In this section, we propose a Halperntype iterative scheme for finding common zeros of an infinite family of accretive operators in a uniformly convex Banach space and prove the following strong convergence theorem.
Theorem 4.1 Let E be a real uniformly convex Banach space which admits the weakly sequentially continuous duality mapping {J}_{\varphi}. Let {\{{A}_{j}\}}_{j=1}^{\mathrm{\infty}} be an infinite family of accretive operators satisfying the range condition, and let C be a nonempty, closed and convex subset of E such that \overline{D({A}_{j})}\subset C\subset {\bigcap}_{r>0}R(I+r{A}_{j}) for each j\in \mathbb{N}. Let {r}_{n}>0 and r>0 be such that {lim}_{n\to \mathrm{\infty}}{r}_{n}=r, and let {J}_{{r}_{n}}^{{A}_{j}}={(I+{r}_{n}{A}_{j})}^{1} be the resolvent of A. Let {\{{\alpha}_{n}\}}_{n\in \mathbb{N}}, {\{{\beta}_{n,j}\}}_{n\in \mathbb{N},j\in \mathbb{N}\cup \{0\}} be sequences in [0,1] satisfying the following control conditions:

(a)
{lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0;

(b)
{\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty};

(c)
{\beta}_{n,0}+{\sum}_{j=1}^{\mathrm{\infty}}{\beta}_{n,j}=1, \mathrm{\forall}n\in \mathbb{N};

