Some results on asymptotically quasi-ϕ-nonexpansive mappings in the intermediate sense and equilibrium problems

Abstract

In this paper, we investigate a common fixed point problem of a finite family of asymptotically quasi-ϕ-nonexpansive mappings in the intermediate sense and an equilibrium problem. Strong convergence theorems of common solutions are established in a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property.

MSC:47H09, 47J25, 90C33.

1 Introduction-preliminaries

Let E be a real Banach space. Recall that E is said to be strictly convex if $\parallel \frac{x+y}{2}\parallel <1$ for all $x,y\in E$ with $\parallel x\parallel =\parallel y\parallel =1$ and $x\ne y$. It is said to be uniformly convex if ${lim}_{n\to \mathrm{\infty }}\parallel x_n-y_n\parallel =0$ for any two sequences $\{x_n\}$ and $\{y_n\}$ in E such that $\parallel x_n\parallel =\parallel y_n\parallel =1$ and ${lim}_{n\to \mathrm{\infty }}\parallel \frac{x_n+y_n}{2}\parallel =1$. Let $U_E=\{x\in E:\parallel x\parallel =1\}$ be the unit sphere of E. Then the Banach space E is said to be smooth if

$$\underset{t\to 0}{lim}\frac{\parallel x+ty\parallel -\parallel x\parallel }{t}$$

exists for each $x,y\in U_E$. It is said to be uniformly smooth if the above limit is attained uniformly for $x,y\in U_E$.

Recall that E has Kadec-Klee property if for any sequence $\{x_n\}\subset E$, and $x\in E$ with $x_n\rightharpoonup x$, 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 of the Kadec-Klee property, the readers can refer to [1] and the references therein. It is well known that if E is a uniformly convex Banach space, then E enjoys the Kadec-Klee property.

Recall that the normalized duality mapping J from E to $2^{E^{\ast }}$ is defined by

$$Jx=\{f^{\ast }\in E^{\ast }:〈x,f^{\ast }〉={\parallel x\parallel }^2={\parallel f^{\ast }\parallel }^2\},$$

where $〈\cdot ,\cdot 〉$ denotes the generalized duality pairing. It is well known that if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E. It is also well known that if E is uniformly smooth if and only if $E^{\ast }$ is uniformly convex.

Next, we assume that E is a smooth Banach space. Consider the functional defined by

$$\varphi (x,y)={\parallel x\parallel }^2-2〈x,Jy〉+{\parallel y\parallel }^2,\mathrm{\forall }x,y\in E.$$

Observe that, in a Hilbert space H, the equality is reduced to $\varphi (x,y)={\parallel x-y\parallel }^2$, $x,y\in H$. 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 [2] recently introduced a generalized projection operator ${\mathrm{\Pi }}_C$ in a Banach space E which is an analog 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 $\varphi (x,y)$, that is, ${\mathrm{\Pi }}_{C}x=\overline{x}$, where $\overline{x}$ is the solution to the minimization problem

$$\varphi (\overline{x},x)=\underset{y\in C}{min}\varphi (y,x).$$

The existence and uniqueness of the operator ${\mathrm{\Pi }}_C$ follow from the properties of the functional $\varphi (x,y)$ and strict monotonicity of the mapping J; see, for example, [1, 2]. In Hilbert spaces, ${\mathrm{\Pi }}_C=P_C$. It is obvious from the definition of the function ϕ that

$${(\parallel x\parallel -\parallel y\parallel )}^2\le \varphi (x,y)\le {(\parallel y\parallel +\parallel x\parallel )}^2,\mathrm{\forall }x,y\in E,$$

(1.1)

and

$$\varphi (x,y)=\varphi (x,z)+\varphi (z,y)+2〈x-z,Jz-Jy〉,\mathrm{\forall }x,y,z\in E.$$

(1.2)

Remark 1.1 If E is a reflexive, strictly convex, and smooth Banach space, then $\varphi (x,y)=0$ if and only if $x=y$; for more details, see [1, 2] and the references therein.

Let C be a nonempty subset of E and let $T:C\to C$ be a mapping. In this paper, we use $F(T)$ to denote the fixed point set of T. T is said to be asymptotically regular on C if for any bounded subset K of C,

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\{\parallel T^{n+1}x-T^{n}x\parallel :x\in K\}=0.$$

T is said to be closed if for any sequence $\{x_n\}\subset C$ 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 weak convergence, respectively.

Recall that a point p in C is said to be an asymptotic fixed point of T [3] iff C contains a sequence $\{x_n\}$ which converges weakly to p such that ${lim}_{n\to \mathrm{\infty }}\parallel x_n-T{x}_n\parallel =0$. The set of asymptotic fixed points of T will be denoted by $\tilde{F}(T)$.

