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Convergence theorems for mixed type asymptotically nonexpansive mappings
Fixed Point Theory and Applications volume 2012, Article number: 224 (2012)
Abstract
In this paper, we introduce a new twostep iterative scheme of mixed type for two asymptotically nonexpansive selfmappings and two asymptotically nonexpansive nonselfmappings and prove strong and weak convergence theorems for the new twostep iterative scheme in uniformly convex Banach spaces.
1 Introduction
Let K be a nonempty subset of a real normed linear space E. A mapping T:K\to K is said to be asymptotically nonexpansive if there exists a sequence \{{k}_{n}\}\subset [1,\mathrm{\infty}) with {lim}_{n\to \mathrm{\infty}}{k}_{n}=1 such that
for all x,y\in K and n\ge 1.
In 1972, Goebel and Kirk [1] introduced the class of asymptotically nonexpansive selfmappings, which is an important generalization of the class of nonexpansive selfmappings, and proved that if K is a nonempty closed convex subset of a real uniformly convex Banach space E and T is an asymptotically nonexpansive selfmapping of K, then T has a fixed point.
Since then, some authors proved weak and strong convergence theorems for asymptotically nonexpansive selfmappings in Banach spaces (see [2–16]), which extend and improve the result of Goebel and Kirk in several ways.
Recently, Chidume et al. [10] introduced the concept of asymptotically nonexpansive nonselfmappings, which is a generalization of an asymptotically nonexpansive selfmapping, as follows.
Definition 1.1 [10]
Let K be a nonempty subset of a real normed linear space E. Let P:E\to K be a nonexpansive retraction of E onto K. A nonselfmapping T:K\to E is said to be asymptotically nonexpansive if there exists a sequence \{{k}_{n}\}\subset [1,\mathrm{\infty}) with {k}_{n}\to 1 as n\to \mathrm{\infty} such that
for all x,y\in K and n\ge 1.
Let K be a nonempty closed convex subset of a real uniformly convex Banach space E.
In 2003, also, Chidume et al. [10] studied the following iteration scheme:
for each n\ge 1, where \{{\alpha}_{n}\} is a sequence in (0,1) and P is a nonexpansive retraction of E onto K, and proved some strong and weak convergence theorems for an asymptotically nonexpansive nonselfmapping.
In 2006, Wang [11] generalized the iteration process (1.3) as follows:
for each n\ge 1, where {T}_{1},{T}_{2}:K\to E are two asymptotically nonexpansive nonselfmappings and \{{\alpha}_{n}\}, \{{\beta}_{n}\} are real sequences in [0,1), and proved some strong and weak convergence theorems for two asymptotically nonexpansive nonselfmappings. Recently, Guo and Guo [12] proved some new weak convergence theorems for the iteration process (1.4).
The purpose of this paper is to construct a new iteration scheme of mixed type for two asymptotically nonexpansive selfmappings and two asymptotically nonexpansive nonselfmappings and to prove some strong and weak convergence theorems for the new iteration scheme in uniformly convex Banach spaces.
2 Preliminaries
Let E be a real Banach space, K be a nonempty closed convex subset of E and P:E\to K be a nonexpansive retraction of E onto K. Let {S}_{1},{S}_{2}:K\to K be two asymptotically nonexpansive selfmappings and {T}_{1},{T}_{2}:K\to E be two asymptotically nonexpansive nonselfmappings. Then we define the new iteration scheme of mixed type as follows:
for each n\ge 1, where \{{\alpha}_{n}\}, \{{\beta}_{n}\} are two sequences in [0,1).
If {S}_{1} and {S}_{2} are the identity mappings, then the iterative scheme (2.1) reduces to the sequence (1.4).
We denote the set of common fixed points of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2} by F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}) and denote the distance between a point z and a set A in E by d(z,A)={inf}_{x\in A}\parallel zx\parallel.
Now, we recall some wellknown concepts and results.
