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

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.

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

$$d(Tx,Ty)\le Au,\mathrm{\forall }x,y\in X,$$

(1)

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., $A(u+v)⪯Au+Av$ for each $u,v\in P$) mapping with $A\theta =\theta$ such that

$$\sum _{i=0}^{\mathrm{\infty }}\parallel A^{i}u\parallel <\mathrm{\infty },\mathrm{\forall }u\in P.$$

(2)

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.

Remark 2 (i) In [1] the authors claim that (see (4) in [1])

$$BA=AB,$$

(3)

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 $b=\frac{1+\sqrt{1+4t_0}}{2}$ by $b>0$. Thus we get $1<\frac{1+\sqrt{1+4t_0}}{2}=b\le \frac{t_0^2}{1-t_0^2}\le 1$, 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.

(iii) Suppose that E is a Banach space and A is a continuous and subadditive mapping such that (2) is satisfied, then by Remark 1 we get

$$\begin{array}{rl}ABu& =A\left(\underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^n{A}^{i}u\right)=\underset{n\to \mathrm{\infty }}{lim}A\left(\sum _{i=0}^n{A}^{i}u\right)\\ ⪯\underset{n\to \mathrm{\infty }}{lim}\left(\sum _{i=0}^n{A}^{i+1}u\right)=\sum _{i=1}^{\mathrm{\infty }}{A}^{i}u=BAu,\mathrm{\forall }u\in P.\end{array}$$

(4)

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 $n\ge 1$,

$$d(x_i,x_j)⪯BAd(x_0,x_1),\mathrm{\forall }1\le i,j\le n.$$

(5)

In the following we shall show this claim by induction.

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

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

$$d(x_{i_0},x_{n+1})⪯BAd(x_0,x_1),\mathrm{\forall }1\le i_0\le n.$$

(6)

By (1),

$$d(x_{i_0},x_{n+1})⪯Au,$$

(7)

where

$$u\in \{d(x_{i_0-1},x_n),d(x_{i_0-1},x_{i_0}),d(x_n,x_{n+1}),d(x_{i_0-1},x_{n+1}),d(x_n,x_{i_0})\}.$$

Consider the case that $i_0=1$.

If $u=d(x_0,x_n)$, then by the triangle inequality, the nondecreasing property, subadditivity of A, the definition of B, (4), (5), and (7)

$$\begin{array}{rl}d(x_{i_0},x_{n+1})& ⪯Ad(x_0,x_n)⪯A[d(x_0,x_1)+d(x_1,x_n)]\\ ⪯A[d(x_0,x_1)+BAd(x_0,x_1)]⪯Ad(x_0,x_1)+ABAd(x_0,x_1)\\ ⪯Ad(x_0,x_1)+B{A}^{2}d(x_0,x_1)=Ad(x_0,x_1)+\sum _{i=2}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)\\ =\sum _{i=1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)=BAd(x_0,x_1),\end{array}$$

i.e., (6) holds.

If $u=d(x_0,x_1)$, then by the definition of B and (7)

$$d(x_{i_0},x_{n+1})⪯Ad(x_0,x_1)⪯\sum _{i=1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)=BAd(x_0,x_1),$$

i.e., (6) holds.

If $u=d(x_0,x_{n+1})$, then by the triangle inequality, the nondecreasing property and subadditivity of A, and (7)

$$\begin{array}{rl}d(x_{i_0},x_{n+1})& ⪯Ad(x_0,x_{n+1})⪯A[d(x_0,x_1)+d(x_{i_0},x_{n+1})]\\ ⪯Ad(x_0,x_1)+Ad(x_{i_0},x_{n+1}).\end{array}$$

Acting on the above inequality with B, by the nondecreasing property and subadditivity of B

$$\begin{array}{rl}Bd(x_{i_0},x_{n+1})& ⪯B[Ad(x_0,x_1)+Ad(x_{i_0},x_{n+1})]\\ ⪯BAd(x_0,x_1)+BAd(x_{i_0},x_{n+1}),\end{array}$$

which together with the definition of B implies that

$$d(x_{i_0},x_{n+1})=Bd(x_{i_0},x_{n+1})-BAd(x_{i_0},x_{n+1})⪯BAd(x_{i_0},x_{n+1}),$$

i.e., (6) holds.

