Erratum to ‘MeirKeeler αcontractive fixed and common fixed point theorems’
 Thabet Abdeljawad^{1, 2}Email author and
 Dhananjay Gopal^{3}
https://doi.org/10.1186/168718122013110
© Abdeljawad and Gopal; licensee Springer 2013
Received: 11 March 2013
Accepted: 11 April 2013
Published: 25 April 2013
The original article was published in Fixed Point Theory and Applications 2013 2013:19
Abstract
In this note we correct some errors that appeared in the article (Abdeljawad in Fixed Point Theory Appl. 2013:19, 2013) by modifying some conditions in the main theorems and by giving an example to support.
MSC:47H10, 54H25.
After examining the calculations in the proof of the uniqueness part in Theorem 8 in [1] and Steps 3 and 4 of Theorem 16, we found that they do not lead to strict inequalities, and hence the proofs failed. In this note, we slightly modify some of the used conditions to achieve our claim.
The following theorem is a modification to Theorem 8 in [1]. The proof is the same as in [1] except the uniqueness part will be proved by using the new modified condition (H) in the statement of the theorem.
 1.
$(f,g)$ is αadmissible and there exists an ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$;
 2.
the pair $(f,g)$ is generalized MeirKeeler αcontractive.
Then the sequence ${d}_{n}=d({x}_{n},{x}_{n+1})$ is monotone decreasing. If, moreover, we assume that
 3.
on the $(f,g)$orbit of ${x}_{0}$, we have $\alpha ({x}_{n},{x}_{j})\ge 1$ for all n even and $j>n$ odd and that f and g are continuous on the $(f,g)$orbit of ${x}_{0}$.
Then either (1) f or g has a fixed point in the $(f,g)$orbit $\{{x}_{n}\}$ of ${x}_{0}$ or (2) f and g have a common fixed point p and $lim{x}_{n}=p$. If, moreover, we assume that the following condition (H) holds:
(H) If for all fixed points x and y of $(f,g)$, $\alpha (x,y)\ge 1$,
then the uniqueness of the fixed point is obtained.
Thus we reach $d(p,q)<d(p,q)$, and hence a contradiction, which implies that $p=q$. □
Using the new modified condition (H) for the pair $(f,f)$, we modify the uniqueness part of Corollary 9 in [1].
The following example shows that we lose the uniqueness if our modified (H) condition is not satisfied.
Notice that f has two common fixed points $x=0$ and $x=\frac{3}{2}$. This is because f satisfies all the hypotheses of the corollary (which is Corollary 9 in [1]) except the condition (H), i.e., $\alpha (0,\frac{3}{2})=0<1$.
The following theorem is a modification of Theorem 16 in [1]. The proofs of Step 3 and Step 4 are given only according to the new modified conditions (H) and (fH).
Theorem 3 Let f, g be continuous selfmaps of a metric space $(X,d)$ such that $g\in {C}_{f}$. Assume that $\alpha ({x}_{n},{x}_{m})\ge 1$ for all $m>n$. If g is a generalized MeirKeeler αfcontractive map such that α satisfies the condition (fH): If $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{m})\ge 1$ for all $m>n$ and $f{x}_{n}\to z$, then $\alpha (z,fz)\ge 1$. Also assume that the condition (H) is satisfied. Then f and g have a unique common fixed point.
Proof

Step 3. We show that $\eta =fz=gz$ is a common fixed point for f and g. Assume that $f\eta \ne \eta $. Then ${f}^{2}z\ne fz$, and by the help of the (fH) condition, we have$\begin{array}{rcl}d(\eta ,f\eta )& =& d(gz,fgz)=d(gz,gfz)\\ \le & \alpha (z,fz)d(gz,gfz)\\ <& max\{d(fz,ffz),d(fz,gz),d(ffz,gfz),\frac{d(fz,gfz)+d(ffz,gz)}{2}\}\\ =& max\{d(\eta ,f\eta ),0,0,d(\eta ,f\eta )\}.\end{array}$
Thus we have $d(\eta ,f\eta )<d(\eta ,f\eta )$, which gives a contradiction and therefore $f\eta =\eta $. Moreover, $g\eta =gfz=f\eta =\eta $.

Step 4. The uniqueness of the common fixed point. Assume that $\eta =fz=gz$ is our common fixed point for f and g, where $f{x}_{n}\to z$, and ω is another common fixed point. Then, by the (H) condition, we have$\begin{array}{rcl}d(\eta ,\omega )& =& d(g\eta ,g\omega )\\ \le & \alpha (\eta ,\omega )d(g\eta ,g\omega )\\ <& max\{d(f\eta ,f\omega ),d(f\eta ,g\eta ),d(f\omega ,g\omega ),\frac{d(f\eta ,g\omega )+d(f\omega ,g\eta )}{2}\},\end{array}$
which gives $d(\eta ,\omega )<d(\eta ,\omega )$, a contradiction, and hence $\eta =\omega $.
□
Instead of the modified condition (fH) above, the following condition can be used (sfH). If $\{{x}_{n}\}$ is a sequence in X such that $\alpha ({x}_{n},{x}_{m})\ge 1$ for all $m>n$ and $f{x}_{n}\to z$, then $\alpha (f{x}_{n},fz)\ge k$ for all n, where $k>1$, and hence Step 3 will be proved as follows.
If we let $n\to \mathrm{\infty}$ above and use the continuity and commutativity of f and g, then we reach $d(\eta ,f\eta )\le {k}^{1}d(\eta ,f\eta )<d(\eta ,f\eta )$, and hence $f\eta =\eta $. Moreover, $g\eta =gfz=f\eta =\eta $.
Finally, according to the modifications above, the (H) condition only in Theorem 18 of [1] is needed to be modified.
Notes
Declarations
Acknowledgements
The second author thanks for the support of CSIR, Govt. of India, Grant No25(0215)/13/EMRII.
Authors’ Affiliations
References
 Abdeljawad T: MeirKeeler α contractive fixed and common fixed point theorems. Fixed Point Theory Appl. 2013., 2013: Article ID 19Google Scholar
Copyright
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