Let E be a real Banach space, {E}^{\ast} be the dual space of E and J:E\to {2}^{{E}^{\ast}} be the normalized duality mapping defined by
for all x\in E, where \u3008\cdot ,\cdot \u3009 denotes duality pairing between E and {E}^{\ast}. A singlevalued normalized duality mapping is denoted by j.
A subset K of a real Banach space E is called a retract of E [10] if there exists a continuous mapping P:E\to K such that Px=x for all x\in K. Every closed convex subset of a uniformly convex Banach space is a retract. A mapping P:E\to E is called a retraction if {P}^{2}=P. It follows that if a mapping P is a retraction, then Py=y for all y in the range of P.
A Banach space E is said to satisfy Opial’s condition [17] if, for any sequence \{{x}_{n}\} of E, {x}_{n}\to x weakly as n\to \mathrm{\infty} implies that
for all y\in E with y\ne x.
A Banach space E is said to have a Fréchet differentiable norm [18] if, for all x\in U=\{x\in E:\parallel x\parallel =1\},
exists and is attained uniformly in y\in U.
A Banach space E is said to have the KadecKlee property [19] if for every sequence \{{x}_{n}\} in E, {x}_{n}\to x weakly and \parallel {x}_{n}\parallel \to \parallel x\parallel, it follows that {x}_{n}\to x strongly.
Let K be a nonempty closed subset of a real Banach space E. A nonselfmapping T:K\to E is said to be semicompact [11] if, for any sequence \{{x}_{n}\} in K such that \parallel {x}_{n}T{x}_{n}\parallel \to 0 as n\to \mathrm{\infty}, there exists a subsequence \{{x}_{{n}_{j}}\} of \{{x}_{n}\} such that \{{x}_{{n}_{j}}\} converges strongly to some {x}^{\ast}\in K.
Lemma 2.1 [15]
Let \{{a}_{n}\}, \{{b}_{n}\} and \{{c}_{n}\} be three nonnegative sequences satisfying the following condition:
for each n\ge {n}_{0}, where {n}_{0} is some nonnegative integer, {\sum}_{n={n}_{0}}^{\mathrm{\infty}}{b}_{n}<\mathrm{\infty} and {\sum}_{n={n}_{0}}^{\mathrm{\infty}}{c}_{n}<\mathrm{\infty}. Then {lim}_{n\to \mathrm{\infty}}{a}_{n} exists.
Lemma 2.2 [8]
Let E be a real uniformly convex Banach space and 0<p\le {t}_{n}\le q<1 for each n\ge 1. Also, suppose that \{{x}_{n}\} and \{{y}_{n}\} are two sequences of E such that
hold for some r\ge 0. Then {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{y}_{n}\parallel =0.
Lemma 2.3 [10]
Let E be a real uniformly convex Banach space, K be a nonempty closed convex subset of E and T:K\to E be an asymptotically nonexpansive mapping with a sequence \{{k}_{n}\}\subset [1,\mathrm{\infty}) and {k}_{n}\to 1 as n\to \mathrm{\infty}. Then IT is demiclosed at zero, i.e., if {x}_{n}\to x weakly and {x}_{n}T{x}_{n}\to 0 strongly, then x\in F(T), where F(T) is the set of fixed points of T.
Lemma 2.4 [16]
Let X be a uniformly convex Banach space and C be a convex subset of X. Then there exists a strictly increasing continuous convex function \gamma :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) with \gamma (0)=0 such that, for each mapping S:C\to C with a Lipschitz constant L>0,
for all x,y\in C and 0<\alpha <1.
Lemma 2.5 [16]
Let X be a uniformly convex Banach space such that its dual space {X}^{\ast} has the KadecKlee property. Suppose \{{x}_{n}\} is a bounded sequence and {f}_{1},{f}_{2}\in {W}_{w}(\{{x}_{n}\}) such that
exists for all \alpha \in [0,1], where {W}_{w}(\{{x}_{n}\}) denotes the set of all weak subsequential limits of \{{x}_{n}\}. Then {f}_{1}={f}_{2}.
