- Erratum
- Open Access

# Erratum to: Nonlinear quasi-contractions in non-normal cone metric spaces

- Zhilong Li
^{1, 2}Email author and - Shujun Jiang
^{3}

**2014**:196

https://doi.org/10.1186/1687-1812-2014-196

© Li and Jiang; licensee Springer. 2014

**Received: **22 September 2014

**Accepted: **22 September 2014

**Published: **24 September 2014

The original article was published in Fixed Point Theory and Applications 2014 2014:165

## Abstract

In the note we correct some errors that appeared in the article (Jiang and Li in Fixed Point Theory Appl. 2014:165, 2014).

**MSC:**06A07, 47H10.

## Keywords

- nonlinear quasi-contraction
- cone metric space

## Correction

Upon critical examination of the main results and their proofs in [1], we note some critical errors under the conditions of the main theorem and its proof in our article [1].

In this note, we would like to supplement some essential conditions, which will ensure that the mapping *B* is well defined, to achieve our claim.

The following theorem is a slight modification of [[1], Theorem 1].

**Theorem 1**

*Let*$(X,d)$

*be a complete cone metric space over a solid cone*

*P*

*of a Banach space*$(E,\parallel \cdot \parallel )$

*and*$T:X\to X$

*a quasi*-

*contraction*(

*i*.

*e*.,

*there exists a mapping*$A:P\to P$

*such that*

*where*$u\in \{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$).

*Assume that*$A:P\to P$

*is a nondecreasing*,

*continuous and subadditive*(

*i*.

*e*., $A(u+v)\u2aafAu+Av$

*for each*$u,v\in P$)

*mapping with*$A\theta =\theta $

*such that*

*If* *B* *is continuous at* *θ*, *where* $Bu={\sum}_{i=0}^{\mathrm{\infty}}{A}^{i}u$ *for each* $u\in P$, *then* *T* *has a unique fixed point* ${x}^{\ast}\in X$, *and for each* ${x}_{0}\in X$, *the Picard iterative sequence* $\{{x}_{n}\}$ *converges to* ${x}^{\ast}$, *where* ${x}_{n}={T}^{n}{x}_{0}$ *for each* *n*.

**Remark 1** In the case that the normed vector space $(E,\parallel \cdot \parallel )$ is complete, if (2) holds then the mapping *B* is well defined. In fact, fix $u\in P$, let ${s}_{n}={\sum}_{i=0}^{n}{A}^{i}u$ and ${S}_{n}={\sum}_{i=0}^{n}\parallel {A}^{i}u\parallel $. By (2), we get ${lim}_{n\to \mathrm{\infty}}{S}_{n}={\sum}_{i=0}^{\mathrm{\infty}}\parallel {A}^{i}u\parallel $ and hence $\{{S}_{n}\}$ is a Cauchy sequence of reals. Note that $\parallel {s}_{m}-{s}_{n}\parallel =\parallel {\sum}_{i=n+1}^{m}{A}^{i}u\parallel \le {\sum}_{i=n+1}^{m}\parallel {A}^{i}u\parallel ={S}_{m}-{S}_{n}$ for each $m>n$, then $\{{s}_{n}\}$ is a Cauchy sequence in *E*. Moreover, by the completeness of *E*, $\{{s}_{n}\}$ is convergent. This implies that $Bu={lim}_{n\to \mathrm{\infty}}{\sum}_{i=0}^{n}{A}^{i}u$ for each *u*, *i.e.*, *B* is well defined. However, in [[1], Theorem 1] the normed vector space *E* is not assumed to be complete, then $\{{s}_{n}\}$ may not be convergent, and consequently, *B* may be not meaningful.

