Abstract
We generalize the results obtained in Cho et al. (Fixed Point Theory Appl. 2013:329, 2013) and give other conditions to prove the existence and uniqueness of a fixed point of α-Geraghty contraction type maps in the context of a complete metric space.
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1 Introduction and preliminaries
The Banach contraction principle [1] is one of the earliest and most important results in fixed point theory. Because of its application in many disciplines such as chemistry, physics, biology, computer science and many branches of mathematics, a lot of authors have improved, generalized and extended this classical result in nonlinear analysis; see, e.g., [2–10] and the references therein.
One of the interesting results was given by Geraghty [6] in the setting of complete metric spaces by considering an auxiliary function. Later, Amini-Harandi and Emami [3] characterized the result of Geraghty in the context of a partially ordered complete metric space, and Caballero et al. [11] discussed the existence of a best proximity point of Geraghty contraction. Gordji et al. [12] defined the notion of ψ-Geraghty type contraction and supposedly improved and extended the results of Amini-Harandi and Emami [3]. Recently, Cho, Bae and Karapınar [13] defined the concept of α-Geraghty contraction type maps in the setting of a metric space and proved the existence and uniqueness of a fixed point of such maps in the context of a complete metric space. Very recently, Karapınar and Samet [14] proved that the results of Gordji et al. [12] and all results inspired by the paper of Gordji et al. [12] are equivalent to existing results in the literature. For other results related to Geraghty contractions, see [13–23].
In this paper, we generalize the results obtained in [13] and give other conditions to prove the existence and uniqueness of a fixed point of α-Geraghty contraction type maps in the context of a complete metric space. Now, we remind some basic definitions and remarkable results on the topic in the literature.
Definition 1 [24]
Let be a map and be a function. Then T is said to be α-admissible if implies .
Definition 2 [16]
A map is said to be triangular α-admissible if:
(T1) T is α-admissible,
(T2) and imply .
Let ℱ be the family of all functions which satisfies the condition
By using such maps, Geraghty [6] proved the following result.
Theorem 1 Let be a complete metric space and let T be a mapping on X. Suppose that there exists such that for all ,
Then T has a unique fixed point and converges to for each .
Amini-Harandi and Emami [3] reconsidered Theorem 1 in the framework of partially ordered metric spaces.
Theorem 2 Let be a partially ordered complete metric space. Let be an increasing mapping such that there exists an element with . If there exists such that
for each with , then f has a fixed point provided that either f is continuous or X is such that if an increasing sequence , then for all n. Besides, if for each there exists which is comparable to x and y, then f has a unique fixed point.
Definition 3 [24]
Let be a metric space and be a function. A map is called a generalized α-Geraghty contraction type map if there exists such that for all ,
where .
Cho et al. [13] proved the following interesting result.
Theorem 3 Let be a complete metric space, be a function, and let be a map. Suppose that the following conditions are satisfied:
-
(1)
T is a generalized α-Geraghty contraction type map;
-
(2)
T is triangular α-admissible;
-
(3)
there exists such that ;
-
(4)
T is continuous.
Then T has a fixed point and converges to .
The continuity of the mapping T can be replaced by a suitable condition (4′) (see Theorem 2.2 [13]):
(4′) If is a sequence in X such that for all n and as , then there exists a subsequence of such that for all k.
Adding the condition (H),
-
(H)
For all , there exists such that and , where denotes the set of fixed points of T,
to the hypotheses of Theorem 3, Cho et al. [13] obtained that is the unique fixed point of T (see Theorem 2.3 [13]). However, we think that this condition is not appropriate. To verify such a condition assumes to know and then the uniqueness of a fixed point is trivial. Also, we think that the following condition is not more appropriate:
-
(iii)
[16] For all , there exists such that and .
This condition implies for all , and then the utility of α is 0. For more details, see Theorem 19 [16].