A mapping T is said to be relatively nonexpansive iff

$$\tilde{F}(T)=F(T)\ne \mathrm{\varnothing },\varphi (p,Tx)\le \varphi (p,x),\mathrm{\forall }x\in C,\mathrm{\forall }p\in F(T).$$

A mapping T is said to be relatively asymptotically nonexpansive iff

$$\tilde{F}(T)=F(T)\ne \mathrm{\varnothing },\varphi (p,T^{n}x)\le (1+{\mu }_n)\varphi (p,x),\mathrm{\forall }x\in C,\mathrm{\forall }p\in F(T),\mathrm{\forall }n\ge 1,$$

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

Remark 1.2 The class of relatively asymptotically nonexpansive mappings were first considered in [4]; see also, [5] and the references therein.

Recall that a mapping T is said to be quasi-ϕ-nonexpansive iff

$$F(T)\ne \mathrm{\varnothing },\varphi (p,Tx)\le \varphi (p,x),\mathrm{\forall }x\in C,\mathrm{\forall }p\in F(T).$$

Recall that a mapping T is said to be asymptotically quasi-ϕ-nonexpansive iff there exists a sequence $\{{\mu }_n\}\subset [0,\mathrm{\infty })$ with ${\mu }_n\to 0$ as $n\to \mathrm{\infty }$ such that

$$F(T)\ne \mathrm{\varnothing },\varphi (p,T^{n}x)\le (1+{\mu }_n)\varphi (p,x),\mathrm{\forall }x\in C,\mathrm{\forall }p\in F(T),\mathrm{\forall }n\ge 1.$$

Remark 1.3 The class of quasi-ϕ-nonexpansive mappings was considered in [6]. The class of asymptotically quasi-ϕ-nonexpansive mappings which was investigated in [7] and [8] includes the class of quasi-ϕ-nonexpansive mappings as a special case.

Remark 1.4 The class of quasi-ϕ-nonexpansive mappings and the class of asymptotically quasi-ϕ-nonexpansive mappings are more general than the class of relatively nonexpansive mappings and the class of relatively asymptotically nonexpansive mappings. Quasi-ϕ-nonexpansive mappings and asymptotically quasi-ϕ-nonexpansive do not require the restriction $F(T)=\tilde{F}(T)$.

Remark 1.5 The class of quasi-ϕ-nonexpansive mappings and the class of asymptotically quasi-ϕ-nonexpansive mappings are generalizations of the class of quasi-nonexpansive mappings and the class of asymptotically quasi-nonexpansive mappings in Banach spaces.

Recall that T is said to be asymptotically quasi-ϕ-nonexpansive in the intermediate sense iff $F(T)\ne \mathrm{\varnothing }$ and the following inequality holds:

$$\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{p\in F(T),x\in C}{sup}(\varphi (p,T^{n}x)-\varphi (p,x))\le 0.$$

(1.3)

Putting

$${\xi }_n=max\{0,\underset{p\in F(T),x\in C}{sup}(\varphi (p,T^{n}x)-\varphi (p,x))\},$$

it follows that ${\xi }_n\to 0$ as $n\to \mathrm{\infty }$. Then (1.3) is reduced to the following:

$$\varphi (p,T^{n}x)\le \varphi (p,x)+{\xi }_n,\mathrm{\forall }p\in F(T),\mathrm{\forall }x\in C.$$

(1.4)

Remark 1.6 The class of asymptotically quasi-ϕ-nonexpansive mappings in the intermediate sense was first considered by Qin and Wang [9].

Remark 1.7 The class of asymptotically quasi-ϕ-nonexpansive mappings in the intermediate sense is a generalization of the class of asymptotically quasi-nonexpansive mappings in the intermediate sense, which was considered by Kirk [10], in the framework of Banach spaces.

Let f be a bifunction from $C\times C$ to ℝ, where ℝ denotes the set of real numbers. Recall the following equilibrium problem. Find $p\in C$ such that

$$f(p,y)\ge 0,\mathrm{\forall }y\in C.$$

(1.5)

We use $EP(f)$ to denote the solution set of the equilibrium problem (1.5). That is,

$$EP(f)=\{p\in C:f(p,y)\ge 0,\mathrm{\forall }y\in C\}.$$

We remark here that the equilibrium problem was first introduced by Fan [11]. Given a mapping $Q:C\to E^{\ast }$, let

$$f(x,y)=〈Qx,y-x〉,\mathrm{\forall }x,y\in C.$$

Then $p\in EP(f)$ if and only if p is a solution of the following variational inequality. Find p such that

$$〈Qp,y-p〉\ge 0,\mathrm{\forall }y\in C.$$

(1.6)

To study the equilibrium problems (1.5), we may assume that F satisfies the following conditions:

(A1) $F(x,x)=0$ for all $x\in C$;

(A2) F is monotone, i.e., $F(x,y)+F(y,x)\le 0$ for all $x,y\in C$;

(A3) for each $x,y,z\in C$,

$$\underset{t\downarrow 0}{lim\hspace{0.17em}sup}F(tz+(1-t)x,y)\le F(x,y);$$

(A4) for each $x\in C$, $y\mapsto F(x,y)$ is convex and weakly lower semi-continuous.