3 Strong convergence theorems
In this section, we prove strong convergence theorems for the iterative scheme given in (2.1) in uniformly convex Banach spaces.
Lemma 3.1 Let E be a real uniformly convex Banach space and K be a nonempty closed convex subset of E. Let {S}_{1},{S}_{2}:K\to K be two asymptotically nonexpansive selfmappings with \{{k}_{n}^{(1)}\},\{{k}_{n}^{(2)}\}\subset [1,\mathrm{\infty}) and {T}_{1},{T}_{2}:K\to E be two asymptotically nonexpansive nonselfmappings with \{{l}_{n}^{(1)}\},\{{l}_{n}^{(2)}\}\subset [1,\mathrm{\infty}) such that {\sum}_{n=1}^{\mathrm{\infty}}({k}_{n}^{(i)}1)<\mathrm{\infty} and {\sum}_{n=1}^{\mathrm{\infty}}({l}_{n}^{(i)}1)<\mathrm{\infty} for i=1,2, respectively, and F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2})\ne \mathrm{\varnothing}. Let \{{x}_{n}\} be the sequence defined by (2.1), where \{{\alpha}_{n}\} and \{{\beta}_{n}\} are two real sequences in [0,1). Then

(1)
{lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}q\parallel exists for any q\in F;

(2)
{lim}_{n\to \mathrm{\infty}}d({x}_{n},F) exists.
Proof (1) Set {h}_{n}=max\{{k}_{n}^{(1)},{k}_{n}^{(2)},{l}_{n}^{(1)},{l}_{n}^{(2)}\}. For any q\in F, it follows from (2.1) that
and so
Since {\sum}_{n=1}^{\mathrm{\infty}}({k}_{n}^{(i)}1)<\mathrm{\infty} and {\sum}_{n=1}^{\mathrm{\infty}}({l}_{n}^{(i)}1)<\mathrm{\infty} for i=1,2, we have {\sum}_{n=1}^{\mathrm{\infty}}({h}_{n}^{2}1)<\mathrm{\infty}. It follows from Lemma 2.1 that {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}q\parallel exists.

(2)
Taking the infimum over all q\in F in (3.2), we have
d({x}_{n+1},F)\le [1+({h}_{n}^{2}1)]d({x}_{n},F)
for each n\ge 1. It follows from {\sum}_{n=1}^{\mathrm{\infty}}({h}_{n}^{2}1)<\mathrm{\infty} and Lemma 2.1 that the conclusion (2) holds. This completes the proof. □
Lemma 3.2 Let E be a real uniformly convex Banach space and K be a nonempty closed convex subset of E. Let {S}_{1},{S}_{2}:K\to K be two asymptotically nonexpansive selfmappings with \{{k}_{n}^{(1)}\},\{{k}_{n}^{(2)}\}\subset [1,\mathrm{\infty}) and {T}_{1},{T}_{2}:K\to E be two asymptotically nonexpansive nonselfmappings with \{{l}_{n}^{(1)}\},\{{l}_{n}^{(2)}\}\subset [1,\mathrm{\infty}) such that {\sum}_{n=1}^{\mathrm{\infty}}({k}_{n}^{(i)}1)<\mathrm{\infty} and {\sum}_{n=1}^{\mathrm{\infty}}({l}_{n}^{(i)}1)<\mathrm{\infty} for i=1,2, respectively, and F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2})\ne \mathrm{\varnothing}. Let \{{x}_{n}\} be the sequence defined by (2.1) and the following conditions hold:

(a)
\{{\alpha}_{n}\} and \{{\beta}_{n}\} are two real sequences in [\u03f5,1\u03f5] for some \u03f5\in (0,1);

(b)
\parallel x{T}_{i}y\parallel \le \parallel {S}_{i}x{T}_{i}y\parallel for all x,y\in K and i=1,2.
Then {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{S}_{i}{x}_{n}\parallel ={lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{T}_{i}{x}_{n}\parallel =0 for i=1,2.
