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A further remark to paper ‘Convergence theorems for the common solution for a finite family of ϕstrongly accretive operator equations’
Journal of Inequalities and Applications volume 2013, Article number: 35 (2013)
Abstract
In this note, we point out several gaps in Gurudwan and Sharma (Appl. Math. Comput. 217(15):67486754, 2011) and Yang (Appl. Math. Comput. 218(21):1036710369, 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 Φstrongly accretive operator equations. Their results are as follows.
Theorem GS [[1], Theorem 3.1]
Let E be an arbitrary real Banach space and let {\{{A}_{i}\}}_{i=1}^{N}:E\to E be uniformly continuous ϕstrongly accretive operators and each range of either {A}_{i} or (I{A}_{i}) be bounded. Let, for i=1,\dots ,N, {\{{u}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}} be sequences in E and {\{{a}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}}, {\{{b}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}}, {\{{c}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}} be real sequences in [0,1] satisfying

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

(ii)
{\sum}_{n=0}^{\mathrm{\infty}}{b}_{n}^{N}=\mathrm{\infty},

(iii)
{\sum}_{n=0}^{\mathrm{\infty}}{c}_{n}<\mathrm{\infty},

(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 {\{{S}_{i}\}}_{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 multistep iterative sequence with errors {\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}} defined by the above converges strongly to the unique solution of the operator equations {\{{A}_{i}x\}}_{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 {\{{A}_{i}\}}_{i=1}^{N}:E\to E be uniformly continuous ϕstrongly accretive operators and each range of either {A}_{i} or (I{A}_{i}) is bounded. Let for i=1,\dots ,N, {\{{u}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}} be bounded sequences in E and {\{{a}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}}, {\{{b}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}}, {\{{c}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}} be real sequences in [0,1] satisfying

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

(ii)
{\sum}_{n=0}^{\mathrm{\infty}}{b}_{n}^{N}=\mathrm{\infty},

(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 {\{{S}_{i}\}}_{i=1}^{N}:E\to E by {S}_{i}x=(I{A}_{i})x+f, \mathrm{\forall}i=1,\dots ,N, \mathrm{\forall}x\in E. Then the multistep iterative sequence with errors {\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}} defined by the above converges strongly to the unique solution of the operator equations {\{{A}_{i}x\}}_{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 xy\parallel \le {\varphi}_{i}^{1}(\parallel {A}_{i}x{A}_{i}y\parallel ), that is, \u3008{A}_{i}x{A}_{i}y,j(xy)\u3009\ge \varphi (\parallel xy\parallel )\parallel xy\parallel \Rightarrow \varphi (\parallel xy\parallel )\le \parallel {A}_{i}x{A}_{i}y\parallel, but we cannot deduce \parallel xy\parallel \le {\varphi}_{i}^{1}(\parallel {A}_{i}x{A}_{i}y\parallel ). The reason is that it is possible \parallel {A}_{i}x{A}_{i}y\parallel does not belong to R(\varphi ) (range of ϕ). A counterexample is as follows. Let us define \varphi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) by \varphi (\alpha )=\frac{{2}^{\alpha}1}{{2}^{\alpha}+1}; then it can be easily seen that ϕ is increasing with \varphi (0)=0, but {lim}_{\alpha \to +\mathrm{\infty}}\varphi (\alpha )=1 and {\varphi}^{1}(2) 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 <\u03f5 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 \{{\gamma}_{n}\} be a real sequence, \{{\gamma}_{{n}_{m}}\} be some infinite subsequence of \{{\gamma}_{n}\} and \{{n}_{m}\} be neither odd nor even sequence, then the conclusions are as follows:
(Ci) {lim}_{n\to \mathrm{\infty}}{\gamma}_{n}=0\iff \mathrm{\forall}\u03f5>0, ∃ nonnegative integer {n}_{0} such that {\gamma}_{{n}_{m}+j}<\u03f5 for {n}_{m}\ge {n}_{0}, j\ge 1.
(Cii) {lim}_{n\to \mathrm{\infty}}{\gamma}_{n}=0\Rightarrow {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 (Ci), it just illustrates the result (Cii). Therefore, the note given by Yang [2] confused the conclusions (Ci) and (Cii).
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
where \u3008\cdot ,\cdot \u3009 denotes the generalized duality pairing. The singlevalued 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(xy)\in J(xy) such that
without loss of generality, we assume that k\in (0,1). The operator T is called ϕstrongly accretive if for any x,y\in E, there exist j(xy)\in J(xy) and a strictly increasing continuous function \varphi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) with \varphi (0)=0 such that
It is obvious that a strongly accretive operator must be the ϕstrongly accretive in the special case in which \varphi (t)=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 ϕstrongly 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 ϕstrongly 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
for all x,y\in E and j(x+y)\in J(x+y).
Lemma 2.3 [5]
Let {\{{\delta}_{n}\}}_{n=0}^{\mathrm{\infty}}, {\{{\lambda}_{n}\}}_{n=0}^{\mathrm{\infty}} and {\{{\gamma}_{n}\}}_{n=0}^{\mathrm{\infty}} be three nonnegative real sequences and \varphi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) be a strictly increasing and continuous function with \varphi (0)=0 satisfying the following inequality:
where {\lambda}_{n}\in [0,1] with {\sum}_{n=0}^{\mathrm{\infty}}{\lambda}_{n}=\mathrm{\infty}, {\gamma}_{n}=o({\lambda}_{n}). 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 {\{{A}_{i}\}}_{i=1}^{N}:E\to E be N uniformly continuous ϕstrongly accretive operators. For i=1,2,\dots ,N, let {\{{u}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}} be bounded sequences in E and {\{{a}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}}, {\{{b}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}}, {\{{c}_{n}^{i}\}}_{n=1}^{\mathrm{\infty}} be real sequences in [0,1] satisfying