which plays an important role in the proof of [[1], Theorem 1]. However, if *A* is a nonlinear mapping, the above claim may not hold. For example, let $E=P=[0,a]$ and $A(t)={t}^{2}$ for each $t\in P$, where $0<a<1$. It is clear that $A:P\to P$ is nonlinear. Note that ${A}^{0}(t)=t$ and ${A}^{i}(t)={t}^{{2}^{i}}$ ($i=1,2,3,\dots $), then ${\sum}_{i=0}^{\mathrm{\infty}}{A}^{i}(t)=t+{\sum}_{i=1}^{\mathrm{\infty}}{t}^{{2}^{i}}\le t+{\sum}_{i=1}^{\mathrm{\infty}}{t}^{2i}=t+\frac{{t}^{2}}{1-{t}^{2}}$ for each $t\in [0,a]$, and hence ${\sum}_{i=0}^{\mathrm{\infty}}{A}^{i}(t)$ is convergent for each $t\in [0,a]$. This implies that the function $B(t)$ is well defined, where $B(t)={\sum}_{i=0}^{\mathrm{\infty}}{A}^{i}(t)$ for each $t\in [0,a]$. For each $t\in [0,a]$, we have $AB(t)=\sqrt{t+{\sum}_{i=1}^{\mathrm{\infty}}{t}^{{2}^{i}}}$ and $BA(t)={\sum}_{i=1}^{\mathrm{\infty}}{t}^{{2}^{i}}$. Suppose that there exists ${t}_{0}\in (0,\frac{\sqrt{2}}{2}]$ such that $AB({t}_{0})=BA({t}_{0})$, and set $b=BA({t}_{0})$. Then we have $0<b\le \frac{{t}_{0}^{2}}{1-{t}_{0}^{2}}$ and ${t}_{0}+b={b}^{2}$. Solve the equation ${t}_{0}+b={b}^{2}$, then $b=\frac{1+\sqrt{1+4{t}_{0}}}{2}$ by $b>0$. Thus we get $1<\frac{1+\sqrt{1+4{t}_{0}}}{2}=b\le \frac{{t}_{0}^{2}}{1-{t}_{0}^{2}}\le 1$, a contradiction. Hence $BA(t)\ne AB(t)$ for each $t\in (0,\frac{\sqrt{2}}{2}]$. This shows that (3) does not hold.

(ii) Note that *A* is not confined to a linear mapping in [[1], Theorem 1], then from (i) we know that (3) may not hold, and consequently, the proof of [[1], Theorem 1] is not finished yet. In order to complete its proof, we add the continuity of *A* to Theorem 1.

*E*is a Banach space and

*A*is a continuous and subadditive mapping such that (2) is satisfied, then by Remark 1 we get

In what follows, we shall complete the proof of Theorem 1 by using (4) instead of (3). Since there are too many changes required for the proof of [[1], Theorem 1], we present the full proof of Theorem 1 as follows.

*Proof of Theorem 1*It follows from (2) and Remark 1 that the mapping

*B*is well defined. Clearly,

*B*is a nondecreasing and subadditive mappings with $B(P)\subset P$ and $B\theta =\theta $ since

*A*is nondecreasing and subadditive, $A(P)\subset (P)$ and $A\theta =\theta $. We claim that for each $n\ge 1$,

In the following we shall show this claim by induction.

If $n=1$, then $i=j=1$, and so the claim is trivial.

*n*. To prove (5) holds for $n+1$, it suffices to show

Consider the case that ${i}_{0}=1$.

*A*, the definition of

*B*, (4), (5), and (7)

*i.e.*, (6) holds.

*B*and (7)

*i.e.*, (6) holds.

*A*, and (7)

*B*, by the nondecreasing property and subadditivity of

*B*

*B*implies that

*i.e.*, (6) holds.

*A*, (4), (5), and (7)

*i.e.*, (6) holds.

Consider the case that $2\le {i}_{0}\le n$.

*A*, (4), (5), and (7)

*i.e.*, (6) holds.

If $u=d({x}_{n},{x}_{n+1})$, or $u=d({x}_{{i}_{0}-1},{x}_{n+1})$, we set ${i}_{1}=n$, or ${i}_{1}={i}_{0}-1\ge 1$, respectively, and then (8) follows.