Proof Set {h}_{n}=max\{{k}_{n}^{(1)},{k}_{n}^{(2)},{l}_{n}^{(1)},{l}_{n}^{(2)}\}. For any given q\in F, {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}q\parallel exists by Lemma 3.1. Now, we assume that {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}q\parallel =c. It follows from (3.2) and {\sum}_{n=1}^{\mathrm{\infty}}({h}_{n}^{2}1)<\mathrm{\infty} that
and
Taking lim sup on both sides in (3.1), we obtain {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}\parallel {y}_{n}q\parallel \le c and so
Using Lemma 2.2, we have
By the condition (b), it follows that
and so, from (3.3), we have
Since
Taking lim inf on both sides in the inequality above, we have
by (3.4) and so
Using (3.1), we have
In addition, we have
and
It follows from Lemma 2.2 that
Now, we prove that
Indeed, since \parallel {x}_{n}{T}_{2}{(P{T}_{2})}^{n1}{x}_{n}\parallel \le \parallel {S}_{2}^{n}{x}_{n}{T}_{2}{(P{T}_{2})}^{n1}{x}_{n}\parallel by the condition (b). It follows from (3.5) that
Since {S}_{2}^{n}{x}_{n}=P({S}_{2}^{n}{x}_{n}) and P:E\to K is a nonexpansive retraction of E onto K, we have
and so
Furthermore, we have
Thus it follows from (3.5), (3.6) and (3.7) that
Since \parallel {x}_{n}{T}_{1}{(P{T}_{1})}^{n1}{x}_{n}\parallel \le \parallel {S}_{1}^{n}{x}_{n}{T}_{1}{(P{T}_{1})}^{n1}{x}_{n}\parallel by the condition (b) and
Using (3.3) and (3.8), we have
and
It follows from
and (3.3) that
In addition, we have
Using (3.3) and (3.11), we obtain that
Thus, using (3.9), (3.10) and the inequality
we have {lim}_{n\to \mathrm{\infty}}\parallel {S}_{1}^{n}{x}_{n}{x}_{n}\parallel =0. It follows from (3.6) and the inequality
that
Since
from (3.8), (3.11) and (3.13), it follows that
Again, since (P{T}_{i}){(P{T}_{i})}^{n2}{y}_{n1}, {x}_{n}\in K for i=1,2 and {T}_{1}, {T}_{2} are two asymptotically nonexpansive nonselfmappings, we have
for i=1,2. It follows from (3.12), (3.14) and (3.15) that
for i=1,2. Moreover, we have
Using (3.4), (3.8) and (3.12), we have
In addition, we have
for i=1,2. Thus it follows from (3.6), (3.10), (3.16) and (3.17) that
Finally, we prove that
In fact, by the condition (b), we have
for i=1,2. Thus it follows from (3.5), (3.6), (3.9) and (3.10) that
This completes the proof. □
Now, we find two mappings, {S}_{1}={S}_{2}=S and {T}_{1}={T}_{2}=T, satisfying the condition (b) in Lemma 3.2 as follows.
Example 3.1 [20]
Let ℝ be the real line with the usual norm \cdot  and let K=[1,1]. Define two mappings S,T:K\to K by
and
Now, we show that T is nonexpansive. In fact, if x,y\in [0,1] or x,y\in [1,0), then we have
If x\in [0,1] and y\in [1,0) or x\in [1,0) and y\in [0,1], then we have
This implies that T is nonexpansive and so T is an asymptotically nonexpansive mapping with {k}_{n}=1 for each n\ge 1. Similarly, we can show that S is an asymptotically nonexpansive mapping with {l}_{n}=1 for each n\ge 1.
Next, we show that two mappings S, T satisfy the condition (b) in Lemma 3.2. For this, we consider the following cases:
Case 1. Let x,y\in [0,1]. Then we have
Case 2. Let x,y\in [1,0). Then we have
Case 3. Let x\in [1,0) and y\in [0,1]. Then we have
Case 4. Let x\in [0,1] and y\in [1,0). Then we have
Therefore, the condition (b) in Lemma 3.2 is satisfied.
