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# Hybrid projection methods for a bifunction and relatively asymptotically nonexpansive mappings

Fixed Point Theory and Applications20132013:294

https://doi.org/10.1186/1687-1812-2013-294

• Received: 8 July 2013
• Accepted: 4 September 2013
• Published:

## Abstract

The purpose of this paper is to investigate a bifunction equilibrium problem and a fixed point problem of relatively asymptotically nonexpansive mappings based on a generalized projection method. A weak convergence theorem for common solutions is established in a uniformly smooth and uniformly convex Banach space.

## Keywords

• bifunction
• equilibrium problem
• fixed point
• generalized projection
• relatively asymptotically nonexpansive mapping

## 1 Introduction and preliminaries

Let E be a real Banach space, ${E}^{\ast }$ be the dual space of E, and C be a nonempty subset of E. Let F be a bifunction from $C×C$ to , where denotes the set of real numbers. Recall the following equilibrium problem: Find $\overline{x}\in C$ such that
$F\left(\overline{x},y\right)\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$
(1.1)
From now on, we use $\mathit{EP}\left(F\right)$ to denote the solution set of equilibrium problem (1.1) and assume that F satisfies the following conditions:
1. (A1)

$F\left(x,x\right)=0$, $\mathrm{\forall }x\in C$;

2. (A2)

F is monotone, i.e., $F\left(x,y\right)+F\left(y,x\right)\le 0$, $\mathrm{\forall }x,y\in C$;

3. (A3)
$\underset{t↓0}{lim sup}F\left(tz+\left(1-t\right)x,y\right)\le F\left(x,y\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y,z\in C;$

4. (A4)

for each $x\in C$, $y↦F\left(x,y\right)$ is convex and weakly lower semi-continuous.

Let ${U}_{E}=\left\{x\in E:\parallel x\parallel =1\right\}$ be the unit sphere of E. Then the Banach space E is said to be smooth iff
$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$

exists for each $x,y\in {U}_{E}$. It is also said to be uniformly smooth iff the above limit is attained uniformly for $x,y\in {U}_{E}$. It is well known that if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E. Recall that E is said to be uniformly convex iff ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{y}_{n}\parallel =0$ for any two sequences $\left\{{x}_{n}\right\}$ and $\left\{{y}_{n}\right\}$ in E such that $\parallel {x}_{n}\parallel =\parallel {y}_{n}\parallel =1$ and ${lim}_{n\to \mathrm{\infty }}\parallel \frac{{x}_{n}+{y}_{n}}{2}\parallel =1$. It is well known that E is uniformly smooth if and only if ${E}^{\ast }$ is uniformly convex.

Recall that a Banach space E enjoys the Kadec-Klee property if for any sequence $\left\{{x}_{n}\right\}\subset E$, and $x\in E$ with ${x}_{n}⇀x$, and $\parallel {x}_{n}\parallel \to \parallel x\parallel$, then $\parallel {x}_{n}-x\parallel \to 0$ as $n\to \mathrm{\infty }$. For more details on the Kadec-Klee property, the readers can refer to  and the references therein. It is well known that if E is a uniformly convex Banach space, then E enjoys the Kadec-Klee property.

Let $T:C\to C$ be a mapping. From now on, we use $F\left(T\right)$ to denote the fixed point set of T. Recall that T is said to be closed if for any sequence $\left\{{x}_{n}\right\}\subset C$ such that ${lim}_{n\to \mathrm{\infty }}{x}_{n}={x}_{0}$ and ${lim}_{n\to \mathrm{\infty }}T{x}_{n}={y}_{0}$, then $T{x}_{0}={y}_{0}$. In this paper, we use → and to denote the strong convergence and the weak convergence, respectively.

