# Erratum to: Fixed point theorems of contractive mappings in cone *b*-metric spacesand applications

- Huaping Huang
^{1}and - Shaoyuan Xu
^{2}Email author

**2014**:55

https://doi.org/10.1186/1687-1812-2014-55

© Xu and Huang; licensee Springer. 2014

**Received: **26 November 2013

**Accepted: **26 November 2013

**Published: **5 March 2014

## Correction

In this note we correct some errors that appeared in the article (Huang and Xu in FixedPoint Theory Appl. 2013:112, 2013) by modifying some conditions in the main theorems andexamples.

After examining the proofs of the main results in [1], we can find that there is something wrong with the proof of the Cauchy sequencein [[1], Theorem 2.1]. This leads to subsequent errors in Theorem 2.3 andrelated examples in [1]. We also find that it is not rigorous to use the corresponding lemmas, and so theproof is inaccurate. The detailed reasons are given in the following.

as $m\to \mathrm{\infty}$. Therefore, it is impossible to utilize [[1], Lemma 1.8, 1.9] and demonstrate that $\{{x}_{n}\}$ is a Cauchy sequence.

In this note, we would like to slightly modify only one of the used conditions to achieveour claim.

The following theorem is a modification to [[1], Theorem 2.1]. The proof is the same as that in [1] except the proof of the Cauchy sequence. We will attain the desired goal by usingthe new modified condition $\lambda \in [0,\frac{1}{s})$ instead of $\lambda \in [0,1)$.

**Theorem 2.1**

*Let*$(X,d)$

*be a complete cone*

*b*-

*metric space with the coefficient*$s\ge 1$.

*Suppose that the mapping*$T:X\to X$

*satisfies the contractive condition*

*where*
$\lambda \in [0,\frac{1}{s})$
*is a constant*. *Then* *T* *has a unique fixed point in* *X*. *Furthermore*, *the iterative sequence*
$\{{T}^{n}x\}$
*converges to the fixed point*.

*Proof*In order to show that $\{{x}_{n}\}$ is a Cauchy sequence, we only need the following calculations.For any $m\ge 1$, $p\ge 1$, it follows that

*p*. Making full use of [[1], Lemma 1.8], we find ${m}_{0}\in \mathbb{N}$ such that

for all $m\ge 1$, $p\ge 1$. So, by [[1], Lemma 1.9], $\{{x}_{n}\}$ is a Cauchy sequence in $(X,d)$. The proof is completed. □

As is indicated in the reviewer’s comments, [[1], Example 2.2] is too trivial. Therefore, [[1], Example 2.2] is withdrawn. Now we give another example as follows.

**Example 2.2**Let $X=[0,0.48]$, $E={\mathbb{R}}^{2}$ and let $1\le p\le 6$ be a constant. Take $P=\{(x,y)\in E:x,y\ge 0\}$. We define $d:X\times X\to E$ as

*b*-metric space with$s={2}^{p-1}$. Let us define $T:X\to X$ as

Hence, by Theorem 2.1, there exists ${x}_{0}\in X$ (in fact, it satisfies $0.472251591454<{x}_{0}<0.472251591479$) such that ${x}_{0}$ is the unique fixed point of *T*.

For the same reason, we need to use the new condition ${\lambda}_{1}+{\lambda}_{2}+s({\lambda}_{3}+{\lambda}_{4})<\frac{2}{1+s}$ instead of the original condition ${\lambda}_{1}+{\lambda}_{2}+s({\lambda}_{3}+{\lambda}_{4})<min\{1,\frac{2}{s}\}$ in [[1], Theorem 2.3]. The correct statement is as follows.

**Theorem 2.3**

*Let*$(X,d)$

*be a complete cone*

*b*-

*metric space with the coefficient*$s\ge 1$.

*Suppose that the mapping*$T:X\to X$

*satisfies the contractive condition*

*where the constant*
${\lambda}_{i}\in [0,1)$
*and*
${\lambda}_{1}+{\lambda}_{2}+s({\lambda}_{3}+{\lambda}_{4})<\frac{2}{1+s}$, $i=1,2,3,4$. *Then* *T* *has a unique fixed point in* *X*. *Moreover*, *the iterative sequence*
$\{{T}^{n}x\}$
*converges to the fixed point*.

*Proof* Following an identical argument that is given in [[1], Theorem 2.3] except substituting $0\le \lambda \le 1$ for $0\le \lambda \le \frac{1}{s}$ in line 26 of p.6 in [1], we obtain the proof of Theorem 2.3. □

In addition, based on the changes of Theorem 2.1, we need to change the condition${h}^{2}<min\{\frac{\delta}{{M}^{2}},\frac{1}{{L}^{2}}\}$ into ${h}^{2}<min\{\frac{\delta}{{M}^{2}},\frac{1}{2{L}^{2}}\}$ for [[1], Example 3.1]. Let us give the corrected example.

where $F:[-h,h]\times [\xi -\delta ,\xi +\delta ]$ is a continuous function.

**Example 2.4**Consider boundary problem (2.1) with the continuous function

*F*,and suppose that $F(x,y)$ satisfies the local Lipschitz condition,

*i.e.*, if$|x|\le h$, ${y}_{1},{y}_{2}\in [\xi -\delta ,\xi +\delta ]$, it induces

Set $M={max}_{[-h,h]\times [\xi -\delta ,\xi +\delta ]}|F(x,y)|$ such that ${h}^{2}<min\{\frac{\delta}{{M}^{2}},\frac{1}{2{L}^{2}}\}$, then there exists a unique solution of (2.1).

*Proof* Let $X=E=C([-h,h])$ and $P=\{u\in E:u\ge 0\}$. Put $d:X\times X\to E$ as $d(x,y)=f(t){max}_{-h\le t\le h}{|x(t)-y(t)|}^{2}$ with $f:[-h,h]\to \mathbb{R}$ such that $f(t)={e}^{t}$. It is clear that $(X,d)$ is a complete cone *b*-metric space with$s=2$.

we speculate that $T:B(\xi ,\delta f)\to B(\xi ,\delta f)$ is a contractive mapping.

*X*. Since$(X,d)$ is complete, there is $x\in X$ such that ${x}_{n}\to x$ ($n\to \mathrm{\infty}$). So, for each $c\in intP$, there exists

*N*, whenever $n>N$, we obtain $d({x}_{n},x)\ll c$. Thus, it follows from

and Lemma 1.12 in [1] that $d(\xi ,x)\le \delta f$, which means $x\in B(\xi ,\delta f)$, that is, $(B(\xi ,\delta f),d)$ is complete. □

Owing to the above statement, all conditions of Theorem 2.1 are satisfied. Hence*T* has a unique fixed point $x(t)\in B(\xi ,\delta f)$. That is to say, there exists a unique solution of (2.1).

**Remark 2.5** Theorem 2.1 and Theorem 2.3 generalize and improve thecorresponding results in [2–4].

## Notes

## Declarations

### Acknowledgements

The authors thank the referees, the editors and the readers including Prof. SriramBalasubramanian and Prof. Reny George. Special thanks are due to Prof. Ravi P. Agarwal andProf. Ljubomir Ciric, who have made a number of valuable comments and suggestions, whichhave improved [1] greatly. The research is partially supported by the PhD Start-up Fund ofHanshan Normal University, Guangdong Province, China (No. QD20110920).

## Authors’ Affiliations

## References

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