#### Abstract

Lakzian and Samet (2010) studied some fixed-point results in generalized metric spaces in the sense of Branciari. In this paper, we study the existence of fixed-point results of mappings satisfying generalized weak contractive conditions in the framework of a generalized metric space in sense of Branciari. Our results modify and generalize the results of Laksian and Samet, as well as, our results generalize several well-known comparable results in the literature.

#### 1. Introduction and Preliminaries

Branciari in [1] initiated the notion of a generalized metric space as a generalization of a metric space in such a way that the triangle inequality is replaced by the “quadrilateral inequality,” for all pairwise distinct points , and of . Afterwards, many authors initiated and studied many existing fixed-point theorems in such spaces. For more details about fixed-point theory in generalized metric spaces, we refer the reader to [1–13].

The following definitions will be needed in the sequel.

*Definition 1.1 (see [1]). *Let be a nonempty set and such that for all and for all distinct points each of them different from and , one has(p1):,(p2):,
(p3):. Then, is called a generalized metric space (or shortly g.m.s).

Any metric space is a generalized metric space, but the converse is not true [1].

*Definition 1.2 (see [1]). *Let be a g.m.s, a sequence in , and . We say that is g.m.s convergent to if and only if as . We denote this by .

*Definition 1.3 (see [1]). *Let be a g.m.s and a sequence in . We say that is a g.m.s Cauchy sequence if and only if for each there exists a natural number such that for all .

*Definition 1.4 (see [1]). *Let be a g.m.s. Then, is called a complete g.m.s if every g.m.s Cauchy sequence is g.m.s convergent in .

Very recently, Lakzian and Samet [9] proved the following nice result.

Theorem 1.5. *Let be a Hausdorff and complete generalized metric space. Suppose that is such that for all **
where is continuous and nondecreasing with if and only if , and is continuous and if and only if . Then, there exists a unique point such that .*

Note that Theorem 1.5 extends a result of Dutta and Choudhury [14] to the set of generalized metric spaces. Moreover, its proof is more technical compared with that of [9].

In this paper, we generalize in some cases Theorem 1.5 by replacing in (1.1) the term by the quantity and the continuity of by lower semicontinuity. Also, we derive some useful corollaries of this result.

#### 2. Main Results

Let be a nonempty set and a given mapping. For all , set Also, let , and . Note that, if , is called an altering distance function [15].

The notion of a periodic point of a given mapping is crucial for proving our main theorem. So we need the following definition.

*Definition 2.1. *Let be a nonempty set. A given mapping admits a periodic point if there exists such that for some . If , is a fixed point.

Hence, each fixed point is also a periodic point of .

Now, in the following, let us prove our main result.

Theorem 2.2. *Let be a Hausdorff and complete generalized metric space. Suppose that is such that for all **
where , , and is defined by (2.1). Then, there exists a unique point such that .*

*Proof. *First, it is obvious that if and only if is a fixed point of . Let an arbitrary point. By induction, we easily construct a sequence such that
*Step* 1. We claim that
Substituting and in (2.2) and using properties of functions and , we obtain
which implies that
Note that
If for some , , then and by a property of , so (2.5) becomes
a contradiction. Thus, for all ,
From (2.9), the sequence is monotone nonincreasing and so bounded below. So there exists such that
Letting in (2.5) and using the above limits with the continuity of and the lower semicontinuity of , we get , which implies that , so by a property of . Thus, (2.4) is proved.*Step* 2. We shall prove that
By (2.2), we have
which implies that
where
Set and . Thus, by (2.12), one can write
which implies that
On the other hand, having in mind that the sequence is monotone nonincreasing, so
From (2.16) and (2.17), we have
Therefore, the sequence is monotone nonincreasing, so it converges to some . Assume that . Now, by (2.4), it is obvious that
Taking the in (2.15) and using (2.19) and the properties of and , we obtain
which implies that , so , a contradiction. Thus, from (2.19),
and hence , so (2.11) is proved.*Step* 3. We claim that has a periodic point.

We argue by contradiction. Assume that has no periodic point. Then, is a sequence of distinct points, that is, for all . We will show that, in this case, is g.m.s Cauchy. Suppose to the contrary. Then, there is a such that for an integer there exist integers such that
For every integer , let be the least positive integer exceeding satisfying (2.22) and such that
Now, using (2.22), (2.23), and the rectangular inequality (because is a sequence of distinct points), we find that
Then, by (2.4) and (2.11), it follows that
Now, by rectangular inequality, we have
Letting in the above inequalities, using (2.4) and (2.25), we obtain
Therefore, by (2.4) and (2.27), we get that
Applying (2.2) with and , we have
Letting in the above inequality and using (2.25) and (2.28), we obtain
which yields that , so , which is a contradiction.