Numerous problems in physics, optimization, and economics reduce to find a solution of (1.5). Recently, many authors have investigated common solutions of fixed point and equilibrium problems in Banach spaces; see, for example, [12–33] and the references therein.

In this paper, we consider a projection algorithm for treating the equilibrium problem and fixed point problems of asymptotically quasi-ϕ-nonexpansive mappings in the intermediate sense.

In order to prove our main results, we need the following lemmas.

Lemma 1.8 [2]

Let E be a reflexive, strictly convex and smooth Banach space. Let C be a nonempty closed convex subset of E and let $x\in E$. Then

$$\varphi (y,{\mathrm{\Pi }}_{C}x)+\varphi ({\mathrm{\Pi }}_{C}x,x)\le \varphi (y,x),\mathrm{\forall }y\in C.$$

Lemma 1.9 [2]

Let C be a nonempty closed convex subset of a smooth Banach space E and let $x\in E$. Then $x_0={\mathrm{\Pi }}_{C}x$ if and only if

$$〈x_0-y,Jx-J{x}_0〉\ge 0,\mathrm{\forall }y\in C.$$

Lemma 1.10 Let C be a closed convex subset of a smooth, strictly convex and reflexive Banach space E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4). Let $r>0$ and $x\in E$. Then

• (a) [34]There exists $z\in C$ such that

  $$f(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉\ge 0,\mathrm{\forall }y\in C.$$

• (b) [6, 24]Define a mapping $T_r:E\to C$ by

  $$S_{r}x=\{z\in C:f(z,y)+\frac{1}{r}〈y-z,Jz-Jx〉,\mathrm{\forall }y\in C\}.$$

Then the following conclusions hold:

• (1) $S_r$ is single-valued;

• (2) $S_r$ 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) $F(S_r)=EP(f)$;

• (4) $S_r$ is quasi-ϕ-nonexpansive;

• (5) $\varphi (q,S_{r}x)+\varphi (S_{r}x,x)\le \varphi (q,x)$, $\mathrm{\forall }q\in F(S_r)$;

• (6) $EP(f)$ is closed and convex.

Lemma 1.11 [35]

Let E be a smooth and uniformly convex Banach space and let $r>0$. Then there exists a strictly increasing, continuous and convex function $g:[0,2r]\to R$ such that $g(0)=0$ and

$${\parallel tx+(1-t)y\parallel }^2\le t{\parallel x\parallel }^2+(1-t){\parallel y\parallel }^2-t(1-t)g(\parallel x-y\parallel )$$

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

2 Main results

Theorem 2.1 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property and let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4) and let N be some positive integer. Let $T_i:C\to C$ an asymptotically quasi-ϕ-nonexpansive mapping in the intermediate sense for every $1\le i\le N$. Assume that $T_i$ is closed asymptotically regular on C and ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)$ is nonempty and bounded. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{ chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ y_n=J^{-1}({\alpha }_{n,0}J{x}_n+{\sum }_{i=1}^N{\alpha }_{n,i}J{T}_i^n{x}_n),\hfill \\ u_n\in C\mathit{\text{ such that }}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)+{\xi }_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,\hfill \end{array}$$

where ${\xi }_n=max\{0,{sup}_{p\in F(T_i),x\in C}(\varphi (p,T_i^{n}x)-\varphi (p,x))\}$, $\{{\alpha }_{n,i}\}$ is a real number sequence in $(0,1)$ for every $1\le i\le N$, $\{r_n\}$ is a real number sequence in $[k,\mathrm{\infty })$, where k is some positive real number. Assume that ${\sum }_{i=0}^N{\alpha }_{n,i}=1$ and ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_{n,0}{\alpha }_{n,i}>0$ for every $1\le i\le N$. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}{x}_1$, where ${\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}$ is the generalized projection from E onto ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)$.

