Comment on ‘Fixed point theorems for contraction mappings in modular metric spaces, Fixed Point Theory and Applications, doi:10.1186/1687-1812-2011-93, 20 pages’
© Dehghan et al.; licensee Springer 2012
Received: 8 January 2012
Accepted: 22 August 2012
Published: 7 September 2012
In this paper, we provide an example to show that some results obtained in [Mongkolkeha et al. in Fixed Point Theory Appl. 2011, doi:10.1186/1687-1812-2011-93] are not valid.
We begin with the definition of a modular metric space.
Definition 1 
Let X be a nonempty set. A function is said to be metric modular on X if for all , the following conditions hold:
(i) for all iff ;
(ii) for all ;
(iii) for all .
Given , the set is called a modular metric space generated by and induced by ω. If its generator does not play any role in the situation, we will write instead of .
We need the following theorems in the proof of the main result of this paper.
Theorem 2 [, Theorem 2.6]
Theorem 3 [, Theorem 2.13]
Let ω be (pseudo) modular on a set X. Given a sequence and , we have as if and only if as for all . A similar assertion holds for Cauchy sequences.
for all and .
Recently, Mongkolkeha et al.  proved the following theorems.
Theorem 4 [, Theorem 3.2]
Let ω be metric modular on X and be a modular metric space induced by ω. If is a complete modular metric space and is a contraction mapping, then T has a unique fixed point in . Moreover, for any , iterative sequence converges to the fixed point.
Theorem 5 [, Theorem 3.4]
Let ω be metric modular on X and be a modular metric space induced by ω. If is a complete modular metric space and is a mapping, which is a contraction mapping for some positive integer N. Then, T has a unique fixed point in .
We show that Theorems 4 and 5 are not correct. To this end, we give the following example.
Example 6 Let and define modular ω by if , and if . It is easy to verify that (see also [, Example 2.7]) and . It follows from Theorem 3 that is a complete modular metric space. Now, define by . We show that T is a contraction while it has no fixed point. Let (for example, ) and . If , then and (1) holds. If , then . Therefore, . Hence T is a contraction. On the other hand, by definition of T, it is easy to see that T has no fixed point. So, Theorems 4 and 5 are not correct.
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- Mongkolkeha C, Sintunavarat W, Kumam P: Fixed point theorems for contraction mappings in modular metric spaces. Fixed Point Theory Appl. 2011. doi:10.1186/1687–1812–2011–93Google Scholar
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