Recall that the normalized duality mapping J from E to ${2}^{{E}^{\ast }}$ is defined by
$Jx=\left\{{f}^{\ast }\in {E}^{\ast }:〈x,{f}^{\ast }〉={\parallel x\parallel }^{2}={\parallel {f}^{\ast }\parallel }^{2}\right\},$
where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. Next, we assume that E is a smooth Banach space. Consider the functional defined by
$\varphi \left(x,y\right)={\parallel x\parallel }^{2}-2〈x,Jy〉+{\parallel y\parallel }^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in E.$
Observe that in a Hilbert space H the equality is reduced to $\varphi \left(x,y\right)={\parallel x-y\parallel }^{2}$, $x,y\in H$. As we all know, if C is a nonempty closed convex subset of a Hilbert space H and ${P}_{C}:H\to C$ is the metric projection of H onto C, then ${P}_{C}$ is nonexpansive. This fact actually characterizes Hilbert spaces and, consequently, it is not available in more general Banach spaces. In this connection, Alber  recently introduced a generalized projection operator ${\mathrm{\Pi }}_{C}$ in a Banach space E which is an analogue of the metric projection ${P}_{C}$ in Hilbert spaces. Recall that the generalized projection ${\mathrm{\Pi }}_{C}:E\to C$ is a map that assigns to an arbitrary point $x\in E$ the minimum point of the functional $\varphi \left(x,y\right)$, that is, ${\mathrm{\Pi }}_{C}x=\overline{x}$, where $\overline{x}$ is the solution to the minimization problem
$\varphi \left(\overline{x},x\right)=\underset{y\in C}{min}\varphi \left(y,x\right).$
Existence and uniqueness of the operator ${\mathrm{\Pi }}_{C}$ follow from the properties of the functional $\varphi \left(x,y\right)$ and strict monotonicity of the mapping J. In Hilbert spaces, ${\mathrm{\Pi }}_{C}={P}_{C}$. It is obvious from the definition of a function ϕ that
${\left(\parallel x\parallel -\parallel y\parallel \right)}^{2}\le \varphi \left(x,y\right)\le {\left(\parallel y\parallel +\parallel x\parallel \right)}^{2},\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in E.$
(1.2)

Remark 1.1 If E is a reflexive, strictly convex, and smooth Banach space, then $\varphi \left(x,y\right)=0$ if and only if $x=y$; for more details, see  and the references therein.

Recall that a point p in C is said to be an asymptotic fixed point of a mapping T iff C contains a sequence $\left\{{x}_{n}\right\}$ which converges weakly to p so that ${lim}_{n\to \mathrm{\infty }}\parallel {x}_{n}-{T}^{n}{x}_{n}\parallel =0$. The set of asymptotic fixed points of T will be denoted by $\stackrel{˜}{F}\left(T\right)$.

Recall that a mapping T is said to be relatively nonexpansive iff
$\stackrel{˜}{F}\left(T\right)=F\left(T\right)\ne \mathrm{\varnothing },\phantom{\rule{2em}{0ex}}\varphi \left(p,Tx\right)\le \varphi \left(p,x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C,\mathrm{\forall }p\in F\left(T\right).$
Recall that a mapping T is said to be relatively asymptotically nonexpansive iff
$\stackrel{˜}{F}\left(T\right)=F\left(T\right)\ne \mathrm{\varnothing },\phantom{\rule{2em}{0ex}}\varphi \left(p,{T}^{n}x\right)\le \left(1+{\mu }_{n}\right)\varphi \left(p,x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C,\mathrm{\forall }p\in F\left(T\right),\mathrm{\forall }n\ge 1,$

where $\left\{{\mu }_{n}\right\}\subset \left[0,\mathrm{\infty }\right)$ is a sequence such that ${\mu }_{n}\to 0$ as $n\to \mathrm{\infty }$.

Remark 1.2 The class of relatively nonexpansive mappings was first considered in Butnariu et al. . The class of relatively asymptotically nonexpansive mappings was first considered in Agarwal et al.  and the references therein.

Recently, many authors investigated fixed point problems of a (relatively) nonexpansive mapping based on hybrid projection methods; for more details, see  and the references therein. However, most of the results are on strong convergence. In this article, we investigate a bifunction equilibrium problem and a fixed point problem of relatively asymptotically nonexpansive mappings based on a generalized projection method. A weak convergence theorem for common solutions is established in a uniformly smooth and uniformly convex Banach space.

The following lemmas play an important role in this paper.