Proof First, we show that ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)$ is closed and convex. From [9], we find that ${\bigcap }_{i=1}^{N}F(T_i)$ is closed and convex, which combines with Lemma 1.10 shows that ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)$ is closed and convex. Next, we show that $C_n$ is closed and convex. It is obvious that $C_1=C$ is closed and convex. Suppose that $C_h$ is closed and convex for some positive integer h. For $z\in C_h$, we see that $\varphi (z,u_h)\le \varphi (z,x_h)+{\xi }_h$ is equivalent to

$$2〈z,J{x}_h-J{u}_h〉\le {\parallel x_k\parallel }^2-{\parallel u_k\parallel }^2+{\xi }_h.$$

It is to see that $C_{h+1}$ is closed and convex. This proves that $C_n$ is closed and convex. This in turn shows that ${\mathrm{\Pi }}_{C_{n+1}}{x}_1$ is well defined. Putting $u_n=S_{r_n}{y}_n$, we from Lemma 1.10 see that $S_{r_n}$ is quasi-ϕ-nonexpansive. Now, we are in a position to prove that ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)\subset C_n$. Indeed, ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)\subset C_1=C$ is obvious. Assume that ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)\subset C_h$ for some positive integer h. Then, for $\mathrm{\forall }w\in {\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)\subset C_h$, we have

$$\begin{array}{rl}\varphi (w,u_h)& =\varphi (w,S_{r_h}{y}_h)\\ \le \varphi (w,y_h)\\ =\varphi (w,J^{-1}({\alpha }_{h,0}J{x}_h+\sum _{i=1}^N{\alpha }_{h,i}J{T}_i^h{x}_h))\\ ={\parallel w\parallel }^2-2〈w,{\alpha }_{h,0}J{x}_h+\sum _{i=1}^N{\alpha }_{h,i}J{T}_i^h{x}_h〉+{\parallel {\alpha }_{h,0}J{x}_h+\sum _{i=1}^N{\alpha }_{h,i}J{T}_i^h{x}_h\parallel }^2\\ \le {\parallel w\parallel }^2-2{\alpha }_{h,0}〈w,J{x}_h〉-2\sum _{i=1}^N{\alpha }_{h,i}〈w,J{T}_i^h{x}_h〉+{\alpha }_{h,0}{\parallel x_h\parallel }^2+\sum _{i=1}^N{\alpha }_{h,i}{\parallel T_i^h{x}_h\parallel }^2\\ ={\alpha }_{h,0}\varphi (w,x_h)+\sum _{i=1}^N{\alpha }_{h,i}\varphi (w,T_i^h{x}_h)\\ \le {\alpha }_{h,0}\varphi (w,x_h)+\sum _{i=1}^N{\alpha }_{h,i}\varphi (w,x_h)+\sum _{i=1}^N{\alpha }_{h,i}{\xi }_h\\ =\varphi (w,x_h)+\sum _{i=1}^N{\alpha }_{h,i}{\xi }_h\\ \le \varphi (w,x_h)+\sum _{i=1}^N{\xi }_h,\end{array}$$

(2.1)

which shows that $w\in C_{h+1}$. This implies that ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)\subset C_n$.

Next, we prove that the sequence $\{x_n\}$ is bounded. Notice that $x_n={\mathrm{\Pi }}_{C_n}{x}_1$. We find from Lemma 1.9 that $〈x_n-z,J{x}_1-J{x}_n〉\ge 0$, for any $z\in C_n$. Since ${\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)\subset C_n$, we find that

$$〈x_n-w,J{x}_1-J{x}_n〉\ge 0,\mathrm{\forall }w\in \bigcap _{i=1}^{N}F(T_i)\cap EF(f).$$

(2.2)

It follows from Lemma 1.8 that

$$\begin{array}{rl}\varphi (x_n,x_1)& \le \varphi ({\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}{x}_1,x_1)-\varphi ({\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}{x}_1,x_n)\\ \le \varphi ({\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}{x}_1,x_1).\end{array}$$

This implies that the sequence $\{\varphi (x_n,x_1)\}$ is bounded. It follows from (1.1) that the sequence $\{x_n\}$ is also bounded. Since the space is reflexive, we may assume, without loss of generality, that $x_n\rightharpoonup \overline{x}$. Next, we prove that $\overline{x}\in {\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)$. Since $C_n$ is closed and convex, we find that $\overline{x}\in C_n$. This implies from $x_n={\mathrm{\Pi }}_{C_n}{x}_1$ that $\varphi (x_n,x_1)\le \varphi (\overline{x},x_1)$. On the other hand, we see from the weakly lower semicontinuity of $\parallel \cdot \parallel$ that

$$\begin{array}{rl}\varphi (\overline{x},x_1)& ={\parallel \overline{x}\parallel }^2-2〈\overline{x},J{x}_1〉+{\parallel x_1\parallel }^2\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}({\parallel x_n\parallel }^2-2〈x_n,J{x}_1〉+{\parallel x_1\parallel }^2)\\ =\underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\varphi (x_n,x_1)\\ \le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (x_n,x_1)\\ \le \varphi (\overline{x},x_1),\end{array}$$

