## 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.

On p.5 in [1], we conclude that

as m\to \mathrm{\infty} for any p\ge 1. This is incorrect. Indeed, note that taking\lambda =\frac{1}{\sqrt{s}}>\frac{1}{s} and p=m+1 leads to

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

Let \theta \ll c be given. Notice that \frac{s{\lambda}^{m}}{1-s\lambda}d({x}_{1},{x}_{0})\to \theta as m\to \mathrm{\infty} for any *p*. Making full use of [[1], Lemma 1.8], we find {m}_{0}\in \mathbb{N} such that

for each m>{m}_{0}. Thus,

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 Therefore, [[1], Example 2.2] is withdrawn.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

Then (X,d) is a complete cone *b*-metric space withs={2}^{p-1}. Let us define T:X\to X as

Thus, for all x,y\in X, we have

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.

We now apply Theorem 2.1 to the first-order periodic boundary problem

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 withs=2.

Note that (2.1) is equivalent to the integral equation

Define a mapping T:C([-h,h])\to \mathbb{R} by Tx(t)=\xi +{\int}_{0}^{t}F(\tau ,x(\tau ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\tau. If

then from

and

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

Finally, we prove that (B(\xi ,\delta f),d) is complete. In fact, suppose that \{{x}_{n}\} is a Cauchy sequence in B(\xi ,\delta f). Then \{{x}_{n}\} is also a Cauchy sequence in *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].