- Research Article
- Open Access
On Mappings with Contractive Iterate at a Point in Generalized Metric Spaces
© Gajić and and Z. Lozanov-Crvenković. 2010
- Received: 7 September 2010
- Accepted: 29 December 2010
- Published: 4 January 2011
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.
- Positive Integer
- Continuous Mapping
- Differential Geometry
- Symmetric Space
- Arbitrary Point
Sehgal in  proved fixed point theorem for mappings with a contractive iterate at a point and therefore generalized a well-known Banach theorem.
Then has a unique fixed point in and , for each .
Guseman  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  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  introduced correct definition of generalized metric space as follows.
Definition 1.2 (see ).
Let be a nonempty set, and let be a function satisfying the following properties:
; 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.
Let be an ordinary metric apace, then can define -metrics on by
Example 1.4 (see ).
and extend to by using the symmetry in the variables. Then it is clear the is a -metric space.
Definition 1.5 (see ).
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 ).
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 ).
Let be a -metric space, then the function is jointly continuous in all three of its variables.
Definition 1.8 (see ).
A -metric space is called symmetric -metric space if , for all .
Proposition 1.9 (see ).
and that in general these inequalities cannot be improved.
Proposition 1.10 (see ).
A -metric space is -complete if and only if is a complete metric space.
for all , we write .
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 .
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 .
so , .
Therefore, , where . For , , .
For the set is called the orbit for .
for all .
If inequality in (2.14) holds for all , then and , .
Moreover, if , then is the fixed point of .
so is Cauchy sequence and there exists , for some , and inequality (2.15) is proved.
implies that .
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).
If (2.24) holds, for all or is orbitally continuous at , then is a fixed point of .
and result follows from Theorem 2 in .
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 .
Let us note that this result is very close to Theorem 2.1 in .
In the statements above does not have to be continuous.
The next theorems are generalizations of Ćirić fixed point results in .
holds for some and all . Then has a unique fixed point . Moreover, for every , .
for all . Then the result follows from Theorem 2.1 in  and it is true for all .
But need not to be less than 1, so we will prove the statement by using -metric.
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.
For we have .
Since , it follows that , which implies that is a fixed point of .
implies that and thus is the unique fixed point in .
If we suppose that is continuous, then we may prove the following theorem.
for some and all . Then has a unique fixed point and , for every .
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 .
The condition that is a continuous mapping can be relaxed by the following condition: is continuous at a point .
Now, we are going to prove Hadžić  fixed point theorem in 2-metric space, in a manner of -metric spaces.
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:
(e.g., there exists a unique common fixed point for , and )
It obvious that .
Now, since that and are continuous we have that so .
which is contradiction!
So, the common fixed point for , and is unique, and proof is completed.
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.
The authors are thankful to professor B. E. Rhoades, for his advice which helped in improving the results. This work was supported by grants approved by the Ministry of Science and Technological Development, Republic of Serbia, for the first author by Grant no. 144016, and for the second author by Grant no. 144025.
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