Theorem 3.1 Under the assumptions of Lemma 3.2, if one of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2} is completely continuous, then the sequence \{{x}_{n}\} defined by (2.1) converges strongly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}.
Proof Without loss of generality, we can assume that {S}_{1} is completely continuous. Since \{{x}_{n}\} is bounded by Lemma 3.1, there exists a subsequence \{{S}_{1}{x}_{{n}_{j}}\} of \{{S}_{1}{x}_{n}\} such that \{{S}_{1}{x}_{{n}_{j}}\} converges strongly to some {q}^{\ast}. Moreover, we know that
and
by Lemma 3.2, which imply that
as j\to \mathrm{\infty} and so {x}_{{n}_{j}}\to {q}^{\ast}\in K. Thus, by the continuity of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}, we have
and
for i=1,2. Thus it follows that {q}^{\ast}\in F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}). Furthermore, since {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{q}^{\ast}\parallel exists by Lemma 3.1, we have {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{q}^{\ast}\parallel =0. This completes the proof. □
Theorem 3.2 Under the assumptions of Lemma 3.2, if one of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2} is semicompact, then the sequence \{{x}_{n}\} defined by (2.1) converges strongly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}.
Proof Since {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{S}_{i}{x}_{n}\parallel ={lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{T}_{i}{x}_{n}\parallel =0 for i=1,2 by Lemma 3.2 and one of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2} is semicompact, there exists a subsequence \{{x}_{{n}_{j}}\} of \{{x}_{n}\} such that \{{x}_{{n}_{j}}\} converges strongly to some {q}^{\ast}\in K. Moreover, by the continuity of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}, we have \parallel {q}^{\ast}{S}_{i}{q}^{\ast}\parallel ={lim}_{j\to \mathrm{\infty}}\parallel {x}_{{n}_{j}}{S}_{i}{x}_{{n}_{j}}\parallel =0 and \parallel {q}^{\ast}{T}_{i}{q}^{\ast}\parallel ={lim}_{j\to \mathrm{\infty}}\parallel {x}_{{n}_{j}}{T}_{i}{x}_{{n}_{j}}\parallel =0 for i=1,2. Thus it follows that {q}^{\ast}\in F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}). Since {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{q}^{\ast}\parallel exists by Lemma 3.1, we have {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{q}^{\ast}\parallel =0. This completes the proof. □
Theorem 3.3 Under the assumptions of Lemma 3.2, if there exists a nondecreasing function f:[0,\mathrm{\infty})\to [0,\mathrm{\infty}) with f(0)=0 and f(r)>0 for all r\in (0,\mathrm{\infty}) such that
for all x\in K, where F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}), then the sequence \{{x}_{n}\} defined by (2.1) converges strongly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}.
Proof Since {lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{S}_{i}{x}_{n}\parallel ={lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{T}_{i}{x}_{n}\parallel =0 for i=1,2 by Lemma 3.2, we have {lim}_{n\to \mathrm{\infty}}f(d({x}_{n},F))=0. Since f:[0,\mathrm{\infty})\to [0,\mathrm{\infty}) is a nondecreasing function satisfying f(0)=0, f(r)>0 for all r\in (0,\mathrm{\infty}) and {lim}_{n\to \mathrm{\infty}}d({x}_{n},F) exists by Lemma 3.1, we have {lim}_{n\to \mathrm{\infty}}d({x}_{n},F)=0.