Lemma 1.3 [37, 38]

Let C be a closed convex subset of a uniformly smooth and uniformly convex Banach space E. Let F be a bifunction from $C×C$ to satisfying (A1)-(A4). Let $r>0$ and $x\in E$. Then there exists $z\in C$ such that $F\left(z,y\right)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0$, $\mathrm{\forall }y\in C$. Define a mapping ${S}_{r}:E\to C$ by ${S}_{r}x=\left\{z\in C:F\left(z,y\right)+\frac{1}{r}〈y-z,Jz-Jx〉,\mathrm{\forall }y\in C\right\}$. Then the following conclusions hold:
1. (a)

${S}_{r}$ is single-valued;

2. (b)
${S}_{r}$ is a firmly nonexpansive-type mapping, i.e., for all $x,y\in E$,
$〈{S}_{r}x-{S}_{r}y,J{S}_{r}x-J{S}_{r}y〉\le 〈{S}_{r}x-{S}_{r}y,Jx-Jy〉;$

3. (c)

$F\left({S}_{r}\right)=\mathit{EP}\left(F\right)$ is closed and convex;

4. (d)

${S}_{r}$ is relatively nonexpansive;

5. (e)

$\varphi \left(q,{S}_{r}x\right)+\varphi \left({S}_{r}x,x\right)\le \varphi \left(q,x\right)$, $\mathrm{\forall }q\in F\left({S}_{r}\right)$.

Lemma 1.4 

Let E be a uniformly smooth and uniformly convex Banach space. Let C be a nonempty closed and convex subset of E. Let $T:C\to C$ be a relatively asymptotically nonexpansive mapping. Then $F\left(T\right)$ is a closed convex subset of C.

Lemma 1.5 

Let E be a reflexive, strictly convex, and smooth Banach space, let C be a nonempty, closed, and convex subset of E, and let $x\in E$. Then
$\varphi \left(y,{\mathrm{\Pi }}_{C}x\right)+\varphi \left({\mathrm{\Pi }}_{C}x,x\right)\le \varphi \left(y,x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$

Lemma 1.6 

Let C be a nonempty, closed, and convex subset of a smooth Banach space E, and let $x\in E$. Then ${x}_{0}={\mathrm{\Pi }}_{C}x$ if and only if
$〈{x}_{0}-y,Jx-J{x}_{0}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }y\in C.$

Lemma 1.7 

Let E be a smooth and uniformly convex Banach space, and let $r>0$. Then there exists a strictly increasing, continuous, and convex function $g:\left[0,2r\right]\to R$ such that $g\left(0\right)=0$ and
${\parallel tx+\left(1-t\right)y\parallel }^{2}\le t{\parallel x\parallel }^{2}+\left(1-t\right){\parallel y\parallel }^{2}-t\left(1-t\right)g\left(\parallel x-y\parallel \right)$

for all $x,y\in {B}_{r}=\left\{x\in E:\parallel x\parallel \le r\right\}$ and $t\in \left[0,1\right]$.

Lemma 1.8 

Let $\left\{{a}_{n}\right\}$, $\left\{{b}_{n}\right\}$, and $\left\{{c}_{n}\right\}$ be three nonnegative sequences satisfying the following condition:
${a}_{n+1}\le \left(1+{b}_{n}\right){a}_{n}+{c}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\ge {n}_{0},$

where ${n}_{0}$ is some nonnegative integer. If ${\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}<\mathrm{\infty }$ and ${\sum }_{n=1}^{\mathrm{\infty }}{c}_{n}<\mathrm{\infty }$, then the limit of the sequence $\left\{{a}_{n}\right\}$ exists. If, in addition, there exists a subsequence $\left\{{\alpha }_{{n}_{i}}\right\}\subset \left\{{\alpha }_{n}\right\}$ such that ${\alpha }_{{n}_{i}}\to 0$, then ${\alpha }_{n}\to 0$ as $n\to \mathrm{\infty }$.

Lemma 1.9 

Let E be a smooth and uniformly convex Banach space, and let $r>0$. Then there exists a strictly increasing, continuous, and convex function $g:\left[0,2r\right]\to R$ such that $g\left(0\right)=0$ and $g\left(\parallel x-y\parallel \right)\le \varphi \left(x,y\right)$ for all $x,y\in {B}_{r}$.

