A further remark to paper ‘Convergence theorems for the common solution for a finite family of ϕ-strongly accretive operator equations’

Abstract

In this note, we point out several gaps in Gurudwan and Sharma (Appl. Math. Comput. 217(15):6748-6754, 2011) and Yang (Appl. Math. Comput. 218(21):10367-10369, 2012) and give the main results under weaker conditions.

MSC:47H10, 47H09, 46B20.

1 Introduction

Recently, Gurudwan, Sharma [1] and Yang [2] studied the strong convergence of the sequence, respectively, which was defined by

for approximation of a common solution of a finite family of uniformly continuous Φ-strongly accretive operator equations. Their results are as follows.

Theorem GS [[1], Theorem 3.1]

Let E be an arbitrary real Banach space and let ${\left\{{A}_{i}\right\}}_{i=1}^{N}:E\to E$ be uniformly continuous ϕ-strongly accretive operators and each range of either ${A}_{i}$ or $\left(I-{A}_{i}\right)$ be bounded. Let, for $i=1,\dots ,N$, ${\left\{{u}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$ be sequences in E and ${\left\{{a}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{b}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{c}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$ be real sequences in $\left[0,1\right]$ satisfying

1. (i)

${a}_{n}^{i}+{b}_{n}^{i}+{c}_{n}^{i}=1$,

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{b}_{n}^{N}=\mathrm{\infty }$,

3. (iii)

${\sum }_{n=0}^{\mathrm{\infty }}{c}_{n}<\mathrm{\infty }$,

4. (iv)

${lim}_{n\to \mathrm{\infty }}{b}_{n}^{i}={lim}_{n\to \mathrm{\infty }}{c}_{n}^{i}={lim}_{n\to \mathrm{\infty }}\frac{{c}_{n}^{i}}{{b}_{n}^{i}+{c}_{n}^{i}}=0$, $\mathrm{\forall }i=1,\dots ,N$, $n\ge 1$.

For any given $f\in E$, define ${\left\{{S}_{i}\right\}}_{i=1}^{N}:E\to E$ by ${S}_{i}x=x-{A}_{i}x+f$, $\mathrm{\forall }i=1,\dots ,N$, $\mathrm{\forall }x\in E$. Then the multi-step iterative sequence with errors ${\left\{{x}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ defined by the above converges strongly to the unique solution of the operator equations ${\left\{{A}_{i}x\right\}}_{i=1}^{N}=f$.

On the basis of the above result, Yang [2] proved the following convergence theorem.

Thoerem Yang [[2], Theorem 2]

Let E be an arbitrary real Banach space and let ${\left\{{A}_{i}\right\}}_{i=1}^{N}:E\to E$ be uniformly continuous ϕ-strongly accretive operators and each range of either ${A}_{i}$ or $\left(I-{A}_{i}\right)$ is bounded. Let for $i=1,\dots ,N$, ${\left\{{u}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$ be bounded sequences in E and ${\left\{{a}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{b}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{c}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$ be real sequences in $\left[0,1\right]$ satisfying

1. (i)

${a}_{n}^{i}+{b}_{n}^{i}+{c}_{n}^{i}=1$,

2. (ii)

${\sum }_{n=0}^{\mathrm{\infty }}{b}_{n}^{N}=\mathrm{\infty }$,

3. (iii)

${lim}_{n\to \mathrm{\infty }}{b}_{n}^{i}={lim}_{n\to \mathrm{\infty }}{c}_{n}^{i}={lim}_{n\to \mathrm{\infty }}\frac{{c}_{n}^{i}}{{b}_{n}^{i}+{c}_{n}^{i}}=0$, $\mathrm{\forall }i=1,\dots ,N$, $n\ge 1$.

For any given $f\in E$, define ${\left\{{S}_{i}\right\}}_{i=1}^{N}:E\to E$ by ${S}_{i}x=\left(I-{A}_{i}\right)x+f$, $\mathrm{\forall }i=1,\dots ,N$, $\mathrm{\forall }x\in E$. Then the multi-step iterative sequence with errors ${\left\{{x}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ defined by the above converges strongly to the unique solution of the operator equations ${\left\{{A}_{i}x\right\}}_{i=1}^{N}=f$.