Now, we show that \{{x}_{n}\} is a Cauchy sequence in K. In fact, from (3.2), we have
for each n\ge 1, where {h}_{n}=max\{{k}_{n}^{(1)},{k}_{n}^{(2)},{l}_{n}^{(1)},{l}_{n}^{(2)}\} and q\in F. For any m, n, m>n\ge 1, we have
where M={e}^{{\sum}_{i=1}^{\mathrm{\infty}}({h}_{i}^{2}1)}. Thus, for any q\in F, we have
Taking the infimum over all q\in F, we obtain
Thus it follows from {lim}_{n\to \mathrm{\infty}}d({x}_{n},F)=0 that \{{x}_{n}\} is a Cauchy sequence. Since K is a closed subset of E, the sequence \{{x}_{n}\} converges strongly to some {q}^{\ast}\in K. It is easy to prove that F({S}_{1}), F({S}_{2}), F({T}_{1}) and F({T}_{2}) are all closed and so F is a closed subset of K. Since {lim}_{n\to \mathrm{\infty}}d({x}_{n},F)=0, {q}^{\ast}\in F, the sequence \{{x}_{n}\} converges strongly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}. This completes the proof. □
4 Weak convergence theorems
In this section, we prove weak convergence theorems for the iterative scheme defined by (2.1) in uniformly convex Banach spaces.
Lemma 4.1 Under the assumptions of Lemma 3.1, for all {q}_{1},{q}_{2}\in F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}), the limit
exists for all t\in [0,1], where \{{x}_{n}\} is the sequence defined by (2.1).
Proof Set {a}_{n}(t)=\parallel t{x}_{n}+(1t){q}_{1}{q}_{2}\parallel. Then {lim}_{n\to \mathrm{\infty}}{a}_{n}(0)=\parallel {q}_{1}{q}_{2}\parallel and, from Lemma 3.1, {lim}_{n\to \mathrm{\infty}}{a}_{n}(1)={lim}_{n\to \mathrm{\infty}}\parallel {x}_{n}{q}_{2}\parallel exists. Thus it remains to prove Lemma 4.1 for any t\in (0,1).
Define the mapping {G}_{n}:K\to K by
for all x\in K. It is easy to prove that
for all x,y\in K, where {h}_{n}=max\{{k}_{n}^{(1)},{k}_{n}^{(2)},{l}_{n}^{(1)},{l}_{n}^{(2)}\}. Letting {h}_{n}=1+{v}_{n}, it follows from 1\le {\prod}_{j=n}^{\mathrm{\infty}}{h}_{j}^{4}\le {e}^{4{\sum}_{j=n}^{\mathrm{\infty}}{v}_{j}} and {\sum}_{n=1}^{\mathrm{\infty}}{v}_{n}<\mathrm{\infty} that {lim}_{n\to \mathrm{\infty}}{\prod}_{j=n}^{\mathrm{\infty}}{h}_{j}^{4}=1. Setting
for each m\ge 1, from (4.1) and (4.2), it follows that
for all x,y\in K and {S}_{n,m}{x}_{n}={x}_{n+m}, {S}_{n,m}q=q for any q\in F. Let
Then, using (4.3) and Lemma 2.4, we have
It follows from Lemma 3.1 and {lim}_{n\to \mathrm{\infty}}{\prod}_{j=n}^{\mathrm{\infty}}{h}_{j}^{4}=1 that {lim}_{n\to \mathrm{\infty}}{b}_{n,m}=0 uniformly for all m. Observe that
Thus we have {lim\hspace{0.17em}sup}_{n\to \mathrm{\infty}}{a}_{n}(t)\le {lim\hspace{0.17em}inf}_{n\to \mathrm{\infty}}{a}_{n}(t), that is, {lim}_{n\to \mathrm{\infty}}\parallel t{x}_{n}+(1t){q}_{1}{q}_{2}\parallel exists for all t\in (0,1). This completes the proof. □
Lemma 4.2 Under the assumptions of Lemma 3.1, if E has a Fréchet differentiable norm, then, for all {q}_{1},{q}_{2}\in F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}), the limit
exists, where \{{x}_{n}\} is the sequence defined by (2.1). Furthermore, if {W}_{w}(\{{x}_{n}\}) denotes the set of all weak subsequential limits of \{{x}_{n}\}, then \u3008{x}^{\ast}{y}^{\ast},j({q}_{1}{q}_{2})\u3009=0 for all {q}_{1},{q}_{2}\in F and {x}^{\ast},{y}^{\ast}\in {W}_{w}(\{{x}_{n}\}).