(d)
{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{\beta}_{n,0}{\beta}_{n,j}>0, \mathrm{\forall}j\in \mathbb{N}.
Let
{\{{x}_{n}\}}_{n\in \mathbb{N}}
be a sequence generated by
\{\begin{array}{l}{x}_{1},u\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\\ {y}_{n}={\beta}_{n,0}{x}_{n}+{\sum}_{j=1}^{\mathrm{\infty}}{\beta}_{n,j}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n},\\ {x}_{n+1}={\alpha}_{n}u+(1{\alpha}_{n}){y}_{n}.\end{array}
(4.1)
If Z:={\bigcap}_{j=1}^{\mathrm{\infty}}{A}_{j}^{1}(0)\ne \mathrm{\varnothing}, then the sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} defined in (4.1) converges strongly to {Q}_{Z}u, where {Q}_{Z} is the sunny nonexpansive retraction from E onto Z.
Proof We divide the proof into several steps.
We first note that Z is closed and convex. Set z={Q}_{Z}u.
Step 1. We prove that the sequences {\{{x}_{n}\}}_{n\in \mathbb{N}}, {\{{y}_{n}\}}_{n\in \mathbb{N}} and {\{{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\}}_{n\in \mathbb{N}} are bounded.
We first show that {\{{x}_{n}\}}_{n\in \mathbb{N}} is bounded. Let p\in Z be fixed. In view of Lemma 2.8, there exists a continuous, strictly increasing and convex function g:[0,\mathrm{\infty})\to [0,\mathrm{\infty}) with g(0)=0 such that for any j\in \mathbb{N},
\begin{array}{rcl}{\parallel {y}_{n}p\parallel}^{2}& =& {\parallel {\beta}_{n,0}{x}_{n}+\sum _{j=1}^{\mathrm{\infty}}{\beta}_{n,j}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}p\parallel}^{2}\\ \le & {\beta}_{n,0}{\parallel {x}_{n}p\parallel}^{2}+\sum _{j=1}^{\mathrm{\infty}}{\beta}_{n,j}{\parallel {J}_{{r}_{n}}^{{A}_{j}}{x}_{n}p\parallel}^{2}{\beta}_{n,0}{\beta}_{n,j}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right)\\ \le & {\beta}_{n,0}{\parallel {x}_{n}p\parallel}^{2}+\sum _{j=1}^{\mathrm{\infty}}{\beta}_{n,j}{\parallel {x}_{n}p\parallel}^{2}{\beta}_{n,0}{\beta}_{n,j}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right)\\ =& {\parallel {x}_{n}p\parallel}^{2}{\beta}_{n,0}{\beta}_{n,j}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right)\\ \le & {\parallel {x}_{n}p\parallel}^{2}.\end{array}
(4.2)
This implies that
\begin{array}{rcl}\parallel {x}_{n+1}p\parallel & =& \parallel {\alpha}_{n}u+(1{\alpha}_{n}){y}_{n}p\parallel \le {\alpha}_{n}\parallel up\parallel +(1{\alpha}_{n})\parallel {y}_{n}p\parallel \\ \le & {\alpha}_{n}\parallel up\parallel +(1{\alpha}_{n})\parallel {x}_{n}p\parallel \le max\{\parallel up\parallel ,\parallel {x}_{n}p\parallel \}.\end{array}
By induction, we obtain
\parallel {x}_{n+1}p\parallel \le max\{\parallel up\parallel ,\parallel {x}_{1}p\parallel \}
for all n\in \mathbb{N}. This implies that the sequence {\{\parallel {x}_{n}p\parallel \}}_{n\in \mathbb{N}} is bounded and hence the sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} is bounded. This, together with (4.1), implies that the sequences {\{{y}_{n}\}}_{n\in \mathbb{N}} and {\{{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\}}_{n\in \mathbb{N}} are bounded too.
Step 2. We prove that for any n\in \mathbb{N},
{\parallel {x}_{n+1}z\parallel}^{2}\le (1{\alpha}_{n}){\parallel {x}_{n}z\parallel}^{2}+2{\alpha}_{n}\u3008uz,{J}_{\varphi}({x}_{n+1}z)\u3009.
(4.3)
Let us show (4.3). For any n,j\in \mathbb{N}, in view of (4.2), we obtain
{\parallel {y}_{n}z\parallel}^{2}\le {\parallel {x}_{n}z\parallel}^{2}{\beta}_{n,0}{\beta}_{n,j}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right).
This implies that
\begin{array}{rcl}{\parallel {x}_{n+1}z\parallel}^{2}& =& {\parallel {\alpha}_{n}u+(1{\alpha}_{n}){y}_{n}z\parallel}^{2}\le {\alpha}_{n}{\parallel uz\parallel}^{2}+(1{\alpha}_{n}){\parallel {y}_{n}z\parallel}^{2}\\ \le & {\alpha}_{n}{\parallel uz\parallel}^{2}+(1{\alpha}_{n})[{\parallel {x}_{n}z\parallel}^{2}{\beta}_{n,0}{\beta}_{n,j}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right)].\end{array}
(4.4)
Let {M}_{2}:=sup\{{\parallel uz\parallel}^{2}{\parallel {x}_{n}z\parallel}^{2}+{\beta}_{n,0}{\beta}_{n,j}g(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel ):n,j\in \mathbb{N}\}. It follows from (4.4) that
{\beta}_{n,0}{\beta}_{n,j}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right)\le {\parallel {x}_{n}z\parallel}^{2}{\parallel {x}_{n+1}z\parallel}^{2}+{\alpha}_{n}{M}_{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall}j\in \mathbb{N}.
(4.5)
In view of Lemma 2.1 and (4.2), we obtain
\begin{array}{rcl}{\parallel {x}_{n+1}z\parallel}^{2}& =& {\parallel {\alpha}_{n}u+(1{\alpha}_{n}){y}_{n}z\parallel}^{2}\\ =& {\parallel {\alpha}_{n}(uz)+(1{\alpha}_{n})({y}_{n}z)\parallel}^{2}\\ \le & {\parallel (1{\alpha}_{n})({y}_{n}z)\parallel}^{2}+2\u3008{\alpha}_{n}(uz),{J}_{\varphi}({x}_{n+1}z)\u3009\\ =& {(1{\alpha}_{n})}^{2}{\parallel {y}_{n}z\parallel}^{2}+2\u3008{\alpha}_{n}(uz),{J}_{\varphi}({x}_{n+1}z)\u3009\\ \le & (1{\alpha}_{n}){\parallel {y}_{n}z\parallel}^{2}+2\u3008{\alpha}_{n}(uz),{J}_{\varphi}({x}_{n+1}z)\u3009\\ =& (1{\alpha}_{n}){\parallel {y}_{n}z\parallel}^{2}+2{\alpha}_{n}\u3008uz,{J}_{\varphi}({x}_{n+1}z)\u3009.\end{array}
Step 3. We prove that {x}_{n}\to z as n\to \mathrm{\infty}.
We discuss the following two possible cases.
Case 1. Suppose that there exists {n}_{0}\in \mathbb{N} such that {\{\parallel {x}_{n}z\parallel \}}_{n={n}_{0}}^{\mathrm{\infty}} is nonincreasing. Then the sequence {\{\parallel {x}_{n}z\parallel \}}_{n\in \mathbb{N}} is convergent. Thus we have {\parallel {x}_{n}z\parallel}^{2}{\parallel {x}_{n+1}z\parallel}^{2}\to 0 as n\to \mathrm{\infty}. This, together with condition (d) and (4.5), implies that