## 2 Main results

Theorem 2.1 Let E be a uniformly smooth and uniformly convex Banach space, and let C be a nonempty closed and convex subset of E. Let F be a bifunction from $C×C$ to satisfying (A1)-(A4). Let $T:C\to C$ be a relatively asymptotically nonexpansive mapping with the sequence $\left\{{\mu }_{n,1}\right\}$, and let $S:C\to C$ be a relatively asymptotically nonexpansive mapping with the sequence $\left\{{\mu }_{n,2}\right\}$. Assume that $\mathrm{\Phi }:=F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)$ is nonempty. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
where $\left\{{\alpha }_{n}\right\}$, $\left\{{\beta }_{n}\right\}$, $\left\{{\gamma }_{n}\right\}$ are real sequences in $\left[0,1\right]$ and $\left\{{r}_{n}\right\}$ is a real number sequence in $\left[r,\mathrm{\infty }\right)$, where $r>0$ is some real number. Assume that J is weakly sequentially continuous and the following restrictions hold:
1. (i)

${\alpha }_{n}+{\beta }_{n}+{\gamma }_{n}=1$;

2. (ii)

${\sum }_{n=1}^{\mathrm{\infty }}{\mu }_{n}<\mathrm{\infty }$;

3. (iii)

${lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}{\beta }_{n}>0$, ${lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}{\gamma }_{n}>0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges weakly to $\overline{x}\in \mathrm{\Phi }$, where $\overline{x}={lim}_{n\to \mathrm{\infty }}{\mathrm{\Pi }}_{\mathrm{\Phi }}{x}_{n}$.

