On Mappings with Contractive Iterate at a Point in Generalized Metric Spaces

Abstract

Using the setting of generalized metric space, the so-called G-metric space, fixed point theorems for mappings with a contractive and a generalized contractive iterate at a point are proved. These results generalize some comparable results in the literature. A common fixed point result is also proved.

1. Introduction

Sehgal in [1] proved fixed point theorem for mappings with a contractive iterate at a point and therefore generalized a well-known Banach theorem.

Theorem 1.1.

Let be a complete metric space and let be a continuous mapping with property that for every there exists so that for every

(1.1)

Then has a unique fixed point in and , for each .

Guseman [2] extended Sehgal's result by removing the condition of continuity of and weakening (1.1) to hold on some subset of such that , where, for some , contains the closure of the iterates of . Further extensions appear in [3, 4]. Our aim in this study is to show that these results are valid in more general class of spaces.

In 1963, S. Gähler introduced the notion of 2-metric spaces but different authors proved that there is no relation between these two function and there is no easy relationship between results obtained in the two settings. Because of that, Dhage [5] introduced a new concept of the measure of nearness between three or more object. But topological structure of so called -metric spaces was incorrect. Finally, Mustafa and Sims [6] introduced correct definition of generalized metric space as follows.

Definition 1.2 (see [6]).

Let be a nonempty set, and let be a function satisfying the following properties:

if ;

; for all , with ;

, for all , with ;

, (symmetry in all three variables);

, for all .

Then the function is called a generalized metric, or, more specifically, a -metric on , and the pair is called a -metric space.

Clearly these properties are satisfied when is the perimeter of triangle with vertices at , , and , moreover taking in the interior of the triangle shows that is the best possible.

Example 1.3.

Let be an ordinary metric apace, then can define -metrics on by

,

.

Example 1.4 (see [6]).

Let . Define on by

(1.2)

and extend to by using the symmetry in the variables. Then it is clear the is a -metric space.

Definition 1.5 (see [6]).

Let be a -metric space, and let be sequence of points of , a point is said to be the limit of the sequence , if , and one says that the sequence is -convergent to

Thus, if in a -metric space , then for any , there exists such that , for all .

Definition 1.6 (see [6]).

Let be a -metric space, a sequence is called -Cauchy if for every , there is such that , for all ; that is, if as .

A -metric space is said to be -complete (or complete -metric) if every -Cauchy sequence in is -convergent in .

Proposition 1.7 (see [6]).

Let be a -metric space, then the function is jointly continuous in all three of its variables.

Definition 1.8 (see [6]).

A -metric space is called symmetric -metric space if , for all .

Proposition 1.9 (see [6]).

Every -metric space will define a metric space by

(1.3)

Note that if is a symmetric -metric space, then

(1.4)

However, if is nonsymmetric, then by the -metric properties it follows that

(1.5)

and that in general these inequalities cannot be improved.

Proposition 1.10 (see [6]).

A -metric space is -complete if and only if is a complete metric space.

In recent years a lot of interesting papers were published with fixed point results in -metric spaces, see [7–18]. This paper is our contribution to the fixed point theory in -metric spaces.

2. Fixed Point Results

Let be a -metric space, a mapping, such that for some and each for there exists a positive integer such that

(2.1)

for all . Then we write . If

(2.2)

for all , we write .

Theorem 2.1.

Let or . Let , with . If there exists such that for , , then is the unique fixed point of in . Moreover, , , for any for , and for if .

Proof.

If is a symmetric space than and (2.1) becomes

(2.3)

and (2.2) becomes

(2.4)

thus the result follows from Theorem 12 in [3] and it is valid for any . Suppose now that is nonsymmetric space. Then by inequality (1.5) we have that (2.1) becomes

(2.5)

and (2.2) becomes

(2.6)

Since need not be less then 1 we can use metric fixed point results only for . On the other side, using the concept of -metric space, we are going to prove the result, if the first case for any , and in the second one for . This means that our results are real generalization in the case of nonsymmetric -metric spaces.

Let . Uniqueness follows from (2.1), since for , it follows that . Now implies that .

Let , and assume for each . For sufficiently large write , , and . Then

(2.7)

so , .

If , uniqueness follows from (2.2) since for , it follows that and further . Now for any

(2.8)

where

(2.9)

For we have

(2.10)

which is a contradiction, and therefore

(2.11)

If then

(2.12)

So

(2.13)

Therefore, , where . For , , .

For the set is called the orbit for .

Theorem 2.2.

Let be a complete -metric space and let be a mapping. Suppose that for some the orbit is complete, and that: for some and each there is an integer such that

(2.14)

for all .

Then , , converges to some and for all ,

(2.15)

If inequality in (2.14) holds for all , then and ,.

Moreover, if , then is the fixed point of .

Proof.

If is a symmetric -metric space the statement easily follows from Guseman fixed point result [2]. Let be nonsymmetric -metric space. Then by inequality (1.5)

(2.16)

Thus, one can use the fixed point result in metric space only for . But here, using the concept of -metric, we prove the result for any . At first let us show that

(2.17)

For any , sufficiently large, there exist , such that . Then

(2.18)

Now, for each

(2.19)

For all , , it follows that

(2.20)

so is Cauchy sequence and there exists , for some , and inequality (2.15) is proved.