\underset{n\to \mathrm{\infty}}{lim}g\left(\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel \right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}j\in \mathbb{N}.
From the properties of g, we conclude that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}j\in \mathbb{N}.
(4.6)
On the other hand, we have
{y}_{n}{x}_{n}=\sum _{j=1}^{\mathrm{\infty}}{\beta}_{n,j}({J}_{{r}_{n}}^{{A}_{j}}{x}_{n}{x}_{n})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}_{n+1}{y}_{n}={\alpha}_{n}(u{y}_{n}).
This implies that
\underset{n\to \mathrm{\infty}}{lim}\parallel {y}_{n}{x}_{n}\parallel =0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n+1}{y}_{n}\parallel =0.
(4.7)
By the triangle inequality, we conclude that
\parallel {x}_{n+1}{x}_{n}\parallel \le \parallel {x}_{n+1}{y}_{n}\parallel +\parallel {y}_{n}{x}_{n}\parallel .
It follows from (4.7) that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n+1}{x}_{n}\parallel =0.
(4.8)
Exploiting Lemma 2.5 and (4.6), we obtain
\parallel {x}_{n}{J}_{r}^{{A}_{j}}{x}_{n}\parallel \le \parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel +\parallel {J}_{{r}_{n}}^{{A}_{j}}{x}_{n}{J}_{r}^{{A}_{j}}{x}_{n}\parallel \le \parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel +\frac{{r}_{n}r}{{r}_{n}}\parallel {x}_{n}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n}\parallel .
This implies that
\underset{n\to \mathrm{\infty}}{lim}\parallel {x}_{n}{J}_{r}^{{A}_{j}}{x}_{n}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}j\in \mathbb{N}.
(4.9)
Since {\{{x}_{n}\}}_{n\in \mathbb{N}} is bounded, there exists a subsequence {\{{x}_{{n}_{i}}\}}_{i\in \mathbb{N}} of {\{{x}_{n}\}}_{n\in \mathbb{N}} such that {x}_{{n}_{i}+1}\rightharpoonup y\in F({J}_{r}^{{A}_{j}}) for all j\in \mathbb{N}. This, together with Lemma 2.1, implies that
\begin{array}{rcl}\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008uz,{J}_{\varphi}({x}_{n+1}z)\u3009& =& \underset{i\to \mathrm{\infty}}{lim}\u3008uz,{J}_{\varphi}({x}_{{n}_{i}+1}z)\u3009\\ =& \u3008uz,{J}_{\varphi}(yz)\u3009\\ \le & 0.\end{array}
(4.10)
Thus we have the desired result by Lemma 2.6.
Case 2. Suppose that there exists a subsequence {\{{n}_{i}\}}_{i\in \mathbb{N}} of {\{n\}}_{n\in \mathbb{N}} such that
\parallel {x}_{{n}_{i}}z\parallel <\parallel {x}_{{n}_{i}+1}z\parallel
for all i\in \mathbb{N}. Then, by Lemma 2.7, there exists a nondecreasing sequence {\{{m}_{k}\}}_{k\in \mathbb{N}}\subset \mathbb{N} such that {m}_{k}\to \mathrm{\infty},
\parallel z{x}_{{m}_{k}}\parallel <\parallel z{x}_{{m}_{k}+1}\parallel \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\parallel z{x}_{k}\parallel \le \parallel {x}_{{m}_{k}+1}\stackrel{\u02c6}{u}\parallel
for all k\in \mathbb{N}. This, together with (4.5), implies that
{\beta}_{{m}_{k},0}{\beta}_{{m}_{k},j}g\left(\parallel {x}_{{m}_{k}}{J}_{r}^{{A}_{j}}{x}_{{m}_{k}}\parallel \right)\le {\parallel {x}_{{m}_{k}}z\parallel}^{2}{\parallel {x}_{{m}_{k}+1}z\parallel}^{2}+{\alpha}_{{m}_{k}}{M}_{2}\le {\alpha}_{{m}_{k}}{M}_{2}
for all k\in \mathbb{N}. Then, by conditions (a) and (c), we get
\underset{k\to \mathrm{\infty}}{lim}g\left(\parallel {x}_{{m}_{k}}{J}_{r}^{{A}_{j}}{x}_{{m}_{k}}\parallel \right)=0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}j\in \mathbb{N}.
From the properties of g, we conclude that
\underset{k\to \mathrm{\infty}}{lim}\parallel {x}_{{m}_{k}}{J}_{r}^{{A}_{j}}{x}_{{m}_{k}}\parallel =0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}j\in \mathbb{N}.
By the same argument as Case 1, we arrive at
\underset{k\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008uz,{J}_{\varphi}({x}_{{m}_{k}}z)\u3009=\underset{k\to \mathrm{\infty}}{lim\hspace{0.17em}sup}\u3008uz,{J}_{\varphi}({x}_{{m}_{k}+1}z)\u3009\le 0.
It follows from (4.3) that
{\parallel {x}_{{m}_{k}+1}z\parallel}^{2}\le (1{\alpha}_{{m}_{k}}){\parallel {x}_{{m}_{k}}z\parallel}^{2}+{\alpha}_{{m}_{k}}\u3008uz,{J}_{\varphi}({x}_{{m}_{k}}z)\u3009.
(4.11)
Since \parallel {x}_{{m}_{k}}z\parallel \le \parallel {x}_{{m}_{k}+1}z\parallel, we have that
\begin{array}{rcl}{\alpha}_{{m}_{k}}{\parallel {x}_{{m}_{k}}z\parallel}^{2}& \le & {\parallel {x}_{{m}_{k}}z\parallel}^{2}{\parallel {x}_{{m}_{k}+1}z\parallel}^{2}+{\alpha}_{{m}_{k}}\u3008uz,{J}_{\varphi}({x}_{{m}_{k}}z)\u3009\\ \le & 2{\alpha}_{{m}_{k}}\u3008uz,{J}_{\varphi}({x}_{{m}_{k}}z)\u3009.\end{array}
(4.12)
In particular, since {\alpha}_{{m}_{k}}>0, we obtain
{\parallel {x}_{{m}_{k}}z\parallel}^{2}\le \u3008uz,{J}_{\varphi}({x}_{{m}_{k}}z)\u3009.
In view of (4.12), we deduce that
\underset{k\to \mathrm{\infty}}{lim}\parallel {x}_{{m}_{k}}z\parallel =0.
This, together with (4.10), implies that
\underset{k\to \mathrm{\infty}}{lim}\parallel {x}_{{m}_{k}+1}z\parallel =0.
On the other hand, we have \parallel {x}_{k}z\parallel \le \parallel {x}_{{m}_{k}+1}z\parallel for all k\in \mathbb{N}, which implies that {x}_{{m}_{k}}\to z as k\to \mathrm{\infty}. Thus, we have {x}_{n}\to z as n\to \mathrm{\infty}, which completes the proof. □
Theorem 4.2 Let E be a real uniformly convex Banach space having a Gâteaux differentiable norm. Let {\{{A}_{j}\}}_{j=1}^{\mathrm{\infty}} be an infinite family of accretive operators satisfying the range condition, and let C be a nonempty, closed and convex subset of E such that \overline{D({A}_{j})}\subset C\subset {\bigcap}_{r>0}R(I+r{A}_{j}) for each j\in \mathbb{N}. Let {r}_{n}>0 and r>0 be such that {lim}_{n\to \mathrm{\infty}}{r}_{n}=r, and let {J}_{{r}_{n}}^{{A}_{j}}={(I+{r}_{n}{A}_{j})}^{1} be the resolvent of A. Let {\{{\alpha}_{n}\}}_{n\in \mathbb{N}}, {\{{\beta}_{n,j}\}}_{n\in \mathbb{N},j\in \mathbb{N}\cup \{0\}} be sequences in [0,1] satisfying the following control conditions:

(a)
{lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=0;

(b)
{\sum}_{n=1}^{\mathrm{\infty}}{\alpha}_{n}=\mathrm{\infty};

(c)
{\beta}_{n,0}+{\sum}_{j=1}^{\mathrm{\infty}}{\beta}_{n,j}=1, \mathrm{\forall}n\in \mathbb{N};

(d)
{lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{\beta}_{n,0}{\beta}_{n,j}>0, \mathrm{\forall}j\in \mathbb{N}.
Let
{\{{x}_{n}\}}_{n\in \mathbb{N}}
be a sequence generated by
\{\begin{array}{l}{x}_{1}\in C\phantom{\rule{1em}{0ex}}\mathit{\text{chosen arbitrarily}},\\ {y}_{n}={\beta}_{n,0}{x}_{n}+{\sum}_{j=1}^{\mathrm{\infty}}{\beta}_{n,j}{J}_{{r}_{n}}^{{A}_{j}}{x}_{n},\\ {x}_{n+1}={\alpha}_{n}u+(1{\alpha}_{n}){y}_{n}.\end{array}
If Z:={\bigcap}_{j=1}^{\mathrm{\infty}}{A}_{j}^{1}(0)\ne \mathrm{\varnothing}, then the sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} converges strongly to {Q}_{F}u, where {Q}_{F} is the sunny nonexpansive retraction from E onto Z.
Remark 4.1 The main results of [22, 23] gave strong convergence theorems for finding common zeros of an accretive operator, while the main results of the present paper (Theorems 4.1 and 4.2) give strong convergence theorems for finding common zeros of an infinite family of accretive operators. So our results extend and improve the corresponding results of [22, 23].
Remark 4.2 The twostep Halpern iteration process is a generalization of the onestep Halpern iteration process. It provides more flexibility in defining the algorithm parameters, which is important from the numerical implementation perspective.