2 Fixed point theorems
In this section, we prove that the results of Cho et al. [13] are still available if we replace condition (2) of Theorem 3 with a weaker condition, and if we extend the notion of generalized α-Geraghty contraction type map.
Definition 4 Let be a complete metric space and let be a function. A map is called a generalized α-Geraghty contraction type map if there exists such that for all ,
where .
Now, we introduce two new concepts.
Definition 5 Let be a map and be a function. Then T is said to be α-orbital admissible if
(T3) implies .
Definition 6 Let be a map and be a function. Then T is said to be triangular α-orbital admissible if T is α-orbital admissible and
(T4) and imply .
Obviously, every α-admissible mapping is an α-orbital admissible mapping and every triangular α-admissible mapping is a triangular α-orbital admissible mapping. The following example shows that there exists a triangular α-admissible mapping which is not triangular α-admissible.
Example 7 Let , , , such that , , , and , if and otherwise. Since and , T is α-orbital admissible. Since , , T is triangular α-orbital admissible. But , , so T is not triangular α-admissible.
Lemma 8 Let be a triangular α-orbital admissible mapping. Assume that there exists such that . Define a sequence by . Then we have for all with .
Proof Since T is α-orbital admissible and , we deduce that . By continuing this process, we get for all . Suppose that and prove that , where . Since T is triangular α-orbital admissible and , we get that . Hence, we have proved that for all with . □
Theorem 4 Let be a complete metric space, be a function, and let be a map. Suppose that the following conditions are satisfied:
-
(1)
T is a generalized α-Geraghty contraction type mapping;
-
(2)
T is a triangular α-orbital admissible mapping;
-
(3)
there exists such that ;
-
(4)
T is continuous.
Then T has a fixed point and converges to .
Proof Let such that . Define a sequence by for . If for some , then obviously T has a fixed point. Hence we suppose that for all . By Lemma 8 we have for all . Then we get, for all ,
where
Since , the case is impossible, so we have . Thus, the sequence is positive and decreasing. Therefore, there exists such that . We will show that . Suppose, to the contrary, that . Then we have
This implies that . Since , we get , and then . This is a contradiction.
Next, we shall show that is a Cauchy sequence. Suppose, to the contrary, that is not a Cauchy sequence. Then there exists such that, for all , there exists with . Let be the smallest number satisfying the conditions above. Hence, we have . Therefore, we get
Letting , we have . Since
we get . Similarly, we obtain
By Lemma 8 we have . Thus, we deduce that
where
Clearly, we deduce that
Hence, we have
Letting , we conclude that , which yields that . Hence, , which is a contradiction. Thus, we get that is a Cauchy sequence. Since X is a complete metric space, it follows that there exists . By the continuity of T, we get , and so , which means that is a fixed point of T. □
Like in [13] we can replace the continuity of the operator T by a suitable condition.
Theorem 5 Let be a complete metric space, be a function, and let be a map. Suppose that the following conditions are satisfied:
-
(1)
T is a generalized α-Geraghty contraction type mapping;
-
(2)
T is triangular α-orbital admissible mapping;
-
(3)
there exists such that ;
-
(4)
if is a sequence in X such that for all n and as , then there exists a subsequence of such that for all k.
Then T has a fixed point and converges to .
Proof Following the lines in the proof of Theorem 4, we have that the sequence defined by for all converges to . By condition (4) of the hypothesis, we deduce that there exists a subsequence of such that for all k. Therefore, we have
where
Now, we suppose that , that is, . Letting in the above equality, we get that
Since for all k, letting , we conclude that . This implies . Hence, . This is a contradiction. Therefore, . □
For the uniqueness of a fixed point of a generalized α-Geraghty contraction type mapping, we consider the following hypothesis:
-
(K)
For all , there exists such that , and .