which implies that ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_1)=\varphi (\overline{x},x_1)$. Hence, we have ${lim}_{n\to \mathrm{\infty }}\parallel x_n\parallel =\parallel \overline{x}\parallel$. In view of the Kadec-Klee property of E, we find that $x_n\to \overline{x}$ as $n\to \mathrm{\infty }$. Since $x_n={\mathrm{\Pi }}_{C_n}{x}_1$, and $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1\in C_{n+1}\subset C_n$, we find that $\varphi (x_n,x_1)\le \varphi (x_{n+1},x_1)$. This shows that $\{\varphi (x_n,x_1)\}$ is nondecreasing. We find from its boundedness that ${lim}_{n\to \mathrm{\infty }}\varphi (x_n,x_1)$ exists. It follows that

$$\begin{array}{rl}\varphi (x_{n+1},x_n)& =\varphi (x_{n+1},{\mathrm{\Pi }}_{C_n}{x}_1)\\ \le \varphi (x_{n+1},x_1)-\varphi ({\mathrm{\Pi }}_{C_n}{x}_1,x_1)\\ =\varphi (x_{n+1},x_1)-\varphi (x_n,x_1).\end{array}$$

This implies that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},x_n)=0.$$

(2.3)

In the light of $x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_1\in C_{n+1}$, we find that

$$\varphi (x_{n+1},u_n)\le \varphi (x_{n+1},x_n)+{\xi }_n.$$

It follows from (2.3) that

$$\underset{n\to \mathrm{\infty }}{lim}\varphi (x_{n+1},u_n)=0.$$

(2.4)

In view of (1.1), we see that ${lim}_{n\to \mathrm{\infty }}(\parallel x_{n+1}\parallel -\parallel u_n\parallel )=0$. This implies that ${lim}_{n\to \mathrm{\infty }}\parallel u_n\parallel =\parallel \overline{x}\parallel$. That is,

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{u}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel u_n\parallel =\parallel J\overline{x}\parallel .$$

(2.5)

This implies that $\{J{u}_n\}$ is bounded. Note that both E and $E^{\ast }$ are reflexive. We may assume, without loss of generality, that $J{u}_n\rightharpoonup u^{\ast }\in E^{\ast }$. In view of the reflexivity of E, we see that $J(E)=E^{\ast }$. This shows that there exists an element $u\in E$ such that $Ju=u^{\ast }$. It follows that

$$\begin{array}{rl}\varphi (x_{n+1},u_n)& ={\parallel x_{n+1}\parallel }^2-2〈x_{n+1},J{u}_n〉+{\parallel u_n\parallel }^2\\ ={\parallel x_{n+1}\parallel }^2-2〈x_{n+1},J{u}_n〉+{\parallel J{u}_n\parallel }^2.\end{array}$$

Taking ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$ on both sides of the equality aboven yields

$$\begin{array}{rl}0& \ge {\parallel \overline{x}\parallel }^2-2〈\overline{x},u^{\ast }〉+{\parallel u^{\ast }\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Ju〉+{\parallel Ju\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Ju〉+{\parallel u\parallel }^2\\ =\varphi (\overline{x},u).\end{array}$$

That is, $\overline{x}=u$, which in turn implies that $u^{\ast }=J\overline{x}$. It follows that $J{u}_n\rightharpoonup J\overline{x}\in E^{\ast }$. Since $E^{\ast }$ enjoys the Kadec-Klee property, we obtain from (2.5) that ${lim}_{n\to \mathrm{\infty }}J{u}_n=J\overline{x}$. Since $J^{-1}:E^{\ast }\to E$ is demicontinuous and E enjoys the Kadec-Klee property, we obtain $u_n\to \overline{x}$, as $n\to \mathrm{\infty }$. Note that

$$\parallel x_n-u_n\parallel \le \parallel x_n-\overline{x}\parallel +\parallel \overline{x}-u_n\parallel .$$

It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel x_n-u_n\parallel =0.$$

(2.6)

Since J is uniformly norm-to-norm continuous on any bounded sets, we have

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{u}_n\parallel =0.$$

(2.7)

On the other hand, we have

$$\begin{array}{rl}\varphi (w,x_n)-\varphi (w,u_n)& ={\parallel x_n\parallel }^2-{\parallel u_n\parallel }^2-2〈w,J{x}_n-J{u}_n〉\\ \le \parallel x_n-u_n\parallel (\parallel x_n\parallel +\parallel u_n\parallel )+2\parallel w\parallel \parallel J{x}_n-J{u}_n\parallel .\end{array}$$

