#### Abstract

In this work, two Geraghty type contractions are introduced in -metric spaces, and some fixed point theorems about the contractions are proved. At the end of this article, a theorem about unique solution of an integral function is proved.

#### 1. Introduction

It is well known that fixed point theorem is an important tool for solving many equations in the study of mathematics, such as integral equations [1] and differential equations [2]. It can be applied in several subjects as well, like game theory [3] and economics [4].

The notion of -metric spaces was introduced by Mustafa and Sims [5] as a generalization of metric spaces. Thereafter, -metric spaces have been studied and applied to obtain different kinds of fixed point theorems, see [612]. Aghajani et al. [13] introduced the notion of -metric spaces based on -metric spaces and -metric spaces introduced by Bakhtin in [14]. Also, some further fixed point theorems were studied after -metric spaces, which related to partial metric spaces and -metric spaces, introduced by Zand and Nezhad in [15]. Ansari et al. [16] proved some common fixed point results in complete -metric spaces with a new approach. More recently, some Geraghty type contraction results were studied in various metric spaces, see [1719].

In this work, we introduce two Geraghty type contractions in -metric spaces and investigate some fixed point theorems about such contractions. In [13], -metric space was introduced as follows.

Definition 1 (see [13]). Let be a nonempty set and be a given real number. Suppose that is a function satisfying the following properties: (1) if ;(2) for all with ;(3) for all with ;(4), where is a permutation of ;(5) for all .

Then the function is called a generalized -metric or a -metric on . The pair is called a -metric space.

It is obvious that -metric space is effectively larger than that of -metric space. Actually, each -metric space is a -metric space with .

Definition 2 (see [13]). A -metric space is said to be symmetric if for all .

Proposition 3 (see [13]). Let be a -metric space. Then for each , it satisfies the following properties: (1)If , then ;(2);(3);(4).

In this paper, we denote as the set of all positive integers and as the set of all real numbers.

Definition 4 (see [13]). Let be a -metric space and be a sequence in such that , then is -convergent; that is, for any , there exists such that , for all .

Proposition 5 (see [13]). Let be a -metric space. The following statements are equivalent: (1) is -convergent to ;(2);(3);(4) as .

Definition 6 (see [13]). Let be a -metric space. A sequence is called a -Cauchy sequence if for any , there is such that for all ; that is, as .

Definition 7 (see [13]). A -metric space is called -complete if every -Cauchy sequence is -convergent in .

Let be the family of all functions which satisfies the condition implying .

Example 8. Let Then .

Let be the family of all functions which satisfies the condition implying .

Example 9. Let Then, .

In [20], Karapınar et al. proved the following result.

Theorem 10. Let be a complete metric-like space and be a mapping. Suppose that there exists such thatfor all , then has a unique fixed point with .

Recently, Aydi et al. [21] introduced a type of Geraghty contraction in metric-like spaces and proved a fixed point theorem about such contraction as follows:

Theorem 11. Let be a complete metric-like space and be a mapping. Suppose that there exists such thatfor all , whereThen has a unique fixed point with .

In our work, enlightened by the preceding works, we introduce the Geraghty contraction to the -metric space and prove fixed point theorems for Geraghty type contractions. At the end, we give an application about a unique solution of an integral function.

#### 2. Main Results

Theorem 12. Let be a complete -metric space and be a given mapping. Suppose there exists such that for all whereThen has a unique fixed point .

Proof. Let . Define a sequence in by for all . Assume that for some , then is the fixed point of ; the proof is completed. Thus, we assume for all . From (6), we havewhere TakeThen, (8) becomesSuppose that there exists such that , then from (11), we have which is a contradiction.
Thus, for all , . From (11), we haveThe real sequence is decreasing; suppose there exists such that . Assume that . Taking in (13), we getThere are two situations that need to be discussed.
(A) When , then (14) becomes We can obtainSince , we getwhich is a contradiction.
(B) When , according to (14) we have which is a contradiction.
In conclusion of the above two conditions, we have , that isNote that thus, we get We shall prove that is a Cauchy sequence in . Equationwill be proved.
Suppose (22) does not hold. Then there exists for which we can find subsequences and of with such that for every Corresponding to , we can find with the smallest index and and satisfying (23), thenBy (23) and (24), we haveTaking in (25) and using (19), we obtainBack to (6), there is Let Taking in the above inequations and by (19), (24), we have Then we obtainFrom (26) and (30), we get We deduce thatSince and by (19), thenFrom (19) and (33), we have which is a contradiction. Thusand is a Cauchy sequence in complete -metric space. So there exists , such thatBy Proposition 3, we have ; therefore by (36) we getWe shall prove that is a fixed point of . Assume that , then . From (6) and (7), we havewhere We also have Taking in the above inequation and using (36), (37), and (38) obtains which impliesSince , thenwhich is a contradiction. Thus and so . Consequently, is a fixed point of .
We shall prove that such is the unique fixed point of . We argue by contradiction. Assume there exists such that . We haveFrom (6), we have which is a contradiction. Thus, has a unique fixed point.

We now present the following result.

Theorem 13. Let be a complete -metric space and be a given mapping. Suppose there exists such thatfor all whereThen has a unique fixed point .

Proof. Let . Define a sequence in by . Assume that for , that is, , then , i.e., . Therefore is a fixed point of .
Suppose for all . From (46) we havewhere TakeThen (48) becomesSuppose there exists such that . From (51), we have which is a contradiction. Thus for all , and the real sequence is decreasing.
Suppose there exists such that . Now we shall prove thatApplying (51), we getTake in (54) to write We obtainSince , thenThusThat isNote thatthen we can getNow we shall prove that is a Cauchy sequence in -metric spaces. We will prove thatSuppose (62) does not hold, then there exists for which we can find subsequences and of with such thatCorresponding to , we can find with the smallest index satisfying (63) and . That isFrom (46) and (63), we havewhereFrom (65), (66), and applying (59), we haveOn the other hand, applying (64) we have Taking in the above inequations, we haveFrom (67) and (69) we obtain which is a contradiction with .
We shall consider the situation which with . From (63) and (64) we haveTaking in (71) and by (59) we haveIn the meantime, By taking and using (59) in the above inequation, we can deduce thatBy (46), we havewhere By using (59), we obtainTaking in (75), we can deduce thatSince , we havewhich is a contradiction. In conclusion of two situations, is a Cauchy sequence in complete -metric spaces. So there exists such thatWe can confirm that . In fact, if , from (46) and (47) we have where We also have Taking in the above inequation and using (80), we obtainWhen , we haveSince , thenwhich is a contradiction.
When , we get a contradiction from (84).
Thus, in conclusion of two situations, and so . is a fixed point of . We shall prove that such is the unique fixed point of . We argue by contradiction. Assume there exists such thatWe haveFrom (46), we have which is a contradiction. Thus, there exists a unique fixed point such that .

#### 3. Application

Let be the set of real continuous functions defined on . Take the -metric given by for all . Then is -metric spaces with . Consider the following integral equationwhere and are two continuous functions and is a function such that for all . Consider the operator defined by

Theorem 14. Suppose that the following conditions are satisfied: (1)there exists for all (2)there exists such that where Then the integral equation (91) has a unique solution in .

Proof. It is clear that any fixed point of (92) is a solution of (91). By conditions (1) and (2), we get