Remark 9 Replacing condition (3) with condition (K) in the hypotheses of Theorem 4 or Theorem 5, we obtain that is the unique fixed point of T. Suppose that and are two fixed points of T such that . Then, by (K), there exists such that , and . Since T is a triangular α-orbital admissible mapping, we get that and for all . Thus we have
for all , where
By Theorem 4 (resp. Theorem 5) we deduce that the sequence converges to a fixed point of T. Letting in the above equality, we get . If we suppose , then we have , and letting , we deduce that . This implies , so , which is a contradiction. Therefore, . Similarly, we get . Hence, , which is a contradiction.
Example 10 Let , , , such that if , if , and , if and otherwise. Then T satisfies the conditions of Theorem 5. Clearly, X is a complete metric space. If , then , so , which implies . Hence T is an α-orbital admissible mapping. For and , we have . Thus, and . Therefore, T is a triangular α-orbital admissible mapping. Also, and if is a sequence in X such that for all n and as , then and for all n. For , we have and . Thus . The case is similar. If and , then , , so . For , we have . Therefore, taking , , we obtain that T is a generalized α-Geraghty contraction type mapping. Hence, T satisfies all the conditions of Theorem 5. However, since , , T is not a triangular α-admissible mapping.
3 α-Orbital attractive mappings
Now we introduce a new concept.
Definition 11 Let be a map and be a function. Then T is said to be α-orbital attractive if
for every .
Theorem 6 Let be a complete metric space, be a function, and let be a map. Suppose that the following conditions are satisfied:
-
(1)
T is a generalized α-Geraghty contraction type mapping;
-
(2)
T is α-orbital admissible;
-
(3)
there exists such that ;
-
(4)
T is α-orbital attractive.
Then T has a unique fixed point and converges to .
Proof Let such that . Define a sequence by for . If for some , then obviously T has a fixed point. Hence we suppose that for all . Since T is α-orbital admissible, we have
Inductively, we get for all . Then we obtain
for all , where
If , we have
which is a contradiction. Thus, we get
Therefore, the sequence is positive and nonincreasing. Hence, there exists such that . We will show that . Suppose, on the contrary, that . Then we have
This implies that . Since , we obtain that , which is a contradiction. Hence .
Now, we shall show that is a Cauchy sequence. Suppose, on the contrary, that is not a Cauchy sequence. Then there exists such that, for all , there exists with . Let be the smallest number satisfying the conditions above. Hence, we have . Therefore, we get
Letting , we have . Since
we get . Similarly, we obtain
Since and T is α-orbital attractive, we have
Hence, we get two cases as follows.
-
(1)
There exists an infinite subset I of N such that for every .
-
(2)
There exists an infinite subset J of N such that for every .
In the first case, we have
where
Letting , , we conclude that
and since
we get that
Since , we obtain that
which is a contradiction.
In the second case, we have
where
Letting , , we conclude that
and since
we get that
Since , we obtain that
which is a contradiction. Thus, we get that is a Cauchy sequence. Since X is a complete metric space, it follows that there exists .
We claim that . Suppose, on the contrary, that . Since T is α-orbital attractive, we have for every that or . Hence, there exists a subsequence of such that or for all . In the first case, we have
for all , where
Letting , we conclude that
and since
we have
Since , we obtain that
which is a contradiction. The second case is similar. Therefore, .
If is another fixed point of T, from the hypothesis we deduce that or . Hence, there exists a subsequence of such that or for all . In the first case, we have
for all , where
Letting , we conclude that
and since
we obtain
Since , we get that
so . This is a contradiction. The second case is similar. □
Example 12 Let , , , such that , , , if and otherwise. Obviously, T is α-orbital admissible and α-orbital attractive. Also, T is a generalized α-Geraghty contraction type mapping if we take , . Hence T satisfies all the conditions of Theorem 6. However, since , , T is not triangular α-orbital admissible. Note that .
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Popescu, O. Some new fixed point theorems for α-Geraghty contraction type maps in metric spaces. Fixed Point Theory Appl 2014, 190 (2014). https://doi.org/10.1186/1687-1812-2014-190
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DOI: https://doi.org/10.1186/1687-1812-2014-190