However, after careful reading of their works, we discovered that there exist some problems in references [1] and [2] as follows.

Problem 1 In the proof course of Theorem 3.1 of Gurudwan and Sharma [1], which happens in line 11 of page 6751. Here, it is defective that they obtained $\parallel x-y\parallel \le {\varphi }_{i}^{-1}\left(\parallel {A}_{i}x-{A}_{i}y\parallel \right)$, that is, $〈{A}_{i}x-{A}_{i}y,j\left(x-y\right)〉\ge \varphi \left(\parallel x-y\parallel \right)\parallel x-y\parallel ⇒\varphi \left(\parallel x-y\parallel \right)\le \parallel {A}_{i}x-{A}_{i}y\parallel$, but we cannot deduce $\parallel x-y\parallel \le {\varphi }_{i}^{-1}\left(\parallel {A}_{i}x-{A}_{i}y\parallel \right)$. The reason is that it is possible $\parallel {A}_{i}x-{A}_{i}y\parallel$ does not belong to $R\left(\varphi \right)$ (range of ϕ). A counterexample is as follows. Let us define $\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ by $\varphi \left(\alpha \right)=\frac{{2}^{\alpha }-1}{{2}^{\alpha }+1}$; then it can be easily seen that ϕ is increasing with $\varphi \left(0\right)=0$, but ${lim}_{\alpha \to +\mathrm{\infty }}\varphi \left(\alpha \right)=1$ and ${\varphi }^{-1}\left(2\right)$ makes no sense (see [3]).

Problem 2 In the paper of Yang [2], he referred to the mistakes of ‘$\parallel {x}_{{n}_{m}+j}^{i}-q\parallel <ϵ$ for $j\ge 1$ to deduce $\parallel {x}_{n}-q\parallel \to 0$ ($n\to \mathrm{\infty }$)’ in [1] and cited an example, i.e.,

Now, we want to clarify the fact. Let $\left\{{\gamma }_{n}\right\}$ be a real sequence, $\left\{{\gamma }_{{n}_{m}}\right\}$ be some infinite subsequence of $\left\{{\gamma }_{n}\right\}$ and $\left\{{n}_{m}\right\}$ be neither odd nor even sequence, then the conclusions are as follows:

(C-i) ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}=0⇔\mathrm{\forall }ϵ>0$, nonnegative integer ${n}_{0}$ such that $|{\gamma }_{{n}_{m}+j}|<ϵ$ for ${n}_{m}\ge {n}_{0}$, $j\ge 1$.

(C-ii) ${lim}_{n\to \mathrm{\infty }}{\gamma }_{n}=0⇒{lim}_{m\to \mathrm{\infty }}{\gamma }_{{n}_{m}}=0$ and ${lim}_{m\to \mathrm{\infty }}{\gamma }_{{n}_{m}+j}=0$ for $\mathrm{\forall }j\ge 1$.

Indeed, the above example () does not satisfy the conclusion (C-i), it just illustrates the result (C-ii). Therefore, the note given by Yang [2] confused the conclusions (C-i) and (C-ii).

The aim of this paper is to generalize the results of papers [1] and [2]. For this, we need the following knowledge.

2 Preliminary

Let E be a real Banach space and ${E}^{\ast }$ be its dual space. The normalized duality mapping $J:E\to {2}^{{E}^{\ast }}$ is defined by

$J\left(x\right)=\left\{f\in {E}^{\ast }:〈x,f〉={\parallel x\parallel }^{2}={\parallel f\parallel }^{2}\right\},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in E,$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. The single-valued normalized duality mapping is denoted by j.

