the function T has exactly one fixed point. Proof. Letting the result follows from Theorem 4. □
If we take
in Theorem 4 we get:
Corollary 6. Let be a complete DMS, a function and . Suppose that for all Suppose also that:
- (i)
T is triangular α-orbital admissible;
- (ii)
there exists in such that ;
- (iii)
either T is continuous, or
- (iv)
the space is regular and the condition (U) is satisfied.
Then the function T has exactly one fixed point.
Example 3. Let a dislocated metric space , where and for any . Let a self mapping T on be defined as follows: Let the functions , and , Let us first notice that for any and . Since we get that assumption of Theorem 4 is satisfied. Also, since by simple calculation we can conclude that the assumptions and are satisfied. We remark that if and then , . Hence,andwhich shows us that T does not satisfy the contraction condition of Banach, neither condition (36) of Corollary 6. We must discuss the next cases: - 1.
If then for we have and . Thus,where - 2.
If we can choose . Then, and - 3.
If and then for we have and - 4.
If , and then and The other cases are not interesting since . Therefore is the unique fixed point for T.
Inspired by Proposition 3 from [
7] we will establish a new fixed point result for a
T function on a DMS, not necessarily complete.
Corollary 7. Let be a DMS and a function . Suppose that for a given such that we can find a positive integer such that the following two conditions hold:andfor any and Suppose also that there exists a point such that , where . Then is fixed point of T.
Proof. Suppose that
. Then we can find a positive integer
such that
Replacing
in (
38), using triangle inequality and keeping in mind (
39) we have
or, since
we get
This is a contradiction. Hence and a is fixed point of T. □
4. Application to Boundary Value Problem
Here we consider the following two point boundary value problems for the second order differential equation.
where
is a continuous function. Recall that the Green’s function associated to (
45) is
Let be the space of all continuous functions defined on We consider on , the dislocated metric given by for all where for each .
Clearly, is a complete DMS.
It is well known that
is a solution of (
45) is equivalent to that
is a solution of the integral equation.
Theorem 6. Let be a complete DMS as defined above. Further, we will assume the following conditions hold:
- 1.
there exists a continuous function such thatfor each and ; - 2.
there exists a continuous function such thatfor each and ; - 3.
;
- 4.
.
The problem (45) has a solution . Proof. Define the mapping
by
for all
and
be a positive integer. Then the Equation (
47) is equivalent to finding
that is a fixed point of
T.
Now let
. We have,
for each
. On the other hand,
Thus
. Similarly, we derive that
Take
. Thus
. By using (
49)–(
51), we get,
Hence
. Thus all the conditions of Corollary 4 are satisfied. Hence
T has exactly one fixed point
i.e., the problem (
45) has a solution