We, therefore, find that

$$\underset{n\to \mathrm{\infty }}{lim}(\varphi (w,x_n)-\varphi (w,u_n))=0.$$

(2.8)

Since E is uniformly smooth, we know that $E^{\ast }$ is uniformly convex. In view of Lemma 1.11, we find that

$$\begin{array}{rl}\varphi (w,u_n)& =\varphi (w,S_{r_n}{y}_n)\\ \le \varphi (w,y_n)\\ =\varphi (w,J^{-1}({\alpha }_{n,0}J{x}_n+\sum _{i=1}^N{\alpha }_{n,i}J{T}_i^n{x}_n))\\ ={\parallel w\parallel }^2-2〈w,{\alpha }_{n,0}J{x}_n+\sum _{i=1}^N{\alpha }_{n,i}J{T}_i^n{x}_n〉+{\parallel {\alpha }_{n,0}J{x}_n+\sum _{i=1}^N{\alpha }_{n,i}J{T}_i^n{x}_n\parallel }^2\\ \le {\parallel w\parallel }^2-2{\alpha }_{n,0}〈w,J{x}_n〉-2\sum _{i=1}^N{\alpha }_{n,i}〈w,J{T}_i^n{x}_n〉+{\alpha }_{n,0}{\parallel x_n\parallel }^2+\sum _{i=1}^N{\alpha }_{n,i}{\parallel T_i^n{x}_n\parallel }^2\\ -{\alpha }_{n,0}{\alpha }_{n,1}g\left(\parallel J{x}_n-J{T}_1^n{x}_n\parallel \right)\\ ={\alpha }_{n,0}\varphi (w,x_n)+\sum _{i=1}^N{\alpha }_{n,i}\varphi (w,T_i^n{x}_n)-{\alpha }_{n,0}{\alpha }_{n,1}g\left(\parallel J{x}_n-J{T}_1^n{x}_n\parallel \right)\\ \le {\alpha }_{n,0}\varphi (w,x_n)+\sum _{i=1}^N{\alpha }_{n,i}\varphi (w,x_n)+\sum _{i=1}^N{\alpha }_{n,i}{\xi }_h\\ -{\alpha }_{n,0}{\alpha }_{n,1}g\left(\parallel J{x}_n-J{T}_1^n{x}_n\parallel \right)\\ =\varphi (w,x_n)+\sum _{i=1}^N{\alpha }_{n,i}{\xi }_n-{\alpha }_{n,0}{\alpha }_{n,1}g\left(\parallel J{x}_n-J{T}_1^n{x}_n\parallel \right)\\ \le \varphi (w,x_n)+{\xi }_n-{\alpha }_{n,0}{\alpha }_{n,1}g\left(\parallel J{x}_n-J{T}_1^n{x}_n\parallel \right).\end{array}$$

It follows that

$${\alpha }_{n,0}{\alpha }_{n,1}g\left(\parallel J{x}_n-J{T}_1^n{x}_n\parallel \right)\le \varphi (w,x_n)-\varphi (w,u_n)+{\xi }_n.$$

In view of the restriction on the sequences, we find from (2.8) that ${lim}_{n\to \mathrm{\infty }}g(\parallel J{x}_n-J{T}_1^n{x}_n\parallel )=0$. It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{T}_1^n{x}_n\parallel =0.$$

In the same way, we obtain

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{x}_n-J{T}_i^n{x}_n\parallel =0,\mathrm{\forall }1\le i\le N.$$

Notice that $\parallel J{T}_i^n{x}_n-J\overline{x}\parallel \le \parallel J{T}_i^n{x}_n-J{x}_n\parallel +\parallel J{x}_n-J\overline{x}\parallel$. It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{T}_i^n{x}_n-J\overline{x}\parallel =0.$$

(2.9)

The demicontinuity of $J^{-1}:E^{\ast }\to E$ implies that $T_i^n{x}_n\rightharpoonup \overline{x}$. Note that

$$|\parallel T_i^n{x}_n\parallel -\parallel \overline{x}\parallel |=|\parallel J{T}_i^n{x}_n\parallel -\parallel J\overline{x}\parallel |\le \parallel J{T}_i^n{x}_n-J\overline{x}\parallel .$$

This implies from (2.9) that ${lim}_{n\to \mathrm{\infty }}\parallel T_i^n{x}_n\parallel =\parallel \overline{x}\parallel$. Since E has the Kadec-Klee property, we obtain ${lim}_{n\to \mathrm{\infty }}\parallel T_i^n{x}_n-\overline{x}\parallel =0$. On the other hand, we have

$$\parallel T_i^{n+1}{x}_n-\overline{x}\parallel \le \parallel T_i^{n+1}{x}_n-T_i^n{x}_n\parallel +\parallel T_i^n{x}_n-\overline{x}\parallel .$$

It follows from the uniformly asymptotic regularity of $T_i$ that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel T_i^{n+1}{x}_n-\overline{x}\parallel =0.$$

That is, $T_i{T}_i^n{x}_n\to \overline{x}$. From the closedness of $T_i$, we find $\overline{x}=T_i\overline{x}$ for every $1\le i\le N$. This proves $\overline{x}\in {\bigcap }_{i=1}^{N}F(T_i)$.