If we suppose that inequality in (2.14) is satisfied for all , then, for all ,

(2.21)

so .

On the other hand,

(2.22)

implies that .

Since is continuous it means that

(2.23)

Hence .

Now, let us suppose that . Since by Theorem 2.1   is the fixed point of in and .

For , in inequality (2.14) independently on , we are going to simplify the proof and to relax the condition in (2.14).

Corollary 2.3.

Let be a complete -metric apace and let . Suppose that there exist a point and with complete and

(2.24)

for each . Then converges to some point and for all , ,

(2.25)

If (2.24) holds, for all or is orbitally continuous at , then is a fixed point of .

Proof.

If is a symmetric space than so (2.24) becomes

(2.26)

and result follows from Theorem 2 in [19].

Now, let be a nonsymmetric -metric space. Then since , ,

(2.27)

so for all , ,

(2.28)

and there exists . If (2.24) holds for all , then by Theorem 2.2, since , it follows that .

The fact that is orbitally continuous at , and that , implies that , and therefore .

Remark 2.4.

Let us note that this result is very close to Theorem 2.1 in [8].

Remark 2.5.

In the statements above does not have to be continuous.

The next theorems are generalizations of Ćirić fixed point results in [4].

Theorem 2.6.

Let be a complete metric space and a mapping. Suppose that for each there exists a positive integer such that

(2.29)

holds for some and all . Then has a unique fixed point . Moreover, for every , .

Proof.

If is a symmetric space then and inequality (2.29) becomes

(2.30)

for all . Then the result follows from Theorem 2.1 in [4] and it is true for all .

Now suppose that is nonsymmetric space. Then, by definition of the metric and inequality (1.5) we have

(2.31)

But need not to be less than 1, so we will prove the statement by using -metric.

First, let us prove prove that

(2.32)

where

(2.33)

Clearly (2.32) is true for . Suppose that , and that (2.32) is true for and let us prove it for . Let . Now

(2.34)

where

(2.35)

If , then (2.34) imply

(2.36)

If , then (2.34) imply

(2.37)

Thus by induction we obtain (2.32).

Let us prove that is a Cauchy sequence. Let , , , and we define inductively a sequence of integers and a sequence of points in as follows: , and , . Evidently, is a subsequence of the orbit . Using this sequence we will prove that is a Cauchy sequence.

Let be any fixed member of and let and be any two members of the orbit which follow after . Then and for some and , respectively. Now, using (2.29) we get

(2.38)

where

(2.39)

Similarly, , where

(2.40)

Repeating this argument times we get

(2.41)

Hence . Similarly , so

(2.42)

Since , it follows that is a Cauchy sequence. Let . We show that is a fixed point of . First, let us prove that , where . For , we now have

(2.43)

and on letting tend to infinity it follows that

(2.44)

For we have .

To show that is a fixed point of , let us suppose that and let . Then

(2.45)

Since , it follows that , which implies that is a fixed point of .

Let us suppose that for some , . Then,

(2.46)

implies that and thus is the unique fixed point in .

If we suppose that is continuous, then we may prove the following theorem.

Theorem 2.7.

Let be a complete -metric space and let be a continuous mapping which satisfies the condition: for each there is a positive integer such that

(2.47)

for some and all . Then has a unique fixed point and , for every .

Proof.

Let be an arbitrary point in . Then, as in the proof of Theorem 2.6, the orbit is bounded and is a Cauchy sequence in the complete -metric space and so it has a limit in . Since by the hypothesis is continuous,

(2.48)

Therefore, is a fixed point of . By the same argument as in the proof of Theorem 2.6, it follows that is a unique fixed point of .

Remark 2.8.

The condition that is a continuous mapping can be relaxed by the following condition: is continuous at a point .

3. A Common Fixed Point Result

Now, we are going to prove Hadžić [20] fixed point theorem in 2-metric space, in a manner of -metric spaces.

Theorem 3.1.

Let be a complete -metric space, and one to one continuous mappings, continuous mapping commutative with and . Suppose that there exists a point such that is complete and that the following conditions are satisfied:

(i)For every there exists so that for all and some

(3.1)

(ii)There exists such that for all

(3.2)

Then there exists one and only one element such that

(3.3)

(e.g., there exists a unique common fixed point for , and )

Proof.

Since starting with we can define the sequence such that

(3.4)

Let

(3.5)

We are going to prove that is Cauchy sequence

(3.6)

Similarly one can prove that , , for all , ,

(3.7)

Thus we proved that is a Cauchy sequence, so there exists such that

(3.8)

It obvious that .

At first we will prove that

(3.9)

so .

Now, since that and are continuous we have that so .

Further, let us prove that .

(3.10)

implies that

(3.11)

and . Similarly one can see that , so we prove that

(3.12)

If we suppose that is some other common fixed point for , and then we have that

(3.13)

which is contradiction!

So, the common fixed point for , and is unique, and proof is completed.

Remark 3.2.

For condition (2.14) is satisfied but the Theorem 2.2 is not just a consequence of Theorem 3.1 since in Theorem 2.2 we do not suppose that is continuous.