Erratum to ‘Common fixed point theorems for expansion mappings in various abstract spaces using concept of weak reciprocal continuity’

On critical examination of the results given in our paper [1], we notice one crucial error. We need to carry out the following correction.

Example 1 given in paper [1] is wrong as $g(X)\not\subset f(X)$ because $f(X)=\{2,6\}$ and $g(X)=[2,20]$. So, Example 1 in paper [1] is replaced by the following example.

Example 1 Let $(X,G)$ be a G-metric space, where $X=[0,1]$ and

$$G(x,y,z)=(|x-y|+|y-z|+|z-x|)$$

for all $x,y,z\in X$.

Define $f,g:X\to X$ by $f(x)=\frac{x}{2}$ and $g(x)=\frac{x}{6}$ for all $x\in X$.

Then, clearly, $g(X)\subset f(X)$as $f(X)=[0,\frac{1}{2}]$ and $g(X)=[0,\frac{1}{6}]$.

Moreover,

$$\begin{array}{rcl}G(fx,fy,fz)& =& (|fx-fy|+|fy-fz|+|fz-fx|)\\ =& \frac{3}{2}(|x-y|)\ge q\left[\frac{1}{2}(|x-y|)\right]=qG(gx,gy,gz)\end{array}$$

for $1<q\le 3$ and hence, the condition (2.2) of Theorem 2 is satisfied. Also, f and g are two weakly reciprocally continuous self-maps by taking the sequence $\{x_n=\frac{1}{n}\}$. However, the maps are compatible. Thus, all the conditions of Theorem 2 are satisfied and $x=0$ is the unique common fixed point of f and g.