An operator $T:E\to E$ is said to be strongly accretive if there exists a constant $k>0$, and for $\mathrm{\forall }x,y\in E$, $\mathrm{\exists }j\left(x-y\right)\in J\left(x-y\right)$ such that

$〈Tx-Ty,j\left(x-y\right)〉\ge k{\parallel x-y\parallel }^{2},$

without loss of generality, we assume that $k\in \left(0,1\right)$. The operator T is called ϕ-strongly accretive if for any $x,y\in E$, there exist $j\left(x-y\right)\in J\left(x-y\right)$ and a strictly increasing continuous function $\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ with $\varphi \left(0\right)=0$ such that

$〈Tx-Ty,j\left(x-y\right)〉\ge \varphi \left(\parallel x-y\parallel \right)\parallel x-y\parallel .$

It is obvious that a strongly accretive operator must be the ϕ-strongly accretive in the special case in which $\varphi \left(t\right)=kt$, but the converse is not true in general. That is, the class of strongly accretive operators is a proper subclass of the class of ϕ-strongly accretive operators.

In order to obtain the main conclusion of this paper, we need the following lemmas.

Lemma 2.1 [1]

Suppose that E is an arbitrary Banach space and $A:E\to E$ is a continuous ϕ-strongly accretive operator. Then the equation $Ax=f$ has a unique solution for any $f\in E$.

Lemma 2.2 [4]

Let E be a real Banach space and let $J:E\to {2}^{{E}^{\ast }}$ be a normalized duality mapping. Then

${\parallel x+y\parallel }^{2}\le {\parallel x\parallel }^{2}+2〈y,j\left(x+y\right)〉,$
(2.1)

for all $x,y\in E$ and $j\left(x+y\right)\in J\left(x+y\right)$.

Lemma 2.3 [5]

Let ${\left\{{\delta }_{n}\right\}}_{n=0}^{\mathrm{\infty }}$, ${\left\{{\lambda }_{n}\right\}}_{n=0}^{\mathrm{\infty }}$ and ${\left\{{\gamma }_{n}\right\}}_{n=0}^{\mathrm{\infty }}$ be three nonnegative real sequences and $\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ be a strictly increasing and continuous function with $\varphi \left(0\right)=0$ satisfying the following inequality:

${\delta }_{n+1}^{2}\le {\delta }_{n}^{2}-{\lambda }_{n}\varphi \left({\delta }_{n+1}\right)+{\gamma }_{n},\phantom{\rule{1em}{0ex}}n\ge 0,$
(2.2)

where ${\lambda }_{n}\in \left[0,1\right]$ with ${\sum }_{n=0}^{\mathrm{\infty }}{\lambda }_{n}=\mathrm{\infty }$, ${\gamma }_{n}=o\left({\lambda }_{n}\right)$. Then ${\delta }_{n}\to 0$ as $n\to \mathrm{\infty }$.

3 Main results

Theorem 3.1 Let E be an arbitrary real Banach space and ${\left\{{A}_{i}\right\}}_{i=1}^{N}:E\to E$ be N uniformly continuous ϕ-strongly accretive operators. For $i=1,2,\dots ,N$, let ${\left\{{u}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$ be bounded sequences in E and ${\left\{{a}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{b}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$, ${\left\{{c}_{n}^{i}\right\}}_{n=1}^{\mathrm{\infty }}$ be real sequences in $\left[0,1\right]$ satisfying

1. (i)

${a}_{n}^{i}+{b}_{n}^{i}+{c}_{n}^{i}=1$, $i=1,2,\dots ,N$;

2. (ii)

${\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}^{N}=+\mathrm{\infty }$;

3. (iii)

${lim}_{n\to \mathrm{\infty }}{b}_{n}^{i}={lim}_{n\to \mathrm{\infty }}{c}_{n}^{i}=0$, $i=1,2,\dots ,N$;

4. (iv)

${c}_{n}^{N}=o\left({b}_{n}^{N}\right)$.