Proof Set ${\mu }_{n}=max\left\{{\mu }_{n,1},{\mu }_{n,2}\right\}$. Fixing $p\in \mathrm{\Phi }$, we find that
$\begin{array}{rcl}\varphi \left(p,{x}_{n+1}\right)& =& \varphi \left(p,{S}_{{r}_{n+1}}{y}_{n+1}\right)\\ \le & \varphi \left(p,{y}_{n+1}\right)\\ =& {\parallel p\parallel }^{2}-2〈p,{\alpha }_{n}J{x}_{n}+{\beta }_{n}J{T}^{n}{x}_{n}+{\gamma }_{n}J{S}^{n}{x}_{n}〉\\ +{\parallel {\alpha }_{n}J{x}_{n}+{\beta }_{n}J{T}^{n}{x}_{n}+{\gamma }_{n}J{S}^{n}{x}_{n}\parallel }^{2}\end{array}$
(2.1)
$\begin{array}{rc}\le & {\parallel p\parallel }^{2}-2{\alpha }_{n}〈p,J{x}_{n}〉-2{\beta }_{n}〈p,J{T}^{n}{x}_{n}〉-2{\gamma }_{n}〈p,J{S}^{n}{x}_{n}〉\\ +{\alpha }_{n}{\parallel {x}_{n}\parallel }^{2}+{\beta }_{n}{\parallel {T}^{n}{x}_{n}\parallel }^{2}+{\gamma }_{n}{\parallel {S}^{n}{x}_{n}\parallel }^{2}\\ =& {\alpha }_{n}\varphi \left(p,{x}_{n}\right)+{\beta }_{n}\varphi \left(p,{T}^{n}{x}_{n}\right)+{\gamma }_{n}\varphi \left(p,{S}^{n}{x}_{n}\right)\\ \le & \varphi \left(p,{x}_{n}\right)+{\beta }_{n}{\mu }_{n}\varphi \left(p,{x}_{n}\right)+{\gamma }_{n}{\mu }_{n}\varphi \left(p,{x}_{n}\right)\\ \le & \left(1+{\mu }_{n}\right)\varphi \left(p,{x}_{n}\right).\end{array}$
(2.2)
In view of Lemma 1.8, we obtain that ${lim}_{n\to \mathrm{\infty }}\varphi \left(p,{x}_{n}\right)$ exits. This implies that the sequence $\left\{{x}_{n}\right\}$ is bounded. In the light of Lemma 1.7, we find that
$\begin{array}{rcl}\varphi \left(p,{x}_{n+1}\right)& =& \varphi \left(p,{S}_{{r}_{n+1}}{y}_{n+1}\right)\\ \le & {\parallel p\parallel }^{2}-2〈p,{\alpha }_{n}J{x}_{n}+{\beta }_{n}J{T}^{n}{x}_{n}+{\gamma }_{n}J{S}^{n}{x}_{n}〉\\ +{\parallel {\alpha }_{n}J{x}_{n}+{\beta }_{n}J{T}^{n}{x}_{n}+{\gamma }_{n}J{S}^{n}{x}_{n}\parallel }^{2}\\ \le & {\parallel p\parallel }^{2}-2{\alpha }_{n}〈p,J{x}_{n}〉-2{\beta }_{n}〈p,J{T}^{n}{x}_{n}〉-2{\gamma }_{n}〈p,J{S}^{n}{x}_{n}〉\\ +{\alpha }_{n}{\parallel {x}_{n}\parallel }^{2}+{\beta }_{n}{\parallel {T}^{n}{x}_{n}\parallel }^{2}+{\gamma }_{n}{\parallel {S}^{n}{x}_{n}\parallel }^{2}-{\alpha }_{n}{\beta }_{n}g\left(\parallel J{T}^{n}{x}_{n}-J{x}_{n}\parallel \right)\\ \le & \varphi \left(p,{x}_{n}\right)+{\beta }_{n}{\mu }_{n}\varphi \left(p,{x}_{n}\right)+{\gamma }_{n}{\mu }_{n}\varphi \left(p,{x}_{n}\right)-{\alpha }_{n}{\beta }_{n}g\left(\parallel J{T}^{n}{x}_{n}-J{x}_{n}\parallel \right)\\ \le & \left(1+{\mu }_{n}\right)\varphi \left(p,{x}_{n}\right)-{\alpha }_{n}{\beta }_{n}g\left(\parallel J{T}^{n}{x}_{n}-J{x}_{n}\parallel \right).\end{array}$
It follows that
${\alpha }_{n}{\beta }_{n}g\left(\parallel J{T}^{n}{x}_{n}-J{x}_{n}\parallel \right)\le \left(1+{\mu }_{n}\right)\varphi \left(p,{x}_{n}\right)-\varphi \left(p,{x}_{n+1}\right).$
This finds from the restrictions (ii) and (iii) that
$\underset{n\to \mathrm{\infty }}{lim}g\left(\parallel J{T}^{n}{x}_{n}-J{x}_{n}\parallel \right)=0.$
This implies that
$\underset{n\to \mathrm{\infty }}{lim}\parallel J{T}^{n}{x}_{n}-J{x}_{n}\parallel =0.$
Since ${J}^{-1}$ is uniformly norm-to-norm continuous on bounded sets, we find that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {T}^{n}{x}_{n}-{x}_{n}\parallel =0.$
In the same way, we find that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {S}^{n}{x}_{n}-{x}_{n}\parallel =0.$
Since $\left\{{x}_{n}\right\}$ is bounded, we see that there exists a subsequence $\left\{{x}_{{n}_{i}}\right\}$ of $\left\{{x}_{n}\right\}$ such that $\left\{{x}_{{n}_{i}}\right\}$ converges weakly to $p\in C$. It follows that $p\in F\left(T\right)\cap F\left(S\right)$. Next, we prove that $p\in \mathit{EP}\left(F\right)$. Let $r={sup}_{n\ge 1}\left\{\parallel {x}_{n}\parallel ,\parallel {y}_{n}\parallel \right\}$. In view of Lemma 1.9, we find that there exists a continuous, strictly increasing and convex function h with $h\left(0\right)=0$ such that
$h\left(x,y\right)\le \varphi \left(x,y\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x,y\in {B}_{r}.$
It follows from (2.1) that
$\begin{array}{rl}h\left(\parallel {x}_{n}-{y}_{n}\parallel \right)& \le \varphi \left({x}_{n},{y}_{n}\right)\\ \le \varphi \left(p,{y}_{n}\right)-\varphi \left(p,{x}_{n}\right)\\ \le \varphi \left(p,{x}_{n-1}\right)-\varphi \left(p,{x}_{n}\right)+{\mu }_{n-1}\varphi \left(p,{x}_{n-1}\right).\end{array}$
This implies that
$\underset{n\to \mathrm{\infty }}{lim}h\left(\parallel {x}_{n}-{y}_{n}\parallel \right)=0.$
It follows from the property of h that
$\underset{n\to \mathrm{\infty }}{lim}\parallel {x}_{n}-{y}_{n}\parallel =0.$