Hence, is g.m.s Cauchy. Since is a complete g.m.s, there exists such that . Applying (2.2) with and , we obtain
which implies that
where
Since , so we obtain that
It follows that
Next, we shall find a contradiction of the fact that has no periodic point in each of the two following cases.(i)If, for all , and , then by rectangular inequality
and, using (2.4), we get that
From (2.35) and (2.37),
Taking the in (2.31) and using (2.34), (2.38), and the properties of and , we obtain
which implies that , so , that is, is a fixed point of , so is a periodic point of . It contradicts the fact that has no periodic point.(ii) Let for some , or . Since has no periodic point, then obviously . Indeed, if , so , that is, is a periodic point of , while if and , so , that is, is a periodic point of .

For all , we have
In the two precedent identities, the integer is fixed, and so and are subsequences from , and since g.m.s. converges to in which is assumed to be Hausdorff, so the two subsequences g.m.s. converge to same unique limit , that is,
Thus,
Again, since is Hausdorff, then by (2.42),
On the other hand, since has no periodic point, it follows that
Using (2.44) and the rectangular inequality, we may write
Letting in the above limit and proceeding as (2.4) (since the point is arbitrary), using (2.43), we obtain
Now, by (2.2),
where
Letting in (2.47) and using (2.46) and the above limit, we get that
which holds only if , that is, , which implies that is a periodic point of . This contradicts the fact that has no periodic point.

Consequently, admits a periodic point, that is, there exists such that for some .*Step* 4. Existence of a fixed point of .

If , then , that is, is a fixed point of . Suppose now that . We will prove that is a fixed point of . Suppose that it is not the case, that is, . Then, and , which implies that . Now, using inequality (2.2), we obtain
which by the monotone nondecreasing property of implies
where
because otherwise we get a contradiction with (2.51). Thus, (2.51) becomes
Again, using (2.2), we have
Again, this implies that
Where
because of (2.55). Thus, from (2.55),
Continuing this process as (2.53) and (2.57), we find that
which is a contradiction. We deduce that is a fixed point of .*Step* 5. Uniqueness of the fixed point of .

Suppose that there are two distinct points such that and . Then, and . By (2.2), we obtain
a contradiction. Thus, has a unique fixed point. This completes the proof of Theorem 2.2.

Now, we state some corollaries of Theorem 2.2, which are given in the following.

Corollary 2.3. *Let be a Hausdorff and complete generalized metric space. Suppose that is such that, for all , there exists and
**
then has a unique fixed point.*

*Proof. *It suffices to take and in Theorem 2.2.

Corollary 2.4. *Let be a Hausdorff and complete generalized metric space. Suppose that is such that, for all , there exists and
**
then has a unique fixed point.*

*Proof. *Let , so . Also, if (2.61) holds, so
Then, it suffices to apply Corollary 2.3.

Another easy consequence of Corollary 2.3 (a Reich contraction type) is the following.

Corollary 2.5. *Let be a Hausdorff and complete generalized metric space. Suppose that is such that, for all , there exists and
**
then has a unique fixed point.*

Corollary 2.6. *Let satisfy the conditions of Theorem 2.2, except that condition (2.2) is replaced by the following: there exist positive Lebesgue integrable functions and on such that and for each and that
**
Then, has a unique fixed point.*

*Proof. *Consider the functions
Then, (2.64) becomes
And, putting and and applying Theorem 2.2, we obtain the proof of Corollary 2.6 (it is easy to verify that and ).

Corollary 2.7. *Let be a Hausdorff and complete generalized metric space. Let . Assume there exist positive Lebesgue integrable functions and on such that and for each and for all , and
**
then has a unique fixed point.*

*Proof. *It follows by taking in Corollary 2.6.

Corollary 2.8. *Let be a Hausdorff and complete generalized metric space. Let . Assume there exist and a positive Lebesgue integrable function on such that for each and for all , and
**
then has a unique fixed point.*

*Proof. *It suffices to take in Corollary 2.7.

Finally, let us finish this paper by noticing the following remark.

*Remark 2.9. *(i) Theorem 2.2 extends Theorem 3.1 of Lakzian and Samet [9].

(ii) Corollary 2.3 extends the results of Branciari [1], Azam and Arshad [2], and Sarma et al. [13].

(iii) Corollary 2.8 extends Theorem 2 of Samet [11].

(iv) Several publications attempting to generalize fixed-point theorems in metric spaces to g.m.s are plagued by the use of some false properties given in [1] (see, e.g., [2–5]). This was observed by Das and Dey [7] who proved a fixed-point theorem without using the false properties. Subsequently, but independently, this was also observed by Samet [12] and Sarma et al. [13] who proved fixed-point theorems assuming that the generalized metric space is Hausdorff. Here, we give a rigorous proof of Theorem 2.2 by taking the same assumption.

#### Acknowledgment

The authors would like to thank the referees for their useful comments and suggestions.