If $u=d(x_n,x_{i_0})$, then by the definition and the nondecreasing property of A, (4), (5), and (7)

$$\begin{array}{rl}d(x_{i_0},x_{n+1})& ⪯Ad(x_{i_0},x_n)⪯ABAd(x_0,x_1)⪯B{A}^{2}d(x_0,x_1)\\ =\sum _{i=2}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)⪯\sum _{i=1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)=BAd(x_0,x_1),\end{array}$$

i.e., (6) holds.

If $u=d(x_n,x_{n+1})$, we set $i_1=n-1$, and then by (7)

$$d(x_{i_0},x_{n+1})⪯Ad(x_{i_1},x_{n+1}).$$

(8)

Consider the case that $2\le i_0\le n$.

If $u=d(x_{i_0-1},x_n)$, or $u=d(x_{i_0-1},x_{i_0})$, or $d(x_n,x_{i_0})$, then by the definition and the nondecreasing property of A, (4), (5), and (7)

$$\begin{array}{rl}d(x_{i_0},x_{n+1})& ⪯Au⪯ABAd(x_0,x_1)⪯B{A}^{2}d(x_0,x_1)\\ =\sum _{i=2}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)⪯\sum _{i=1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)=BAd(x_0,x_1),\end{array}$$

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.

From the above discussions of both cases, we find the result that either (6) holds, and so the proof of our claim is complete, or there exists $i_1\in \{1,2,\dots ,n\}$ such that (8) holds. For the latter situation, continuing in a similar way, it will be found as a result that either

$$d(x_{i_1},x_{n+1})⪯BAd(x_0,x_1),$$

which together with the definition and the nondecreasing property of A, (4), and (8), forces that

$$\begin{array}{rl}d(x_{i_0},x_{n+1})& ⪯ABAd(x_0,x_1)⪯B{A}^{2}d(x_0,x_1)\\ =\sum _{i=2}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)⪯\sum _{i=1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)=BAd(x_0,x_1),\end{array}$$

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

$$d(x_{i_1},x_{n+1})⪯Ad(x_{i_2},x_{n+1}).$$

If the above procedure ends by the 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

$$\begin{array}{c}d(x_{i_0},x_{n+1})⪯Ad(x_{i_1},x_{n+1}),\hfill \\ d(x_{i_1},x_{n+1})⪯Ad(x_{i_2},x_{n+1}),\dots ,\hfill \\ d(x_{i_{k-1}},x_{n+1})⪯Ad(x_{i_k},x_{n+1}),\hfill \\ d(x_{i_k},x_{n+1})⪯BAd(x_0,x_1),\hfill \end{array}$$

then by the nondecreasing property of A and (4)

$$\begin{array}{rl}d(x_{i_0},x_{n+1})& ⪯A^{k}BAd(x_0,x_1)⪯B{A}^{k+1}d(x_0,x_1)=\sum _{i=k+1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)\\ ⪯\sum _{i=1}^{\mathrm{\infty }}{A}^{i}d(x_0,x_1)=BAd(x_0,x_1),\end{array}$$

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

If the above procedure continues more than n steps, then there exist $n+1$ integers $i_0,i_1,i_n\in \{1,2,\dots ,n\}$ such that

$$\begin{array}{c}d(x_{i_0},x_{n+1})⪯Ad(x_{i_1},x_{n+1}),\hfill \\ d(x_{i_1},x_{n+1})⪯Ad(x_{i_2},x_{n+1}),\dots ,\hfill \\ d(x_{i_{n-1}},x_{n+1})⪯Ad(x_{i_n},x_{n+1}).\hfill \end{array}$$

(9)

It is clear that $i_0,i_1,i_n\in \{1,2,\dots ,n\}$ implies there exist two integers $k,l\in \{0,1,2,\dots ,n\}$ with $k<l$ such that $i_k=i_l$, then by the nondecreasing property of A and (9)

$$d(x_{i_k},x_{n+1})⪯A^{l-k}d(x_{i_l},x_{n+1})=A^{l-k}d(x_{i_k},x_{n+1}).$$

(10)

Acting on (10) with B, by the nondecreasing property of B we get

$$Bd(x_{i_k},x_{n+1})⪯B{A}^{l-k}d(x_{i_k},x_{n+1}),$$

which together with the definition of B implies that

$$d(x_{i_k},x_{n+1})⪯\sum _{j=0}^{l-k-1}{A}^{j}d(x_{i_k},x_{n+1})=Bd(x_{i_k},x_{n+1})-B{A}^{l-k}d(x_{i_k},x_{n+1})⪯\theta ⪯BAd(x_0,x_1),$$