Since J is uniformly norm-to-norm continuous on bounded sets, one has
$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_{n}-J{y}_{n}\parallel =0.$
Since $\left\{{r}_{n}\right\}$ is a real number sequence in $\left[r,\mathrm{\infty }\right)$, where $r>0$ is some real number, one finds that
$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel J{x}_{n}-J{y}_{n}\parallel }{{r}_{n}}=0.$
Notice that ${x}_{n}={S}_{{r}_{n}}{y}_{n}$, one sees that
$F\left({x}_{n},x\right)+\frac{1}{{r}_{n}}〈x-{x}_{n},J{x}_{n}-J{y}_{n}〉\ge 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
By replacing n by ${n}_{i}$, one finds from (A2) that
$\begin{array}{rl}\parallel x-{x}_{{n}_{i}}\parallel \frac{\parallel J{x}_{{n}_{i}}-J{y}_{{n}_{i}}\parallel }{{r}_{{n}_{i}}}& \ge \frac{1}{{r}_{{n}_{i}}}〈x-{x}_{{n}_{i}},J{x}_{{n}_{i}}-J{y}_{{n}_{i}}〉\\ \ge F\left(x,{x}_{{n}_{i}}\right).\end{array}$
Letting $i\to \mathrm{\infty }$ in the above inequality, one obtains from (A4) that
$F\left(x,p\right)\le 0,\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in C.$
For $0 and $y\in C$, define ${x}_{t}=tx+\left(1-t\right)p$. It follows that ${x}_{t}\in C$, which yields that $F\left({x}_{t},p\right)\le 0$. It follows from (A1) and (A4) that
$0=F\left({x}_{t},{x}_{t}\right)\le tF\left({x}_{t},x\right)+\left(1-t\right)F\left({x}_{x},p\right)\le tF\left({x}_{t},x\right).$
That is,
$F\left({x}_{t},x\right)\ge 0.$
Letting $t↓0$, we obtain from (A3) that $F\left(p,x\right)\ge 0$, $\mathrm{\forall }x\in C$. This implies that $p\in \mathit{EP}\left(F\right)$. This completes the proof that $p\in F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)$. Define ${z}_{n}={\mathrm{\Pi }}_{F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)}{x}_{n}$. It follows from (2.1) that
$\varphi \left({z}_{n},{x}_{n+1}\right)\le \left(1+{\mu }_{n}\right)\varphi \left({z}_{n},{x}_{n}\right).$
(2.3)
This in turn implies from Lemma 1.5 that
$\begin{array}{rl}\varphi \left({z}_{n+1},{x}_{n+1}\right)& =\varphi \left({\mathrm{\Pi }}_{F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)}{x}_{n+1},{x}_{n+1}\right)\\ \le \varphi \left({z}_{n},{x}_{n+1}\right)-\varphi \left({z}_{n},{\mathrm{\Pi }}_{F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)}{x}_{n+1}\right)\\ \le \varphi \left({z}_{n},{x}_{n+1}\right)-\varphi \left({z}_{n},{z}_{n+1}\right)\\ \le \varphi \left({z}_{n},{x}_{n+1}\right).\end{array}$
It follows from (2.3) that
$\varphi \left({z}_{n+1},{x}_{n+1}\right)\le \left(1+{\mu }_{n}\right)\varphi \left({z}_{n},{x}_{n}\right).$
This finds from Lemma 1.8 that the sequence $\left\{\varphi \left({z}_{n},{x}_{n}\right)\right\}$ is a convergence sequence. It follows from (2.1) that
$\varphi \left(p,{x}_{n+m}\right)\le \varphi \left(p,{x}_{n}\right)+L\left(\sum _{i=1}^{m}{\mu }_{n+m-i}\right),$
(2.4)
where $L={sup}_{n\ge 1}\varphi \left(p,{x}_{n}\right)$. Since ${z}_{n}\in F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)$, we find that
$\varphi \left({z}_{n},{x}_{n+m}\right)\le \varphi \left({z}_{n},{x}_{n}\right)+M\left(\sum _{i=1}^{m}{\mu }_{n+m-i}\right),$
where $M={sup}_{n\ge 1}\varphi \left({z}_{n},{x}_{n}\right)$. Since ${z}_{n+m}={\mathrm{\Pi }}_{F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)}{x}_{n+m}$, we find from Lemma 1.5 that
$\varphi \left({z}_{n},{z}_{n+m}\right)+\varphi \left({z}_{n+m},{x}_{n+m}\right)\le \varphi \left({z}_{n},{x}_{n+m}\right)\le \varphi \left({z}_{n},{x}_{n}\right)+M\left(\sum _{i=1}^{m}{\mu }_{n+m-i}\right).$
It follows that
$\varphi \left({z}_{n},{z}_{n+m}\right)\le \varphi \left({z}_{n},{x}_{n}\right)-\varphi \left({z}_{n+m},{x}_{n+m}\right)+M\left(\sum _{i=1}^{m}{\mu }_{n+m-i}\right).$
In view of Lemma 1.9, we find that there exists a continuous, strictly increasing, and convex function g with
$g\left(\parallel {z}_{n}-{z}_{m}\parallel \right)\le \varphi \left({z}_{n},{z}_{m}\right)\le \varphi \left({z}_{n},{x}_{n}\right)-\varphi \left({z}_{n+m},{x}_{n+m}\right)+M\left(\sum _{i=1}^{m}{\mu }_{n+m-i}\right).$
This shows that $\left\{{z}_{n}\right\}$ is a Cauchy sequence. Since $F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)$ is closed, one sees that $\left\{{z}_{n}\right\}$ converges strongly to $z\in F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)$. Since $p\in F\left(T\right)\cap F\left(S\right)\cap \mathit{EP}\left(F\right)$, we find from Lemma 1.6 that
$〈{z}_{{n}_{k}}-p,J{x}_{{n}_{k}}-J{z}_{{n}_{k}}〉\ge 0.$