Next, we show that $\overline{x}\in EF(f)$. In view of Lemma 1.8, we find that

$$\begin{array}{rl}\varphi (u_n,y_n)& \le \varphi (w,y_n)-\varphi (w,u_n)\\ \le \varphi (w,x_n)+{\mu }_n-\varphi (w,u_n).\end{array}$$

It follows from (2.8) that ${lim}_{n\to \mathrm{\infty }}\varphi (u_n,y_n)=0$. This implies that ${lim}_{n\to \mathrm{\infty }}(\parallel u_n\parallel -\parallel y_n\parallel )=0$. It follows from (2.6) that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel y_n\parallel =\parallel \overline{x}\parallel .$$

It follows that

$$\underset{n\to \mathrm{\infty }}{lim}\parallel J{y}_n\parallel =\underset{n\to \mathrm{\infty }}{lim}\parallel y_n\parallel =\parallel \overline{x}\parallel =\parallel J\overline{x}\parallel .$$

This shows that $\{J{y}_n\}$ is bounded. Since $E^{\ast }$ is reflexive, we may assume that $J{y}_n\rightharpoonup v^{\ast }\in E^{\ast }$. In view of $J(E)=E^{\ast }$, we see that there exists $v\in E$ such that $Jv=v^{\ast }$. It follows that

$$\begin{array}{rl}\varphi (u_n,y_n)& ={\parallel u_n\parallel }^2-2〈u_n,J{y}_n〉+{\parallel y_n\parallel }^2\\ ={\parallel u_n\parallel }^2-2〈u_n,J{y}_n〉+{\parallel J{y}_n\parallel }^2.\end{array}$$

Taking ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}$ the both sides of equality above yields that

$$\begin{array}{rl}0& \ge {\parallel \overline{x}\parallel }^2-2〈\overline{x},v^{\ast }〉+{\parallel v^{\ast }\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Jv〉+{\parallel Jv\parallel }^2\\ ={\parallel \overline{x}\parallel }^2-2〈\overline{x},Jv〉+{\parallel v\parallel }^2\\ =\varphi (\overline{x},v).\end{array}$$

That is, $\overline{x}=v$, which in turn implies that $v^{\ast }=J\overline{x}$. It follows that $J{y}_n\rightharpoonup J\overline{x}\in E^{\ast }$. Since $E^{\ast }$ enjoys the Kadec-Klee property, we obtain $J{y}_n-J\overline{x}\to 0$ as $n\to \mathrm{\infty }$. Note that $J^{-1}:E^{\ast }\to E$ is demicontinuous. It follows that $y_n\rightharpoonup \overline{x}$. Since E enjoys the Kadec-Klee property, we obtain $y_n\to \overline{x}$ as $n\to \mathrm{\infty }$. Note that

$$\parallel u_n-y_n\parallel \le \parallel u_n-\overline{x}\parallel +\parallel \overline{x}-y_n\parallel .$$

This implies that ${lim}_{n\to \mathrm{\infty }}\parallel u_n-y_n\parallel =0$. Since J is uniformly norm-to-norm continuous on any bounded sets, we have ${lim}_{n\to \mathrm{\infty }}\parallel J{u}_n-J{y}_n\parallel =0$. From the assumption $r_n\ge k$, we see that

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\parallel J{u}_n-J{y}_n\parallel }{r_n}=0.$$

(2.10)

Since $u_n=S_{r_n}{y}_n$, we find that

$$f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C.$$

It follows from (A2) that

$$\parallel y-u_n\parallel \frac{\parallel J{u}_n-J{y}_n\parallel }{r_n}\ge \frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge f(y,u_n),\mathrm{\forall }y\in C.$$

In view of (A4), we find from (2.10) that

$$f(y,\overline{x})\le 0,\mathrm{\forall }y\in C.$$

For $0<t<1$ and $y\in C$, define $y_t=ty+(1-t)\overline{x}$. It follows that $y_t\in C$, which yields $f(y_t,\overline{x})\le 0$. It follows from (A1) and (A4) that

$$0=f(y_t,y_t)\le tf(y_t,y)+(1-t)f(y_t,\overline{x})\le tf(y_t,y).$$

That is,

$$f(y_t,y)\ge 0.$$

Letting $t\downarrow 0$, we obtain from (A3) that $f(\overline{x},y)\ge 0$, $\mathrm{\forall }y\in C$. This implies that $\overline{x}\in EP(f)$.