For any given $f\in E$, define ${\left\{{S}_{i}\right\}}_{i=1}^{N}:E\to E$ with ${\bigcap }_{i=1}^{N}F\left({S}_{i}\right)\ne \mathrm{\varnothing }$ by ${S}_{i}x=x-{A}_{i}x+f$, $\mathrm{\forall }i=1,2,\dots ,N$, $\mathrm{\forall }x\in E$, where $F\left({S}_{i}\right)=\left\{x\in E:{S}_{i}x=x\right\}$. Then, for some ${x}_{0}\in E$, the multi-step iterative sequence with errors ${\left\{{x}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ defined by

converges strongly to the unique solution of the operator equations ${\left\{{A}_{i}x\right\}}_{i=1}^{N}=f$.

Proof Since ${\left\{{A}_{i}\right\}}_{i=1}^{N}:E\to E$ is ϕ-strongly accretive operator, we obtain that each equation ${A}_{i}x=f$ has the unique solution by Lemma 2.1, denote ${q}_{i}$, i.e., ${q}_{i}$ is the unique fixed point of ${S}_{i}$ by ${S}_{i}x=x-{A}_{i}x+f$. Since ${\bigcap }_{i=1}^{N}F\left({S}_{i}\right)\ne \mathrm{\varnothing }$, then ${\bigcap }_{i=1}^{N}F\left({S}_{i}\right)$ is a single set, let q. Meanwhile, there exists a strictly increasing continuous function $\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,+\mathrm{\infty }\right)$ with $\varphi \left(0\right)=0$ such that

$〈{A}_{i}x-{A}_{i}q,j\left(x-q\right)〉\ge \varphi \left(\parallel x-q\parallel \right),$

for $x\in E$, $q\in F\left(T\right)$, that is,

Choose some ${x}_{0}\in E$ and ${x}_{0}\ne {S}_{i}{x}_{0}$ such that ${r}_{0}\in R\left(\mathrm{\Phi }\right)$, where

${r}_{0}=max\left\{\parallel {x}_{0}-{S}_{1}{x}_{0}\parallel \cdot \parallel {x}_{0}-q\parallel ,\parallel {x}_{0}-{S}_{2}{x}_{0}\parallel \cdot \parallel {x}_{0}-q\parallel ,\dots ,\parallel {x}_{0}-{S}_{N}{x}_{0}\parallel \cdot \parallel {x}_{0}-q\parallel \right\},$

$R\left(\mathrm{\Phi }\right)$ is the range of Φ. Indeed, if $\mathrm{\Phi }\left(r\right)\to +\mathrm{\infty }$ as $r\to +\mathrm{\infty }$, then ${r}_{0}\in R\left(\mathrm{\Phi }\right)$; if $sup\left\{\mathrm{\Phi }\left(r\right):r\in \left[0,+\mathrm{\infty }\right)\right\}={r}_{1}<+\mathrm{\infty }$ with ${r}_{1}<{r}_{0}$, then for $q\in E$, there exists a sequence $\left\{{w}_{n}\right\}$ in E such that ${w}_{n}\to q$ as $n\to \mathrm{\infty }$ with ${w}_{n}\ne q$. Since ${A}_{i}$ is uniformly continuous, so is ${S}_{i}$. Furthermore, we obtain that ${S}_{i}{w}_{n}\to {S}_{i}q$ as $n\to \mathrm{\infty }$, then $\left\{{w}_{n}-{S}_{i}{w}_{n}\right\}$ is the bounded sequence for $i=1,2,\dots ,N$. Hence, there exists the common natural number ${n}_{0}$ such that $\parallel {w}_{n}-{S}_{i}{w}_{n}\parallel \cdot \parallel {w}_{n}-q\parallel <\frac{{r}_{1}}{2}$ for $n\ge {n}_{0}$ and $i=1,2,\dots ,N$, then we redefine ${x}_{0}={w}_{{n}_{0}}$ and $\parallel {x}_{0}-{S}_{i}{x}_{0}\parallel \cdot \parallel {x}_{0}-q\parallel <\frac{{r}_{1}}{2}$. Thus, ${max}_{1\le i\le N}\left\{\parallel {x}_{0}-{S}_{i}{x}_{0}\parallel \cdot \parallel {x}_{0}-q\parallel \right\}\in R\left(\varphi \right)$. It is to ensure that ${\mathrm{\Phi }}^{-1}\left({r}_{0}\right)$ is defined well.

