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

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

$$\frac{s^p{\lambda }^{m+1}}{s-\lambda }d(x_1,x_0)+s^{p-1}{\lambda }^{m}d(x_1,x_0)\to \theta$$

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

$$\frac{s^p{\lambda }^{m+1}}{s-\lambda }d(x_1,x_0)+s^{p-1}{\lambda }^{m}d(x_1,x_0)=\frac{s^{\frac{m+2}{2}}}{s^{\frac{3}{2}}-1}d(x_1,x_0)+s^{\frac{m}{2}}d(x_1,x_0)\nrightarrow \theta$$

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

$$d(Tx,Ty)\le \lambda d(x,y)\mathit{\text{for }}x,y\in X,$$

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

$$\begin{array}{rcl}d(x_m,x_{m+p})& \le & s[d(x_m,x_{m+1})+d(x_{m+1},x_{m+p})]\\ \le & sd(x_m,x_{m+1})+s^2[d(x_{m+1},x_{m+2})+d(x_{m+2},x_{m+p})]\\ \le & sd(x_m,x_{m+1})+s^{2}d(x_{m+1},x_{m+2})+s^{3}d(x_{m+2},x_{m+3})\\ +\cdots +s^{p-1}d(x_{m+p-2},x_{m+p-1})+s^{p-1}d(x_{m+p-1},x_{m+p})\\ \le & s{\lambda }^{m}d(x_1,x_0)+s^2{\lambda }^{m+1}d(x_1,x_0)+s^3{\lambda }^{m+2}d(x_1,x_0)\\ +\cdots +s^{p-1}{\lambda }^{m+p-2}d(x_1,x_0)+s^p{\lambda }^{m+p-1}d(x_1,x_0)\\ =& s{\lambda }^m[1+s\lambda +s^2{\lambda }^2+\cdots +{(s\lambda )}^{p-1}]d(x_1,x_0)\le \frac{s{\lambda }^m}{1-s\lambda }d(x_1,x_0).\end{array}$$

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

$$\frac{s{\lambda }^m}{1-s\lambda }d(x_1,x_0)\ll c$$

for each $m>m_0$. Thus,

$$d(x_m,x_{m+p})\le \frac{s{\lambda }^m}{1-s\lambda }d(x_1,x_0)\ll c$$

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

$$d(x,y)=({|x-y|}^p,{|x-y|}^p)\text{for all }x,y\in X.$$

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

$$Tx=\frac{1}{2}(cos\frac{x}{2}-|x-\frac{1}{2}|)\text{for all }x\in X.$$

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

$$\begin{array}{rcl}d(Tx,Ty)& =& ({|Tx-Ty|}^p,{|Tx-Ty|}^p)\\ =& \frac{1}{2^p}({|(cos\frac{x}{2}-cos\frac{y}{2})-(|x-\frac{1}{2}|-|y-\frac{1}{2}|)|}^p,\\ {|(cos\frac{x}{2}-cos\frac{y}{2})-(|x-\frac{1}{2}|-|y-\frac{1}{2}|)|}^p)\\ \le & \frac{1}{2^p}({(|cos\frac{x}{2}-cos\frac{y}{2}|+|x-y|)}^p,{(|cos\frac{x}{2}-cos\frac{y}{2}|+|x-y|)}^p)\\ \le & \frac{1}{2^p}({(\frac{|x+y|}{8}|x-y|+|x-y|)}^p,{(\frac{|x+y|}{8}|x-y|+|x-y|)}^p)\\ \le & {0.56}^p({|x-y|}^p,{|x-y|}^p)<\frac{1}{2^{p-1}}({|x-y|}^p,{|x-y|}^p).\end{array}$$

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

$$d(Tx,Ty)\le {\lambda }_{1}d(x,Tx)+{\lambda }_{2}d(y,Ty)+{\lambda }_{3}d(x,Ty)+{\lambda }_{4}d(y,Tx)\mathit{\text{for }}x,y\in X,$$

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 $h^2<min\{\frac{\delta }{M^2},\frac{1}{2L^2}\}$ for [[1], Example 3.1]. Let us give the corrected example.

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

$$\{\begin{array}{c}\frac{\mathrm{d}x}{\mathrm{d}t}=F(t,x(t)),\hfill \\ x(0)=\xi ,\hfill \end{array}$$

(2.1)

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

$$|F(x,y_1)-F(x,y_2)|\le L|y_1-y_2|.$$

Set $M={max}_{[-h,h]\times [\xi -\delta ,\xi +\delta ]}|F(x,y)|$ such that $h^2<min\{\frac{\delta }{M^2},\frac{1}{2L^2}\}$, 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$s=2$.

Note that (2.1) is equivalent to the integral equation

$$x(t)=\xi +{\int }_0^{t}F(\tau ,x(\tau ))\mathrm{d}\tau .$$

Define a mapping $T:C([-h,h])\to \mathbb{R}$ by $Tx(t)=\xi +{\int }_0^{t}F(\tau ,x(\tau ))\mathrm{d}\tau$. If

$$x(t),y(t)\in B(\xi ,\delta f)\triangleq \{\phi (t)\in C([-h,h]):d(\xi ,\phi )\le \delta f\},$$

then from

$$\begin{array}{rcl}d(Tx,Ty)& =& f(t)\underset{-h\le t\le h}{max}{|{\int }_0^{t}F(\tau ,x(\tau ))\mathrm{d}\tau -{\int }_0^{t}F(\tau ,y(\tau ))\mathrm{d}\tau |}^2\\ =& f(t)\underset{-h\le t\le h}{max}{\left|{\int }_0^t[F(\tau ,x(\tau ))-F(\tau ,y(\tau ))]\mathrm{d}\tau \right|}^2\\ \le & h^{2}f(t)\underset{-h\le \tau \le h}{max}{|F(\tau ,x(\tau ))-F(\tau ,y(\tau ))|}^2\\ \le & h^2{L}^{2}f(t)\underset{-h\le \tau \le h}{max}{|x(\tau )-y(\tau )|}^2\\ =& h^2{L}^{2}d(x,y),\end{array}$$

and

$$d(Tx,\xi )=f(t)\underset{-h\le t\le h}{max}{\left|{\int }_0^{t}F(\tau ,x(\tau ))\mathrm{d}\tau \right|}^2\le h^{2}f\underset{-h\le \tau \le h}{max}{\left|F(\tau ,x(\tau ))\right|}^2\le h^2{M}^{2}f\le \delta f,$$

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

$$d(\xi ,x)\le d(x_n,\xi )+d(x_n,x)\le \delta f+c$$

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. HenceT 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].