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

Note that 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 $\{u_n\}\in P$,

$$B{u}_n\stackrel{w}{\to }\theta ,A{u}_n\stackrel{w}{\to }\theta ,BA{u}_n\stackrel{w}{\to }\theta ,$$

(11)

provided that $u_n\stackrel{w}{\to }\theta$. By (2), we get

$$\underset{n\to \mathrm{\infty }}{lim}\parallel A^{n}u\parallel =0,\mathrm{\forall }u\in P.$$

(12)

Then in analogy to the proof of [[1], Theorem 1], by (5), (11), (12) we can show that

$$d(x_m,x_n)\stackrel{w}{\to }\theta (n>m\to \mathrm{\infty }),$$

(13)

and there exists some $x^{\ast }\in X$ such that

$$d(x_n,x^{\ast })\stackrel{w}{\to }\theta (n\to \mathrm{\infty }).$$

(14)

By (1),

$$d(T{x}^{\ast },x^{\ast })⪯d(x_{n+1},T{x}^{\ast })+d(x_{n+1},x^{\ast })⪯Au+d(x_{n+1},x^{\ast }),\mathrm{\forall }n,$$

(15)

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).

If $u=d(x^{\ast },T{x}^{\ast })$, then by (15)

$$d(x^{\ast },T{x}^{\ast })⪯Ad(x^{\ast },T{x}^{\ast })+d(x_{n+1},x^{\ast }),\mathrm{\forall }n.$$

Acting on the above inequality with B, by the nondecreasing and subadditivity of B we get

$$Bd(x^{\ast },T{x}^{\ast })⪯B[Ad(x^{\ast },T{x}^{\ast })+d(x_{n+1},x^{\ast })]⪯BAd(x^{\ast },T{x}^{\ast })+Bd(x_{n+1},x^{\ast }),\mathrm{\forall }n,$$

which together with the definition of B implies that

$$d(x^{\ast },T{x}^{\ast })=Bd(x^{\ast },T{x}^{\ast })-BAd(x^{\ast },T{x}^{\ast })⪯Bd(x_{n+1},x^{\ast }),\mathrm{\forall }n,$$

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

If $u=d(x_n,T{x}^{\ast })$, then, by the triangle inequality, the nondecreasing property, and subadditivity of A and (15), we get

$$\begin{array}{rl}d(T{x}^{\ast },x^{\ast })& ⪯d(x_{n+1},x^{\ast })+Ad(x_n,T{x}^{\ast })\\ ⪯d(x_{n+1},x^{\ast })+A[d(x_n,x^{\ast })+d(x^{\ast },T{x}^{\ast })]\\ ⪯d(x_{n+1},x^{\ast })+Ad(x_n,x^{\ast })+Ad(x^{\ast },T{x}^{\ast }),\mathrm{\forall }n.\end{array}$$

Acting on the above inequality with B, then by the nondecreasing property and subadditivity of B

$$\begin{array}{rl}Bd(x^{\ast },T{x}^{\ast })& ⪯B[d(x_{n+1},x^{\ast })+Ad(x_n,x^{\ast })+Ad(x^{\ast },T{x}^{\ast })]\\ ⪯Bd(x_{n+1},x^{\ast })+BAd(x_n,x^{\ast })+BAd(x^{\ast },T{x}^{\ast }),\mathrm{\forall }n,\end{array}$$

which together with the definition of B implies that

$$d(x^{\ast },T{x}^{\ast })=Bd(x^{\ast },T{x}^{\ast })-BAd(x^{\ast },T{x}^{\ast })⪯Bd(x_{n+1},x^{\ast })+BAd(x_n,x^{\ast }),\mathrm{\forall }n,$$

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.

If x is another fixed point of T, then by (1)

$$d(x,x^{\ast })=d(Tx,T{x}^{\ast })⪯Au,$$

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 $d(x,x^{\ast })⪯Ad(x,x^{\ast })$. Acting on it with B, by the nondecreasing property of B we get $Bd(x,x^{\ast })⪯BAd(x,x^{\ast })$. Moreover, by the definition of B, we have $d(x,x^{\ast })=Bd(x,x^{\ast })-BAd(x,x^{\ast })⪯\theta$ and hence $d(x,x^{\ast })=\theta$. This shows $x^{\ast }$ is the unique fixed point of T. The proof is complete. □