Step I. We show that $\left\{{x}_{n}\right\}$ is a bounded sequence.

Set $R={\mathrm{\Phi }}^{-1}\left({r}_{0}\right)$, then from the above formula (@), we obtain that $\parallel {x}_{0}-q\parallel \le R$. Denote

${B}_{1}=\left\{x\in E:\parallel x-q\parallel \le R\right\},\phantom{\rule{2em}{0ex}}{B}_{2}=\left\{x\in E:\parallel x-q\parallel \le 2R\right\}.$

Since ${S}_{i}$ is uniformly continuous, then ${S}_{i}$ is bounded. We let

$M=\underset{1\le i\le N}{max}\left\{\underset{x\in {B}_{2}}{sup}\left\{\parallel {S}_{i}x-q\parallel +1\right\}\right\}+\underset{1\le i\le N}{max}\left\{\underset{n}{sup}\left\{\parallel {u}_{n}^{i}-q\parallel \right\}\right\}.$

Next, we want to prove that ${x}_{n}\in {B}_{1}$. If $n=0$, then ${x}_{0}\in {B}_{1}$. Now, assume that it holds for some n, i.e., ${x}_{n}\in {B}_{1}$. We prove that ${x}_{n+1}\in {B}_{1}$. Suppose it is not the case, then $\parallel {x}_{n+1}-q\parallel >R>\frac{R}{2}$. Since ${S}_{i}$ is uniformly continuous for $i=1,2,\dots ,N$, then for ${ϵ}_{0}=\frac{\mathrm{\Phi }\left(\frac{R}{2}\right)}{8R}$, there exists common $\delta >0$ such that $\parallel {S}_{i}x-{S}_{i}y\parallel <{ϵ}_{0}$ when $\parallel x-y\parallel <\delta$. Denote

${\tau }_{0}=min\left\{1,\frac{R}{M},\frac{\mathrm{\Phi }\left(\frac{R}{2}\right)}{8R\left(M+2R\right)},\frac{\delta }{2M+5R}\right\}.$

Since ${b}_{n}^{i},{c}_{n}^{i}\to 0$ as $n\to \mathrm{\infty }$ for $i=1,2,\dots ,p$. Without loss of generality, we let $0\le {b}_{n}^{i},{c}_{n}^{i}\le {\tau }_{0}$ for any $n\ge 0$ and $i=1,2,\dots ,N$. Since ${c}_{n}^{N}=o\left({b}_{n}^{N}\right)$, let ${c}_{n}^{N}<{b}_{n}^{N}{\tau }_{0}$. Now, estimate $\parallel {x}_{n}^{i}-q\parallel$ for $i=1,2,\dots ,N$. From the multi-step iteration, we have

(3.1)

then ${x}_{n}^{1}\in {B}_{2}$. Similarly, we have

(3.2)

then ${x}_{n}^{2}\in {B}_{2}$.  , we have

(3.3)

then ${x}_{n}^{N-1}\in {B}_{2}$. Therefore, we get

(3.4)

And we also have

(3.5)

and

(3.6)

By the uniform continuity of ${S}_{N}$, we have

$\parallel {S}_{N}{x}_{n+1}-{S}_{N}{x}_{n}^{N-1}\parallel <\frac{\mathrm{\Phi }\left(\frac{R}{2}\right)}{8R}.$

Using Lemma 2.2 and the above formulas, we have

(3.7)

which is a contradiction. So, ${x}_{n+1}\in {B}_{1}$, i.e., $\left\{{x}_{n}\right\}$ is a bounded sequence, from which it follows that $\left\{{x}_{n}^{1}\right\},\left\{{x}_{n}^{2}\right\},\dots ,\left\{{x}_{n}^{N-1}\right\}$ are all bounded sequences as well.