Proof This follows basically as in the proof of Lemma 3.2 of [12] using Lemma 4.1 instead of Lemma 3.1 of [12]. □
Theorem 4.1 Under the assumptions of Lemma 3.2, if E has a Fréchet differentiable norm, then the sequence \{{x}_{n}\} defined by (2.1) converges weakly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}.
Proof Since E is a uniformly convex Banach space and the sequence \{{x}_{n}\} is bounded by Lemma 3.1, there exists a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} which converges weakly to some q\in K. By Lemma 3.2, we have
for i=1,2. It follows from Lemma 2.3 that q\in F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}).
Now, we prove that the sequence \{{x}_{n}\} converges weakly to q. Suppose that there exists a subsequence \{{x}_{{m}_{j}}\} of \{{x}_{n}\} such that \{{x}_{{m}_{j}}\} converges weakly to some {q}_{1}\in K. Then, by the same method given above, we can also prove that {q}_{1}\in F. So, q,{q}_{1}\in F\cap {W}_{w}(\{{x}_{n}\}). It follows from Lemma 4.2 that
Therefore, {q}_{1}=q, which shows that the sequence \{{x}_{n}\} converges weakly to q. This completes the proof. □
Theorem 4.2 Under the assumptions of Lemma 3.2, if the dual space {E}^{\ast} of E has the KadecKlee property, then the sequence \{{x}_{n}\} defined by (2.1) converges weakly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}.
Proof Using the same method given in Theorem 4.1, we can prove that there exists a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} which converges weakly to some q\in F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}).
Now, we prove that the sequence \{{x}_{n}\} converges weakly to q. Suppose that there exists a subsequence \{{x}_{{m}_{j}}\} of \{{x}_{n}\} such that \{{x}_{{m}_{j}}\} converges weakly to some {q}^{\ast}\in K. Then, as for q, we have {q}^{\ast}\in F. It follows from Lemma 4.1 that the limit
exists for all t\in [0,1]. Again, since q,{q}^{\ast}\in {W}_{w}(\{{x}_{n}\}), {q}^{\ast}=q by Lemma 2.5. This shows that the sequence \{{x}_{n}\} converges weakly to q. This completes the proof. □
Theorem 4.3 Under the assumptions of Lemma 3.2, if E satisfies Opial’s condition, then the sequence \{{x}_{n}\} defined by (2.1) converges weakly to a common fixed point of {S}_{1}, {S}_{2}, {T}_{1} and {T}_{2}.
Proof Using the same method as given in Theorem 4.1, we can prove that there exists a subsequence \{{x}_{{n}_{k}}\} of \{{x}_{n}\} which converges weakly to some q\in F=F({S}_{1})\cap F({S}_{2})\cap F({T}_{1})\cap F({T}_{2}).
Now, we prove that the sequence \{{x}_{n}\} converges weakly to q. Suppose that there exists a subsequence \{{x}_{{m}_{j}}\} of \{{x}_{n}\} such that \{{x}_{{m}_{j}}\} converges weakly to some \overline{q}\in K and \overline{q}\ne q. Then, as for q, we have \overline{q}\in F. Using Lemma 3.1, we have the following two limits exist:
Thus, by Opial’s condition, we have
which is a contradiction and so q=\overline{q}. This shows that the sequence \{{x}_{n}\} converges weakly to q. This completes the proof. □
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Acknowledgements
The project was supported by the National Natural Science Foundation of China (Grant Number: 11271282) and the second author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (Grant Number: 20120008170).
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Guo, W., Cho, Y.J. & Guo, W. Convergence theorems for mixed type asymptotically nonexpansive mappings. Fixed Point Theory Appl 2012, 224 (2012). https://doi.org/10.1186/168718122012224
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DOI: https://doi.org/10.1186/168718122012224
Keywords
 mixed type asymptotically nonexpansive mapping
 strong and weak convergence
 common fixed point
 uniformly convex Banach space