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

(ii)
{\sum}_{n=1}^{\mathrm{\infty}}{b}_{n}^{N}=+\mathrm{\infty};

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

(iv)
{c}_{n}^{N}=o({b}_{n}^{N}).
For any given f\in E, define {\{{S}_{i}\}}_{i=1}^{N}:E\to E with {\bigcap}_{i=1}^{N}F({S}_{i})\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({S}_{i})=\{x\in E:{S}_{i}x=x\}. Then, for some {x}_{0}\in E, the multistep iterative sequence with errors {\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}} defined by
converges strongly to the unique solution of the operator equations {\{{A}_{i}x\}}_{i=1}^{N}=f.
Proof Since {\{{A}_{i}\}}_{i=1}^{N}:E\to E is ϕstrongly 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({S}_{i})\ne \mathrm{\varnothing}, then {\bigcap}_{i=1}^{N}F({S}_{i}) is a single set, let q. Meanwhile, there exists a strictly increasing continuous function \varphi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) with \varphi (0)=0 such that
for x\in E, q\in F(T), that is,
Choose some {x}_{0}\in E and {x}_{0}\ne {S}_{i}{x}_{0} such that {r}_{0}\in R(\mathrm{\Phi}), where
R(\mathrm{\Phi}) is the range of Φ. Indeed, if \mathrm{\Phi}(r)\to +\mathrm{\infty} as r\to +\mathrm{\infty}, then {r}_{0}\in R(\mathrm{\Phi}); if sup\{\mathrm{\Phi}(r):r\in [0,+\mathrm{\infty})\}={r}_{1}<+\mathrm{\infty} with {r}_{1}<{r}_{0}, then for q\in E, there exists a sequence \{{w}_{n}\} 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 \{{w}_{n}{S}_{i}{w}_{n}\} 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}\{\parallel {x}_{0}{S}_{i}{x}_{0}\parallel \cdot \parallel {x}_{0}q\parallel \}\in R(\varphi ). It is to ensure that {\mathrm{\Phi}}^{1}({r}_{0}) is defined well.
Step I. We show that \{{x}_{n}\} is a bounded sequence.
Set R={\mathrm{\Phi}}^{1}({r}_{0}), then from the above formula (@), we obtain that \parallel {x}_{0}q\parallel \le R. Denote
Since {S}_{i} is uniformly continuous, then {S}_{i} is bounded. We let
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 {\u03f5}_{0}=\frac{\mathrm{\Phi}(\frac{R}{2})}{8R}, there exists common \delta >0 such that \parallel {S}_{i}x{S}_{i}y\parallel <{\u03f5}_{0} when \parallel xy\parallel <\delta. Denote
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({b}_{n}^{N}), 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 multistep iteration, we have
then {x}_{n}^{1}\in {B}_{2}. Similarly, we have
then {x}_{n}^{2}\in {B}_{2}. ⋯⋯ , we have
then {x}_{n}^{N1}\in {B}_{2}. Therefore, we get
And we also have
and
By the uniform continuity of {S}_{N}, we have
Using Lemma 2.2 and the above formulas, we have
which is a contradiction. So, {x}_{n+1}\in {B}_{1}, i.e., \{{x}_{n}\} is a bounded sequence, from which it follows that \{{x}_{n}^{1}\},\{{x}_{n}^{2}\},\dots ,\{{x}_{n}^{N1}\} 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 \{{x}_{n}\}, \{{x}_{n}^{N1}\} are bounded. From (3.5) and (3.6), we obtain
By (3.7), we have
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:

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

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

(c)
The proof method of our theorem differs from that of [1] and [2].
Theorem 3.3 Let E, \{{u}_{n}^{i}\}, \{{a}_{n}^{i}\}, \{{b}_{n}^{i}\}, \{{c}_{n}^{i}\} (i=1,2,\dots ,N) be as in Theorem 3.1 and let {\{{T}_{i}\}}_{i=1}^{N}:E\to E be N uniformly continuous ϕstrongly pseudocontractive mappings. Then, for some {x}_{0}\in E, the multistep iterative sequence with errors {\{{x}_{n}\}}_{n=1}^{\mathrm{\infty}} defined by
converges strongly to the unique common fixed point of {\{{T}_{i}\}}_{i=1}^{N}.
Proof See [1]. □
References
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
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
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
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Acknowledgements
The authors are very grateful to Professor YeolJe 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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Xue, Z., Zhou, H. A further remark to paper ‘Convergence theorems for the common solution for a finite family of ϕstrongly accretive operator equations’. J Inequal Appl 2013, 35 (2013). https://doi.org/10.1186/1029242X201335
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DOI: https://doi.org/10.1186/1029242X201335
Keywords
 uniformly continuous
 Φstrongly accretive
 multistep iteration with errors
 Banach space