*A*, (4), and (8), forces that

*i.e.*, (6) holds, and so the proof of our claim is complete; or there exists ${i}_{2}\in \{1,2,\dots ,n\}$ such that

*k*th step with $k\le n-1$, that is, there exist $k+1$ integers ${i}_{0},{i}_{1},\dots ,{i}_{k}\in \{1,2,\dots ,n\}$ such that

*A*and (4)

*i.e.* (6) holds, and so the proof of our claim is complete.

*n*steps, then there exist $n+1$ integers ${i}_{0},{i}_{1},{i}_{n}\in \{1,2,\dots ,n\}$ such that

*A*and (9)

*B*, by the nondecreasing property of

*B*we get

*B*implies that

*i.e.*, (6) holds. The proof of our claim is complete.

*B*and

*A*are nondecreasing and continuous at

*θ*, $B\theta =\theta $ and $A\theta =\theta $, then it follows from Lemma 3 of [1] that for each $\{{u}_{n}\}\in P$,

where $u\in \{d({x}_{n},{x}^{\ast}),d({x}_{n},{x}_{n+1}),d({x}^{\ast},T{x}^{\ast}),d({x}_{n},T{x}^{\ast}),d({x}^{\ast},{x}_{n+1})\}$.

If $u=d({x}_{n},{x}^{\ast})$, or $u=d({x}_{n},{x}_{n+1})$, or $u=d({x}^{\ast},{x}_{n+1})$, then it follows from (11), (13), and (14) that $Au\stackrel{w}{\to}\theta $ and hence $d(T{x}^{\ast},{x}^{\ast})=\theta $ by (15).

*B*, by the nondecreasing and subadditivity of

*B*we get

*B*implies that

and hence $d({x}^{\ast},T{x}^{\ast})=\theta $ since $Bd({x}_{n+1},{x}^{\ast})\stackrel{w}{\to}\theta $ by (11) and (14).

*A*and (15), we get

*B*, then by the nondecreasing property and subadditivity of

*B*

*B*implies that

and hence $d({x}^{\ast},T{x}^{\ast})=\theta $ since $Bd({x}_{n+1},{x}^{\ast})\stackrel{w}{\to}\theta $ and $BAd({x}_{n},{x}^{\ast})\stackrel{w}{\to}\theta $ by (11) and (14). This shows that ${x}^{\ast}$ is a fixed point of *T*.

*x*is another fixed point of

*T*, then by (1)

where $u\in \{d(x,{x}^{\ast}),d(x,Tx),d({x}^{\ast},T{x}^{\ast}),d(x,T{x}^{\ast}),d({x}^{\ast},Tx)\}$. If $u=d(x,Tx)$, or $u=d({x}^{\ast},T{x}^{\ast})$, then $u=\theta $, and hence $d(x,{x}^{\ast})=\theta $ since $A\theta =\theta $. If $u=d(x,{x}^{\ast})$, or $u=d(x,T{x}^{\ast})$ or $u=d({x}^{\ast},Tx)$, then we must have $u=d(x,{x}^{\ast})$, and hence $d(x,{x}^{\ast})\u2aafAd(x,{x}^{\ast})$. Acting on it with *B*, by the nondecreasing property of *B* we get $Bd(x,{x}^{\ast})\u2aafBAd(x,{x}^{\ast})$. Moreover, by the definition of *B*, we have $d(x,{x}^{\ast})=Bd(x,{x}^{\ast})-BAd(x,{x}^{\ast})\u2aaf\theta $ and hence $d(x,{x}^{\ast})=\theta $. This shows ${x}^{\ast}$ is the unique fixed point of *T*. The proof is complete. □

## Notes

## Declarations

### Acknowledgements

The work was supported by Natural Science Foundation of China (11161022, 11461029, 71462015), Natural Science Foundation of Jiangxi Province (20122BAB201015, 20142BCB23013, 20143ACB21012), Educational Department of Jiangxi Province (KJLD14034), and Program for Excellent Youth Talents of JXUFE (201201).

## Authors’ Affiliations

## References

- Jiang S, Li Z: Nonlinear quasi-contraction in non-normal cone metric spaces.
*Fixed Point Theory Appl.*2014., 2014: Article ID 165Google Scholar

## Copyright

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.