Finally, we turn our attention to proving that $\overline{x}={\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}{x}_1$.

Letting $n\to \mathrm{\infty }$ in (2.2), we obtain

$$〈\overline{x}-w,J{x}_1-J\overline{x}〉\ge 0,\mathrm{\forall }w\in \bigcap _{i=1}^{N}F(T_i)\cap EF(f).$$

In view of Lemma 1.9, we find that $\overline{x}={\mathrm{\Pi }}_{{\bigcap }_{i=1}^{N}F(T_i)\cap EF(f)}{x}_1$. This completes the proof. □

From the definition of quasi-ϕ-nonexpansive mappings, we see that every quasi-ϕ-nonexpansive mapping is asymptotically quasi-ϕ-nonexpansive in the intermediate sense. We also know that every uniformly smooth and uniformly convex space is a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property (note that every uniformly convex Banach space enjoys the Kadec-Klee property).

Remark 2.2 Theorem 2.1 can be viewed an extension of the corresponding results in Qin et al. [6], Kim [12], Qin et al. [22], Takahashi and Zembayashi [24], respectively. The space $L^p$, where $p>1$, satisfies the restriction in Theorem 2.1.

3 Applications

Theorem 3.1 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property and let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4). Let $T:C\to C$ an asymptotically quasi-ϕ-nonexpansive mapping in the intermediate sense. Assume that T is closed asymptotically regular on C and $F(T)\cap EF(f)$ is nonempty and bounded. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{ chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ y_n=J^{-1}({\alpha }_{n,0}J{x}_n+{\alpha }_{n,1}J{T}^n{x}_n),\hfill \\ u_n\in C\mathit{\text{ such that }}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{y}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)+{\xi }_n\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,\hfill \end{array}$$

where ${\xi }_n=max\{0,{sup}_{p\in F(T),x\in C}(\varphi (p,T^{n}x)-\varphi (p,x))\}$, $\{r_n\}$ is a real number sequence in $[k,\mathrm{\infty })$, where k is some positive real number, $\{{\alpha }_{n,0}\}$ and $\{{\alpha }_{n}n,1\}$ are two real number sequence in $(0,1)$. Assume that ${lim\hspace{0.17em}inf}_{n\to \mathrm{\infty }}{\alpha }_{n,0}{\alpha }_{n,1}>0$. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{F(T)\cap EF(f)}{x}_1$, where ${\mathrm{\Pi }}_{F(T)\cap EF(f)}$ is the generalized projection from E onto $F(T)\cap EF(f)$.

Proof Putting $N=1$, we draw from Theorem 2.1 the desired conclusion immediately. □

Remark 3.2 If the mapping T in Theorem 3.1 is quasi-ϕ-nonexpansive, then the restrictions that T is closed asymptotically regular on C and $F(T)\cap EF(f)$ is bounded will not be required anymore.

If $T_i=I$, where I is the identity for every $1\le i\le N$, then we find from Theorem 2.1 the following.

Theorem 3.3 Let E be a uniformly smooth and strictly convex Banach space which also enjoys the Kadec-Klee property and let C be a nonempty closed and convex subset of E. Let f be a bifunction from $C\times C$ to ℝ satisfying (A1)-(A4). Assume that $EF(f)$ is nonempty. Let $\{x_n\}$ be a sequence generated in the following manner:

$$\{\begin{array}{c}{x}_0\in E\mathit{\text{ chosen arbitrarily}},\hfill \\ C_1=C,\hfill \\ x_1={\mathrm{\Pi }}_{C_1}{x}_0,\hfill \\ u_n\in C\mathit{\text{ such that }}f(u_n,y)+\frac{1}{r_n}〈y-u_n,J{u}_n-J{x}_n〉\ge 0,\mathrm{\forall }y\in C,\hfill \\ C_{n+1}=\{z\in C_n:\varphi (z,u_n)\le \varphi (z,x_n)\},\hfill \\ x_{n+1}={\mathrm{\Pi }}_{C_{n+1}}{x}_0,\hfill \end{array}$$

where $\{r_n\}$ is a real number sequence in $[k,\mathrm{\infty })$, where k is some positive real number. Then the sequence $\{x_n\}$ converges strongly to ${\mathrm{\Pi }}_{EF(f)}{x}_1$, where ${\mathrm{\Pi }}_{EF(f)}$ is the generalized projection from E onto $EF(f)$.