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’

Abstract

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.

MSC:47H09, 47H10.

We begin with the definition of a modular metric space.

Definition 1 [1]

Let X be a nonempty set. A function $\omega :(0,\mathrm{\infty })\times X\times X\to [0,\mathrm{\infty }]$ is said to be metric modular on X if for all $x,y,z\in X$, the following conditions hold:

(i) ${\omega }_{\lambda }(x,y)=0$ for all $\lambda >0$ iff $x=y$;

(ii) ${\omega }_{\lambda }(x,y)={\omega }_{\lambda }(y,x)$ for all $\lambda >0$;

(iii) ${\omega }_{\lambda +\mu }(x,y)\le {\omega }_{\lambda }(x,z)+{\omega }_{\mu }(z,y)$ for all $\lambda ,\mu >0$.

Given $x_{\star }\in X$, the set $X_{\omega }(x_{\star })=\{x\in X:{lim}_{\lambda \to \mathrm{\infty }}{\omega }_{\lambda }(x,x_{\star })=0\}$ is called a modular metric space generated by $x_{\star }$ and induced by ω. If its generator $x_{\star }$ does not play any role in the situation, we will write $X_{\omega }$ instead of $X_{\omega }(x_{\star })$.

We need the following theorems in the proof of the main result of this paper.

Theorem 2 [[1], Theorem 2.6]

If ω is metric (pseudo) modular on X, then the modular set $X_{\omega }$ is a (pseudo) metric space with (pseudo) metric given by

$$d_{\omega }^{\circ }(x,y)=inf\{\lambda >0:{\omega }_{\lambda }(x,y)\le \lambda \},x,y\in X_{\omega }.$$

Theorem 3 [[1], Theorem 2.13]

Let ω be (pseudo) modular on a set X. Given a sequence $\{x_n\}\subset X_{\omega }$ and $x\in X_{\omega }$, we have $d_{\omega }^{\circ }(x_n,x)\to 0$ as $n\to \mathrm{\infty }$ if and only if ${\omega }_{\lambda }(x_n,x)\to 0$ as $n\to \mathrm{\infty }$ for all $\lambda >0$. A similar assertion holds for Cauchy sequences.

Let ω be modular on a set X. A mapping $T:X_{\omega }\to X_{\omega }$ is said to be contraction [[2], Definition 3.1] if there exists $k\in [0,1)$ such that

$${\omega }_{\lambda }(Tx,Ty)\le k{\omega }_{\lambda }(x,y)$$

(1)

for all $\lambda >0$ and $x,y\in X_{\omega }$.

Recently, Mongkolkeha et al. [2] proved the following theorems.

Theorem 4 [[2], Theorem 3.2]

Let ω be metric modular on X and $X_{\omega }$ be a modular metric space induced by ω. If $X_{\omega }$ is a complete modular metric space and $T:X_{\omega }\to X_{\omega }$ is a contraction mapping, then T has a unique fixed point in $X_{\omega }$. Moreover, for any $x\in X_{\omega }$, iterative sequence $\{T^n(x)\}$ converges to the fixed point.

Theorem 5 [[2], Theorem 3.4]

Let ω be metric modular on X and $X_{\omega }$ be a modular metric space induced by ω. If $X_{\omega }$ is a complete modular metric space and $T:X_{\omega }\to X_{\omega }$ is a mapping, which $T^N$ is a contraction mapping for some positive integer N. Then, T has a unique fixed point in $X_{\omega }$.

We show that Theorems 4 and 5 are not correct. To this end, we give the following example.

Example 6 Let $X=\mathbb{R}$ and define modular ω by ${\omega }_{\lambda }(x,y)=\mathrm{\infty }$ if $\lambda \le |x-y|$, and ${\omega }_{\lambda }(x,y)=0$ if $\lambda >|x-y|$. It is easy to verify that (see also [[1], Example 2.7]) $X_{\omega }=\mathbb{R}$ and $d_{\omega }^{\circ }(x,y)=|x-y|$. It follows from Theorem 3 that $\mathbb{R}$ is a complete modular metric space. Now, define $T:\mathbb{R}\to \mathbb{R}$ by $Tx=x+1$. We show that T is a contraction while it has no fixed point. Let $k\in [0,1)$ (for example, $k=1/2$) and $x,y\in \mathbb{R}$. If $\lambda \le |x-y|$, then ${\omega }_{\lambda }(x,y)=\mathrm{\infty }$ and (1) holds. If $|x-y|<\lambda$, then $|Tx-Ty|=|x-y|<\lambda$. Therefore, ${\omega }_{\lambda }(Tx,Ty)={\omega }_{\lambda }(x,y)=0$. 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.

Remark 7 In [[2], Example 3.7], the authors mentioned that ‘Thus, T is not a contraction mapping and then the Banach contraction mapping cannot be applied to this example.’ It is true that T is not contraction with the Euclidean metric, but one can easily verify that

$$d_{\omega }^{\circ }(Tx,Ty)\le \frac{\sqrt{3}}{2}{d}_{\omega }^{\circ }(x,y).$$

Thus, the Banach contraction guarantees the existence of a fixed point. Note that

$$d_{\omega }^{\circ }((a_1,0),(a_2,0))=\sqrt{\frac{4|a_1-a_2|}{3}},d_{\omega }^{\circ }((0,b_1),(0,b_2))=\sqrt{|b_1-b_2|}$$

and

$$d_{\omega }^{\circ }((a,0),(0,b))=\sqrt{\frac{4a}{3}+b}.$$