Step II. We want to prove $\parallel {x}_{n}-q\parallel \to 0$ as $n\to \mathrm{\infty }$.

Since ${b}_{n}^{i},{c}_{n}^{i}\to 0$ as $n\to \mathrm{\infty }$ for $i=1,2,\dots ,N$ and $\left\{{x}_{n}\right\}$, $\left\{{x}_{n}^{N-1}\right\}$ are bounded. From (3.5) and (3.6), we obtain

$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}\parallel =0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n+1}-{x}_{n}^{N-1}\parallel =0,\phantom{\rule{2em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}\parallel {S}_{N}{x}_{n+1}-{S}_{N}{x}_{n}^{N-1}\parallel =0.$

By (3.7), we have

(3.8)

where

By Lemma 2.3, we obtain ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-q\parallel =0$. This completes the proof. □

Remark 3.2 Theorem 3.1 generalizes Theorem 3.1 of [1] and Theorem 2 of [2] in the following cases:

1. (a)

It is not necessary for each range of ${A}_{i}$ or $I-{A}_{i}$ to be bounded in [1] and [2].

2. (b)

The condition of $\left\{{c}_{n}^{i}\right\}$ is weakened to ${c}_{n}^{N}=o\left({b}_{n}^{N}\right)$ from ${lim}_{n\to \mathrm{\infty }}\frac{{c}_{n}^{i}}{{b}_{n}^{i}+{c}_{n}^{i}}=0$ ($i=1,2,\dots ,N$).

3. (c)

The proof method of our theorem differs from that of [1] and [2].

Theorem 3.3 Let E, $\left\{{u}_{n}^{i}\right\}$, $\left\{{a}_{n}^{i}\right\}$, $\left\{{b}_{n}^{i}\right\}$, $\left\{{c}_{n}^{i}\right\}$ ($i=1,2,\dots ,N$) be as in Theorem  3.1 and let ${\left\{{T}_{i}\right\}}_{i=1}^{N}:E\to E$ be N uniformly continuous ϕ-strongly pseudocontractive mappings. Then, for some ${x}_{0}\in E$, the multi-step iterative sequence with errors ${\left\{{x}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ defined by

converges strongly to the unique common fixed point of ${\left\{{T}_{i}\right\}}_{i=1}^{N}$.

Proof See [1]. □

References

1. Gurudwan N, Sharma BK: Convergence theorem for the common solution for a finite family of ϕ -strongly accretive operator equations. Appl. Math. Comput. 2011, 217(15):6748–6754. 10.1016/j.amc.2011.01.093

2. Yang L: A note on a paper ‘Convergence theorem for the common solution for a finite family of ϕ -strongly accretive operator equations’. Appl. Math. Comput. 2012, 218(21):10367–10369. 10.1016/j.amc.2012.04.037

3. Rafiq A: On iterations for families of asymptotically pseudocontractive mappings. Appl. Math. Lett. 2011, 24(1):33–38. 10.1016/j.aml.2010.08.005

4. Deimling K: Nonlinear Functional Analysis. Springer, Berlin; 1985.

5. Moore C, Nnoli BVC: Iterative solution of nonlinear equations involving set-valued uniformly accretive operators. Comput. Math. Appl. 2001, 42(1–2):131–140. 10.1016/S0898-1221(01)00138-9

Acknowledgements

The authors are very grateful to Professor Yeol-Je Cho for good suggestions which helped to improve the manuscript. This work is supported by the Hebei Natural Science Foundation No. A2011210033.

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All authors contributed equally in writing this paper, and read and approved the final manuscript.

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Xue, Z., Zhou, H. A further remark to paper ‘Convergence theorems for the common solution for a finite family of ϕ-strongly accretive operator equations’. J Inequal Appl 2013, 35 (2013). https://doi.org/10.1186/1029-242X-2013-35