Notice that J is weakly sequentially continuous. Letting $k\to \mathrm{\infty }$, we find that $〈z-p,Jp-Jz〉\ge 0$. It follows from the monotonicity of J that $〈z-p,Jp-Jz〉\le 0$. Since the space is uniformly convex, we find that $z=p$. This completes the proof. □

Remark 2.2 Theorem 2.1 improves Theorem 2.5 in Qin et al.  on the mappings from the class of relatively nonexpansive mappings to the class of relatively asymptotically nonexpansive mappings.

If $T=S$, then Theorem 2.1 is reduced to the following.

Corollary 2.3 Let E be a uniformly smooth and uniformly convex Banach space, and let C be a nonempty closed and convex subset of E. Let F be a bifunction from $C×C$ to satisfying (A1)-(A4). Let $T:C\to C$ be a relatively asymptotically nonexpansive mapping with the sequence $\left\{{\mu }_{n}\right\}$. Assume that $\mathrm{\Phi }:=F\left(T\right)\cap \mathit{EP}\left(F\right)$ is nonempty. Let $\left\{{x}_{n}\right\}$ be a sequence generated in the following manner:
where $\left\{{\alpha }_{n}\right\}$ is a real sequence in $\left[0,1\right]$ and $\left\{{r}_{n}\right\}$ is a real number sequence in $\left[r,\mathrm{\infty }\right)$, where $r>0$ is some real number. Assume that J is weakly sequentially continuous and the following restrictions hold:
1. (i)

${\sum }_{n=1}^{\mathrm{\infty }}{\mu }_{n}<\mathrm{\infty }$;

2. (ii)

${lim inf}_{n\to \mathrm{\infty }}{\alpha }_{n}\left(1-{\alpha }_{n}\right)>0$.

Then the sequence $\left\{{x}_{n}\right\}$ converges weakly to $\overline{x}\in \mathrm{\Phi }$, where $\overline{x}={lim}_{n\to \mathrm{\infty }}{\mathrm{\Pi }}_{\mathrm{\Phi }}{x}_{n}$.

Remark 2.4 Corollary 2.3 is an improvement of Theorem 4.1 in Zembayashi and Takahashi  on the nonlinear mapping.

## Declarations

### Acknowledgements

This work is supported by the Fundamental Research Funds for the Central Universities (Grant No.: 9161013002). The authors are grateful to the editor and the anonymous reviewers’ suggestions which improved the contents of the article.

## Authors’ Affiliations

(1)
Department of Applied Mathematics and Physics, North China Electric Power University, Baoding, 071003, China
(2)
Department of Mathematics and Sciences, Shijiazhuang University of Economics, Shijiazhuang, Hebei, 050031, China

## References

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